()2 4 (2)(1) (1)(4) (0)(1) 6ˆˆ ˆ ˆ ()...ˆˆˆ xy z yz z y zx x z x y yx ij k aa a b cbc bc bc...
TRANSCRIPT
CHAPTER 1 FUNDAMENTAL CONCEPTS: VECTORS
1.1 (a) ˆ ˆˆ ˆ ˆ ˆ ˆ( ) ( ) 2A B i j j k i j k+ = + + + = + +
v v
12(1 4 1) 6A B+ = + + =
v v
(b) ˆ ˆˆ ˆ ˆ ˆ ˆ3 2 3( ) 2( ) 3 2A B i j j k i j− = + − + = + −
v vk
(c) (1)(0) (1)(1) (0)(1) 1A B⋅ = + + =
v v
(d)
ˆˆ ˆˆ ˆˆ ˆ ˆ ˆ1 1 0 (1 0) (0 1) (1 0)
0 1 1
i j kA B i j k i j k× = = − + − + − = − +v v
12(1 1 1) 3A B× = + + =
v v
1.2 (a) ( ) ( ) ( )ˆˆ ˆ ˆ ˆ2 4 (2)(1) (1)(4) (0)(1) 6A B C i j i j k⋅ + = + ⋅ + + = + + =
v vv
( ) ( )ˆˆ ˆ ˆ3 4 (3)(0) (1)(4) (1)(0) 4A B C i j k j+ ⋅ = + + ⋅ = + + =
v vv
(b) ( )2 1 01 0 1 80 4 0
A B C⋅ × = = −v vv
( ) ( ) 8A B C A B C× ⋅ = ⋅ × = −
v v v vv v
(c) ( ) ( ) ( ) ( ) ( )ˆ ˆˆ ˆ ˆ ˆ4 2 4 4 8A B C A C B A B C i k j i j k× × = ⋅ − ⋅ = + − = − +
v v v v v vv v v4
( ) ( ) ( ) ( )
( ) ( )ˆ ˆˆ ˆ ˆ ˆ0 2 4 4 4
A B C C A B C B A C A B
i j i k i k
⎡ ⎤× × = − × × = − ⋅ − ⋅⎣ ⎦⎡ ⎤= − + − + = +⎣ ⎦
v v v v v v v vv v v v
1
1.3 2
22 2
( )( ) (2 )(2 ) (0)(3 ) 5cos5 145 14
A B a a a a a aAB aa a
θ ⋅ + += = =v v
1 5cos 5314
θ −= ≈ °
1.4 (a) ˆˆ ˆA i j k= + +
v : body diagonal
ˆ ˆˆ ˆ ˆ ˆ 3A A A i i j j k k= ⋅ = ⋅ + ⋅ + ⋅ =v v
(b) ˆ ˆB i j= +v
: face diagonal 2B B B= ⋅ =
v v
(c)
ˆˆ ˆ
1 1 11 1 0
i j kC A B= × =v v v
(d) 1 1cos 03 2
A BAB
θ ⋅ −= = =v v
90θ∴ = o
1.5
sinB B A C AC θ= = × =v vv
sinyBC CA
θ∴ = =
cosA C AC uθ⋅ = =v v
cosxuC CA
θ∴ = =
2x yA B A u B A BC C C AA A AB A
× × ⎛ ⎞= + = + ⎜ ⎟× ⎝ ⎠
v v v
B A
v vv v
vv
2 2
1u A B AA A
= + ×v vv
1.6 ( ) ( ) ( )2 3ˆ ˆˆ ˆ ˆ ˆ2 3 2d d di t j t k t i j t k tdt dt dt dtdA α β γ α β γ= + + = + +v
2
2ˆˆ2 6d A j k t
dtβ γ= +
v
2
1.7 ( )( ) ( )( ) ( )( ) 20 3 1 2A B q q q q q= ⋅ = + − + = − +v v
3 2
0
( ) , q( )2 1q q− − = 1= or 2 1.8 ( ) ( )2 2 2 2A B A B A B A B A B+ = + ⋅ + = + + ⋅
v v v vv v v v
2 2 2 2A B A B AB+ = + +v v
⎡ ⎤⎣ ⎦
Since cosA B AB ABθ⋅ = ≤
v v, A B A B+ ≤ +
v vv v
cos cosA B AB A B A Bθ θ⋅ = = ≤v v vv v v
cosB Bθ ≤
1.9 Show ( ) ( ) ( )A B C A C B A B C× × = ⋅ − ⋅
v v v v v vv v v
or ( ) (ˆˆ ˆ
x y z x x y y z z x x y y z z
x y z
i j k)B B A C A C A C B A B A B A B C
C C C× = + + − + +v vvA B
( ) ˆx x x y x y z x z x x x y y x z z xA B C A B C A B C A B C A B C A B C i= + + − − −
( ) ˆx y x y y y z y z x x y y y y z z yA B C A B C A B C A B C A B C A B C j+ + + − − −
( ) ˆx z x y z y z z z x x z y y z z z zA B C A B C A B C A B C A B C A B C k+ + + − − −
( )( )( )
ˆ
ˆ
ˆ
y x y z x z y y x z z x
x y x z y z x x y z z y
x z x y z y x x z y y z
A B C A B C A B C A B C i
A B C A B C A B C A B C j
A B C A B C A B C A B C k
+ − −
= + + − −
+ + − −
3
ˆˆ ˆ
x y z
y z z y z x x z x y y x
i j kA A A
B C B C B C B C B C B C− − −
( )( )( )
ˆ
ˆ
ˆ
y x y y y x z z x z x z
z y z z z y x x y x y x
x z x x x z y y z y z y
i A B C A B C A B C A B C
j A B C A B C A B C A B C
k A B C A B C A B C A B C
− − +
= + − − +
+ − − +
1.10
siny A θ=
( )12 s2
xy y B x xy yB xy AB inθ⎛ ⎞Α = + − = + − =⎜ ⎟⎝ ⎠
A B×
v vΑ =
1.11
( )ˆˆ ˆ
x y z
x y z x y z ( )x y z
x y z
x y z
B B B
x y z x y
i j k A AC A B B B B B B
C C C C C C⋅ × = ⋅ =v v vv
z
AA B A A A B A C
C C C− = − ⋅ ×
v vv=
1.12
x z
Let = (Ax,Ay,Az), A
vBv
= (0,By,0) and Cr
= (0,Cy,Cz) Cz is the perpendicular distance between the plane A
v, Bv
and its opposite. u B is
directed along the x-axis since the vectors
xC=rrr
Br
, Cr
are in the y,z plane. xu B = BxC=rr
yCz
is the area of the parallelogram formed by the vectors Br
, Cr
. Multiply that area times the height of plane
v,A Bv
= Ax to get the volume of the parallopiped ( )x x x y zV A u A B C A B xC= = = •
r rr
Br
Ar
Cr
ur
y
4
1.13 For rotation about the z axis:
ˆ ˆ ˆ ˆcos ,i i j jφ′ ′⋅ = = ⋅ ˆ ˆ 1k k ′⋅ = ˆ ˆ sini j φ′⋅ = − ˆ ˆ sinj i φ′⋅ =
For rotation about the axis: y′
ˆ ˆˆ ˆ cos ,i i k kθ′ ′⋅ = = ⋅ ˆ ˆ 1j j′⋅ = ˆˆ sini k θ′⋅ = ˆ ˆ sink i θ′⋅ = −
cos 0 sin cos sin 0 cos cos cos sin sin
0 1 0 sin cos 0 sin cos 0sin 0 cos 0 0 1 sin cos sin sin cos
Tθ θ φ φ θ φ θ φ θ
φ φ φ φθ θ θ φ θ φ θ
− −⎛ ⎞⎛ ⎞ ⎛⎜ ⎟⎜ ⎟ ⎜= − = −⎜ ⎟⎜ ⎟ ⎜⎜ ⎟⎜ ⎟ ⎜⎝ ⎠⎝ ⎠ ⎝
t⎞⎟⎟⎟⎠
1.14
3ˆ ˆ cos302
1ˆ ˆ sin 302
ˆˆ 0
i i
i j
i k
′⋅ = =
′⋅ = − = −
′⋅ =
o
o
1ˆ ˆ sin 302
3ˆ ˆ cos302
ˆˆ 0
j i
j j
j k
′⋅ = =
′⋅ = =
′⋅ =
o
o
ˆ ˆ 0
ˆ ˆ 0
ˆ ˆ 1
k i
k j
k k
′⋅ =
′⋅ =
′⋅ =
3 1 30 32 2 221 3 30 3 3 12 2 2
10 0 1 1
x
y
z
AAA
′
′
′
⎡ ⎤ ⎡ ⎤⎢ ⎥ +⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎣ ⎦ ⎣ ⎦ −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦
ˆˆ ˆ3.232 1.598A i j k′ ′ ′= + −v
5
1.15 1. Rotate thru φ about z-axis 45φ = o Rφ
2. Rotate thru θ about x’-axis 45θ = o Rθ 3. Rotate thru ψ about z’-axis 45ψ = o Rψ
1 1 02 21 1 0Rφ
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= −⎜ ⎟2 2
0 0 1⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
1 0 01 102 21 102 2
Rθ
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠
1 1 02 21 1 02 2
0 0 1
Rψ
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
( )
1 1 1 1 12 22 2 2 21 1 1 1 1, ,2 22 2 2 2
1 12 2 2
R R R Rψ θ φψ θ φ
⎛ ⎞− +⎜ ⎟⎜ ⎟⎜ ⎟= = − − − +⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠
2
21
or ( )10 , ,0
Rα
ψ θ φ βγ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Condition is: x Rx′ =v v where and 100
x⎛ ⎞⎜ ⎟′ = ⎜ ⎟⎜ ⎟⎝ ⎠
v xαβγ
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
v
Since 1x x⋅ =v v we have: 2 2 2 1ψ β α+ + =
After a lot of algebra: 1 2 ,2 4
α = − 1 2 ,2 4
β = + 12
γ =
1.16 ˆ ˆv v ctτ τ= =v
2 2
ˆ ˆ ˆv ca v n c nb
τ τρ
= + = +v&
2
ˆt
6
at btc
= , ˆv bτ=v c and ˆ ˆa c cnτ= +v
2
1cos22
v a c bcva bc c
θ ⋅= = =v v
45θ = o
1.17 ( ) ( ) ( )ˆ ˆsin 2 cosv t ib t j b tω ω ω= − +v ω
( ) ( )1 1
2 2 2 2 2 2 22 2sin 4 cos 1 3cosv b t b t b tω ω ω ω ω ω= + = +v
( ) 2 2ˆ ˆcos 2 sina t ib t j b tω ω ω= − −v ω
( )1
2 2 21 3sina b tω ω= +v
at , 0t = 2v bω=v ; at 2
t πω
= , v bω=v
1.18 ( ) ˆˆ ˆcos sin 2v t ib t jb t k ctω ω ω ω= − +v
( ) 2 2 ˆˆ ˆsin cos 2a t ib t jb t k cω ω ω ω= − − +v
( ) ( )1 1
2 4 2 2 4 2 2 2 4 22 2sin cos 4 4a b t b t c b cω ω ω ω ω= + + = +v 1.19 ˆ ˆ ˆkt kt
r rv re r e bke e bce eθ θθ= + = +v &&
( ) ( ) ( )2 2 2ˆ ˆ ˆ2 2kt ktr ra r r e r r e b k c e e bcke eθ θθ θ θ= − + + = − +v & && &&& &
( )
( ) ( )
2 2 2 2 2 2 2
11 2 22 2 2 2 2 22
2cos
4
kt kt
kt kt
b k k c e b c kev ava
be k c be k c c k
φ− +⋅
= =⎡ ⎤+ − +⎢ ⎥⎣ ⎦
v v
( )
( ) ( ) ( )
2 2
1 12 2 2 2 2 22 2
cosk k c k
k c k c k cφ
+= =
+ + +, a constant
1.20 ( ) ( )2 ˆ ˆ2R za R R e R R e zeφφ φ φ= − + + +v & & &&&& & &&ˆ
2 ˆ ˆ2R za b e ceω= − +v
( )1
2 4 2 24a b cω= +v
7
1.21 ( ) ( )ˆ ˆ1 kt ktr t i e je−= − +v
( ) ˆ ˆkt ktr t ike jke−= +v
( ) 2 2ˆ ˆkt ktr t ik e jk e−= − +v
1.22 ˆ ˆ ˆsinrv e r e r e rφ θφ θ θ= + +v &&
( ) ( )1ˆ ˆsin 1 cos 4 sin 42 4 2
v e b t e b tφ θπ πω ω ω⎧ ⎫⎡ ⎤= + −⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭
v ω
( ) ( )ˆ ˆcos cos 4 sin 48 2
v e b t e b tφ θπ πω ω ω⎡ ⎤= −⎢ ⎥⎣ ⎦
v ω
12 2
2 2cos cos 4 sin 48 4
v b t tπ πω ω⎡ ⎤⎛ ⎞= +⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
v ω
Path is sinusoidal oscillation about the equator. 1.23 2v v v⋅ =v v
2dv dvv v vvdt dt
⋅ + ⋅ =v vv v
&
2 2v a vv⋅ =v v
& v a vv⋅ =v v
&
8
1.24 ( ) ( ) ( )d dr dr v a v a r v adt dt dt
⋅ × = ⋅ × + ⋅ ×⎡ ⎤⎣ ⎦
vv v v v v v v v
( ) dv dav v a r a vdt dt
⎡ ⎤⎛ ⎞ ⎛ ⎞× + ⋅ × + ×⎜ ⎟ ⎜ ⎟= ⋅v
⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
vv v v v v v
( )0 0r v a⎡ ⎤= + ⋅ + ×⎣ ⎦v v v&
( ) ( )d r v a r v adt
⋅ × = ⋅ ×⎡ ⎤⎣ ⎦v v v v v v&
1.25 ˆv vτ=v and ˆ ˆna a a nττ= +v
v a vaτ⋅ =v v , so v aavτ⋅
=v v
2 2
na a aτ= + 2 , so ( )1
2 2 2na a aτ= −
1.26 For 1.14, ( )
2 3 2 3 2
12 2 2 2 2 2 2 2 2
cos sin sin cos 4
cos sin 4
b t t b t tab t b t c t
τω ω ω ω ω ω
ω ω ω ω
− ⋅ + ⋅ +=
+ +
c t
( )
2
12 2 2 2 2
4
4
c tab c t
τ
ω=
+
14 2 2
2 2 22 2 2 2
1644n
c ta b cb c t
ωω
⎛ ⎞= + −⎜ ⎟+⎝ ⎠
For 1.15, ( )
( )( )
2 2 2 2 2 2 2 12 2 2
12 2 2
2kt ktkt
kt
b k k c e b c kea b
be k cτ
− += =
+ke k c+
( ) ( ) ( )1 12 22 2 2 2 2 2 2 2 2 2 2 2kt kt kt
na b e k c b k e k c bce k c⎡ ⎤= + − + = +⎢ ⎥⎣ ⎦
1.27
n
τ ˆˆd dθ=−τ n
dθ
ρ
9
The figure above shows the unit vectors and n τ which are normal and tangent to the curve. The vectors are shown at 2 different points along the trajectory. Since the particle is moving tangentially to the curve and in a direction perpendicular to ρ , the local radius of curvature, we have …
ˆv=vv τ 2
ˆ ˆˆ ˆ vv v vθρ
= + =− = −v & &&a nτ τ n (since 0v=& and from the above figure)
2 2
ˆˆ v vv v3v
ρ ρ ρ⎛ ⎞
× = × − = =⎜ ⎟⎝ ⎠
v vv a nτ
1.28
ˆ ˆsin cosPr ib jbθ θ= +o
v ˆ ˆcos sinrelv ib jbθ θ θ= −v & & θ
( ) ( )2 2ˆ ˆcos sin sin cosrela ib jbθ θ θ θ θ θ θ θ= − − +v && & && &
at the point θ π= , relv v= −v v
So, relv bθ= =v & v
vb
θ =& v ab b
θ = = o&&&
Now, 2 2
ˆ ˆ ˆrelrel rel
v va v n ab
τ τρ
= + = +o
v& n
14 2
22rel
va ab
⎛ ⎞= +⎜ ⎟⎝ ⎠
o
v
P relv v v= +v v v and P rela a a= +ov v v
2 2
2 2ˆ ˆcos sin sin cosP
a v a va i a b jbb b b b
θ θ θ⎡ ⎤⎛ ⎞ ⎛
= + − − +⎢ ⎥⎜ ⎟ ⎜⎝ ⎠ ⎝⎣ ⎦
o oo
v θ⎞⎟⎠
14 2 2
2 2
22 2cos sinPv va a
a b a bθ θ
⎛ ⎞= + + −⎜ ⎟
⎝ ⎠o
o o
v
is a maximum at Pav 0θ = , i.e., at the top of the wheel.
10
222sin cos 0v
a bθ θ− − =
o
2
1tan va b
θ − ⎛ ⎞= −⎜ ⎟
⎝ ⎠o
Comments : Note that a point on the bottom of the wheel is instantaneously at rest, i.e., there is no relative motion between the ground and the bottom of the wheel assuming no slipping.
1.29 Therefore,
2
2
0 2 0 00 0 2
0 0 1 0 0 1
x -x 0 x x x= x x 0 x x x
0 0 1
⎛ ⎞⎛ ⎞⎛ ⎞⎜⎜ ⎟⎜ ⎟− = ⎜⎜ ⎟⎜ ⎟
⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
%RR 0⎟⎟12
x =
The transformation represents a rotation of 45o about the z-axis (see Example 1.8.2)
1.30 (a) ˆ ˆcos sinθ θ+a = i j
a
b θ φ
ˆ ˆcos sinϕ ϕ= +b i j
( ) ( ) ( )ˆ ˆ ˆ ˆcos cos sin cos sinθ ϕ θ θ ϕ⋅ = − = + ⋅ +a b i j i j ϕ
( )cos cos cos sin sinθ ϕ θ ϕ θ− = + ϕ
(b) ( ) ( ) ( )ˆ ˆ ˆ ˆ ˆsin cos sin cos sinθ ϕ θ θ ϕ× = − = + × +b a k i j i j ϕ
( )sin sin cos cos sinθ ϕ θ ϕ θ− = − ϕ --------------------------------------------------------------------------------------------------------
11
CHAPTER 2 NEWTONIAN MECHANICS:
RECTILINEAR MOTION OF A PARTICLE
2.1 (a) ( )1x F ctm
= +o&&
( ) 2
0
12
t F cx F ct dt t tm m
= + = +∫ oo
&m
2 2
0 2 6t F Fc c 3x t t dt t
m m m m⎛ ⎞= + = +⎜ ⎟⎝ ⎠∫ o o t
(b) sinFx ctm
= o&&
( )00sin cos 1 cos
t tF F Fx ct dt ct ctm cm cm
= = − = −∫ o o o&
( )0
11 cos 1 sint F Fx ct dt ctcm cm c
⎛ ⎞= − = −⎜ ⎟⎝ ⎠∫ o o
(c) ctFx em
= o&&
( )0
1tct ctF Fx e e
cm cm= =o o& −
( )2
1 1 1ct ctF Fx e t ecm c c c m
⎛ ⎞= − − = −⎜ ⎟⎝ ⎠
o o ct−
2.2 (a) dx dx dx dxx xdt dx dt dx
= = ⋅ =& &
&& &&
( )1dxx F cxdx m
= +o&
&
( )1xdx F cx dxm
= +o& &
2
21 12 2
cxx F xm⎛ ⎞
= +⎜ ⎟⎝ ⎠
o&
( )12
2xx F cxm⎡ ⎤= +⎢ ⎥⎣ ⎦
o&
(b) 1 cxdxx x Fdx m
−= = o
&&& & e
1
1 cxxdx F e dxm
−= o& &
( ) ( )21 1 12
cx cxF Fx e ecm cm
− −= − − = −o o&
( )122 1 cxFx e
cm−⎡ ⎤= −⎢ ⎥⎣ ⎦
o&
(c) ( )1 cosdxx x Fdx m
cxo
&&= =&&
cosFxdx cx dxm
= o& &
21 sin2
Fx cxcm
= o&
122 sinFx cx
cm⎛ ⎞= ⎜ ⎟⎝ ⎠
o&
2.3 (a) ( ) ( )2
2x
x
cxV x F cx dx F x C= − + = − − +∫o
o o
(b) ( )x cx cx
x
FV x F e dx e Cc
− −= − = +∫o
oo
(c) ( ) cos sinx
x
FV x F cx dx cx Cc
= − = − +∫o
oo
2.4 (a) ( ) ( )dV xF x kx
dx= − = −
( ) 2
0
12
xV x kx dx kx= =∫
x V x= +o(b) T T ( ) ( )
( ) ( ) ( )212
T x T V x k A x= − = −o
(c) 212
E T k= =o A
→
(d) turning points @ T x ( )1 0 1x A∴ = ±
2.5 (a) ( )3
2
kxF x kxA
− + so ( )3 4
22 20
1 12 4
x kx kxV x kx dx kxA A
⎛ ⎞= − = −⎜ ⎟
⎝ ⎠∫
(b) ( ) ( )4
22
1 12 4
kxT x T V x T kxA
= − = − +o o
(c) E T= o
2
(d) ( )V x has maximum at ( ) 0mF x →
3
2 0mm
kxkxA
− = mx A= ±
( )4
2 22
1 1 12 4 4m
kAV x kA kAA
= − =
If ( m )E V x< turning points exist.
Turning points @ let ( )1 0T x → 21u x=
2
2
1 1 02 4
kuE kuA
− + =
solving for u , we obtain
122
2
41 1 Eu AkA
⎡ ⎤⎛ ⎞⎢ ⎥= ± −⎜ ⎟⎢ ⎥⎝ ⎠
⎣ ⎦
or
12
1 2
41 1 Ex AkA
⎡ ⎤⎛ ⎞= ± − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
2.6 ( )x v xxα
= =& 2
2 3x xx xα α
= − = −&& &
( )2
3
mF x mxxα
= = −&&
2.7
sinF Mg θ≥
2.8 dxF mx mxdx
= =&
&& &
3x bx−=&
43dx bxdx
−= −&
( )(3 43F m bx bx− −= − ) 2 73F mb x−= −
3
2.9 (a) ( ) ( )2.145 9.8 1250 .3048 541m mV mgx kg ft Js ft
⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
(b) 2
222
2
1 1 1 12 2 2 2 .22t
mg m gT mv mv mc D
⎛ ⎞= = = =⎜ ⎟
⎝ ⎠
( )
( )( ) ( )( )
22
2
.145 9.887
2 .22 2 .0366
mkgsT Jkg
m
⎛ ⎞⎜ ⎟⎝ ⎠= =
⎡ ⎤⎣ ⎦
(c) 3
2 3 tanhttFdx cv dx c v dt c v dtτ
⎛ ⎞⎛ ⎞= − = − = − − ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
∫ ∫ ∫ ∫
3 21 tanh tanh2t
t tcv dτ tτ τ τ
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦∫
3 21 tanh ln cosh2t
t tcv ττ τ
⎡ ⎤⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
Now 2tanh 1tτ⎛ ⎞ ≅⎜ ⎟⎝ ⎠
for t τ>>
Meanwhile tanh ln cosht tt tx vdt v dt vττ τ
⎛ ⎞⎛ ⎞ ⎛ ⎞= = − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠∫ ∫
ln cosht
t xvτ τ
⎛ ⎞ =⎜ ⎟⎝ ⎠
( )1250 .3048 381mx ft mft
⎛ ⎞= =⎜ ⎟
⎝ ⎠
( )
( )( )
12
122
22
.145 9.834.72
.22 .0732t
mkgmg msv kgc s
m
⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥= = =⎜ ⎟
⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦
( )
( )( )
12
12
222
.1453.543
.22 .0732 9.8
kgm skg mc gm s
τ
⎡ ⎤⎢ ⎥⎛ ⎞⎢ ⎥= = =⎜ ⎟ ⎛ ⎞⎢ ⎥⎝ ⎠ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
( )( ) ( ) ( ) ( )( )2 3 3.81.22 .0732 34.72 3.543 .5 454
34.72 3.54Fdx J
⎡ ⎤= − +⎢ ⎥
⎣ ⎦∫ =
541 87 454V T J J J− = − = 2.10 Going up: sin 30 cos30xF mg mgμ= − −o o
4
( ) 2sin 30 0.1cos30 5.749 mx gs
= − + = −o o&&
v v at= +o
at the highest point so 0v = 0.174upvt va
= − =oo s
2 221 0.174 .087 0.0872up up up
2x v t at v v v m= + = − =o o o o
Going down: 20.087x v′ =o o , 0v ′ =o , ( )9.8 0.5 0.0866a′ = − −
2 210 0.087 4.05132down downx v t= = −o
0.207downt v= o ss 0.381total up downt t t v= + = o
2.11 32dv dv dx dv ca v
dt dx dt dx m= = ⋅ = ⋅ = − v
12 cv dv dx
m−
= −
max12
0
v x
v
cv dv dxm
−= −∫ ∫
o
12 max2 cv x
m− = −o
12
max2mvx
c= o
2.12 Going up: 2
2xF mg c v= − −
2dva v g kvdx
= = − − , 2
2ckm
=
2 0
v
v
vdv dxg kv
=− −∫ ∫
o
x (Letting 0 0x = … starting from ground)
( )21 ln2
v
vg kv x
k− − − =
o
2
22
kxg kv eg kv
−+=
+ o
22 2kxg gv v ek k
−⎛ ⎞= + −⎜ ⎟⎝ ⎠
o
5
Going down: 22xF mg c v= − +
2dvv gdx
= − + kv
20 0
v xvdv dxg kv
=− +∫ ∫
( )2
0
1 ln2
vg kv x x
k− + = − o
22 21 kxkxk v e eg
−− = o
22 2kx kxg gv ek k
−⎛ ⎞= − ⎜ ⎟⎝ ⎠
o e
2.13 At the top so 0v = max2
2
kx
gke g v
k
− =+ o
Coming down maxx x=o and at the bottom 0x =
( )2
22
22
1 1
g vg g kv ggk k vv
kk
⎛ ⎞⎜ ⎟⎛ ⎞ ⎝ ⎠= − =⎜ ⎟ ⎛ ⎞⎝ ⎠ ++⎜ ⎟
⎝ ⎠
o
oo
( )1
2 2 2
t
t
v vvv v
=+
o
o
, 2
tg mvk c
= =g
2.14 2dv ka v xdx m
−= = −
20
v x
b
kdxvdvmx
= −∫ ∫
21 12
kvm x b⎛ ⎞= −⎜ ⎟⎝ ⎠
1
1 12 22 1 1 2dx k k b xv
dt m x b mb x⎡ ⎤ ⎡ −⎛ ⎞ ⎛ ⎞= = − =⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣
⎤⎥⎦
1211
3 220 0
0 12 2 1
t
b
xmb x mb xbdt dx dxk b x k b
b
⎛ ⎞⎜ ⎟⎛ ⎞⎡ ⎤⎛ ⎞ ⎛= = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜⎢ ⎥−⎝ ⎠ ⎝⎣ ⎦ ⎝ ⎠ ⎜ ⎟−⎜ ⎟⎝ ⎠
∫ ∫ ∫ ⎞⎟⎠
6
Since x b≤ , say 2sinxb
θ=
( )1 1
3 32 20 0 2
2 2
sin 2sin cos 2 sin2 cos
dmb mbt dk kπ π
θ θ θ θθ θ
θ− −
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
∫ ∫
13 2
8mbt
kπ
⎛ ⎞= ⎜ ⎟⎝ ⎠
2.15 21 2
dvm mg c v cdt
= − − v
20 01 2
t vdt dvm mg c v c v=
− −∫ ∫
Using 2
2 2 2
1 2 4ln4 2 4
dx cx b b aca bx cx b ac cx b b a
+ − −=
+ + − + + −∫ c
,
22 1 1 2
2 21 2 2 1 1 2 0
2 41 ln4 2 4
v
c v c c mgctm c mgc c v c c mgc
− − − +=
+ − − + +
( ) ( )( )( )( )
2 21 2 1 1 2 1 1 22 2
1 2 2 22 1 1 2 1 1 2
2 4 44 ln
2 4 4
c v c c mgc c c mgct c mgcm c v c c mgc c c mgc
+ + + − ++ =
+ − + + +
as t , →∞ 22 1 1 22 4tc v c c mgc+ − + = 0
12 2
1 1
2 2 22 2tc c mgvc c c
⎡ ⎤⎛ ⎞⎢ ⎥= − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
Alternatively, when , tv v=
21 20 t t
dvm mg c v cdt
= = − − v
12 2
1 1
2 2 22 2tc c mgvc c c
⎡ ⎤⎛ ⎞⎢ ⎥= − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
2.16 ( )( )4 2 61 1.55 10 10 1.55 10 kgc
s− − −= × = ×
( )( )22 52 0.22 10 2.2 10 kgc
s− −= = ×
7
( )( )
12 7 26 6
5 5
10 9.81.55 10 1.55 102 2.2 10 2 2.2 10 2.2 10tv
−− −
− −
⎡ ⎤⎛ ⎞× ×⎢ ⎥= − + +⎜ ⎟× × × × ×⎢ ⎥⎝ ⎠⎣ ⎦
5−
0.179tmvs
=
Using Equation 2.4.10 … ( )( )
172
52
10 9.80.211
2.2 10tmg mvc s
−
−
⎛ ⎞= = =⎜ ⎟ ×⎝ ⎠
2.17 ( ) ( )dv dvm mv f x gdt dx
= = v
( ) ( )mvdv f x dx
g v=
By integration, get ( ) dxv v xdt
= =
If ( ) ( ) ( ),F x t f x g t= :
( ) ( )2
2
d x d dxm m f x gdt dt dt
⎛ ⎞= =⎜ ⎟⎝ ⎠
t
This cannot, in general, be solved by integration. If ( ) ( ) ( ),F v t f v g t= :
( ) ( )dvm f v gdt
= t
( ) ( )mdv g t dt
f v=
Integration gives ( )v v t=
( )dx v tdt
=
( )dx v t dt=
A second integration gives ( )x x t=
2.18 dvF kxv mdt
= =
m dv dvkx mv dt dx
= =
kxdx mdv=∫ ∫
2
2xk mv= +C
At x = 0, t = 0 0v v= 0C mv= −
8
2
02x dxk mv m
dt+ =
2 202
dx dxdtkx m v A B x
= =+ +∫ ∫ ∫ 2 2 where 2 2
0 2kA v Bm
= =
11 tan Bt xAB A
−=
Solving …
1 12 2
0 02 tan2
mv kvx tk m
⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2.19 ( ) xF x Ae mα= − =& &&x or ( ) vF v Ae mα= − = &v v
dv A dte mα = −
Let u veα= vdu e dvαα= v
du dudve uαα α
= = 2
du A dtu m
α∴ = −
Integrating
1 1 A tu u m
α− =o
and substituting ve uα =
(a) 1 ln 1 vAv v e tm
α αα
⎡ ⎤= − +⎢ ⎥⎣ ⎦o
o
v= =
(b) t T @ 0
ln 1 vAv em
α Tα α⎡ ⎤= +⎢ ⎥⎣ ⎦o
o
1v vAe em
α α Tα= +o o 1 vmT eA
α
α−⎡ ⎤= −⎣ ⎦
o
(c) vdv Av v edx m
α= = − v
vdv A dxe mα = −
again, let u veα= du udvα= or dudvuα
= 1 lnv u α
=
1 ln duuAu dx
u mα α⎡ ⎤⎢ ⎥⎣ ⎦ = − Integrating and solving
( )2 1 1 vmx vA
ααα
−⎡ ⎤= − +⎣ ⎦o
o e
2.20
( )d mvF mv vm
dt= = + = mg
9
but 343
m rρ π= o 21m rρ π= v
so (1) 3 2 2 21
4 43 3
r v r v r gπρ πρ π πρ+ = =o o34
3
Now 31 10ρρ
−
o
≈ so, second term is negligible-small
hence v and g≈ v gt≈ speed t∝ but
or 2 214m r rρ π ρ π= =o
& & r v 114
r vρρ
≅o
& Hence 114
r ρρ
≈o
gt and rate of
growth t∝ The exact differential equation from (1) above is:
2
11 1
4 4 4 43 3
rr r rρ ρ gπρ πρ πρ ρ
+ =o oo o
&&& ρ
which reduces to: 2
134
rr gr
ρρ
+ =o
&&&
Using Mathcad, solve the above non-linear d.e. letting 31 10ρ
ρ−≈
o
and 0.01R mm≈o (small raindrop). Graphs
show that v and r t∝ ∝& 2r t∝
10
CHAPTER 3 OSCILLATIONS
3.1 ( ) [ ]10.002sin 2 512x s t mπ −⎡ ⎤= ⎣ ⎦
( )( )( )max 0.002 2 512 6.43m mxs s
π ⎡ ⎤ ⎡= = ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣
&⎦
( )( ) ( )2 2 4max 2 20.002 2 512 2.07 10m mx
s sπ ⎡ ⎤ ⎡= = ⎤×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣
&&⎦
3.2 0.1sinx tω= o [m] 0.1 cos mx ts
ω ω ⎡ ⎤= ⎢ ⎥⎣ ⎦o o
&
When t = 0, x = 0 and 0.5 0.1mxs
ω⎡ ⎤= =⎢ ⎥⎣ ⎦o
&
15 sω −=o2 1.26T sπω
= =o
3.3 ( ) cos sinxx t x t tω ωω
= + oo o
o
&o and 2 fω π=o
( ) ( )[ ]0.25cos 20 0.00159sin 20x t tπ π= + m 3.4 ( )cos cos cos sin sinα β α β α− = + β
( )cos cos cos sin sinx A t A t A tω φ φ ω φ= − = +o o ωo cos sinx t tω ω= Α +Βo o , cosA φΑ = , sinA φΒ =
3.5 2 2 2 21 1 2
1 1 1 12 2 2 2
mx kx mx kx+ = +& & 2
( ) ( )2 2 2 21 2 2 1k x x m x x− = −& &
12 2 22 12 21 2
k x xm x x
ω⎛ ⎞−
= = ⎜ ⎟−⎝ ⎠o
& &
2 21 1
1 1 12 2 2
kA mx kx= +& 2
2 2 2 2
2 2 2 1 1 2 11 1 2 2
2 1
m x x x xA x xk x x
−= + = +
−& &
&& &
21x
12 2 2 2 21 2 2 1
2 22 1
x x x xAx x
⎛ ⎞−= ⎜ ⎟−⎝ ⎠
& &
& &
1
3.6 1 1 2.59.826
lT sg
π π= = ≈o s
3.7 For springs tied in parallel:
( ) ( )1 2 1 2sF x k x k x k k= − − = − + x
( )12
1 2k km
ω+⎡ ⎤
= ⎢ ⎥⎣ ⎦
For springs tied in series: The upward force is m eqk x . Therefore, the downward force on spring is 2k eqk x . The upward force on the spring is 2k 1k x′ where x′ is the displacement of P, the point at which the springs are tied. Since the spring is in equilibrium, 2k 1 eqk x k′ = x . Meanwhile, The upward force at P is 1k x′ . The downward force at P is ( )2k x x′− . Therefore, ( )1 2k x k x x′ ′= −
2
1 2
k xxk k
′ =+
And eqk 21
1 2
k xx kk k
⎛ ⎞= ⎜ ⎟+⎝ ⎠
( )
12
1 2
1 2
eqk k km k k m
ω⎡ ⎤
= = ⎢ ⎥+⎣ ⎦
3.8 For the system ( )M m+ , ( )kX M m X− = + &&
The position and acceleration of m are the same as for ( )M m+ :
m mkx x
M m= −
+&&
cos cosmk kx A t d t
M m Mδ
⎛ ⎞= + =⎜ ⎟⎜ ⎟+ +⎝ ⎠ m
r
The total force on m, m mF mx mg F= = −&&
2
cosr mmk mkd kF mg x mg t
M m M m M= + = +
+ + m+
For the block to just begin to leave the bottom of the box at the top of the vertical oscillations, 0rF = at mx d= − :
0 mkdmgM m
= −+
( )g M md
k+
=
3.9 ( )cost
dx e A tγ ω φ−= −
( ) ( )sin cost td d d
dx e A t e A tdt
γ γω ω φ γ ω φ− −= − − − −
maxima at ( ) (0 sin cosd d ddx t tdt
)ω ω φ γ ω φ= = − + −
( )tan dd
t γω φω
− = −
thus the condition of relative maximum occurs every time that t increases by 2
d
πω
:
12
i id
t t πω+ = +
For the i th maximum: ( )cositi dx e A tγ
iω φ−= −
( )1
2
1 1cosi dti d i ix e A t e x
πγγ ωω φ+
−−
+ += − =
2
1
d dTi
i
x e ex
πγω γ
−
+
= =
3.10 (a) 132c sm
γ −= = 2 225k sm
ω −= =o
2 2 2 16d sω ω γ 2−= − =o 2 2 2 7r d sω ω γ 2−= − =
17r sω −∴ =
(b) max48 0.2
60.4d
FA m mCω
= = =o
(c) ( ) 22 2
2 2tan2 3
r r r
r
γω γω ωφγ γω ω
= = = =−o
7 41.4φ∴ ≈ o
3
3.11 (a) 2173 02
mx mx mxβ β+ + =&& &
32
γ β= and 2 2172
ω β=o
2 2 22 4r2ω ω γ β= − =o 2rω β∴ =
(b) max 2 d
FAmγω
= o 2 2 2 254d
2ω ω γ β= − =o 52dω β∴ =
2
215
Aβ
=
3.12 12
dTe γ− =
1 ln 2 ln 2dd
fT
γ = =
(a) 1
2 2 2( )dω ω γ= −o
So, 1
2 2 2( )dω ω γ= +o
1 12 22 2
2 ln 212 2d df f fγπ π
⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞= + = +⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢⎣ ⎦ ⎣
o
⎤⎥⎥⎦
100.6f Hz=o
(b) ( )1
2 2 2r dω ω γ= −
1 12 22 2
2 ln 212 2r d df f fγπ π
⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞= − = −⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢⎣ ⎦ ⎣
⎤⎥⎥⎦
99.4rf Hz= 3.13 Since the amplitude diminishes by dTe γ− in each complete period,
( ) 11d
nTe ee
γ− −= =
1dT nγ = 1
2d
dT n nωγπ
= =
Now ( )1
2 2 2dω ω γ= −o
So ( )1
1 22 2 22 2
114d d n
ω ω γ ωπ
⎛ ⎞= + = +⎜ ⎟⎝ ⎠
o
4
12
2 2
2112 4
d d
d
TT n
πω ωπ ω πω
⎛ ⎞= = = +⎜ ⎟⎝ ⎠
o
o
o
For large n, 2 2
118
dTT nπ
≈ +o
3.14 (a) ( )1
21 22 2 222 2 0.707
2rωω ω γ ω
⎡ ⎤⎛ ⎞= − = − =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
oo o ωo
(b) ( )
121
2 2 2114 0.866
2 2 22
dQωω γω
ωγ γ
⎛ ⎞−⎜ ⎟− ⎝ ⎠= = = =⎛ ⎞⎜ ⎟⎝ ⎠
oo
o
(c) ( )
2 2 2 2
2 22 22tan
4 3
ω ωγωφ
ω ω ω ω
⎛ ⎞⎜ ⎟⎝ ⎠= = =
− −
oo
o o o
−
1 2tan 146.33
φ − ⎛ ⎞= − =⎜ ⎟⎝ ⎠
o
(d) ( ) ( ) ( )1
2 222 2 24 4 4 3.6062
D ω 2ω ω ω ω ω⎡ ⎤⎛ ⎞= − + =⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
oo o o o
( ) ( ) 20.277
FFmA
D mω
ω ω= =
o
o
o
3.15 ( )( )
max1
2 2 2
AA γωω ω γ
≈⎡ ⎤− +⎣ ⎦o
for ( ) max12
A Aω = , ( )
12 2 2
12
γ
ω ω γ=⎡ ⎤− +⎣ ⎦o
( )2 2 24ω ω γ γ− + =o
3ω ω γ− = ±o 3ω ω γ= ±o
5
3.16 (b) 2 2
2 2dQ
ω γωγ γ
−= = o
2 1LC
ω =o , 2RL
γ =
2
2
2
14 1
422
RLC L LQ
R R CL
⎛ ⎞⎛ ⎞ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎛ ⎞ ⎛⎝ ⎠= = ⎜ ⎟ ⎜⎛ ⎞ ⎝ ⎠ ⎝
⎜ ⎟⎝ ⎠
⎞− ⎟⎠
(c) 2
LRCQ
R Rωγ
⎛ ⎞= = =⎜ ⎟⎝ ⎠
o o
3.17 sin Im i t
extF F t F e ωω ⎡ ⎤= = ⎣ ⎦o o
and ( )x t is the imaginary part of the solution to:
i tmx cx kx F e ω+ + = o&& &
i.e. ( ) ( ) ( )Im sini tx t Ae A tω φ ω φ−⎡ ⎤= =⎣ ⎦ −
where, as derived in the text,
( )1
2 22 2 2
FA
k m cω ω
=⎡ ⎤− +⎢ ⎥⎣ ⎦
o
and
2 2
2tan γωφω ω
=−o
3.18 Using the hint, , where ( )Re t
extF F eβ= o iβ α ω= − + , and x(t) is the real part of the solution to:
tmx cx kx F eβ+ + = o&& & .
Assuming a solution of the form: t ix Aeβ φ−=
( )2 iFm c k x xeA
φβ β ⎛ ⎞+ + = ⎜ ⎟⎝ ⎠
o
( )2 22 cFm im m c ic k iA
os sinα αω ω α ω φ φ− − − + + = +o
( )2 2 cosFm c kA
α ω α− − + = o φ
( )2 sFm cA
inω α φ− + = o
6
( )( )
12 2
2tan
c mm c
ω αφ
α ω α− −
=− − + k
Using , 2 2sin cos 1φ φ+ =
( ) ( )2 2 22 2 22 2F m c k c
Amα ω α ω α⎡ ⎤= − − + + −⎣ ⎦
o
( ) ( ) 1
2 222 2 2 2
FA
m c k c mα ω α ω α
=
⎡ ⎤− − + + −⎣ ⎦
o
and ( ) ( )costx t Ae tα ω φ−= − + the transient term.
3.19 (a)
12 2
2 18
l ATg
π−
⎛ ⎞≈ −⎜ ⎟
⎝ ⎠
for 4
A π= , 2 1.041lT
gπ≈ ×
(b) 2
2
4 1.084lgTπ
= ×
Using 2 lTg
π=o gives 2
24 lgTπ
=o
, approximately 8% too small.
(c) 3
232AB λω
= −o
and 2
6ωλ = o
2
192B AA=
for 4
A π= , 0.0032B
A=
3.20 ( ) in t
nn
f t c e ω=∑ . . . 0, 1, 2,n = ± ±
( ) cos sinn nn n
f t c n t c i n tω ω= +∑ ∑ , 0, 1, 2,n = ± ± . . .
and ( )2
2
1 Tin t
Tnc f t e dtT
ω−
−= ∫ 0, 1, 2,, n = ± ± . . .
( ) ( ) ( ) ( )2 2
2 2
1 cos sinT T
T Tnic f t n t dt f t n t
T Tω ω
− −= −∫ ∫ dt
The first term on is the same for and nc n n− ; the second term changes sign for vs. . The same holds true for the trigonometric terms in n n− ( )f t . Therefore, when terms that cancel in the summations are discarded:
7
( ) ( ) ( )2
2
1 cos cosT
Tn
f t c f t n t dt nT
tω ω−
⎛ ⎞= + ⎜ ⎟
⎝ ⎠∑ ∫o
( ) ( )2
2
1 sin sinT
Tn
f t n t dt nT
tω ω−
⎛ ⎞+ ⎜ ⎟
⎝ ⎠∑ ∫ ,
. . ., and 1, 2,n = ± ± ( )2
2
1 T
Tc f tT −
= ∫o dt
Now, due to the equality of terms in n± :
( ) ( ) ( )2
2
2 cos cosT
Tn
f t c f t n t dt n tT
ω ω−
⎛ ⎞= + ⎜ ⎟
⎝ ⎠∑ ∫o
( ) ( )2
2
2 sin sinT
Tn
f t n t dt nT
tω ω−
⎛ ⎞+ ⎜ ⎟
⎝ ⎠∑ ∫ ,
. . . 1,2,3,n = Equations 3.9.9 and 3.9.10 follow directly.
3.21 ( ) in tn
nf t c e ω=∑ , ( )2
2
1 Tin t
Tnc f t eT
ω−
−= ∫ dt , and 0, 1, 2,n = ± ± …
2T πω
= so ( )2
in tnc f t e
πωω
πω
ωπ
−
−= ∫ dt
( )0
02in t in te dt e dt
πω ωω
πω
ωπ
− −
−
⎡ ⎤= − +⎢ ⎥
⎣ ⎦∫ ∫
0
0
1 12
in t in te ein in
πωω ω
πω
ωπ ω ω
− −
−
⎡ ⎤⎢ ⎥= −⎢ ⎥⎣ ⎦
1 1 12
in ine ein
π π
π+ −⎡ ⎤= − − +⎣ ⎦
For n even, and the term in brackets is zero. 1in ine eπ π−= = For n odd, 1in ine eπ π−= = −
42nc
inπ= , . . . 1, 3,n = ± ±
( ) 42
in t
n
f t ein
ω
π=∑ , . . . 1, 3,n = ± ±
( )4 1 12
in t in t
n
e en i
ω ω
π−= −∑ , 1,3,5,n = . . .
( )4 1 sinn
n tn
ωπ
=∑ , 1,3,5,n = . . .
( ) 4 1 1sin sin 3 sin 53 5
f t t t tω ω ωπ⎡ ⎤= + + +⎢ ⎥⎣ ⎦
K
8
3.22 In steady state, ( ) ( )ni n t
nn
x t A e ω φ−=∑
( )1
2 22 2 2 2 2 24
n
n
FmA
n nω ω γ ω
=⎡ ⎤− +⎢ ⎥⎣ ⎦o
Now 4n
FFnπ
= o , . . . and 1,3,5,n = 3ω ω=o
1002
Q ωγ
= ≈ o so 2
2 940,000ωγ ≈
( )1 1
2 2 222 2
4 1
99 440000
FAmπ ω ωω ω
= ⋅⎡ ⎤⋅
− +⎢ ⎥⎣ ⎦
o
1 22FA
mπω≈ o
( )3 1
2 2222 2
4 13
99 9 4200
FAmπ
ωω ω
= ⋅⎡ ⎤⎛ ⎞
− +⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
o
3 2
40027
FAmπω
≈ o
( ) ( )5 1
2 22 22 2
4 15
39 25 4 5200
FAmπ
ωω ω ω
= ⋅⎡ ⎤⎛ ⎞− +⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
o
5 220FA
mπω≈ o
i.e., 1A : : = 1 : 29.6 : 0.1 3A 5A 3.23 (a) Thus 2 0x xω+ =o
&& y x= & 2y xω= − o& x y=&
divide these two equations: 2y dy x
x dx yω
= = − o&
&
(b) Solving 2 0ydy xdxω
+ =o
and Integrating 2 2
22 2y x Cω
+ =o
Let 22C A=
2 2
2 2 2 1y xA Aω
+ =o
an ellipse →
9
3.24 The equation of motion is ( ) 3F x x x mx= − = && . For simplicity, let m=1. Then
3x x x= −&& . This is equivalent to the two first order equations … x y=& and 3y x x= −& (a) The equilibrium points are defined by ( )( )3 1 1x x x x x− = − + = 0 Thus, the points are: (-1,0), (0,0) and (+1,0). We can tell whether or not
the points represent stable or unstable points of equilibrium by examining the phase space plots in the neighborhood of the equilibrium points. We’ll do this in part (c).
(b) The energy can be found by integrating 3dy y x x
dx x y−
= =&
& or
or ( )3y dy x x dx C= − +∫ ∫
2 2 4
2 2 4y x x C= − +
` In other words … 2 4 2
2 4 2y x xE T V C
⎡ ⎤= + = + − =⎢ ⎥
⎣ ⎦. The total energy C is
constant. (c) The phase space trajectories are given by solutions to the above equation
14 2
2 22xy x C
⎛ ⎞= ± − +⎜ ⎟
⎝ ⎠.
The upper right quadrant of the trajectories is shown in the figure below. The trajectories are symmetrically disposed about the x and y axes. They form closed paths for energies C<0 about the two points (-1,0) and (+1,0). Thus, these are points of stable equilibrium for small excursions away from these points. The trajectory passes thru the point (0,0) for C=0 and is a saddle point. Trajectories never pass thru the point (0,0) for positive energies C>0. Thus, (0,0) is a point of unstable equilibrium.
0 0.5 1 1.5 20
0.5
1
10
3.25 sin 0θ θ+ =&& ⇒2
cos 02
ddt
θ θ⎧ ⎫
− =⎨ ⎬⎩ ⎭
Integrating: 2
0
cos2
θθ
θ
θ θ=o
&&
or ( )2 2 cos cosθ θ θ= − o&
( )1
0 2
42 cos cos
dTθ θ
θ θ∴ =
−⎡ ⎤⎣ ⎦∫o
o
Time for pendulum to swing from 0θ = to θ θ= o is 4T
Now—substitute sin
2sinsin
2
θ
φ θ=o
so 2πφ = at θ θ= o
and use the identity 2cos 1 2sin2θθ = − 1
0 22 2
4
4 sin sin2 2
dTθ θ
θ θ∴ =
⎡ ⎤⎛ ⎞−⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
∫o
o
and after some algebra … 1 12 22 2 21 sin 4 sin sin
2 2 2
d dφ θ
θ θ θ=
⎡ ⎤ ⎡ ⎤⎛ ⎞− −⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦o
or
(a) 2
120 2
41 sin
dT
π
φ
α φ=
⎡ ⎤−⎣ ⎦∫ where 2sin
2θα = o
(b) ( )1
2 2 22 1 31 sin 1 sin sin2 8
α φ α φ α φ−
− ≈ + + K4 +
2
2 2 4
0
1 34 1 sin sin2 8
T d
π
φ α φ α φ⎡ ⎤= + + +⎢ ⎥⎣ ⎦∫ K 292 14 64
T απ α⎡ ⎤= + + +⎢ ⎥⎣ ⎦K
(c) 23 2
2sin2 2 48 4θ θ θ θα
⎡ ⎤= ≈ − +⎢ ⎥
⎣ ⎦o o o K o . . .
2
2 116
T θπ⎡ ⎤
= + +⎢ ⎥⎣ ⎦
o K
------------------------------------------------------------------------------------------------------
11
1
CHAPTER 4 GENERAL MOTION OF A PARTICLE IN THREE
DIMENSIONS --------------------------------------------------------------------------------------------------------- Note to instructors … there is a typo in equation 4.3.14. The range of the projectile is …
2
0 sin 2vR xg
α= = … NOT
2 20 sin 2... v
gα
---------------------------------------------------------------------------------------------------------
4.1 (a) ˆˆ ˆV V VF V i j kx y z
∂ ∂ ∂= −∇ = − − −
∂ ∂ ∂
v v
( )ˆˆ ˆF c iyz jxz kxy= − + +v
(b) ˆˆ ˆ2 2 2F V i x j y k zα β γ= −∇ = − − −v v
(c) ( ) ( )ˆˆ ˆx y zF V ce i j kα β γ α β γ− + += −∇ = + +v v
(d) 1 1ˆ ˆ ˆsinr
V V VF V e e er r rθ φθ θ φ
∂ ∂ ∂= −∇ = − − −
∂ ∂ ∂
v v
1ˆ nrF e cnr −= −
v
4.2 (a)
ˆˆ ˆ
0
i j k
Fx y zx y z
∂ ∂ ∂∇× = =
∂ ∂ ∂
v v conservative
(b)
( )2
ˆˆ ˆ
ˆ 1 1 0
i j k
F kx y zy x z
∂ ∂ ∂∇× = = − − ≠
∂ ∂ ∂−
v v SconservativeS
(c)
( )3
ˆˆ ˆ
ˆ 1 1 0
i j k
F kx y zy x z
∂ ∂ ∂∇× = = − =
∂ ∂ ∂
v v conservative
(d)
2
2
ˆ ˆ ˆ sin1 0sin
0 0
r
n
e e r e r
Fr r
kr
θ φ θ
θ θ φ−
∂ ∂ ∂∇× = =
∂ ∂ ∂−
v v conservative
4.3 (a)
( )2 3
ˆˆ ˆ
2
i j k
F k cx xx y z
xy cx z
∂ ∂ ∂∇× = = −
∂ ∂ ∂
v v
2 0cx x− = 12
c =
(b)
2
ˆˆ ˆi j k
Fx y zz cxz xy y y
∂ ∂ ∂∇× =
∂ ∂ ∂
v v
2 2 2 2
1 1 ˆˆ ˆx cx cz zi j ky y y y y y
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − + − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 2 0x cxy y
− − = 1c = −
also 2 2 0cz zy y
+ = implies that 1c = − as it must
4.4 (a) E = constant ( ) 21, ,2
V x y z mv= +
at the origin 2102
E mv= + o
at ( )1,1,1 2 21 12 2
E mv mvα β γ= + + + = o
3
( )2 2 2v vm
α β γ= − + +o
( )122 2v v
mα β γ⎡ ⎤= − + +⎢ ⎥⎣ ⎦
o
(b) ( )2 2 0vm
α β γ− + + =o
( )122v
mα β γ⎡ ⎤= + +⎢ ⎥⎣ ⎦
o
(c) xVmx Fx
∂= = −
∂&&
mx α= −&&
2Vmy yy
β∂= − = −
∂&&
23Vmz zz
γ∂= − = −
∂&&
4.5 (a) ˆ ˆF ix jy= +
r
on the path x y= : ˆ ˆdr idx jdy= +v
( )
( )1,1 1 1 1 1
0,0 0 0 0 01x yF dr F dx F dy xdx ydy⋅ = + = + =∫ ∫ ∫ ∫ ∫
v v
on the path along the x-axis: ˆdr idx=v and on the line 1x = : ˆdr jdy=v
( )
( )1,1 1 1
0,0 0 01x yF dr F dx F dy⋅ = + =∫ ∫ ∫
v v
Fv
is conservative.
(b) ˆ ˆF iy jx= −r
on the path x y= :
( )
( )1,1 1 1 1 1
0,0 0 0 0 0x yF dr F dx F dy ydx xdy⋅ = + = −∫ ∫ ∫ ∫ ∫v v
and, with x y= ( )
( )1,1 1 1
0,0 0 00F dr xdx ydy⋅ = − =∫ ∫ ∫
v v
on the x-axis: ( )
( )1,0 1 1
0,0 0 0xF dr F dx ydx⋅ = =∫ ∫ ∫v v
and, with 0y = on the x-axis ( )
( )1,0
0,00F dr⋅ =∫
v v
on the line 1x = : ( )
( )1,1 1 1
1,0 0 0yF dr F dy xdy⋅ = =∫ ∫ ∫v v
4
and, with 1x = ( )
( )1,1 1
1,0 01F dr dy⋅ = =∫ ∫
v v
on this path ( )
( )1,1
0,00 1 1F dr⋅ = + =∫
v v
Fv
is not conservative.
4.6 From Example 2.3.2, ( ) ( )2
e
e
rV z mgr z
= −+
( )1
1ee
zV z mgrr
−⎛ ⎞
= − +⎜ ⎟⎝ ⎠
From Appendix D, ( ) 1 21 1x x x−+ = − + +K
( )2
21ee e
z zV z mgrr r
⎛ ⎞= − − + +⎜ ⎟
⎝ ⎠K
( )2
ee
mgzV z mgr mgzr
= − + − +K
With emgr− an additive constant,
( ) 1e
zV z mgzr
⎛ ⎞≈ −⎜ ⎟
⎝ ⎠
( )ˆF V k V zz∂
= −∇ = −∂
v v
1ˆ 1e e
zkmg zr r
⎡ ⎤⎛ ⎞= − − + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
2ˆ 1e
zF kmgr
⎛ ⎞= − −⎜ ⎟
⎝ ⎠
v
0xmx F= =&& , 0ymy F= =&& 21e
zmz mgr
⎛ ⎞= − −⎜ ⎟
⎝ ⎠&&
21e
dz zmz mgdz r
⎛ ⎞= − −⎜ ⎟
⎝ ⎠
&&
0
0
21z
h
ve
zzdz g dzr
⎛ ⎞= − −⎜ ⎟
⎝ ⎠∫ ∫
o
&
2
212 z
e
hv g hr
⎛ ⎞− = − −⎜ ⎟
⎝ ⎠o
2
2 02e z
er vh r h
g− + =o
5
2
2 212 2e e z
er r vh r
g= − − o ( ), eh z r
221
2 2e e z
e
r r vhgr
= − − o
From Appendix D, ( )21
21 12 8x xx+ = + − +K
2 4
2 2 2 4e e z z
e
r r v vhg gr
= − + + +o o K
2 2
12 2
z z
e
v vhg gr⎛ ⎞
≈ +⎜ ⎟⎝ ⎠
o o
From Example 2.3.2, 2
2vhg
= o
12
12 e
vgr
−⎛ ⎞−⎜ ⎟
⎝ ⎠o
And with ( ) 11 1x x−− ≈ + , 2 2
12 2 e
v vhg gr⎛ ⎞
≈ +⎜ ⎟⎝ ⎠
o o
4.7
For a point on the rim measured from the center of the wheel: ˆ ˆcos sinr ib jbθ θ= −v
v ttb
θ ω= = o , so ˆ ˆsin cosr iv jvθ θ= − −o o
v&
Relative to the ground, ( )ˆ ˆ1 sin cosv iv jvθ θ= − −o o
v For a particle of mud leaving the rim:
siny b θ= −o and cosyv v θ= −o o So cosy yv v gt v gtθ= − = − −o o
and 21sin cos2
y b v t gtθ θ= − − −o
At maximum height, 0yv = :
cosvtgθ
= − o
2
cos 1 cossin cos2
v vh b v gg gθ θθ θ
⎛ ⎞ ⎛ ⎞= − − − − −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠o o
o
2 2cossin
2vh b
gθθ= − + o
Maximum h occurs for 22 cos sin0 cos
2dh vbd g
θ θθθ= = − − o
6
2sin gbv
θ = −o
4 2 2
2 24cos 1 sin v g b
vθ θ −= − = o
o
2 4 2 2 2 2
max 2 2 22 2 2gb v g b gb vhv gv v g
−= + = +o o
o o o
Measured from the ground,
2 2
max 22 2gb vh bv g
′ = + + o
o
The mud leaves the wheel at 12sin gb
vθ − ⎛ ⎞= −⎜ ⎟
⎝ ⎠o
4.8 cosx R φ= and ( )cosxx v t v tα= =o o
so coscos
Rtv
φα
=o
siny R φ= and ( )2 21 1sin2 2yy v t gt v t gtα= − = −o o
( )2
cos 1 cossin sincos 2 cos
R RR v gv v
φ φφ αα α
⎛ ⎞= − ⎜ ⎟
⎝ ⎠o
o o
2
2 2
cossin tan cos2 cosgRv
φφ α φα
= −o
( ) ( )2 2 2
2 2
2 cos 2 costan cos sin sin cos cos sincos cos
v vRg g
α αα φ φ α φ α φφ φ
= − = −o o
From Appendix B, ( )sin sin cos cos sinθ φ θ φ θ φ+ = +
( )2
2
2 cos sincos
vRg
α α φφ
= −o
R is a maximum for ( ) ( )2
2
20 sin sin cos coscos
dR vd g
α α φ α α φα φ= = − − + −⎡ ⎤⎣ ⎦
o
Implies that ( ) ( )cos cos sin sin 0α α φ α α φ− − − =
From appendix B, ( )cos cos cos sin sinθ φ θ φ θ φ+ = −
so ( )cos 2 0α φ− =
22πα φ− =
4 2π φα = +
2
max 2
2 cos sincos 4 2 4 2
vRg
π φ π φφ
⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
o
φ α
7
Now sin cos cos4 2 2 4 2 4 2π φ π π φ π φ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − − = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
2
2max 2
2 coscos 4 2
vRg
π φφ
⎛ ⎞= +⎜ ⎟⎝ ⎠
o
Again using Appendix B, 2 2 2cos 2 cos sin 2cos 1θ θ θ θ= − = −
2
max 2
2 1 1coscos 2 2 2
vRg
π φφ⎡ ⎤⎛ ⎞= + + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
o2
2 cos 1cos 2v
gπ φ
φ⎡ ⎤⎛ ⎞+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
o
Using cos sin2π θ θ⎛ ⎞+ = −⎜ ⎟⎝ ⎠
,
( ) ( )
2
max 21 sin
1 sinvR
gφ
φ= −
−o
( )
2
max 1 sinvR
g φ=
+o
4.9
(a) Here we note that the projectile is launched “downhill” towards the target, which is located a distance h below the cannon along a line at an angle φ below the horizon. α is the angle of projection that yields maximum range, RBmaxB. We can use the results from problem 4.8 for this problem. We simply have to replace the angle φ in the above result with the angle -φ, to account for the downhill slope. Thus, we get for the downhill range …
( )20
2
cos sin2cos
vRg
α α ϕϕ+
=
The maximum range and the angle is α are obtained from the problem above again by
replacing φ with the angle -φ … ( )20
max 2
1 sincos
vRg
ϕϕ
+= and … 2
2πα ϕ= − .
We can now calculate α … ( )( )
2 20 0
max 2
1 sinsin cos 1 sin
v vhRg g
ϕϕ ϕ ϕ
+= = =
−
Solving for sinϕ … 2 20 0
sin 1gh ghv v
ϕ⎛ ⎞
= +⎜ ⎟⎝ ⎠
But, from the above … 2sin sin 2 cos 2 1 2sin2πϕ α α α⎛ ⎞= − = = −⎜ ⎟⎝ ⎠
Thus … 22 2
0 0
1 2sin 1gh ghv v
α⎛ ⎞
− = +⎜ ⎟⎝ ⎠
h RBmaxB
α
φ
8
22 2 2
0 02
0
122sin 1 1csc 1
gh ghghv vv
αα
⎛ ⎞= = − + =⎜ ⎟
⎝ ⎠ +
Finally … 22
0
csc 2 1 ghv
α⎛ ⎞
= +⎜ ⎟⎝ ⎠
(b)
Solving for RBmaxB … max 2 2sin 1 2sin 1 2 csch h hRϕ α α
= = =− −
Substituting for 2csc α and solving …
2
0max 2
0
1v ghRg v⎛ ⎞
= +⎜ ⎟⎝ ⎠
4.10 We can again use the results of problem 4.8. The maximum slope range from problem 4.8 is given by …
( )
20
max 1 sin sinv hR
g ϕ ϕ= =
+
Solving for sinϕ …
2 20 0
sin 1gh ghv v
ϕ⎛ ⎞
= −⎜ ⎟⎝ ⎠
Thus …
max maxcoscossin
x R h ϕϕϕ
= =
We can calculate cosϕ from the above relation for sinϕ …
( )1
1 22 2
2 20 0
cos 1 sin 1 2 1gh ghv v
ϕ ϕ⎛ ⎞ ⎛ ⎞
= − = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Inserting the results for sinϕ and cosϕ into the above …
12 2
0max 2
0
cos 1 2sin
v ghx hg v
ϕϕ
⎛ ⎞= = −⎜ ⎟
⎝ ⎠
4.11 We can simplify this problems somewhat by noting that the trajectory is symmetric about a vertical line that passes through the highest point of the trajectory. Thus we have the following picture …
9
We have “reversed the trajectory so that hB0 B (= 9.8 ft), and xB0 B , the height and range within which Mickey can catch the ball represent the starting point of the trajectory. hB1B (=3.28 ft) is the height of the ball when Mickey strikes it at home plate. δ is the distance behind home plate where the ball would be hypothetically launched at some angle α to achieve the total range R. xB1B (=328 ft) is the distance the ball actually would travel from home plate if not caught by Mickey. (Note, because of the symmetry, vB0B is the speed of the ball when it strikes the ground … also at the same angle α at which was launched. We will calculate the value of xB0 B assuming a time-reversed trajectory!)
(1) The range of the ball … 2 2
0 0sin 2 2 sin cosv vRg g
α α α= =
(2) The maximum height … 2
max 2 20
tan2 2 cos 2R g Rz
vα
α⎛ ⎞= − ⎜ ⎟⎝ ⎠
(3) The height at xB1B … ( )21 1 12 2
0
tan2 cos
gh x xv
αα
= −
From (1) … 2 20
tan2 cos
gv R
αα= and inserting this into (2) gives …
max tan tan tan2 4 4R R Rz α α α= − =
Thus, max4tan
zRα
= and inserting this expression and the first previously derived into (3)
(4) ( )21
1 1max
tantan
4x
h xzα
α= −
Let 1 tanu x α= and we obtain the following quadratic … 2
max max 14 4 0u z u z h− + = and solving for u …
( )12
max 1 max2 1 1u z h z⎡ ⎤= ± −⎢ ⎥⎣ ⎦ and letting 1
max
hz
ε = , we get …
R
xB1B
δxB0B
hB0 B hB1 B
zBmaxB
z
x
vB0B
α
10
max 1u z hε≈ = or ( ) ( )max max max2 2 2 2 .0475 3.9u z z zε≈ − = − = . This result is the correct one …
Thus, 0max
1
3.9tan 0.821 39.4zx
α α= = ∴ =
Now solve for xB0 B using a relation identical to (4) …
( )20
0 0max
tantan
4x
h xzα
α= −
Again we obtain a quadratic expression for 0 tanu x α= which we solve as before. This time, though, the first result for u is the correct one to use … max 0u z hε= ≈ and we obtain …
00 11.9
tanhx ftα
= =
4.12 The x and z positions of the ball vs. time are
1cos2
x v t θ= o o 21 1cos sin2 2
z v t gtθ θ= −o o o
Since 1cos2xv v θ= o o
The horizontal range is 2
2 1cos sin 22
vRg
θ θ= oo o
The maximum range occurs @ 0dRdθ
=o
2
2 1 1 12cos cos 2 cos sin sin 2 02 2 2
vdRd g
θ θ θ θ θθ
⎛ ⎞= − =⎜ ⎟⎝ ⎠
oo o o o o
o
Thus, 2 1 1 12cos cos 2 cos sin sin 22 2 2θ θ θ θ θ=o o o o o
Using the identities: 22cos 1 cos 2θ θ= +o o and sin 2 2sin cosθ θ θ=o o o We get: ( )( ) ( )2 21 cos 2cos 1 sin sin cos 1 cos cosθ θ θ θ θ θ θ+ − = = −o o o o o o o
or ( )( )21 cos 3cos cos 1 0θ θ θ+ − − =o o o
Thus cos 1θ = −o , ( )1cos 1 136
θ = ±o
Only the positive root applies for the θo -range: 02πθ≤ ≤o
( )1cos 1 13 0.76766
θ = + =o 39 51θ ′= oo
Thus (b) for 125v m s−=o max 55.4 @ 39 51R m θ ′= = oo
11
(a) The maximum height occurs at 0dzdt
=
1cos sin2
v gTθ θ =o o o or at
1cos sin2
vT
g
θ θ=
o o o
or 2
2 21cos sin2 2vH
gθ θ= oo o maximum at fixed θo
The maximum UpossibleU height occurs @ 0dHdα
=
2
2 21 1 12cos sin cos cos sin sin 02 2 2 2vdH
d gθ θ θ θ θ θ
α⎛ ⎞= − =⎜ ⎟⎝ ⎠
oo o o o o o
Using the above trigonometric identities, we get
( ) ( )2 21 11 cos sin cos sin sin sin 1 cos2 2
θ θ θ θ θ θ θ+ = = −o o o o o o o
or ( )( )sin 1 cos 3cos 1 0θ θ θ+ − =o o o
There are 3-roots: sin 0θ =o , cos 1θ = −o , 1cos3
θ =o
The first two roots give minimum heights; the last gives the maximum
Thus, 1max
118.9 @ cos 70 323
H m θ − ′= = = oo
4.13 The trajectory of the shell is given by Eq. 4.3.11 with r replacing x
222
z gz r rr r
= −o
o o
&
& & where cosr v θ=o o o
& sinz v θ=o o o&
Thus, 2
22tan sec
2g rz r
vθ θ= −o o
o
Since 2 2sec 1 tanθ θ= +o o We have:
2 2
22 2tan tan 0
2 2g r g rr z
v vθ θ− + + =o o
o o
(r,z) are target coordinates. The above equation yields two possible roots:
( )1
2 4 2 2 2 21tan 2v v gzv g rgr
θ⎡ ⎤
= ± − −⎢ ⎥⎣ ⎦
o o o o
The roots are only real if 4 2 2 22 0v gzv g r− − ≥o o The UcriticalU surface is therefore: 4 2 2 22 0v gzv g r− − =o o 4.14 If the velocity vector, of magnitude s& , makes an angle θ with the z-axis, and its
12
projection on the xy-plane make an angle φ with the x-axis: sin cosx s θ φ=& & , and sin cosx rF F mxθ φ= = && sin siny s θ φ=& & , and sin siny rF F myθ φ= = && cosz s θ= && , and cosz rF mg F mzθ= − + = && Since ( )2 2 2 2
2 2rF c s c x y z= − = − + +& & & & , the differential equations of motion are not separable. 2
2 2sin cosmx c s c sxθ φ= − = −&& & &&
2dx dx ds dxm m ms c sxdt ds dt ds
= ⋅ = = −& & &
& &&
2dx c ds dsx m
γ= − = −&
&, where 2c
mγ =
ln ln ln xx x sx
γ− = = −o
o
&& &
&
sx x e γ−= o& &
Similarly sy y e γ−= o& &
4.15 From eqn 4.3.16, max max2 2 ln 1 0x xz g g
x xγ γ
γ γ γ⎛ ⎞⎛ ⎞
+ + − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
o
o o
&
& &
From Appendix D: ( )2 3
ln 12 3u uu u− = − − − −K for 1u <
2 2 3 3
max max max max2 3ln 1
2 3x x x xx x x x
γ γ γ γ⎛ ⎞− = − − − +⎜ ⎟
⎝ ⎠o o o o& & & &
terms in 4γ
2 3
max max max max max2 32 3
z x gx gx gx g xx x x x x
γγ γ
+ − − − +o
o o o o o
&
& & & & & terms in 2 0γ =
2
2max max
3 3 02x x zx x
gγ γ+ − ≈o o o
& & &
12 2 2
max 2
3 9 34 16x x x zx
gγ γ γ⎛ ⎞
≈ − ± +⎜ ⎟⎝ ⎠
o o o o& & & &
12
max3 3 1614 4 3x x zx
gγ
γ γ⎛ ⎞
≈ − ± +⎜ ⎟⎝ ⎠
o o o& & &
Since max 0x > , the + sign is used. From Appendix D:
1 2216 8 1611 1
3 3 8 3z z z
g g gγ γ γ⎛ ⎞ ⎛ ⎞
+ = + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
o o o& & &
terms in 3γ
2
max 2
3 3 2 84 4 3x x x z x zx
g gγ
γ γ= − + + − +o o o o o o
& & & && & terms in 2γ
θ
φ
y
x
z
sr&
13
2
max 2
2 83
x z x zxg g
γ= − +o o o o& && &
K
For sinz v α=o o& and 22 sin 2x z v α=o o o& & :
2 3
max 2
sin 2 4 sin 2 sin3
v vxg g
α α α γ= − +o o K
4.16
( )cosx A tω α= + , ( )sinx A tω ω α= − +& from 0x =o& , 0α = from x A=o , cosx A tω=
( )cosy B tω β= + , ( )siny B tω ω β= − +&
2 2 21 1 12 2 2
kB ky my= +o o&
with 4y A=o , 3y Aω=o& and km
ω = :
( )2 2 2 2 22
116 9 25B A A Aωω
= + =
5B A= Then 4 5 cosA A β= and 3 5 sinA Aω ω β= −
1 14 3cos sin 36.95 5
β − −⎛ ⎞ ⎛ ⎞= = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
o
( )5 cos 36.9y A tω= − o Since maximum x and y displacements are A± and 5A± , respectively, the motion takes place entirely within a rectangle of dimension 2A and 10A . 36.9 0 36.9β αΔ = − = − − = −o o
From eqn 4.4.15, 2 2
2 costan 2 ABA B
ψ Δ=
−
( )( ) ( )
( )22
4102 5 cos 36.9 15tan 224 35
A A
A Aψ
⎛ ⎞⎜ ⎟− ⎝ ⎠= = = −−−
o
11 1tan 9.22 3
ψ − ⎛ ⎞= − = −⎜ ⎟⎝ ⎠
o
4.17 2x
Vmx F kx mxx
π∂= = − = − = −
∂&&
( )cos coskx A t A tm
α π α⎛ ⎞
= + = +⎜ ⎟⎜ ⎟⎝ ⎠
&
y
x
14
24Vmy mxy
π∂= − = −
∂&&
( )cos 2y B tπ β= +
29Vmz mzz
π∂= − = −
∂&&
( )cos 3z C tπ γ= +
Since 0x y z= = = at 0t = , 2πα β γ= = = −
cos sin2
x A t A tππ π⎛ ⎞= − =⎜ ⎟⎝ ⎠
cosx A tπ π=& Since 2 2 2 2v x y z= + +o o o o
& & & and x y z= =o o o& & & ,
3
vx Aπ= =oo&
3
vAπ
= o
sin3
vx tππ
= o
sin 2y B tπ= , 2 cos 2y B tπ π=&
23
vy Bπ= =oo&
2 3
vBπ
= o
sin 22 3
vy tππ
= o
sin 3z C tπ= , 3 cos3z C tπ π=&
33
vz Cπ= =oo&
3 3
vCπ
= o
sin 33 3
vz tππ
= o
Since xω π= , 2yω π= , and 3zω π= the ball does retrace its path.
31 2min
22 2
x y z
nn nt ππ πω ω ω
= = =
The minimum time occurs at 1 1n = , 2 2n = , 3 3n = .
15
min2 2t ππ
= =
4.18 Equation 4.4.15 is 2 2
2 costan 2 ABA B
ψ Δ=
−
Transforming the coordinate axes xyz to the new axes x y z′ ′ ′ by a rotation about
the z-axis through an angleψ given, from Section 1.8: cos sinx x yψ ψ′ = + , sin cosy x yψ ψ′ = − + or, cos sinx x yψ ψ′ ′= − , and sin cosy x yψ ψ′ ′= +
From eqn. 4.4.10: 2 2
22 2
2cos sinx yxyA AB B
Δ− + = Δ
Substituting:
( )2 2 2 22
1 cos 2 cos sin sinx x y yA
ψ ψ ψ ψ′ ′ ′ ′− +
( )2 2 2 22cos cos sin cos sin cos sinx x y yAB
ψ ψ ψ ψ ψ ψΔ ⎡ ⎤′ ′ ′ ′− + − −⎣ ⎦
( )2 2 2 2 22
1 sin 2 cos sin cos sinx x y yB
ψ ψ ψ ψ′ ′ ′ ′+ + + = Δ
For x′ to be a major or minor axis of the ellipse, the coefficient of x y′ ′ must vanish.
( )2 22 2
2cos sin 2cos 2cos sincos sin 0A AB Bψ ψ ψ ψψ ψΔ
− − − + =
From Appendix B, 2cos sin sin 2ψ ψ ψ= and 2 2cos sin cos 2ψ ψ ψ− =
2 2
sin 2 2cos cos 2 sin 2 0A AB Bψ ψ ψΔ
− − + =
2 2
1 1 2costan 2B A AB
ψ Δ⎛ ⎞− =⎜ ⎟⎝ ⎠
2 2
2 costan 2 ABA B
ψ Δ=
−
4.19 Shown below is a face-centered cubic lattice. Each atom in the lattice is centered
within a cube on whose 6 faces lies another adjacent atom. Thus each atom is surrounded by 6 nearest neighbors at a distance d. We neglect the influence of atoms that lie at further distances. Thus, the potential energy of the central atom can be approximated as
6
1i
i
V cr α−
=
=∑
2d
16
( )1
2 2 2 21r d x y z⎡ ⎤= − + +⎣ ⎦
( )2 2 2 2
2 2 2 2 21 2
22 1 x x y zr d dx x y z dd d
αα
α α−
−− − ⎛ ⎞+ += − + + + = − +⎜ ⎟
⎝ ⎠
From Appendix D, ( ) ( ) 211 1 12
nx nx n n x+ = + + − +K
2 2 2
1 2
2 11 12 2 2 2
x x y zr dd d
α α α α α− − ⎧ ⎛ ⎞+ +⎪ ⎛ ⎞⎛ ⎞= − − + + − − −⎨ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎪ ⎝ ⎠⎩
22 2 2 2 2 2 2
2 2
2 22x x x y z x y zd d d d
⎡ ⎤⎛ ⎞ ⎛ ⎞+ + + +⎛ ⎞ ⎛ ⎞− − +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
+ terms in 3
3
xd
⎫⎬⎭
( )2
2 2 21 2 2
41 12 4 2
x xr d x y zd d d
α α α α α α− − ⎡⎧ ⎛ ⎞= + − + + + + +⎨ ⎜ ⎟ ⎢⎩ ⎝ ⎠ ⎣
terms in 3
3
xd
⎫⎤⎪⎬⎥⎪⎦⎭
( )2 2 2 21 2 21 1
2 2xr d x y z x
d d dα α α α α α− − ⎡ ⎤⎛ ⎞≈ + − + + + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
( )1 1
2 2 2 2 2 2 22 22 2r d x y z d dx x y z⎡ ⎤ ⎡ ⎤= − − + + = + + + +⎣ ⎦⎣ ⎦
2 2 2 2
2 2
21 x x y zr dd d
α
α α−
− − ⎛ ⎞+ += + +⎜ ⎟
⎝ ⎠
( )2 2 2 22 2 21 1
2 2xr d x y z x
d d dα α α α α α− − ⎡ ⎤⎛ ⎞≈ − − + + + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
( ) ( )2 2 2 21 2 2 22 2r r d x y z x
d dα α α α α α− − − ⎡ ⎤+ ≈ − + + + +⎢ ⎥⎣ ⎦
Similarly:
( ) ( )2 2 2 23 4 2 22 2r r d x y z y
d dα α α α α α− − − ⎡ ⎤+ ≈ − + + + +⎢ ⎥⎣ ⎦
( ) ( )2 2 2 25 6 2 22 2r r d x y z z
d dα α α α α α− − − ⎡ ⎤+ ≈ − + + + +⎢ ⎥⎣ ⎦
( ) ( )2
2 2 2 2 2 22 2 2
3 26V cd x y z x y zd d d
α α α α− ⎡ ⎤⎛ ⎞≈ − + + + + + +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
( )( )2 2 2 2 26cd cd x y zα α α α− − −≈ + − + + ( )2 2 2V A B x y z≈ + + + 4.20
17
( )F q E v B= + ×
r r rr
( )ˆ ˆˆ ˆ ˆ ˆv B ix jy kz kB iyB jxB× = + + × = −rr
& & & &&
( )ˆ ˆF iqyB jq E xB= + −r
& & xmx F qyB= =&& &
qBx x ym
− =o& &
yqBmy F qE qxB qE qB x ym
⎛ ⎞= = − = − +⎜ ⎟⎝ ⎠
o&& & &
2 2qE qBx qB eE eBx eBy y y
m m m m m m⎛ ⎞ ⎛ ⎞= − − = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
o o& &
&&
2 eEy y xm
ω ω+ = − + o&& & , eB
mω =
( )2
1 coseEy x A tm
ω ω θω
⎛ ⎞= − + + +⎜ ⎟⎝ ⎠
o o
( )siny A tω ω θ= − + o&
0y =o& , so 0θ =o
0y =o , so 2
1 eEA xm
ωω
⎛ ⎞= − − +⎜ ⎟⎝ ⎠
o&
( )1 cosy a tω= − , 2
1 eEa xm
ωω
⎛ ⎞= − +⎜ ⎟⎝ ⎠
o&
( )1 cosqBx x y x y x a tm
ω ω ω= + = − = − −o o o& & & &
( ) cosx x a a tω ω ω= − +o& &
( ) sinx x a t a tω ω= − +o&
sinx a t btω= + , b x aω= −o& 0zmz F= =&& 0z z t z= + =o o& 4.21
212 2
bmv mqh mg+ =
( )2 2v g b h= −
2
cosrmvF mg R
bθ= − = − +
x
z
0iv
jE
kB
y
b
h
y
x
18
cos hb
θ =
( ) ( )2
2 3h mv mg mgR mg h b h h bb b b b
= − = − − = −⎡ ⎤⎣ ⎦
the particle leaves the side of the sphere when 0R =
3bh = , i.e.,
3b above the central plane
4.22 21 02
mv mgh+ =
at the bottom of the loop, h b= −
so 212
mv mgb= ,
2v gb=
2
rmvF mg R
b= − + =
2
2 3mvR mg mg mg mgb
= + = + =
4.23 From the equation for the energy as a function of s in Example 4.6.2,
2 21 12 2 4
mgE ms sA
⎛ ⎞= + ⎜ ⎟⎝ ⎠
& ,
s is undergoing harmonic motion with:
" " 14 2
k g gm A A
ω = = =
Since 4 sins A φ= , φ increases by 2π radians during the time interval:
2 2 2 ATg
π πω
⎛ ⎞′ = = ⎜ ⎟⎜ ⎟
⎝ ⎠
For cycloidal motion, x and z are functions of 2φ so they undergo a complete cycle every time φ changes by π . Therefore, the period for the cycloidal motion is one-half the period for s .
1 22
AT Tg
π′= =
------------------------------------------------------------------------------------------------------
v
h b
CHAPTER 5 NONINERTIAL REFERENCE SYSTEMS
5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net force acting on him is zero. The scale exerts an upward force, N
r, whose value is equal to the
scale reading --- the “weight,” W’, of the observer in the accelerated frame. Thus
0 0N mg mA+ − =rr r
0Ar
(a) (b)
Nr
mgr
05 0
4 4gN mg mA N mg m N mg− − = − − = − =
5 54 4
W N mg W′ = = =
150 .W lb′ = (b) The acceleration is downward, in the same direction as gr
04gN mg m⎛ ⎞− + =⎜ ⎟
⎝ ⎠ 3
4 4WW W W′ = − =
b
r
90 .W l′ = 5.2 (a) ( )centF mω ω ′= − × ×
r r r r
For rω ′⊥r r , 2 ˆcent rF m r eω ′′=
r
1 1500 1000s sω π− −= =
( )26 2ˆ10 1000 5 5cent rF eπ π−= × × =r
dynes outward
(b) ( )2241000 5
5.04 10980
cent
g
F m rF mg
πω ′= = = ×
5.3 (See Figure 5.1.2) 0mg T mA+ − =o
rrr
ˆ ˆ ˆ ˆcos sin 010gmg j T j T i m iθ θ ⎛ ⎞− + + − =⎜ ⎟
⎝ ⎠
cosT mgθ = , and sin10mgT θ =
1tan10
θ = , 5.71θ = o
1.005cosmgT mθ
= = g
5.4 The non-inertial observer thinks that g′r points downward in the direction of the hanging plumb bob… Thus
1
ˆ ˆ10gg g A g j′ = − = −o
rr r i
For small oscillations of a simple pendulum:
12T g
π=′
2
2 1.00510gg g ⎛ ⎞′ = + =⎜ ⎟
⎝ ⎠g
1 12 1.9951.005
T g g
π π= =
5.5 (a) f mgμ= − is the frictional force acting on the
box, so
0f mA ma′− =r r
0Ar
fr
r ( a′r is the acceleration of the box relative to the truck. See
Equation 5.1.4b.) Now, fr
the only real force acting horizontally, so the acceration relative to the road is
(b) 3
f mg ga gm m
μ μ= = − = − = −
(For + in the direction of the moving truck, the – indicates that friction opposes the forward sliding of the box.)
2gA = −o (The truck is decelerating.)
from above, 0ma mA ma′− = so
(a) 3 2 6g g ga a A′ = − = − + =o
5.6 (a) ( )ˆ ˆcos sinr i x R t jR t= + Ω + Ωo
v
ˆ ˆsin cosr i R t j R= − Ω Ω + Ω Ωv& t
2
2 2r r v R⋅ = = Ω
v v& & v R∴ =Ω circular motion of radius R
(b) r r where rω′ ′= − ×
v v v v& & ˆ ˆr ix jy′ ′= +v ′
( )ˆˆ ˆ ˆsin cosi R t j R t k ix jyω ˆ′ ′= − Ω Ω + Ω Ω − × +
2
ˆ ˆ ˆsin cosi R t j R t j x i yˆω ω′ ′= − Ω Ω + Ω Ω − + sinx y R tω′ ′= −Ω Ω&
cosy x R tω′ ′= − +Ω Ω& (c) Let here u x iy′ ′= + ′ 1i = − !
sin cosu x iy y R t i x i R tω ω′ ′ ′ ′ ′= + = −Ω Ω − + Ω Ω& & & i uω ′ = yω ′= − i xω ′+
sin cos Rei tu i u R t i R t iω Ω′ ′∴ + = −Ω Ω + Ω Ω = Ω& Try a solution of the form
i t i tu Ae Beω− Ω′ = + i t i tu i Ae i Beωω − Ω′ = − + Ω&
i t i ti u i Ae i Beωω ω ω− Ω′ = +
( ) i tu i u i Be ωω ω′ ′∴ + = +Ω& so RB Ω=ω +Ω
=
Also at t the coordinate systems coincide so 0 ( ) ( )0 0u A B x iy x R′ ′ ′= + = + = +o
RA x R B x RωΩ
∴ = + − = + −+Ωo o so, RA x ω
ω= +
+Ωo
Thus, i t i tR Ru x e eωωω ω
− ΩΩ⎡ ⎤′ = + +⎢ ⎥+Ω +Ω⎣ ⎦o
5.7 The x, y frame of reference is attached to the Earth, but the x-axis always points away from the Sun. Thus, it rotates once every year relative to the fixed stars. The X,Y frame of reference is fixed inertial frame attached to the Sun.
(a) In the x, y rotating frame of reference ( ) ( )cosx t R t Rεω= Ω− −
( ) ( )siny t R tω= − Ω− where R is the radius of the asteroid’s orbit and RE is the radius of the Earth’s orbit. Ω is the angular frequency of the Earth’s revolution about the Sun and ω is the angular frequency of the asteroid’s orbit.
(b) ( ) ( ) ( )sin 0x t R tω ω= − Ω− Ω− →& at 0t =
( ) ( ) ( ) ( )cosy t R t Rω ω= − Ω− Ω− Ω−→ −& ω at t 0= (c) 2a A A r r rε= − −Ω× − Ω× −Ω×Ω×
vv v v v v vv v& &v
3
Where a is the acceleration of the asteroid in the x, y frame of reference, v
,A Aε
v v are the accelerations of the asteroid and the Earth in the fixed, inertial frame
of reference. 1st : examine: A A rε− −Ω×Ω×
v v v v v
( )R R R Rε εω ω= × × −Ω×Ω× −Ω×Ω× −v v v v v v v vv v
= ( ) ( )2 2R Rω ω ω× −Ω×Ω × = − −Ωv v v vv v note: kω ω=v , kΩ = Ω
v
Thus: a R ( )2 2 2 vω= Ω − − Ω×
v vv v
⎡ ⎤+ = Ω − Ω− − Ω−⎣ ⎦&& &&
Therefore: ( ) ( ) ( ) ( )2 2ˆ ˆ ˆ ˆcos sinix jy iR t jR tω ω ω ˆ ˆ2 2j x i y− Ω + Ω& &
Thus: ( ) ( )2 2 cos 2x R tω ω= Ω − Ω− + Ω&& &y
x
( ) ( )2 2 sin 2y R tω ω= − Ω − Ω− − Ω&& &
Let ( )x yω= Ω−&& & and ( )y xω= Ω−&& & Then, we have
( ) ( ) ( )2cosy R tω ωω
yΩ= Ω− Ω− +
Ω−& & which reduces to
( ) ( )cosy R tω ω= − Ω− Ω−& Integrating … ( )sin 0y R tω= − Ω− → at t 0=
Also, ( ) ( ) ( )2 2 sin 2x R tω ω ω x− = − Ω − Ω− − Ω& &− Ω
or ( ) ( )sin 2x R tω ω= Ω+ Ω− + Ω& &x
)
( ) (sinx R tω ω= − Ω− Ω−&
Integrating … ( )cosx R tω= Ω− + const
( )cosx R t R R Rε εω− − → − 0t= Ω at =
5.8 Relative to a reference frame fixed to the turntable the cockroach travels at a
constant speed v’ in a circle. Thus 2
ˆrva eb ′
′′ = −r .
Since the center of the turntable is fixed. 0A =o
r
4
b
ωr
x′
y′
The angular velocity, ω, of the turntable is constant, so kω ω ′=
r , with 0ω =r&
, so ˆrr be ′′ =r ( ) 2 ˆrr bω ω ω e ′′× × = −
r r r ˆv v eθ ′′ ′=
r , so ˆrv v eω ω ′′ ′× = −r r
From eqn 5.2.14, 2 (a a v r )ω ω ω′ ′= + × + × ×r r r r r r r′ and putting in terms from above
222r
va vb
bω ω′
′′= − − −
For no slipping sF mμ≤r
g , so sa gμ≤r
2
22 sv v b gb
ω ω μ′
′+ + ≤
2 2 22 0m m sv bv b b gω ω μ′ ′+ + − =
2 2 2 2m sv b b b b gω ω ω μ′ = − ± − +
Since v was defined positive, the +square root is used. ′ m sv b b gω μ′ = − + (b) ˆv v eθ ′′ ′= −
r ˆrv v eω ω ′′ ′× = +
r r
2
22rva vb
bω ω′
′′= − + −
2
22 sv v b gb
ω ω μ′
′− + ≤
m sv b b gω μ′ = +
5.9 As in Example 5.2.2, ˆV jωρ
′= or and 2
ˆVA iρ
′= oo
r
For the point at the front of the wheel:
2
ˆVr jb
′ ′= or&& and ˆv V k′ ′= − o
r
0ω =r&
( )ˆ ˆ ˆV Vr k bjω b iρ ρ
′ ′ ′× = × − =o or r ′
( )2
ˆ ˆ ˆV V b V br k iω ωρ ρ ρ
′ ′ ′× × = × =o o or r r j′
( )ˆ ˆ 0Vv k V kωρ
′ ′ ′× = × − =oo
r r
( )2 2 2
2ˆ ˆV V V ba r r A i j
bω ω
ρ ρ⎛ ⎞
′ ′ ′= + × × + = + +⎜ ⎟⎝ ⎠
o o oo
rr r r r r&& ′
5
5.10 (See Example 5.3.3)
2m x mxω ′ ′= && ( ) t tx t Ae Beω ω−′ = +
( ) t tx t Ae Beω ωω ω −′ = −& Boundary Conditions:
( )2 02lx A B= = +
( ) ( )0 0x A Bω′ = = −&
4lA∴ =
4lB =
(a) ( ) cosh2lx t tω′ = ( ) sinh
2lx t tω ω′ =&
(b) ( ) cosh2 2 2l l lx T Tω′ = + = when the bead reaches the end of the rod
cosh 2Tω∴ = or 11 1.317cosh 2Tω ω
−= =
(c) ( ) sinh2lx T Tω ω′ =&
( )1sinh cosh 2 1.732 0.8662 2l l lω ω ω−⎡ ⎤= = =⎣ ⎦
or 1
2 2cosh 1 3 0.8662 2l lT lωω ω ω⎡ ⎤− = =⎣ ⎦
5.11 mphˆ400v j′ ′=
r ˆ586.67 j′= ft ⋅ s 1−
( )5 ˆˆ7.27 10 cos 41 sin 41j kω − ′ ′= × +o or s 1−
( )( )( )5 ˆ7.27 10 586.67 sin 41v iω −′ ′× = − × or r ft ⋅ s 2−
2cor
grav
m vFF mg
ω ′− ×=
r r
( )( )( )52 7.27 10 586.67 sin 41
32
−×=
o
0.0017cor
grav
FF
=
The Coriolis force is in the vω ′− ×r r direction, i.e., i′+ or east.
5.12 (See Figure 5.4.3)
ˆˆy zj kω ω ω′ ′′ ′= +
r
6
ˆ ˆx yv v i v j′ ′′ ′= +
r ′
k ˆˆ ˆz y z x y xv v i v j vω ω ω ω′ ′ ′ ′ ′ ′′ ′ ′× = − + −
r r ′
v j′
( ) ˆ ˆz y z xhoriz
v v iω ω ω′ ′ ′ ′′ ′ ′× = − +r r
( ) ( ) ( )1 1
2 2 2 2 2 22 2z y z x z x y zhoriz
v v v v vω ω ω ω′ ′ ′ ′ ′ ′ ′ ′′ ′× = + = + =r r vω
2corF m vω ′= − ×r r r
( ) ( )2cor zhorizhoriz2F m v m vω ω′= × =
r r r ′ ′ , independent of the direction of v . ′r
5.13 From Example 5.4.1 …
13 21 8 cos
3hhxg
ω λ⎛ ⎞
′ = ⎜ ⎟⎝ ⎠
and 0hy′ = .
( )1
3 3 25 1
2
1 8 12507.27 10 cos 413 32h
ftx sft s
− −−
⎛ ⎞×′ = × ⎜ ⎟⋅⎝ ⎠o
0.404hx ft′ = to the east. 5.14 From Example 5.4.2:
2
sinHv
ω λΔ ≈o
is the deflection of the baseball towards the south since it
was struck due East at Yankee Stadium at latitude (problem 5.13). v41 Nλ = o0 is the
initial speed of the baseball whose range is H. From eqn 4.3.18b, without air resistance in an inertial reference frame, the horizontal range is …
2 sin 2vH
gα
= o
Solving for v0 … 12
sin 2gHv
α⎛ ⎞= ⎜ ⎟⎝ ⎠
o
12 2
132 200 113sin 30
ft s ftv f−
t s−⎛ ⎞⋅ ×
= =⎜ ⎟⎝ ⎠
o o⋅
( )( )5 1 2 2
1
7.27 10 200sin 41 0.0169 0.2
113s ft
ft ift s
− −
−
×n∴ Δ ≈ = =
⋅o
A deflection of 0.2 inches should not cause the outfielder any difficulty. 5.15 Equation 5.2.10 gives the relationship between the time derivative of any vector in a fixed and rotating frame of reference. Thus …
fixed rot
da dar adt dt
ω⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
r rr r&&& ×
r
7
( )2a r r r rω ω ω ω′ ′ ′= + × + × + × ×r r r r r r r r r&& & & ′
2 2rot
da r r r rdt
ω ω ω ω⎛ ⎞ ′ ′ ′ ′= + × + × + × + ×⎜ ⎟⎝ ⎠
rr r r r r r r r&&& && & & & & &&r′r
( ) ( ) ( )r rω ω ω ω ω ω′ ′ ′+ × × + × × + × ×r r r r r r r r r& & r&
( ) ( ) ( )2a r r r rω ω ω ω ω ω ω ω ω′ ′ ′× = × + × × + × × + × × × ′⎡ ⎤⎣ ⎦r r r r r r r r r r r r r r&& & &
Now ( ) is ⊥ to rω ′×r r ωr and r′r . Let this define a direction : n
ˆr rω ω′ ′× = ×r r nr r
Since nω ⊥r , is in the plane defined by ( rω ω ′× ×
r r ) ωr and r′r and ˆ( )r n rω ω ω ω ω ω′ ′× × = × × = ×
r r r r r r r rr′ .
Since … ( )rω ω ω ′⊥ × ×r r r r
( ) 2r rω ω ω ω ω′ ′× × × = ×⎡ ⎤⎣ ⎦r r r r r r
And is in the direction ( rω ω ω ′× × ×⎡⎣r r r r )⎤⎦ n−
Thus ( ) ( )2r rω ω ω ω ω′ ′× × × = − ×⎡ ⎤⎣ ⎦r r r r r r
( ) ( ) ( )22a r r r rω ω ω ω ω ω ω ω′ ′ ′× = × + × × + × × − ×r r r r r r r r r r r&& & & ′
( )3 3r r r r rω ω ω ω ω′ ′ ′ ′= + × + × + × + × ×r r r r r r r r r r r&&& &&& && & & && & r′
( ) ( ) ( )22 3r rω ω ω ω ω ω′ ′+ × × + × × − ×r r r r r r r r& & r′
5.16 With , and 0x y z x y′ ′ ′ ′ ′= = = = =o o o o o
& & z v′ ′=o& o Equations 5.4.15a – 5.4.15c become:
( ) 3 21 cos cos3
x t gt t vω λ ω λ′ ′= − o
( ) 0y t′ =
( ) 212
z t gt v t′ ′= − + o
When the bullet hits the ground ( ) 0z t′ = , so
2vtg′
= o
3 2
3 2
8 41 cos cos3
v vx g vg g
ω λ ω λ′ ′⎛ ⎞ ⎛ ⎞
′ ′= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
o oo
3
2
4 cos3
vxgω λ
′′ = − o
x′ is negative and therefore is the distance the bullet lands to the west. 5.17 With and 0x y z′ ′ ′= = =o o o cosx v α′ =o o
& 0y′ =o& sinz v′ = αo o& we can solve equation 5.4.15c to find the time it takes the projectile
8
to strike the ground …
( ) 2 21 sin cos cos 02
z t gt v t v tα ω α λ′ ′= − + + =o o
or 2 sin 2 sin2 cos cos
v vtg v g
α αω α λ
′ ′= ≈
′−o o
o
We have ignored the second term in the denominator—since v′o would have to be impossibly large for the value of that term to approach the magnitude g For example, for 41 45o oandλ α= = 2 cos cosg v g vω α λ ω′ ′− ≈ −o o
or 144g kmsω
′ ≈ ≈ov !
Substituting t into equation 5.4.15b to find the lateral deflection gives
( ) [ ]3
2 22
4cos sin sin sin cosvy t v tgωω α λ λ α
′= − = − o
o& α
5.18 Let … ao
r = acceleration of object relative to Earth
ωo
r = kωo = its angular speed
Rr
x y
0Rr
rr Ao
r = acceleration of satellite
kω ω=r = its angular speed
( )2a a v r Aω ω ω= + × + × × +o o
rr r r r r r r (Equation 5.2.14)
2a a A v rω ω ω= − − × − × ×o o
rr r r r r r r As in problem 5.7 Evaluate the term …
( )a a A r R R R Rω ω ω ω ω ω ω ωΔ = − − × × = × × − × × − × × −o o o o o o
rr r r r rr r r r r r r r r r
( )2 2a Rω ωΔ = −o o
r r
… given the condition that 2 3 2 3R Rω ω=o o
3
231a
RRR
ω⎛ ⎞
Δ = − −⎜ ⎟⎝ ⎠
o
o
r r
but ( ) ( ) 2 2 2 cosR R R r R r R r rR θ⋅ = + ⋅ + = + +o o
r r r rr r
Letting x cosθ= 2 2 2 22 1 xR R R r Rx RR
⎛⋅ = + + ≈ +⎜⎝ ⎠
o o
r r ⎞⎟ for small r
and 323 3 21 xR R
R⎛ ⎞≈ +⎜ ⎟⎝ ⎠
o or 3
3 2
3
21R xR R
−⎛ ⎞= +⎜ ⎟⎝ ⎠o
322 22 ˆ1 1 3 3a
Rx 2R xR R
ω ω−⎡ ⎤
⎛ ⎞⎢ ⎥Δ = − − + ≈ − ≈ −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
oo
rr r
x iω for small r
9
Hence: ( )2 ˆˆ ˆ2 3 2aa v x i k ixω ω ω= Δ − × = − − × + jy
rr r r& &
2ˆ ˆ ˆ ˆ3 2 2a ix jy xi yi xjω ω ω= + = − + −r
&& && & & So 22 3x y xω ω− − =&& & 0 2 0y xω+ =&& & 5.19 ( )mr qE q v B= + ×
r rr r&&
Equation 5.2.14 ( )2r r r v rω ω ω ω′ ′ ′= + × + × + × ×r r r r r r r r r&& && & ′
Equation 5.2.13 v v rω′ ′= + ×r r r r
2q Bm
ω = −rr so 0ω =
r&
( ) ( ) ( )2qmr q B v B r qE q v r Bω ω⎡ ⎤′ ′ ′ ′ ′− × − × × = + + × ×⎣ ⎦
r r rr r r r r r r&&r
( ) ( ) ( ) ( )2qmr q v B r B qE q v B q r Bω ω′ ′ ′ ′ ′+ × + × × = + × + × ×
r r r rr r r r r r r&&r
( )2qmr qE r Bω′ ′= + × ×
r rr r r&&
( ) ( )( )( ) 2sin2 2 2q q qBr B r B B
mω θ⎛ ⎞′ ′× × = ∝⎜ ⎟
⎝ ⎠
rr r
Neglecting terms in 2B , mr qE′ =rr&&
5.20 For cos sinx x t y tω ω′ ′= + ′
t
sin cosy x t yω ω′ ′= − + ′ cos sin sin cosx x t x t y t y tω ω ω ω ω ω′ ′ ′ ′ ′ ′= − + +& & & ′
t
sin cos cos siny x t x t y t yω ω ω ω ω ω′ ′ ′ ′ ′ ′= − − + −& & & ′ cos sinx x t y t yω ω ω′ ′ ′= + +& & & ′ ′
x
sin cosy x t y tω ω ω′ ′ ′= − + −& & & ′ ′ cos sin sin cosx x t x t y t y t yω ω ω ω ω ω ω′ ′ ′ ′ ′ ′ ′= − + + +&& && & && & &′ ′
x
sin cos cos siny x t x t y t y tω ω ω ω ω ω ω′ ′ ′ ′ ′ ′ ′= − − + − −&& && & && & &′ ′ 2cos sin 2x x t y t y xω ω ω ω′ ′ ′ ′ ′= + + +&& && && & ′ ′
2sin cos 2y x t y t xω ω ω ω′ ′ ′ ′ ′= − + − +&& && && & y′ ′ Substituting into Eqns 5.6.3:
2cos sin 2x t y t y xω ω ω ω′ ′ ′ ′+ + +&& && & ′ ′
cos sin 2g gx t y tl l
yω ω ω′ ′= − − + &′ ′ 2sin cos 2x t y t x yω ω ω ω′ ′ ′ ′+ − +&& && & ′ ′
10
sin cos 2g gx t y tl l
xω ω ω′ ′= + − − &′ ′
Collecting terms and neglecting terms in 2ω′ :
cos sin 0g gx x t y y tl l
ω ω⎛ ⎞ ⎛ ⎞′ ′+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠&& && =
sin cos 0g gx x t y y tl l
ω ω⎛ ⎞ ⎛ ⎞′ ′+ − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠&& && =
5.21 24sin
Tλ
= hours
24 73.7sin19
T = =o
hours
5.22 Choose a coordinate system with the origin at the center of the wheel, the x′ and
axes pointing toward fixed points on the rim of the wheel, and the y′ z′ axis pointing toward the center of curvature of the track. Take the initial position of the x′ axis to be horizontal in the V− o
r direction, so the initial position of the y′
axis is vertical.
The bicycle wheel is rotating with angular velocity Vbo about its axis, so …
ˆl
Vkb
ω ′= or
A unit vector in the vertical direction is:
ˆ ˆˆ sin cosV t V tn i jb b
′ ′= +o o
At the instant a point on the rim of the wheel reaches its highest point:
ˆ ˆˆ sin cosV t V tr bn b i jb b
⎛ ⎞′ ′ ′= = +⎜ ⎟⎝ ⎠
o or
Since the coordinate system is moving with the wheel, every point on the rim is fixed in that coordinate system. and 0r′ =r& 0r′ =r&& The x y z′ ′ ′ coordinate system also rotates as the bicycle wheel completes a circle around the track:
2ˆ ˆˆ sin cosV V V t V tn i j
b bω
ρ ρ⎛ ⎞′ ′= = +⎜ ⎟⎝ ⎠
o o o or
The total rotation of the coordinate axes is represented by:
1 2ˆˆ ˆsin cosV V t V ti j k
b bω ω ω
ρ⎛ ⎞ V
b′ ′ ′= + = + +⎜ ⎟
⎝ ⎠o o or r r o
2
ˆ ˆcos sinV V t Vi jb b
ωρ
⎛ ⎞′ ′= −⎜ ⎟⎝ ⎠
o or& tbo
11
2 2
2 2ˆ ˆcos sinV V t V t Vr k kb b
ω kρ ρ⎛ ⎞′ ′ ′× = + =⎜ ⎟⎝ ⎠
o o or r& ′o
0rω ′× =r r&
ˆ ˆ ˆ ˆsin cos sin cos sin cosV b V t V t V t V t V t V tr k k V j ib b b b b b
ωρ⎛ ⎞ ⎛′ ′ ′ ′ ′× = − + −⎜ ⎟ ⎜⎝ ⎠ ⎝
o o o o o oo
r r ⎞⎟⎠
o
( )2 2
2 2ˆ ˆ ˆ ˆsin cos sin cosV V t V t V V t V tr k k i jb b b b
ω ωρ⎛ ⎞ ⎛′ ′ ′ ′ ′× × = + + − −⎜ ⎟ ⎜⎝ ⎠ ⎝
o o o o or r r
b⎞⎟⎠
o
( )2 2
ˆ ˆV Vr k nb
ω ωρ
′ ′× × = −o or r r
Since the origin of the coordinate system is traveling in a circle of radius ρ :
2
ˆ VA kρ
′= oo
r
( )2r r r r r Aω ω ω ω′ ′ ′ ′= + × + × + + × × + o
rr r r r r r r r r&& && & &
2 2 2
ˆ ˆ ˆˆV V V Vr k k n kb
2
ρ ρ ρ′ ′ ′= + − +o o or&& o
2 2
ˆ ˆ3V Vr kbρ
′= −o or&& n
With appropriate change in coordinate notation, this is the same result as obtained in Example 5.2.2. --------------------------------------------------------------------------------------------------
12
CHAPTER 6 GRAVITATIONAL AND CENTRAL FORCES
6.1 343 sm V rρ ρ π= =
133
4smrπρ
⎛ ⎞= ⎜ ⎟⎝ ⎠
( )
2 242 3 33
24 4
4 3 4 32 s
Gmm Gm GF mmrπρ πρ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2
12 334
4 3F F Gm mW mg g
πρ⎛ ⎞= = ⎜ ⎟⎝ ⎠
( )2
11 2 2 3 6 3 133
2 3
6.672 10 4 11.35 1 10 14 9.8 3 10 1
F N m kg g cm kg cm kgW m s g m
π− − −
−
⎛ ⎞× ⋅ ⋅ × ⋅= ×⎜ ⎟× ⋅ ⎝ ⎠
3× ×
92.23 10FW
−= ×
6.2 (a) The derivation of the force is identical to that in Section 6.2 except here r < R. This means that in the last integral equation, (6.2.7), the limits on u are R – r to R + r.
2 2
2 214
R r
R r
GmM r RF dsRr s
+
−
⎛ ⎞−= +⎜ ⎟
⎝ ⎠∫
( )2 2 2 2
24GmM R r R rR r R r
Rr R r R r⎡ ⎤− −
= + − − + −⎢ ⎥+ −⎣ ⎦
( )2 2 04GmMF r R r R r
Rr= + − − + =⎡ ⎤⎣ ⎦
(b) Again the derivation of the gravitational potential energy is identical to that in Example 6.7.1,
except that the limits of integration on s are ( ) ( )R r R
ψ θ P
Q
R
r
s
r− → + .
22 R r
R r
RG drRπρφ
+
−= − ∫ s
( )22 RG R r R
rRπρ
= − + − −⎡ ⎤⎣ ⎦r
24 R MG G
R Rπ ρφ = − = −
For , r R< φ is independent of r. It is constant inside the spherical shell.
6.3 2ˆr
GMmF er
= −r
1
The gravitational force on the particle is due only to the mass of the earth that is inside the particle’s instantaneous displacement from the center of the earth, r. The net effect of the mass of the earth outside r is zero (See Problem 6.2).
343
M rπ ρ=
4 ˆ ˆ3 r rF G mre krπρ= − = − e
r
The force is a linear restoring force and induces simple harmonic motion.
r Fr
2 32 24
mTk G
π π πω πρ
= = =
The period depends on the earth’s density but is independent of its size. At the surface of the earth,
32 2
43 e
e e
GMm Gmmg RR R
π ρ= = ⋅
43 e
G gR
πρ=
6
2
6.38 10 12 2 1.49.8 3600
eR m hrT hg m s s
π π −
×= = × ≈
⋅r
6.4 2ˆg r
GMmF er
= −r
, where 343
M rπ ρ=
The component of the gravitational force perpendicular to the tube is balanced by the normal force arising from the side of the tube. The component of force along the tube is cosx gF F θ=
The net force on the particle is … 4ˆ cos3
F i G mrπρ θ= −r
cosr xθ =
4ˆ ˆ3
F i G mx ikπρ= − = −r
x
As in problem 6.3, the motion is simple harmonic with a period of 1.4 hours.
2
6.5 2
2
GMm mvr r
= so 2 GMvr
=
for a circular orbit r, v is constant.
2 rTvπ
=
2 2 2
2 32
4 4rT rv GMπ π
= = ∝ 3r
6.6 (a) 2 rTvπ
=
From Example 6.5.3, the speed of a satellite in circular orbit is … 1
2 2egRv
r⎛ ⎞
= ⎜ ⎟⎝ ⎠
32
12
2
e
rTg R
π=
12 2 3
24eT gRr
π⎛ ⎞
= ⎜ ⎟⎝ ⎠
1 12 2 2 2 2 23 3
2 2 6
24 3600 9.84 4 6.38 10e e
r T g hr s hr m sR R mπ π
− −⎛ ⎞ ⎛ ⎞× ⋅ × ⋅= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠
2
6.62 7e
rR
= ≈
(b) ( )3 3
32 2
1 12 2
2 60 602 2e e
e e
R RrTg
g R g R
ππ π= = =
13 6 2
2 2 2 2 2 2 2
60 6.38 1029.8 3600 24
mm s s hr hr day
π − −
⎛ ⎞× ×= ⎜ ⎟⋅ × ⋅ × ⋅⎝ ⎠
−
27.27 27T day day= ≈ 6.7 From Example 6.5.3, the speed of a satellite in a circular orbit just above the
earth’s surface is … ev gR=
2 2e eR RTv gπ π= =
3
This is the same expression as derived in Problem 6.3 for a particle dropped into a hole drilled through the earth. 1.4T ≈ hours.
6.8 The Earth’s orbit about the Sun is counter-clockwise as seen from, say, the north star. It’s coordinates on approach at the latus rectum are ( ) ( ), ,x y aε α= − .
The easiest way to solve this problem is to note that 160
ε = is small. The
orbit is almost circular!
2
2SGM m mv
r r∴ = and 2 SGMv
r=
with r a bα= ≈ ≈ when 0ε ≈
43 10SGM mvsα
⎛ ⎞≈ = ⋅⎜ ⎟⎝ ⎠
More exactly
cosr v v lα β× = =r r , but
2mlk
α = (equation 6.5.19)
Since Sk GM m=2
S
lGM
α = , hence ( )12cos Sl v GMα β α= =
Or 12 1
cosSGMv
α β⎛ ⎞= ⎜ ⎟⎝ ⎠
The angle β can be calculated as follows: 2 2
2 2 1y xb a
+ = (see appendix C)
2
2
dy b xdx a y
∴ = − and at ( ) ( ), ,x y aε α= −
so ( )
2 2
2 2 21dy b a bdx a a
ε ε εα ε
= = =−
since 2
221 b
aε− =
here tandydx
β ε= = or β ε≈ (small ε)
and 1 12 21
cosS SGM GMv
α ε α⎛ ⎞ ⎛= ≈⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠
⎞⎟ as before.
6.9 ( ) s dF r F F= +
2sGMmF
r= −
2d
dGM mF
r= −
The net effect of the dust outside the planet’s radius is zero (Problem 6.2). The mass of the dust inside the planet’s radius is:
4
34
3dM rπ ρ=
( ) 2
43
GMmF r mGrr
πρ= − −
6.10 1 1 ku er r
θ−= =o
kdu k ed r
θ
θ−= −
o
2 2
22
kd u k e kd r
θ
θ−= =
o
u
From equation 6.5.10 …
( )2
2 12 2
1d u u k u u f udu ml u
−+ = + = − 2
3
( ) ( )1 2 2 1f u ml k− = − + u
( ) ( )2 2
3
1ml kf r
r+
= −
The force varies as the inverse cube of r. From equation 6.5.4, 2r lθ =&
22
kd l edt r
θθ −=o
22
k le d dtr
θ θ =o
22
12
k lte Ck r
θ = +o
2
1 2ln2
klt Ck r
θ⎛ ⎞
′= +⎜ ⎟⎝ ⎠o
θ varies logarithmically with t.
6.11 ( ) 33
kf r kr
= = u
From equation 6.5.10 …
2
32 2 2
1d u kuu kud ml uθ
+ = − ⋅ = − 2ml
2
2 21 0d u k ud mlθ
⎛ ⎞+ + =⎜ ⎟⎝ ⎠
5
If 21 0kml
⎛ ⎞+ <⎜ ⎟⎝ ⎠
, 2
2 0d u cudθ
− = , , for which 0c > bu ae θ= is a solution.
If 21 0kml
⎛ ⎞+ =⎜ ⎟⎝ ⎠
, 2
2 0d udθ
=
1du Cdθ
=
1 2u c cθ= +
1 2
1rc cθ
=+
If 21 0kml
⎛ ⎞+ >⎜ ⎟⎝ ⎠
, 2
2 0d u cudθ
+ = , 0c >
( )cosu A cθ δ= +
1
2cos 1 kr Aml
θ δ−
⎡ ⎤⎛ ⎞= + +⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
6.12 1 1cos
ur r θ
= =o
2
sincos
dud r
θθ θ=
o
2 2
2 3
1 1 2sincos cos
d ud r
θθ θ θ
⎛ ⎞= + =⎜ ⎟
⎝ ⎠o
2
2 2
1 2 2cos 1 21 1cos cos cos cosr r
θθ θ θ⎛ ⎞− ⎛ ⎞
θ+ = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠o o
( )2
2 2 2 32 2 1 2d u u r u r u u
dθ= − =o o −
Substituting into equation 6.5.10 …
( )2 3 12 2
12r u u u f uml u
−− + = −o
( )1 22 2 5f u r ml− = − o u
( )2 2
5
2r mlf rr
= − o
6.13 From Chapter 1, the transverse component of the acceleration is … 2a r rθ θ θ= +&& && If this term is nonzero, then there must be a transverse force given by … ( ) ( 2 )f m r rθ θ θ= +&& && For r aθ= , and btθ = ( ) 22 0f mabθ = ≠
Since , the force is not a central field. ( ) 0f θ ≠
6
For r aθ= , and the force to be central, try nbtθ = ( ) ( )2 2 2 2 2 2 22 1n nf m ab n t ab n n tθ − −⎡ ⎤= + −⎣ ⎦
For a central field … ( ) 0f θ =
( )2 1n n+ − = 0
13
n =
13btθ =
6.14 (a)
Calculating the potential energy
( )2
3 5
4dv af r kdr r r
⎛ ⎞− = = − +⎜ ⎟
⎝ ⎠
Thus, 2
2 4
24aV k
r r⎛ ⎞
= − +⎜ ⎟⎝ ⎠
The total energy is … 2
2 2 2 2
1 2 1 1 9 9 02 4 2 2 4
k kE T V v ka a a a
⎛ ⎞ ⎛ ⎞= + = − + = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
o o o
Its angular momentum is … 2 2 2 49
2kl a v constant r 2θ= = = =o
&
Its KE is …
( )2 2 2
2 2 2 2 2 24
1 1 12 2 2
dr dr lT r r r rd d
θ θθ θ
⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= + = + = +⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
& &&r
The energy equation of the orbit is … 2 2 2
24 2
1 202 4
dr l aT V r kd r rθ
⎡ ⎤ ⎛ ⎞⎛ ⎞+ = = + − +⎢ ⎥ ⎜ ⎟⎜ ⎟⎝ ⎠⎢ ⎥ ⎝ ⎠⎣ ⎦
4r
2 22
4 2
9 24 4
dr k ar kd r rθ
⎡ ⎤ ⎛ ⎞⎛ ⎞= + − +⎢ ⎥ ⎜ ⎟⎜ ⎟⎝ ⎠⎢ ⎥ ⎝ ⎠⎣ ⎦
4r
or ( )2
2 219
dr a rdθ
⎛ ⎞ = −⎜ ⎟⎝ ⎠
Letting cosr a φ= then sindr dad d
φφθ
= − θ
So 2 1
9ddφθ
⎛ ⎞ =⎜ ⎟⎝ ⎠
13
φ θ∴ =
7
Thus 1cos3
r a θ= ( )@ 0r a θ= = o
(b) at 32πθ = the origin of the force. To find how long it takes … 0r →
22 2 21 1cos cos
3 3
av vlr a a
θθ θ
= = =o o&
2 1cos3
adt dv
θ θ=o
32 2
2 2
0 0
1 3 3cos cos3 4
a aT d dv v
π π
avπθ θ φ φ= =∫ ∫
o o
=o
Since 12
2
92
kva
⎛ ⎞= ⎜ ⎟⎝ ⎠
o
1 122 223 2 2
4 9 4aT a
k kππ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
(c) Since the particle falls into the center of the force
v →∞ (since ) l vr const⊥= =
6.15 From Example 6.5.4 …
12
1
1
2
c
v rv r r
⎛ ⎞= ⎜ ⎟+⎝ ⎠
o
o
Letting c
vVv
= o we have
12
1
2
1V r
r
⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟+⎜ ⎟⎝ ⎠
o
So: 2
21 1
1 2 12
r rdVdr V r r
−⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
o o
1
Thus 21 11
1 11
1
1 1 1 12
112
1
dVr rVr rdr rr
r rrrr
= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ++⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟ ⎝ ⎠
⎜ ⎟+⎜ ⎟⎝ ⎠
o o
oo
o
8
(a) 11
1
12
dVrVrdr
r
⎛ ⎞⎜ ⎟⎝ ⎠ ≈⎛ ⎞⎜ ⎟⎝ ⎠
o (b) ( )1 1
1
2 2 60 1% 120%!dr r dVr r V
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ o
The approximation of a differential has broken down – a correct result can be obtained by calculating finite differences, but the implication is clear – a 1% error in boost causes rocket to miss the moon by a huge factor --- ∼ 2! 6.16 From section 6.5, 0.967ε = and mi. 655 10r = ×o
From equations 6.5.21a&b, 111
r r εε
+=
−o
( )6
1 6
1 55 102 1 1 0.967 93 10
r mi AUa r rmiε
×= + = = ×
− − ×o
o
1
U
17.92a A=
From equation 6.6.5, 32caτ =
3 32 21 17.92yr AU AUτ
−= ⋅ ×
32
75.9 yrτ = From equation 6.5.21a and 6.5.19 …
2
0 0
1 1mlr krαε = − = −
2
1mr vk
ε = o o − and k GMm=
( )12
1GMvr
ε⎡ ⎤
= +⎢ ⎥⎣ ⎦
o
o
From Example 6.5.3 we can translate the factor GM into the more convenient 2
e eGM a v= … with ae the radius of a circular orbit and ve the orbital speed …
( ) ( )1 1
2 62 2
6
93 101 1.955 10
e ee
a v miv vr mi
ε⎡ ⎤ ⎡ ⎤×
= + =⎢ ⎥ ⎢ ⎥×⎣ ⎦⎣ ⎦o
o
67
0 1.824 ev v= Since l is constant … 1 1r v r v= o o
11
1 1 .967 1.824 0.03061 1.967 e e
rv v v vr
vεε
− −= = = × =
+o
o o
6
1 1
2 2 93 10 66,7051 365 24
ee
a miv myr day yr hr day
π πτ − −
× ×≈ = =
× ⋅ × ⋅ph
mph ph
51.22 10v = ×o and 31 2.04 10v m= ×
6.17 From Example 6.10.1 …
9
( )1222 21 sinq qd
dε ⎡ ⎤⎛ ⎞= + −⎜ ⎟⎢ ⎝ ⎠⎣ ⎦
φ ⎥ where e
vqv
= and e
rda
=
are dimensionless ratios of the comet’s speed and distance from the Sun in terms of the Earth’s orbital speed and radius, respectively (q and d are the same as the factors V and R in Example 6.10.1). φ is the angle between the comet’s orbital velocity and direction vector towards the Sun (see Figure 6.10.1). The orbit is hyperbolic, parabolic, or elliptic as ε is > , = , or < 1 …
i.e., as 2 2qR
⎛ −⎜⎝ ⎠
⎞⎟ is > , = , or < 0.
2 2qR
⎛ −⎜⎝ ⎠
⎞⎟
vv
is > , = , or < 0 as is > , = , or < 2. 2q d
6.18 Since l is constant, occurs at r and occurs at , i.e. and
and form the constancy of l … maxv o minv 1r maxv = o
min 1v = 1 1v r v r= o o
2min max 1
1
rv v v v vr
= = oo o
( )2 1kr vm
ε= +o o (See Example 6.5.4)
From equation 6.6.5 … 22k aGM a
mπτ
⎛ ⎞= = ⎜ ⎟⎝ ⎠
( )2
min max1
12 aav vrεπ
τ+⎛ ⎞= ⋅⎜ ⎟
⎝ ⎠
From equation 6.5.21a&b … 111
r r εε
+=
−o . With 12a r r= +o :
( ) ( )( ) ( )1
1 1 1
1 11 1 1 11 1 1 ( 1)2 2 2 1
a r r rr r rε ε εε ε
ε+ + + ⎛ ⎞ ⎡ − ⎤⎛ ⎞= = + + = +⎜ ⎟ ⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦⎝ ⎠
o o 1+ =
2
min max2 av v πτ
⎛ ⎞= ⎜ ⎟⎝ ⎠
6.19 As a result of the impulse, the speed of the planet instantaneously changes; its orbital radius does not, so there is no change in its potential energy V. The instantaneous change in its total orbital energy E is due to the change in its kinetic energy, T, only, so
2 21 22
v vE T mv mv v mv Tv vδ δδ δ δ δ⎛ ⎞= = = = =⎜ ⎟
⎝ ⎠
2E vT vδ δ
=
But the total orbital energy is
10
2kEa
= − So 22kE aa
δ δ=
Since planetary orbits are nearly circular
~ kVa
− and ~2kTa
Thus, aE Taδδ ≅ and E a
T aδ δ
=
We obtain 2a va vδ δ
=
6.20 (a) 0
1V Vτ
τ= ∫ dt
( ) kV rr
= −
From equation 6.5.4, 2l r θ= &
2
ddt r
lθ= or
2r ddtlθ
=
2
0 0
krVdt dl
τ πθ= −∫ ∫
From equation 6.5.18a …
( )21
1 cosa
rε
ε θ−
=+
( )2
2
0 0
11 cos
ka dVdtl
τ πε θε θ
−= −
+∫ ∫
From equation 6.6.4 … 2
22 1alπτ ε= −
2 2
0
12 1 co
k dVa
πε θsπ ε θ
−= −
+∫
2
20
21 cos 1
dπ θ πε θ ε
=+ −
∫ , 2 1ε < kVa
∴ = −
(b) This problem is an example of the virial theorem which, for a bounded, periodic
system, relates the time average of the quantity 0
ii
p rτ
⋅∑∫ & to its kinetic energy T. We will
derive it for planetary motion as follows:
0 0 0
1 1 1p rdt mr rdt F rdtτ τ τ
τ τ τ⋅ = ⋅ = ⋅∫ ∫ ∫& &&
Integrate LHS by parts
11
20
0 0
1 1 1mr r mr dt F rdtτ τ
τ
τ τ τ⎡ ⎤⋅ − = ⋅⎣ ⎦ ∫ ∫& &
The first term is zero – since the quantity has the same value at 0 and τ . Thus 2 T F= − ⋅ r where denote time average of the quantity within brackets.
but dV kr F r r Vdr r
V− ⋅ = = = = −⋅∇
hence 2 T V= −
but 2 2V V
E T V V= + = − + =
hence 2V = E but 2kE constanta
= − =
and 0
12kE Edt Ea
τ
τ= = = −∫ so 2 kE
a= −
Thus: kVa
= − as before and therefore 12 2
kT Va
= − =
6.21 The energy of the initial orbit is
212 2
k kmv Er a
− = = −
(1) 2 2 1kvm r a⎛ ⎞= −⎜ ⎟⎝ ⎠
Since ( )1ar a ε= + at apogee, the speed , at apogee is 1v
( )
( )( )
21
12 11 1
k kvm a a ma
εε ε
⎛ ⎞ −= − =⎜ ⎟⎜ ⎟+ +⎝ ⎠
To place satellite in circular orbit, we need to boost its speed to such that cv
212 2c
a a
kmvr r
− = −k since the radius of the orbit is ar
( )
2
1ca
k kvmr ma ε
= =+
Thus, the boost in speed 1 1cv v vΔ = −
(2) ( ) ( )
12 1
21 1 1
1kv
maε
ε⎡ ⎤ ⎡ ⎤Δ = − −⎢ ⎥ ⎢ ⎥+ ⎣ ⎦⎣ ⎦
Now we solve for the semi-major axis a and the eccentricity ε of the first orbit. From (1) above, at launch v at , so v= o Er R=
2 2 1
E
kvm R a⎛ ⎞
= −⎜ ⎟⎝ ⎠
o
12
and solving for a
22E
E
Ra Rmvk
=− o
noting that …
(3) EE
E E
k GM gRmR R
= =
32
4.49 101.426
2
E E
E
R Ra kv
gR
= = = ⋅⎛ ⎞
−⎜ ⎟⎝ ⎠
o
m
The eccentricity ε can be found from the angular momentum per unit mass, l, equation 6.5.19, and the data on ellipses defined in figure 6.5.1 …
( ) ( )1
1 2 222
1sinE
kakl r v Rm m
εαθ θ⎡ ⎤−⎡ ⎤ ⎢ ⎥= = = =⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
o o&
where v , o θo are the launch velocity, angle Solving for ε (using (3) above)
2 22 21 2 sin 0.795
E E
v vgR gR
ε θ⎛ ⎞
= − − =⎜ ⎟⎝ ⎠
o oo
0.892ε∴ = Inserting these values for a, ε into (2) and using (3) gives
(a) ( )1
123 12
1 1 1 4.61 101
EE
R av gR km sεε
−⎡ ⎤⎛ ⎞ ⎡ ⎤Δ = − − = ⋅ ⋅⎜ ⎟⎢ ⎥ ⎢ ⎥+⎝ ⎠ ⎣ ⎦⎣ ⎦
(b) ( ) 31 2.09 1Eh a R kmε= + − = ⋅ 0 altitude above the Earth … at perigee
6.22 ( ) 2 3 2
22br br brbe e ef r k k b
r r r
− − −⎛ ⎞− ⎛ ⎞′ = − − = +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ r
( )( )
2f ab
f a a′ ⎛ ⎞= − +⎜ ⎟
⎝ ⎠
From equation 6.14.3, ( )( ) ( )
11223 3
f aa a
f aψ π π
−
2b−⎡ ⎤′
= + = − +⎡ ⎤⎢ ⎥ ⎣ ⎦⎣ ⎦
1 abπψ =−
6.23 From Problem 6.9, ( ) 2
43
GMmf r mGrr
πρ= − −
( ) 3
2 43
GMmf r mr
πρ′ = − G
13
( )( )
33
32
4 42 23 34 413 3
aGMma mGf a Mf a aGMma mGa a
M
πρπρ
πρπρ
−
−
− − +′= =
⎛ ⎞− − +⎜ ⎟⎝ ⎠
From equation 6.14.3, ( )( )
12
3f a
af a
ψ π−
⎡ ⎤′= +⎢ ⎥
⎣ ⎦
11 3 23 2
3 3
44 1 42 3334 41 13 3
aaMM
a aM M
πρπρ
ψ π ππρ πρ
−− ⎡ ⎤⎛ ⎞⎡ ⎤ +⎢ ⎥⎜ ⎟− +⎢ ⎥ ⎝ ⎠⎢ ⎥= + =⎢ ⎥ ⎢ ⎥⎢ ⎥+ +⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
121
1 4cc
ψ π +⎛ ⎞= ⎜ ⎟+⎝ ⎠,
343
acM
πρ=
6.24 We differentiate equation 6.11.1b to obtain ( )dU rmr
dr= −&&
For a circular orbit at r a , so = 0r =&&
0r adUdr = =
For small displacements x from r a= , and r x a= + r x=&& &&From Appendix D …
( ) ( ) ( ) ( )2
2xf x a f a xf a f a′ ′′+ = + + +K
Taking ( )f r to be dUdr
, ( )2
2
d Uf rdr
′ =
Near … r a=
2
2r a r adU dU d Uxdr dr dr= == + +K
2
2 r ad Umx xdr == −&&
This represents a “restoring force,” i.e., stable motion, so long as 2
2 0d Udr
> at . r a=
6.25 ( ) 3 5
2 4kf rr r
ε′ = +
From equation 6.13.7, the condition for stability is ( ) ( ) 03af a f a′+ <
14
2 4 3 5
2 4 03
k a ka a a a
ε ε⎛ ⎞− − + + <⎜ ⎟⎝ ⎠
2 4 03 3ka a
ε− + <
2 kaε<
12
akε⎛ ⎞> ⎜ ⎟
⎝ ⎠
6.26 (a) ( ) 2
bref r kr
−
= −
( ) 2 3 2
2 2brbr b ef r ke k b
r r r r
−− ⎛ ⎞ ⎛′ = − − − = +⎜ ⎟ ⎜
⎝ ⎠ ⎝⎞⎟⎠
From equation 6.13.7, the condition for stability is ( ) ( ) 03af a f a′+ <
2 2
2 03
ba bae a ek k ba a a
− − ⎛ ⎞− + + <⎜ ⎟⎝ ⎠
2 03 3
ba bae bek ka a
− −
− + <
1ba
< 1ab
<
(b) ( ) 3
kf rr
= −
( ) 4
3kf rr
′ =
( ) ( ) 3 4
3 03 3a k a kf a f a
a a⎛ ⎞′+ = − + ⎜ ⎟⎝ ⎠
=
Since ( ) ( )3af a f ′+ a is not less than zero, the orbit is not stable.
6.27 (See Figure 6.10.1) From equation 6.5.18a 2(1 )
1 cosar εε θ−
=+
and the data on ellipses in Figure 6.5.1 (1 )p a ε= − so
(1 )1 cos
r p εε θ+
=+
15
For a parabolic orbit, 1ε = The comet intersects earth’s orbit at r a= .
21 cos
paθ
=+
2cos 1 pa
θ = − +
6.28 (See Figure 6.10.1) T along the comet’s trajectory inside earth’s orbit d= ∫ t
From equation 6.5.4, 2 2 dr rdt
lθθ = =& so 2r ddtlθ
=
2r dTlθ
= ∫
From equation 6.5.18a 2(1 )
1 cosar εε θ−
=+
and the data on ellipses in Figure 6.5.1 (1 )p a ε= − so
(1 )1 cos
r p εε θ+
=+
From equation 6.5.18b, with 1ε = for a parabolic orbit:
21 cos
prθ
=+
At 2πθ = the distance to the comet is 2 2
1 cos2
pr pα π= = =+
From equation 6.5.19, 2ml
kα = , where k GMm= , so
2
2lpGM
=
As shown in Example 6.5.3, 2eGM av=
For a circular orbit, 21e
avyrπ
=
( ) ( ) ( )11 1 3
122 2 22 2 2el GMp ap v a p yrπ −= = =
( )
( )12 2 3
1222
4 21 cos
r d pT al
θ θ
θ θ
θ π θθ
+ + −− −
− −= = ⋅
+∫ ∫o o
o o
p d yr
where 1 2cos 1 pa
θ − ⎛= − +⎜⎝ ⎠
o
⎞⎟ from Problem 6.27
( )
32
3 22
21 cos
p dT ya
θ
θ
θθπ
+
−=
+∫o
o
r
From a table of integrals, ( )
32
1 1tan tan2 2 61 cos
dx x xx
= ++∫ 2
16
32 32 1tan tan
2 3 2pT ya
θ θπ
⎡ ⎤⎛ ⎞= +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦o o r
121 costan
2 1 cosx x
x−⎛ ⎞= ⎜ ⎟+⎝ ⎠
12 1
222
tan 22
pa pa
p pa
θ⎛ ⎞−⎜ ⎟ ⎛ ⎞−
= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟
⎝ ⎠
o
1 332 222 1
3p a p a pT ya p pπ
⎡ ⎤⎛ ⎞ ⎛ ⎞− −⎛ ⎞ ⎢ ⎥= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
r
13222 1
3p a p a p yra p pπ
⎛ ⎞ ⎛ ⎞− −⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
122 2 1 1
3p pT y
a aπ⎛ ⎞⎛ ⎞= + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
r
)
T is a maximum when ( ) (122 p a a p+ − is a maximum.
( ) ( ) ( )( )( ) ( ) (2 22 2 2 2 2d p a a p p a a p p adp
⎡ ⎤+ − = + − + + −⎣ ⎦ )1
( )( )2 3 6p a a p= + −
T is a maximum when 2ap = .
( )122 1 22 77.5
3 2 3T yr day
π π⎛ ⎞= = =⎜ ⎟⎝ ⎠
When 0.6p a=
( )2 2.2 .04 0.2088 76.23
T yr dayπ
= = =
6.29 ( ) 3
k kV rr r
ε= − −
( ) ( )22 4 4
3 3dV k k kf r rdr r r r
ε ε= − = + = +
( ) ( )23 5 5
2 12 2 6k k kf r rr r r
ε ε′ = − − = − +
( )( )
2
2
2 63
f a rf a a r
εε
′ ⎛ ⎞+= − ⎜ ⎟+⎝ ⎠
17
From equation 6.14.3, ( )( )
12
3f a
af a
ψ π−
⎡ ⎤′= +⎢ ⎥
⎣ ⎦
12 22
2 2
6 33 23 3
r rr r
ε εψ π πε ε
−⎡ ⎤⎛ ⎞+ +
= − =⎢ ⎥⎜ ⎟+ −⎝ ⎠⎣ ⎦
For 25
R Rε = Δ , , 4000R mi= 13R miΔ =
( )( ) 4 22 4000 13 2.08 105
miε = = ×
For , r R≈ 2 71.6 10r m= × 2i 1.0039 180.7ψ π= = o
6.30 2
2
12rel
k kV Er m c r
⎛ ⎞= − − +⎜ ⎟⎝ ⎠o
( ) 2 2
22
dV k k kf r Edr r m c r r
⎛ ⎞⎛= − = − + + −⎜ ⎟⎜⎝ ⎠⎝o
2⎞⎟⎠
( ) 2 2
11k kf r Er m c r
⎡ ⎤⎛ ⎞= − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦o
( ) 3 2 2 2
2 1 11k k kf r Er m c r r m c r
⎡ ⎤ ⎛ ⎞⎛ ⎞ ⎛′ = + + − −⎜ ⎟⎜ ⎟ ⎜⎢ ⎥⎝ ⎠ ⎝⎣ ⎦ ⎝ ⎠o o
2
k ⎞⎟⎠
( ) 3 2
1 32 2k kf r Er m c r⎡ ⎤⎛ ⎞′ = + +⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦o
( )( )
2
2
1 32 21
11
kEf a m c a
kf a a Em c a
⎡ ⎤⎛ ⎞+ +⎜ ⎟⎢ ⎥′ ⎝ ⎠⎢ ⎥= −⎛ ⎞⎢ ⎥+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
o
o
( )( )
2 2
2
3 13 2 23
11
k kE Ef a m c a m c a
af a kE
m c a
⎡ ⎤⎛ ⎞ ⎛+ + − − +⎜ ⎟ ⎜⎢ ⎥′ ⎝ ⎠ ⎝⎣ ⎦+ =⎡ ⎤⎛ ⎞+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
o o
o
3 ⎞⎟⎠
2
2
1
11
Em c
kEm c a
⎡ ⎤+⎢ ⎥
⎣ ⎦=⎡ ⎤⎛ ⎞+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
o
o
18
( )( )
2
23
f a m c Ea kf a m c Ea
⎡ ⎤′ ⎢ ⎥+
+ = ⎢ ⎥⎢ ⎥+ +⎣ ⎦
o
o
( )( )
12
3f a
af a
ψ π−
⎡ ⎤′= +⎢ ⎥
⎣ ⎦
12
21
ka
m c Eψ π
⎡ ⎤⎢ ⎥
= +⎢ ⎥+⎢ ⎥⎣ ⎦
o
6.31 From equation 6.5.18a 2(1 )
1 cosar εε θ−
=+
… (Here θ is the polar angle of conic
section trajectories as illustrated by the coordinates in Figure 6.5.1) … and the data on ellipses in Figure 6.5.1 0 (1 )r a ε= − so
11 coscomr r ε
ε θ+
=+o
From equation 6.5.18b 1 cos
r αε θ
=+
and at 0θ = o 0 1r α
ε=
+
And from equation 6.5.19 2ml
kα = so
2
0 (1 )mlr
k ε=
+
1m mk GMm GM= =
From Example 6.5.3, and 2e eGM a v= 2 2 2 2sincom coml r v φ=
( )( )( )
2 2 2
2
sin 11 1 cos
com comcom
e e
r vr
a vφ ε
ε ε θ+
=+ +
2 2 11 sin1 cos
RV φε θ
=+
( )2 21cos sin 1RVθ φε
= −
( ) ( )1
1 222 2 222
1sin 1 cos 1 sin 1RVθ θ φε
⎡ ⎤= − = − −⎢ ⎥⎣ ⎦
( ) ( )1
2 22 2 2 2 21sin sin 2 sin 1RV RVθ ε φ φε⎡ ⎤= − + −⎢ ⎥⎣ ⎦
Again from Example 6.5.3 …
( )1222 21 siV RV
Rε φ⎡ ⎤⎛ ⎞= + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
n
19
( )22 2 21 sin 2 sinRV RV 2ε φ φ= + −
( ) ( )1
2 2 22 2 21sin sin sinRV RVθ φε⎡ ⎤= −⎢ ⎥⎣ ⎦
φ
21sin sin cosRVθ φ φε
=
2 2
2
cos sin 1sin sin cos
RVRV
θ φθ φ φ
−=
2
2cot tansin 2RV
θ φφ
= −
12
2cot tan csc 2RV
θ φ φ− ⎛ ⎞= −⎜ ⎟⎝ ⎠
For , , : 0.5V = 4R = 30φ = o
( )
12
2cot tan 30 csc604 .5
θ −⎛ ⎞
= −⎜ ⎟⎜ ⎟⎝ ⎠
o o
( )1 11 2cot cot 33 3
2
− −
⎛ ⎞⎜ ⎟⎜ ⎟ = −⎜ ⎟⎜ ⎟⎝ ⎠
= −
30θ = − o
6.32
The picture at left shows the orbital transfer and the position of the two satellites at the moment the transfer is initiated. Satellite B is “ahead” of satellite A by the angle θ0
θ0
r1
r2
1 2
2r ra +
= is the semi-major axis of the elliptical transfer orbit.
From Kepler’s 3rd law (Equation 6.6.5) applied to objects in orbit about Earth …
22 3
2
4
E
aGMπτ =
The time to intercept is … 3 32 21
2tE E
T aGM R g
τ ππ= = = a since 2E
E
GM gR
=
Letting and where h1 1Er R h= + 2 Er R h= + 2 1 and h2 are the heights of the 2 satellites above the ground. Inserting these into the above gives …
3 33 2 2
1 2 1 22 2 2t EE EE E
h h h hT R
R RR g R gπ π+ +⎛ ⎞ ⎛ ⎞
= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
20
From Example 6.6.2, RE = 6371 km, h1 = 200 mi = 324 km and h2 = r2 - RE = 42,400 km - 36,029 km. Putting in the numbers … Tt = 4.79 hr
(b) Thus, 0 00 180 1 108
12tTθ ⎛ ⎞= − =⎜ ⎟
⎝ ⎠
6.33
The potential for the inverse-cube force law
is … ( ) 22kV rr
=
Letting 1u r−= , we have (Equation 6.9.3)
( )2
2 2 112
duml u V u Edθ
−⎡ ⎤⎛ ⎞ + + =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
( ) 22
2 E Vdu ud mlθ
−= −
( ) ( ) 2 2 22
2
2 2mld du
E V m E V m l uuml
θ = =− − −
−
du
)
θ0
θ0θS
b
θ r
Now, integrating from up to ( 0r u= ∞ = ( )min maxr r u u= = …
( )
max
0 2 2 20 2
u ml dum E V m l u
θ =− −
∫
But 20 0
1 ,2
E mv l bv= = , so …
max
00
0 2 2 2 2 2 20 0
1 122 2
u mbv dum mv ku m b v u
θ =⎛ ⎞− −⎜ ⎟⎝ ⎠
∫
max
00
0 2 2 2 2 2 20 0
1 122 2
u mbv dum mv ku m b v u
θ =⎛ ⎞− −⎜ ⎟⎝ ⎠
∫
Before evaluating this integral, we need to find ( )1
max minu , in other words, the distance of closest approach to the scattering center.
r −=
( ) ( ) 2 2min min 02
min
1 1 12 2 2
kE T r V r mv mvr
= + = + =
But, the angular momentum per unit mass l is … and substituting for v into the above gives … 0 minl bv r v= =
21
2
202 2
min min
ml k mvr r
+ = so … 2
20max2 2
min
1 mv ur ml k
= =+
Solving for … maxu
max2
20
1ukb
mv
=+
Now we evaluate the integral for θ0 … maxmax
10
2 2max0 02 22 22
0 00
sin2
1
uu b b uduuk kk b bb u mv mvmv
b πθ −= = =⎛ ⎞ ⎛ ⎞⎛ ⎞ + +− + ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠
∫
Solving for b …
( ) 00 2 2 2
0 0
24
kbmv
θθπ θ
=−
But (012 S )θ π θ= − . Thus, we have …
( )( )2
0 2S
SS S
kbmv
π θθθ π θ
−=
−
We can now compute the differential cross section …
( ) ( )( )
2
22 20
sin 2 sinS
SS S S S
kb dbd mv
π π θσ θ
θ θSθ π θ θ
−= =
−
Since 2 sin s sd dπ θ θΩ = we get …
( ) ( )( )
3
2 22
2S
S SS S
kd bdbE
π θπ dσ θ ππ θ θ
⎡ ⎤−Ω = = ⎢ ⎥
−⎢ ⎥⎣ ⎦θ
---------------------------------------------------------------------------------------------------
22
CHAPTER 7 DYNAMICS OF SYSTEMS AND PARTICLES
7.1 From eqn. 7.1.1, 1cm i i
ir m
m= r∑r r
( ) ( )1 2 31 1 ˆ ˆˆ ˆ ˆ3 3cmr r r r i j j k= + + = + + + +
r r r r k
( )1 ˆˆ ˆ2 23cmr i j= + +
r k
( ) ( )1 2 31 1 ˆˆ ˆ ˆ ˆ23 3cm cm
dv r v v v i j i jdt
= = + + = + + + +r r r r r k
3
2
1
z
y
x
( )1 ˆˆ ˆ3 23cmv i j= + +
r k
3
From eqn 7.1.3, 1 2i ii
p m v v v v= = +∑r r r r+r
ˆˆ ˆ3 2p i j= + +r k
7.2 (a) From eqn. 7.2.15, 212 i i
iT m v=∑
( )2 2 2 2 21 2 1 1 1 1 42
T ⎡ ⎤= + + + + =⎣ ⎦
(b) From Prob. 7.1, ( )1 ˆˆ ˆ3 23cmv i j= + +
r k
( )2 2 2 21 1 13 3 2 1 22 2 9cmmv = × × + + =
13
v
(c) From eqn.7.2.8, i ii
L r m= ×∑r r r
( ) ( ) ( )ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ2 ˆL i j i j k j k i j k⎡ ⎤ ⎡⎡ ⎤= + × + + × + × + +⎣ ⎦ ⎣ ⎦ ⎣r ⎤
⎦2k
( ) ( ) ( )ˆ ˆˆ ˆ ˆ ˆ ˆ2 2L k i j i i j= − + − + − = − + −r
7.3 b gv v v= −o
r r r
Since momentum is conserved and the bullet and gun were initially at rest: 0 b gmv Mv= +
r r
g bv vγ= −r r , m
Mγ =
( )1 bv vγ= +o
r r
1b
vvγ
=+o
rr
1
1g
vv γγ
= −+
o
rr
7.4 Momentum is conserved: 2 blkvmv m Mv⎛ ⎞= +⎜ ⎟
⎝ ⎠o
o
12blkv vγ= o m
Mγ =
2 221 1 1
2 2 2 2 2i fv vT T mv m M γ⎡ ⎤⎛ ⎞ ⎛ ⎞− = − +⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
o oo
2
21 1 112 4
mv Mm 4
γ⎛ ⎞= − −⎜ ⎟
⎝ ⎠o
34 4
i f
i
T TT
γ−= −
v0
60 o
θ
v0 / 2
7.5 At the top of the trajectory: ˆ ˆcos60
2vv iv i= =o o
o
r
Momentum is conserved:
2ˆ ˆ
2 2 2 2v m v mim j v⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠o o r
2ˆ ˆ
2vv iv j= − o
o
r
Direction: 1 2tan 26.6
v
vθ −
⎛ ⎞−⎜ ⎟= =⎜ ⎟
⎜ ⎟⎝ ⎠
o
o
o
below the horizontal.
Speed:
12 2
22 1.118
2vv v
⎡ ⎤⎛ ⎞= + =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
oo ov
7.6 When a ball reaches the floor, 212
mv mgh= .
As a result of the bounce, vv
ε′= .
The height of the first bounce: 212
mgh mv′ ′=
2 2 2
2
2 2v vh h
g gε ε
′′ = = =
Similarly, the height of the second bounce, h h2 4hε ε′′ ′= =
2
Total distance 2 4 2
0
2 2 1 2 n
n
h h h hε ε ε∞
=
⎛ ⎞= + + + = − +⎜ ⎟⎝ ⎠
∑K
Now 0 1
n
n
aarr
∞
=
=−∑ , 1r < .
2
22 2
0
2 11 2 11 1
n
n
εεε ε
∞
=
+⎛ ⎞− + = − + =⎜ ⎟ − −⎝ ⎠∑
total distance2
2
11
h εε
⎛ ⎞+= ⎜ ⎟−⎝ ⎠
For the first fall, 212
gt h=o , so 2htg
=o
For the fall from height h′ : 12 2h htg g
ε′
= =
Accounting for equal rise and fall times:
Total time ( )2
0
2 21 2 2 1 2 n
n
h hg g
ε ε ε∞
=
⎛ ⎞= + + + = − +⎜ ⎟⎝ ⎠
∑K
Total time 2 11
hg
εε
+⎛ ⎞= ⎜ ⎟−⎝ ⎠
v0m -v0/2 4m 7.7 From eqn. 7.5.5:
( ) ( )1 2 1 2 21
1 2
m m x m m xx
m mε ε− + +
′ =+
& && 2
)
( ) (1 1 1 2 12
1 2
m m x m m xx
m mε ε+ + −
′ =+
& && vc vt2
1 14 4 4 0 54 4 2 2
4 5c
v vm m v m m mv
m m m
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− + + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠= =
+
o oo
2v
= − o
1 144 4
4t
vm m v m mv
m m
⎛ ⎞ ⎛ ⎞⎛+ + − −⎜ ⎟ ⎜ ⎟⎜⎝ ⎠ ⎝ ⎠⎝=
+
oo 2
⎞⎟⎠
5 154 4 2
5 8
vmv mv
m
⎛ ⎞+ −⎜ ⎟⎝ ⎠= = −
oo
o
Both car and truck are traveling in the initial direction of the truck
with speeds 2vo and
8vo , respectively.
3
7.8 From eqn. 7.2.15, 2 2 21 1 2 2
1 1 12 2 2i i
iT m v m v m= = +∑ v
Meanwhile:
( )2
22 2 1 1 2 2 1 21 2
1 2
1 1 1 12 2 2 2cm
m v m v m mmv v m v vm m m
μ +⎛ ⎞+ = + −⎜ ⎟ +⎝ ⎠
r rr r
( )2 2 2 2 2 21 1 2 2 1 2 1 2 1 2 1 2 1 2
1 2 22
m v m v m m v v m m v v v vm⎡ ⎤= + + ⋅ + + − ⋅⎣ ⎦
r r r r
2 21 1 2 2
1 12 2
m v m v= +
Therefore, 2 21 12 2cmT mv vμ= +
7.9 From Prob. 7.8, 2 21 12 2cmT mv vμ= +
and since Q T T ′= − cm cmv v′= :
2 21 12 2
Q v vμ μ ′= −
From eqn. 7.5.4, vv
ε′
=
( )2 21 12
Q vμ ε= −
7.10 Conservation of momentum: 1 1 1 1 2 2m v m v m v′ ′= +
21 1
1
mv v vm
′ ′= − 2
2
2 2 2 21 1 1 2 2
1 1
2m mv v v v vm m
′ ′= − + 22′
Conservation of energy:
2 21 1 1 1 2 2
1 1 12 2 2
m v m v m v′ ′= + 2
2
2 2 221 1 1 1 2 1 2 2 2 2
1
1 1 12 2 2 2
mm v m v m v v v m vm
2′ ′ ′= − + +
22 22 2 1 2
1
1 02
m m v m v vm
⎛ ⎞′ ′+ − =⎜ ⎟
⎝ ⎠
1 12
1 2
2m vvm m
′ =+
( )
22 2 2 1 2
1 1 1 1 1 1 2 2 121 2
1 1 1 22 2 2
m mT T m v m v m v vm m
′ ′ ′− = − = =+
2
4
211
1 1
211 1
24
12
m vT T mT mm v
μμ′−
= =
7.11 From eqn. 7.2.14, cm cm i i i
i
L r mv r m v= × + ×∑r r rr r
1 1 1 2 2i i ii
r m v r m v r m v× = × + ×∑r r r r r r
2
From eqn. 7.3.2, 1 1 21 1
2 2
1 m m m m11R r r
m mr
μ⎛ ⎞ ⎛ ⎞+
= + = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
r r r r
Since from eqn. 7.3.1, 21 2
1
mr rm
= −r r
22
mR rμ
= −r r
1 1 2 21 2
i i ii
r m v R m v R m vm mμ μ⎛ ⎞
× = × + − ×⎜ ⎟⎝ ⎠
∑r rr r r r
( )1 2R v v R vμ μ= × − = ×r rr r r
cm cmL r mv R vμ= × + ×r rr r r
7.12 Let ms = mass of Sun and ae = semi-major axis of Earth’s orbit
then from eqn. 7.3.9c,
1 32 2
1 2
s s e
m m a yrm m a
τ−
⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
16 ms 20 ms
12 3
1 23
e s
a m ma m m
τ⎛ ⎞
= +⎜ ⎟⎝ ⎠s
( )2
21333
15.6 20 16365
yrda yrda
−⎛ ⎞+⎜ ⎟
⎝ ⎠= ×
10.205e ea a= = a
7.13 (Ignore primes in notation)
a) The coordinates of the two primaries, P1 and P2, are shown at left – along with the coordinates of 4L and . 5L
5
b) ( ) ( )
( ) ( )
2 2
1 12 22 22 2
1,
21
x yV x yx y x y
α α
α α
− += − − −
⎡ ⎤ ⎡ ⎤− + + − +⎣ ⎦ ⎣ ⎦
(7.4.13)
( )( )[ ]
( )[ ]
3 32 2
1 1x xV xx
α α α α− − + −∂= +
∂−
Now x 0.5α= − at 4L and 5L also, each bracket term in the denominator equals 1 at 4L , 5L
( )( ) ( ) ( )1 0.5 0.5 1 0.5Vx
α α α α α α α∂= − − − + − + − − −
∂
0.5 0.5 0.5 0.5 0α α α= − + + − + ≡
( )[ ] [ ]
3 32 2
1 yV y yy
α α−∂= +
∂−
Again, the denominator in brackets equals 1 @ 4L , 5L
So, ( ) 3 312 2
Vy
α α⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂
= − ± + ± − ±⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟3
2⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂
3 3 3 3 02 2 2 2
α α= ± ± ≡m m
Thus ( ) ˆ ˆ, 0V VV x y i jx y
∂ ∂+
∂ ∂∇ = ≡ at 4L , . 5L
7.14 Conservation of momentum:
45 o
φ
v’
vo
v’p
4p p p pm v m v m vα′ ′= +o
r r r
cos 45 4 cospv v vα φ′ ′= +oo
4 cos2pv
v vα φ′
′ = −o
0 sin 45 4 sinpv vα φ′ ′= −o
4 sin2pv
vα φ′
′ =
2 216 2 p pv v v v vα′ ′= − +o o2′
Conservation of energy:
2 21 1 1 42 2 2p p pm v m v m v 2
p α′ ′= +o
16 2 24 4 pv v vα′ ′= −o
Subtracting: 2 20 3 2 5p pv v v v′ ′= − − +o o
6
( )2 22 2 60
2 6210 10p
v v v vv± +
′ = =o o o o ±
, so the positive square root is used. 0pv′ > 0.9288pv v′ = o
0.6572p
px py
vv v
′′ ′= = = ov
( ) ( )1 1
2 2 22 21 1 .92882 2p
vv v vα′ ′= − = −oo
0.1853v vα′ = o
.92882tan2 2 .9288
2
p
p
p p
vv
v v vvφ
′′
= = =′ ′− −− oo
1tan 1.9134 62.41φ −= = o
cos 0.086xv vα α vφ′ ′= = o sin 0.164yv vα α vφ′ ′= − = − o 7.15 Conservation of energy:
2 2 21 1 1 1 142 2 2 4 2p p p p pm v m v m v m vα
⎛ ⎞′ ′= + + ⎜ ⎟⎝ ⎠
o o2
2
2 216 3 4 pv v vα′ ′= −o From the conservation of momentum eqn of Prob. 7.14: 2 216 2 p pv v v v vα′ ′= − +o o
2′
Subtracting: 2 20 2 2 5p pv v v v′ ′= − − +o o
( )2 22 2 40
2 4210 10p
v v v vv± +
′ = =o o o o ±
Using the positive square root, since 0pv′ > : 0.7895pv v′ = o
0.5582p
px py
vv v v
′′ ′= = = o
( )1
122 2 2 23 1 .75 .789516 4 2p
vv v vα⎛ ⎞′ ′= − = −⎜ ⎟⎝ ⎠
oo
0.1780v vα′ = o
From the conservation of momentum eqns of Prob. 7.14:
.7895tan2 2 .7895
p
p
vv v
φ′
= =′− −o
7
1tan 1.2638 51.65φ −= = o
cos 0.110xv vα α vφ′ ′= = o sin 0.140yv vα α vφ′ ′= − = − o
7.16 From eqn. 7.6.14, 1sintan
cosθφ
γ θ=
+ φ1 and θ are the scattering angles in
the Lab and C.M. frames respectively.
From eqn. 7.6.16, for Q = 0, 1
2
mm
γ =
sintan 45 1 1 cos4
θ
θ= =
+
o
1 cos sin4
θ θ+ = and squaring …
2 21 1 cos cos 1 cos16 2
θ θ θ+ + = −
2 1 152cos cos 02 16
θ θ+ − =
1 1 152 4 2cos .125 .696
4θ
− ± += = − ±
Since 02πθ< < , 1cos .571 55.2θ −= ≈ o
7.17 From eqn. 7.6.14, 1sintan
cosθφ
γ θ=
+
From eqn. 7.6.18,
12
1 1
2 2
1 1m Q mm T m
γ−
⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
121 1 11 1 0.3015
4 4 4γ
−⎡ ⎤⎛ ⎞= − + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
sintan 45.3015 cos
θθ
=+
o
.3015 cos sinθ θ+ = (since sin cosθ θ> , ) 45θ > o
2 2 2.3015 sin 2sin cos cosθ θ θ= − + θ Using the identity 2sin cos sin 2θ θ θ= 2sin 2 1 .3015 0.9091θ = − = Since 45θ > , : o 2 90θ > o 12 sin .9091 114.62θ −= = o
57.3θ = o
8
φ
P1’
ψ 7.18 Conservation of momentum:
P1( )1 1 2cos cosP P Pφ ψ φ′ ′= + −
( )1 20 sin sinP Pφ ψ φ′ ′= − −
From Appendix B for ( )sin α β+ and ( )cos α β+ : P2’ ( )1 1 2cos cos cos sin sinP P Pφ ψ φ ψ φ′ ′= + +
( )1 20 sin sin cos cos sinP Pφ ψ φ ψ φ′ ′= − −
( )2 2 2 2 2 2 2 21 1 2cos cos cos 2cos cos sin sin sin sinP P Pφ ψ φ ψ φ φ ψ ψ φ′ ′= + + +
( )21 22 cos cos cos sin sinP P φ ψ φ ψ′ ′ φ+ +
( )2 2 2 2 2 2 21 20 sin sin cos 2sin cos cos sin cos sinP Pφ ψ φ ψ φ ψ φ ψ φ′ ′= + − +
( )21 22 sin sin cos cos sinP P φ ψ φ ψ φ′ ′− −
Adding: P P2 2 21 1 2 1 22 cosP P P ψ′ ′′ ′= + +
Conservation of energy:
2 2 2
1 1 2
2 2 2P P P Qm m m
′ ′= + +
( ) ( )2 2 21 1 2 1 2
1 1 2 cos2 2
Q P P P P Pm m
ψ′ ′′ ′= − − =
1 2 cosP PQ ψm
′ ′=
7.19 21 1
12
T m= 1v 21 1
12
T m v′1′=
let 2
1 12
1 1
T vrT v′ ′
= = … ratio of scattered particle to incident particle energy
Looking at Figure 7.6.2 … ( ) ( )1 1 1 1cm cmv v v v v v′ ′ ′ ′⋅ = − ⋅ −
2 2 21 1 12 coscm cmv v v v v 1φ′ ′ ′= + −
hence 2 2 21 1 12cm cmv v v v v γ′ ′ ′= − + where 1cosγ φ=
22
112 2 21 1 1
2cm cmv v vvrv v v
γ′′∴ = − +
but 1 1v v …the center of mass speeds of the incident and scattered particle are the same.
′ =
1 2
1 2 1 1v mv m m
αα
′= =
+ + …from equation 7.6.12 where 2
1
mm
α =
9
1
1 2 1
11
cmv mv m m α
= =+ +
Equation 7.6.11
Thus
( ) ( ) ( )
2
2 2
1 211 1
r α γαα α
= − +++ +
( ) ( ) ( )
121 2
2 21
1 211 1
v rv
α γαα α
′= − +
++ +
Simplifying
122 1 0
1 1r rγ α
α α−⎛ ⎞− + ⎜ ⎟+ +⎝ ⎠
=
Let 2x r= and solving the resulting quadratic for x
( )1
2 2 21 11 1
x γ γ αα α
⎡ ⎤= + − −⎣ ⎦+ +
Squaring
( )
( )1
2 2 2 2 2 22
1 2 1 2 11
r x γ α γ γ αα
⎡ ⎤= = + − + + −⎢ ⎥
+ ⎣ ⎦
Now ( )
( )1
2 2 2 21 22
1
11 1 2 1 2 11
T rT
γ α γ γ αα
⎡ ⎤Δ= − = − + − + + −⎢ ⎥
+ ⎣ ⎦
And, after a little algebra, we get the desired solution
( )
2 212
1
2 2 11 1
TT
γ γ γ αα α
Δ ⎡ ⎤= − + + −⎣ ⎦+ +
7.20 From Equation 7.6.15 … ( ) ( )
1 1 1 1
1 1 2 2 1 2 11m v m v
v m m m m m vγ = =
′ ′+ +
Now we solve for 1
1
vv′
…
12
11
2Tvm
⎛ ⎞= ⎜ ⎟⎝ ⎠
and now solving for 1v′ starting with Equation 7.6.9 …
2 21 1
1 12 2
v vμ μ′ = −Q and using 21
1 2
mvm m
′ =+ 1v we get …
( ) ( )
21 1
1 2 1 2
1 12 1 1
Tm v Q Qm m m m
= − =+ +
−
( ) ( )
21
1 1 2 1 2
21 1
Tv Qm m m m m
⎡ ⎤′ = −⎢ ⎥+ +⎣ ⎦
.
Thus, solving for γ …
10
( )
( ) ( ) ( )
12
111
2 21 12 2
1 1 21 2
2
2 11
T mmm
Tm m m Qm m
γ =⎡ ⎤
+ −⎢ ⎥+⎣ ⎦
Finally…
( )
11
2 21 2
1
11
mm Q m m
T
γ =+⎡ ⎤
−⎢ ⎥⎣ ⎦
7.21 The time of flight, τ = constant—so 1
rv
τ =′
but from problem 7.19 above
2 211 1
1vr v ττ γ γ αα⎡ ⎤′= = + + −⎣ ⎦+
As an example, let 1 1v τ = and we have 1r γ= 1α = pp scattering
22
1 33
r γ γ⎡ ⎤= + +⎣ ⎦ 2α = p – D
23
1 155
r γ γ⎡ ⎤= + +⎣ ⎦ 4α = p – He
24
1 14313
r γ γ⎡ ⎤= + +⎣ ⎦ 12α = p – C
Below is a polar plot of these four curves. 7.22 From eqn. 7.7.6, u gF F mv vm− = +& &
since v = constant, 0v =& m z vλ λ= =& & , λ =mass per unit length ( )gF z gλ=
( )2
uvF zg v v g zg
λ λ λ⎛ ⎞
= + = +⎜ ⎟⎝ ⎠
is equal to the weight of a length uF2vz
g+ of chain.
11
7.23 343
m rπ ρ=
24m r rπ ρ=& & ∝ where v z2r zπ & = & k a constant of proportionality r kz=& &
r r kz= +o From eqn. 7.7.6, mg mv vm= +& &
( )3 3 24 4 43 3
r g r z r kz zπ ρ π ρ π ρ= +&& & &
23kzg z
r= +
&&&
23zz g rzk
= −+ o
&&&
For 0r = , o
23zz gz
= −&
&&
A series solution is used for this differential equation:
2
0
nn
n
z a z∞
=
=∑&
21 ( )
2dz dz dz dz d zz zdt dz dt dz dz
= = ⋅ = =& & & &
&& &
2
1( ) nn
n
d z a nzdz
−= ∑&
2
1nn
n
z a zz
−= ∑&
1 11 32
n nn n
n n
z a nz g a z− −∴ = = −∑ ∑&&
For 1n = : 1 11 32
a g a= −
127
a g=
For 1n ≠ : 1 32 n nna a= −
Since n is an integer, for 0na = 1n ≠
2 27
z g z=&
3 27 7
gz g g zz⎛ ⎞= − =⎜ ⎟⎝ ⎠
&&
12
7.24 From eqn. 7.7.6, , where m and v refer to the portion of the chain hanging over the edge of the table.
mg mv vm= +& &
z m λ= and v where λ is the mass per unit length of chain z= & m zλ=& & and v z =& &&
21 ( )
2dz dz dz dz d zz zdt dz dt dz dz
= = ⋅ = =& & & &
&& &
( )21 ( )
2d zz g z z z
dzλ λ λ
⎛ ⎞= +⎜ ⎟
⎝ ⎠
&& &
2 21 ( )
2d z zz g
dz z= = −
& &&&
Because of the initial condition 0z b= ≠o , a normal power series solution to this differential equation (…as in Prob. 7.22) does not work. Instead, we use the Method of Frobenius …
2
0
n sn
n
z a z∞
+
=
=∑&
( )2
1( ) n sn
n
d z a n s zdz
+ −= +∑&
2
1n sn
n
z a zz
+ −= ∑&
( ) 1 112
n s n sn n
n n
z a n s z g a z+ − += + = −∑ ∑&& −
Equality can be attained for 0na ≠ at n = 0 and n = 3
…otherwise 0 0,3na n= ≠
For , 0n =12
a s a= −o o
2s = −
( ) 3 31 22
n nn n
n n
z a n z g a z− −= − = −∑ ∑&&
For 3n = , 3 312
a g a= −
323
a g=
For all 0n ≠ , 3: ( )1 22 n na n a− = −
, . 0na = 0,3n ≠
2 2 23
z a z gz−= +o&
At 0t = , , and 0z =& z b=
13
2
203
a gb
= +o b
323
a g b= −o
3
22
2 23 3
bz g gz
= − +& z
At z a , = ( )3
2 32 2
2 23 3
b gz g a a ba a
⎛ ⎞= − = −⎜ ⎟
⎝ ⎠& 3
( )123 3
2
23
gz a ba
⎡ ⎤= −⎢ ⎥⎣ ⎦&
7.25 Initially, the upward buoyancy force balances the weight of the balloon and sand.
(1) ( ) 0BF M m g− + =o
Let ( )m m the mass of sand at time t where 0 t= − t t≤ ≤ o
1 tm mt
⎛ ⎞= −⎜
⎝ ⎠o
o
⎟ (2)
The velocity of sand relative to the balloon is zero upon release so 0V in equation 7.7.5 … there is no upward “rocket-thrust.” =
As sand is released, the net upward force is the difference between the initial buoyancy force, FB, and the weight of the balloon and remaining sand. Let be the subsequent displacement of the balloon, so equation 7.7.5 reduces to F = ma
B y
( ) ( )BdvF M m g M mdt
− + = +
and using (1) and (2) above we get
( )
( )( )
M m gtdv m gt gdt M m t m t M m t m t
+= = − +
+ − + −o oo
o o o o o o
whose solution is:
( )( )
ln 1M m gt m tdyv g t
dt m M m t⎛ ⎞+
= = − − −⎜ ⎟⎜ ⎟+⎝ ⎠
o o o
o o o
( )ln 1gy C gt kt dtk
⎛ ⎞= − + −⎜ ⎟⎝ ⎠∫ ,
( )mk
t M m=
+o
o o
( )21 ln 12 1
gt tdtC gt kt gk kt
− − −= −−∫
Integrating by parts
14
( )21 1ln 1 12 1
gt gC gt kt dk k
⎛ ⎞= − − − − − +⎜ ⎟−⎝ ⎠∫ tkt
( ) ( )22
1 ln 1 ln 12
gt gt gC gt kt ktk k k
= − − − + + −
= ( ) ( )22
1 1 ln 12
g gC gt kt ktk k
− + − − +t
but 0y = at so 0t = 0C =
( ) ( )22
1 1 ln 12
gt gy gt ktk k
= − + − − kt
and at t t = o
(a) ( ) ( )2
2 2 2 ln2gt MH M m m M M mm M
⎡ ⎤⎛ ⎞= + + +⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦
oo o o
o om
(b) ( ) ( )lnM mgtv M m
m M+⎡ ⎤
= + −⎢ ⎥⎣ ⎦
ooo o
o
m
(c) letting 1mM
ε = <<o we have
( ) ( ) ( )2
2 2 2 1 ln 12gtH ε ε εε
= + − + +⎡ ⎤⎣ ⎦o ε
( )2 2
22 2 2 1
2 2gt ε εε ε ε εε
⎡ ⎤⎛ ⎞= + − + − + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦o K
3
3
2
6gtH ε≅ o
Similarly:
( ) ( )1 ln 1gtv ε ε εε
= + + −⎡ ⎤⎣ ⎦o
( )2 3
12 3
gt ε εε ε εε⎡ ⎤⎛ ⎞
= + − + − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
o K
12
gt ε≈ o
(d) 327H m= ; 19.8v ms−= 7.26 or m m m = −& k k t= −o
Burn-out occurs at time mTk
ε= o
So – the rocket equation (7.7.7) becomes
dvm Vdt
= − &m (-) since V is oppositely directed to v&
15
dv Vkdt m k t
= +−o
Thus
( ) 1lndtv Vk V m k t Cm k t
= = − −−∫ o
o
+
t= =
where C is a constant. 1
Now, v so C V m0@ 0 1 ln o =
Hence ( )lnm kt
v V , m−⎡ ⎤
= − ⎢ ⎥⎣ ⎦
o
o
0 mtkε
≤ ≤ o
Let y be the displacement at the time t so
( )2ln
m kty V dt C
m−⎡ ⎤
= − +⎢ ⎥⎣ ⎦
∫ o
o
Integrating the above expression by parts
( )2ln
m k t t dty Vt Vk Cm m k t−⎡ ⎤
= − − +⎢ ⎥ −⎣ ⎦∫o
o o
( )2ln= − 1
m k t mVt V dt Cm m k t−⎡ ⎤ ⎛ ⎞
+ − +⎜ ⎟⎢ ⎥ −⎝ ⎠⎣ ⎦∫o o
o o
( ) ( ) 2ln= − lnm k t VmVt Vt m k t C
m k−⎡ ⎤
+ + − +⎢ ⎥⎣ ⎦
o oo
o
= since 0y = at t , 0 2lnVm mC
k−
= o o and we have
( ) ( )lnm k tVy Vt m k t
k m−⎡ ⎤
= + − ⎢ ⎥⎣ ⎦
oo
o
At burn-out mt T εk
= = o so
(a) ( ) ( ) ( )1 ln 1m Vy Dk
ε ε ε ε= = + − −⎡ ⎤⎣ ⎦o
(b) ε cannot exceed 1.0 although it can approach 1.0 for small payloads
Thus ( )max
lim1
m Vy yk
εε
= =→
o
7.27 From eqn. 7.7.5, kv mv Vm− = −
rr r& & r
Since V is opposite in direction from vr − = kv mv Vm+& &
m mv v Vk k
= − −&
& vV
mk
α =&
and since 0m <& , mk
α = −&
16
m m dv m dv dm dvv mm m dt m dm dt dm
α α α α α− = = = =&& & &
v V
dm dvm v Vα α=
−
0
1 m v
m
dm dvm v Vα α
=−∫ ∫
o
1 ln lnm v Vm V
αα α
⎛ ⎞ −⎛ ⎞=⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠o
1
1m vm V
α
α⎛ ⎞
= − +⎜ ⎟ ⎝ ⎠o
1
1 mv V m
αα⎡ ⎤
⎛ ⎞⎢ ⎥= − ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦o
7.28 From eqn. 7.7.5, mg mv Vm− = −
rr r& &
V
v since V is opposite in direction from
rvr ,
− = mg mv Vm+& &
+ − = mgdt mdv Vdm
dmm so dt
=&dmdt = m&
dmmg mdv Vdmm
− = +&
g Vdv dmm m
⎛ ⎞= − +⎜ ⎟⎝ ⎠&
0
e pv m
m
g Vdv dmm m
⎛= − +⎜⎝ ⎠∫ ∫
o &
⎞⎟ mp =payload mass
( ) lne pp
mgv m m Vm m
= − + oo&
o
f pm m m m= − ≈o mf = fuel mass
ln 1 fe f
p
mgv m Vm m
⎛ ⎞= + +⎜⎜
⎝ ⎠&⎟⎟
ln 1 f e
p
m v g mm V V m
⎛ ⎞+ = −⎜ ⎟⎜ ⎟
⎝ ⎠o
&
exp 1f e
p
m v g mm V V m
⎛ ⎞= −⎜ ⎟⎝ ⎠
o
&−
17
For , eV kv=1exp 1f
p e
m g mm k kv m
⎛ ⎞= −⎜ ⎟
⎝ ⎠o
&−
From chap. 2, Section 2.3 … 11ekmvs
≈
For 10.01m m −= o& s and 1
4k = :
( )( )
9.8exp 4 11 11,000 .014
f
p
mm
⎡ ⎤⎢ ⎥
= −⎢ ⎥⎢ ⎥−⎣ ⎦
−
77f
p
mm
=
7.29 We can use Equation 7.7.9 to calculate the final velocity attained by the ion rocket during the 100 hour burn. Assuming the rocket starts from rest (even if the ion rocket is turned on while in Earth orbit, the initial rocket speed 4
0 10 0v c−≈ ≈ . Thus …
0lnp
mv Vm
≈ and 0 2 3F p p pm m m m m m= + = + = p
. The final rocket velocity is a little more than 10% c. ln 3 0.1099v V c≈ =
4 36.40.1099 0.1099
L LYT yc LY yr
= = = r
7.30 We again use Equation 7.7.9 …
0ln ln ln 2F pion ion ion
p p
m mmv V V Vm m
+≈ = = for the ion rocket. For the chemical
rocket …
ln Fchem
p
m mv V
m+
≈ p . Setting these two equations equal …
ln ln 2F pchem ion
p
m mV
m+
=V . Solving for mF …
( ) ( )5 40.1 10 10 5002 2 2 1ion chemc cV VF p
p
m mm
−+= = = ≈ 0 which demonstrates the virtue of
ejecting mass at high velocity! ----------------------------------------------------------------------------------------------------------
18
1
CHAPTER 8 MECHANICS OF RIGID BODIES:
PLANAR MOTION
8.1 (a) For each portion of the wire having a mass3m and centered at
,2 2b b⎛ ⎞−⎜ ⎟
⎝ ⎠, ( )0,0 , and ,
2 2b b⎛ ⎞
⎜ ⎟⎝ ⎠
…
b
b x
y
1 0 02 3 2 3cmb m b mx
m⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦
=
1 02 3 2 3 3cmb m b m by
m⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= + + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦
(b) ( )1
2 2 2ds xdy b y dy= = −
( )1
2 2 20
1 b
cmy y b y dym
ρ= −∫
( ) ( )
12 2 2 22
0
2
214
y b
ycm
b y d b yy
b
ρ
π ρ
=
=− − −
=∫
4y3cm
bπ
=
From symmetry, 4x3cm
bπ
=
(c) The center of mass is on the y-axis.
( )122 2ds xdy by dy= =
( )
( )
3122
0 01 12 2
0 0
2
2
b b
cm b b
y by dy y dyy
by dy y dy
ρ
ρ= =∫ ∫∫ ∫
3y = 5cmb
(d) The center of mass is on the z-axis.
( )2 2 2dv r dz x y dz bzdzπ π π= = + =
1
2
2
0 0
0 0
b b
cm b b
z bzdz z dzz
bzdz zdz
ρ π
ρπ= =∫ ∫∫ ∫
23cmz b=
(e) The center of mass is on the z-axis.
α is the half-angle of the apex of the cone. is the radius of the base at z = 0 and is radius of a circle at some arbitrary z in a plane parallel to the base.
ror
tan r rb b z
α = =−
o , a constant
( )22 2tandv r dz b z dzπ π α= = −
2 31 1 tan3 3
m r b b 2ρ π πρ= =o α
( )( )
2 22 2 30
3 03 2
tan 3 21 tan3
bb
cm
z b z dzz b
bb
ρ π α
πρ α
−= = −∫
∫ z bz z dz+
4cmbz =
8.2 2
0
0
b
cm b
cx dxxdxx
dx cxdx
ρ
ρ= = ∫∫∫ ∫
23cmbx =
8.3 The center of mass is on the z-axis. Consider the sphere with the cavity to be made
of a (i) solid sphere of radius and mass a sM , with its center of mass at , and (ii) a solid sphere the size of the cavity, with
mass
0z =
cM− and center of mass at 2az = − . The actual sphere with
the cavity has a mass s cm M M− and center of mass . cmz=
102c c
s
amM m z
M⎡ ⎤⎛ ⎞= − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
343sM aπ ρ= ,
343 2c
aM π ρ⎛ ⎞= ⎜ ⎟⎝ ⎠
3 33
3
102 2 2 cma a aa z
a
⎧ ⎫⎡ ⎤⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + −⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
2
3
14cmaz =
8.4 (a) 2 2
2 03 2 2z i i
i
m b bI m R⎡ ⎤⎛ ⎞ ⎛ ⎞= = + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
∑
2
6zmbI =
(b) ds rd drθ= , sinR r θ= 2
zI R dsρ= ∫
2 240
4
sinr b
z rI r rdr d
πθ
πθρ θ θ
= =
= =−= ∫ ∫
4
24
4
sin4zbI d
π
πρ θ θ
−= ∫
2 sin 2sin2 4
d θ θθ θ⎡ ⎤= −⎢ ⎥⎣ ⎦∫
4 1
4 4 2zbI ρ π⎛ ⎞= −⎜ ⎟
⎝ ⎠
214
m bρπ=
( )2
24z
mbI ππ
= −
(c) 2xds hdx b dx
b⎛ ⎞
= = −⎜ ⎟⎝ ⎠
Where the parabola intersects the line , y b=
( )12x by b= = ±
2 4
2 2b b
y b b
x xI x b dx bx dxb b
ρ ρ− −
⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∫ ∫
4415yI bρ=
2
243
b
b
xm b dxb
bρ ρ−
⎛ ⎞= − =⎜ ⎟
⎝ ⎠∫
215yI mb=
(d) dv 2 RhdRπ=
h b z= −
3
4
( ) ( )1 1
2 2 2 2R x y bz= + =
121
2bdR dzz
⎛ ⎞= ⎜ ⎟⎝ ⎠
( ) ( )1
1 22 20
122
b
zbI R dv bz bz b z dzz
ρ ρ π ⎛ ⎞= = − ⎜ ⎟⎝ ⎠∫ ∫
( )2 2
0
16
b
z5I b bz z dz bπρ π= − =∫ ρ
( ) ( )1
1 22
0
122
b bm dv bz b zz
ρ ρ π ⎛ ⎞= = − ⎜ ⎟⎝ ⎠∫ ∫ dz
( ) 3
0
12
bm b b z dz bπρ π= − =∫ ρ
213zI mb=
(e)
α is the half-angle of the apex of the cone. r is the radius of the base at z = 0 and is radius of a circle at some arbitrary z in a plane parallel to the base.
o
r
tan R Rb b z
α = =−
o , a constant
( )2 2b z R
dv RhdR zdRb
π π−
= = o
Since ( )b z RR = ,
b− o RdR dz
b= − o , and the limits of integration for 0R R= → o
correspond to z b 0= →
( ) ( )2 202
2 2z b
b z R b z R RI R dv z dzb b
ρ ρ π− − ⎛ ⎞= = −⎜ ⎟
⎝ ⎠∫ ∫ o o o
b
( )4
3 2 2 3 44 0
12 3 310
b
zR 4I b z b z bz z dz R bb
πρ πρ= + − + − =∫oo
213
m Rρ π= o b
2310zI mR= o
4
5
8.5 Consider the sphere with the cavity to be made of a (i) solid sphere of radius and and mass
asM , with its center of mass at 0z = , and (ii) a solid sphere the size of the
cavity, with mass cM− and center of mass at 2az = − . The actual sphere with the cavity
has a mass s cm M M= − and center of mass . cmz
3
3 34 4 7 43 3 2 8 3s c
am M M a aπ ρ π ρ π⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
ρ
87sM m= and 1
7cM m=
From eqn. 8.3.2, s cI I I= +
2
22 25 5 2s c
aM a M ⎛ ⎞ I= +⎜ ⎟⎝ ⎠
2
2 22 8 2 1 315 7 5 7 4 70
aI m a m ma⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
8.6 The moment of inertia about one of the straight edges is
2zI R dvρ= ∫ where 2 2 2R x y= + . From Appendix F …
2 sindv r dr d dθ θ φ= z
x
y
R
r θ
φ
2 2 2 2 2sinR x y r θ= + = Let a = radius of sphere
2 2 22 20 0 0
sin sinr a
z rI r r dr d
π πθ φ
θ φdθρ θ θ
= = =
= = == ∫ ∫ ∫ φ
4 320 0
sin2
r a
z rI r drd
πθ
θ
πρ θ θ= =
= == ∫ ∫
5 320
1 sin10zI a d
π
ρπ θ= ∫ θ
33 cossin cos
3d θθ θ θ
⎡ ⎤= −⎢ ⎥
⎣ ⎦∫
5230zI aρπ=
3 31 4 18 3 6
m a aπ ρ π= = ρ
225zI ma=
5
6
8.7 For a rectangular parallelepiped:
dv hdxdy= h is the length of the box in the z-direction
2a
2b
y
x
z
( )1
2 2 2R x y= +
( )2 2 2x a y b
z x a y bI R dv x y hdxdyρ ρ
= =
=− =−= = +∫ ∫ ∫
( )3
2 22 423 3
a
z a
b 2I h bx dx hab a bρ ρ−
⎛ ⎞= + =⎜ ⎟
⎝ ⎠∫ +
( )( )2 2 4m a b h abhρ ρ= =
( )2 2
3zmI a b= +
For an elliptic cylinder:
Again dv hdxdy= , and ( )1
2 2 2R
x
y
z
2b
2a
x y= +
On the surface, 2 2
2 2 1x ya b
+ =
12 2
21 yx a b
⎛ ⎞= ± −⎜ ⎟
⎝ ⎠
( )1
2 2
2
12 2
2
12 2
1
yx ay b bz y b yx a
b
2I R dv x y hdxdyρ⎛ ⎞
= −⎜ ⎟= ⎜ ⎟⎝ ⎠
=− ⎛ ⎞=− −⎜ ⎟⎜ ⎟
⎝ ⎠
= = +∫ ∫ ∫
3 12 22 2
3 22 2
2 1 2 13
b
z b
y yI h a a yb b
ρ−
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
∫ dy
( ) ( )3 12
2 2 2 2 22 222
3b b
b b
a ah b y dy b y yb b
ρ− −
dy⎡ ⎤
= − + −⎢ ⎥⎣ ⎦
∫ ∫
From a table of integrals:
( ) ( ) ( )3 3 1
2 2 2 2 2 2 2 4 12 2 23 3 sin4 8 8y yb y dy b y b y b y b
b−− = − + − +∫
( ) ( ) ( )1 3 12 4
2 2 2 2 2 2 2 12 2 2 sin4 8 8y b y b yb y y dy b y b y
b−− = − − + − +∫
( )2 4
4 22
3 123 8 8 4z
a a b 2I h b h ab ab b
ρ π π ρ π⎡ ⎤⎛ ⎞= + =⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦b+
6
7
( )m h abρ π=
( )2 2
4zmI a b= +
For an ellipsoid: , dv hdxdy= 2 2 2R x y= + and on the surface, 2 2 2
2 2 2 1x y za b c
+ + = 1
2 2 2
2 21 x yz ca b
⎛ ⎞= ± − −⎜ ⎟
⎝ ⎠
x
y
z 2a
2b
12 2 2
2 22 2 1 x yh z ca b
⎛ ⎞= = − −⎜ ⎟
⎝ ⎠
In the xy plane 2 2
2 2 1x ya b
+ = 1
2 2
21 yx ab
⎛ ⎞= ± −⎜ ⎟
⎝ ⎠
( )1
2 2
2
12 2
2
12 21 2
2 2 22 2
1
2 1yx ay b b
z y b yx ab
x yI R dv x y c dxdya b
ρ ρ⎛ ⎞
= −⎜ ⎟= ⎜ ⎟⎝ ⎠
=− ⎛ ⎞=− −⎜ ⎟⎜ ⎟
⎝ ⎠
⎛ ⎞= = + − −⎜ ⎟
⎝ ⎠∫ ∫
1 12 2 2 22 2
2 2 2 2 2 22 2
2 y b
z y b
c a y a yI a x x dx y a xa b bρ =
=−
⎧ ⎫⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞⎪ ⎪= − − + − −⎨ ⎬⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭
∫ ∫ ∫ dx
From a table of integrals:
( ) ( ) ( )3
1 12 422 2 2 2 2 2 2 12 2 sin
4 8 8x k x k xk x x dx k x k x
k−− = − − + − +∫
( ) ( )1 1 2
2 2 2 2 12 2 sin2 2x k xk x dx k x
k−− = − +∫
24 2 2 2
22 2
2 1 18 2
b
z b
c a y a yI y da b bρ π π
−
⎡ ⎤⎛ ⎞ ⎛ ⎞= − + −⎢ ⎥⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∫ y
2 2 4 4
22 4 2
214
b
z b
a y y yI ac y dyb b b
ρπ−
⎡ ⎤⎛ ⎞= − + + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦∫
( )2 2415zI abc a bρπ= +
For an ellipsoid, 43
m abcρ π= , so ( )2 2
5zmI a b= +
7
8
8.8 (See Figure 8.4.1) Note that l l′ + is the distance from 0′ to 0 , defined as dFrom eqn. 8.4.13, 2
cmk ll′= ( )2 2 2
cmk l ll l l l l′ ′+ = + = +
2 2cmk l l+ = d
2
d=
From eqn. 8.4.9b, 2 2cmk k l= +
k l 2
From eqn. 8.4.6, 2
2 kT gl
π=o
2 dT g
π=o
8.9 Period of a simple pendulum: 2 aTM
π=
Period of real pendulum: 2 ITMgl
π=o (eqn. 8.4.5)
Where I =moment of inertia distance to CM of physical pendulum l = distance to CM of bob a =
radius of bob b = location of CM of physical pendulum:
( ) ( )( ) ( )2
2 2 2
a bm M m a m a b ml a a
m M m M M
−+ − ⎛ ⎞= = − + = −⎜ ⎟+ − ⎝ ⎠
a b+
12 2m m bl aM M a
⎡ ⎤= − −⎢ ⎥⎣ ⎦
Moment of inertia:
( ) ( ) ( )2 2 22 25 5bob
2I M m a M m b M m a b⎛ ⎞= − + − = − +⎜ ⎟⎝ ⎠
2
22
21 15
m bMaM a
⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
( )2
2 21 1 13 3rod
m bI m a b MaM a⎡ ⎤⎛ ⎞= − = −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
22
22
2 11 1 15 3bob rod
m b m bI I I MaM a M a
⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞∴ = + = − + + −⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠ ⎦
⎣
letting mM
α = and ba
β =
8
9
( ) ( )
1222 22 11 1 1
5 321
2 2
MaT
Mga
α β α βπ
α αβ
⎧ ⎫⎡ ⎤⎛ ⎞− + + −⎜ ⎟⎪ ⎪⎢ ⎥⎪ ⎪⎝ ⎠⎣ ⎦= ⎨ ⎬⎡ ⎤⎪ ⎪− −⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
o
(a) ( ) ( )222 11 1 1
15 3 1121
2 2
TT
α β α βα
α αβ
⎛ ⎞− + + −⎜ ⎟⎝ ⎠= ≈⎛ ⎞− −⎜ ⎟⎝ ⎠
o − to 1st order inα
(b) 10m g= 1M kg= 1.27a m= 5b cm= 0.01α = 0.0394β =
11 0.999212
TT
α≈ − =o (actually 0.9994 using complete expression)
8.10 The period of the “seconds” pendulum is
2 2 2IT sMgl
π= =
The period of the modified pendulum is
2 220
I nTM gl n
π′ ⎛ ⎞′ = = ⎜ ⎟′ ′ −⎝ ⎠
where I ′ , M ′ , l refer to parameters of pendulum with m attached and ′ n ( )24 60 60= × × is the number of seconds in a day.
2mI I ml′ = + mMl mll
M+′ =
where is the distance of the attached mass from the pivot point. ml m
So ( )
2 2 22201 m
m
2I mlIn M gl Ml ml
π ππ− ′ +⎛ ⎞− = =⎜ ⎟ ′ ′ +⎝ ⎠ g
( )
2 2m
m
Mgl mlMl ml g
π+=
+
Thus ( )( )
2
2 2
201 m
m
Ml ml gn Mgl mlπ
+⎛ ⎞− =⎜ ⎟ +⎝ ⎠
Or 2 2
1401
1
m
m
ll
n lgl
α
π α
⎛ ⎞+⎜ ⎟⎝ ⎠− ≈⎛ ⎞+⎜ ⎟
⎝ ⎠
mM
α =
Solving for α gives the approximate result
9
10
2
40
1
m
m
ln lm
M lg
απ
= ≅⎛ ⎞
−⎜ ⎟⎝ ⎠
Letting ; ; we obtain 1.3ml m= 1.0l = m 31.15 10α −≅ ⋅
2cmI ma ( l mass8.11 (a) ⊥ = al in rim)
22 2 2rimI ma ma ma⊥ = + =
22 2rimI aTmga g
π π⊥∴ = =
(b) z x y cm
22I I ma= + = I= ( )cmI⊥= I2
2cmmaI∴ =
hence 2
2 232 2rim
maI ma ma= + =
32 22
rimI aTmga g
π π∴ = =
8.12
cmmx mg T= −&&
a
x
Tr
mgr
cmI aTω =& cmx aω=&& &
225cmI ma=
21 2 25 5
cm cmcm cm
I xmx mg mg ma mg mxa a aω ⎛ ⎞= − = − = −⎜ ⎟
⎝ ⎠
& &&&& &&
57cmx g=&&
8.13 When two men hold the plank, each supports 2
mg .
When one man lets go: cmmg R mx− = && and 2 cmlR I ω= &
From table 8.3.1, 2
12cmmlI =
2
12 62Rl R
ml mlω = ⋅ =&
10
11
32cml Rx
mω= =&& &
3 3Rmg R m Rm
⎛ ⎞− = =⎜ ⎟⎝ ⎠
4
mgR =
6 64end
R mgx l lml m
ω ⎛ ⎞= = = ⎜ ⎟⎝ ⎠
&& &
32endx g=&&
8.14 For a solid sphere:
343sM aπ ρ= and 22
5s sI M a=
( )2 225cm s
k a=
For subscript c representing a solid sphere the size of the cavity, from eqn. 8.3.2: s cI I I= +
3 2
3 2 52 4 2 4 2 4 315 3 5 3 2 2 5 3 32
a aI a a aπ ρ π ρ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = ⋅ ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦π ρ
3
3 34 4 4 73 3 2 3 8
am a aπ ρ π ρ π⎛ ⎞= − = ⋅⎜ ⎟⎝ ⎠
ρ
2 22 31 8 315 32 7 70cm
Ik am
= = ⋅ =⋅ 2a
From eqn. 8.6.11, for a sphere rolling down a rough inclined plane:
2
2
sin
1cm
cm
gxka
θ=
⎛ ⎞+ ⎜ ⎟⎝ ⎠
&&
( )2
2
2
2
1
1
cm s
cms
kx a
kxa
+=
+
&&
&&
2 715 531 101170 70
+= =
+
98101s
xx=
&&
&&
11
12
8.15 Energy is conserved:
m1
m2
x
a
22 2
1 2 2 11 1 12 2 2 2
xm x m x I m gx m gx E⎛ ⎞+ + + −⎜ ⎟⎝ ⎠
&& & =
1 2 2 12 0Im xx m xx xx m gx m gxa
+ + + − =&&& &&& &&& & &
( )1 2
1 2 2
m m gx Im m
a
−=
+ +&&
8.16 While the cylinders are in contact:
2
coscmr
mvf mg Rr
θ= = −
vcm
a
b
Rr
mgr
r a b= + , so 2
coscmmv mg Ra b
θ= −+
From conservation of energy:
( ) ( )2 21 1 cos2 2cmmg a b mv I mg a bω θ+ = + + +
From table 8.3.1, 212
I ma=
cmva
ω =
( )(2
2 22
1 1 1 1 cos2 2 2
cmcm
vmv ma mg a ba
)θ⎛ ⎞⎛ ⎞+ = +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
−
( )2 4 1 cos
3cmmv mg
a bθ= −
+
When the rolling cylinder leaves, 0R = :
( )4cos 1 cos3
mg mg θ= − θ
7 4cos3 3
θ =
1 4cos7
θ −=
8.17 2mx N=&&
1my N mg= −&&
12
13
2 21 12 2cmmv I mgy mgyω+ + = o
cos2lx θ= , sin
2lx θ θ= − && , ( )2 cos sin
2lx θ θ θ θ= +& &&&&
sin2ly θ= , cos
2ly θ θ= && , ( )2 sin cos
2ly θ θ θ θ= − +& &&&&
2 2 2 2
2 2 2 sin cos2 2cml lv x y
4l θθ θ θ θ⎛ ⎞ ⎛ ⎞= + = − + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
&& && &
2
12mlI = , and ω θ= &
2 2 2
21 1 sin sin2 4 2 12 2 2
l ml l lm mg mgθ θ θ θ+ + = o
&&
( )2 sin sin3l gθ θ θ= −o&
( )123 sin sing
lθ θ⎡ ⎤= −⎢ ⎥⎣ ⎦
o& θ
( ) ( )121 3 3 3sin sin cos cos
2 2g gl l
gl
θ θ θ θ θ−
⎡ ⎤ ⎛ ⎞= − − = −⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠o
&& & θ
( )23 3cos sin sin sin cos
2 2ml g gN mx
l lθ θ θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞= = − − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
o&& θ
Separation occurs when 2 0N = :
1sin sin sin 02
θ θ θ− − =o , 1 2sin sin3
θ θ− ⎛ ⎞= ⎜ ⎟⎝ ⎠
o
8.18 xR mx= && yR mg my− = &&
sin2lx θ= , cos
2lx θ θ= && , ( )2 sin cos
2lx θ θ θ θ= − +& &&&&
cos2ly θ= , sin
2ly θ θ= − && , ( )2 cos sin
2ly θ θ θ θ= − +& &&&&
2 21 1 cos2 2 2cm
l lmv I mg mgθ θ+ + =&2
2cmlv θ= & ,
2
12mlI =
13
14
( )2 2 2
21 1 cos2 4 2 12 2m l ml lmgθ θ θ+ = −
&&
( )2
1 cos3
l gθ θ= −&
( )123 1 cosg
lθ θ⎡ ⎤= −⎢ ⎥⎣ ⎦&
( )121 3 3 31 cos sin sin
2 2g gl l
gl
θ θ θθ−
⎡ ⎤ ⎛ ⎞= − =⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠&& & θ
( ) ( )3 3sin 1 cos cos sin2 2x
ml g gRl l
θ θ θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞= − − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
( )3 sin 3cos 24xmgR θ θ= −
( )3 3cos 1 cos sin sin2 2y
ml g gR mgl l
θ θ θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞= − − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
2
23 scos cos2 2ymgR mg in θθ θ
⎛ ⎞= − − +⎜ ⎟
⎝ ⎠
( )23cos 14y
mgR θ= −
The reaction force constrains the tail of the rocket from sliding backward for : 0xR > 3cos 2 0θ − >
1 2cos3
θ −<
The rocket is constrained from sliding forward for 0xR < :
1 2cos3
θ −>
8.19 sin cosmx mg mgθ μ θ= − −&&
( )sin cosx g θ μ θ= − +&&
Since acceleration is constant, 212
x
θ mgr
fr
x x t xt= +o& && :
( )2
sin cos2
gtx v t θ μ θ= − +o
Meanwhile ( ) 22cos5
mg a I maμ θ ω= =& &ω
5 cos2
ga
μ θω =&
5 cos2
g ta
μ θω =
14
15
The ball begins pure rolling when v aω= …
( ) 5 cossin cos2
gv v xt v g t a ta
μ θθ μ θ= + = − + =o o&&
( )
22sin 7 cos
vtg θ μ θ
=+o
At that time:
( )
( )( )
22
22
4 sin cos22sin 7 cos 2 2sin 7 cos
vv gxg g
θ μ θθ μ θ θ μ θ
+= −
+ +oo
( )( )
2
2
sin 6 cos22sin 7 cos
vxg
θ μ θ
θ μ θ
+=
+o
8.20 mx mgμ=&& x gμ=&&
x gtμ=& , and 212
x gtμ=
225
I ma mgaω ω μ= = −& &
52
gaμω = −&
52
g taμω ω= −o
Slipping ceases to occur when v aω= …
52
gt a gtμ ω μ= −o
27
atgω
= o
2
1 22 7
ax ggωμμ
⎛ ⎞= ⎜ ⎟
⎝ ⎠o
2 22
49ax
gωμ
= o
8.21 Let the moments of inertia of A and be B 212a aI M a⎛ ⎞=⎜ ⎟
⎝ ⎠and 21
2b bI M b⎛ ⎞=⎜ ⎟⎝ ⎠
.
The angular velocity of A is α& while that of is B β α φ− +& & & (remember that in two dimensions, angular velocity is the rate of change of an angle between an line or direction fixed to the body and one fixed in space). For rolling contact, lengths traveled along the perimeters of the disks A and must be equal to the arc length traveled along the track C.
B
( )( )2a b a bφ β α= = + −φ
15
16
so that ( )( )
22a b
a bα
φ+
=+
and ( )( )
22
a a bb a b
αβ
+=
+
After some algebra … the angular velocity of is found to be … B
2Ba
bαω β α φ= − + =&& &&
For A , we take moments about O and for we take moments about its center. Call
BAT and the components of the reaction forces tangent to BT A and (the
“upward-going” B
AT acts on disk . The “downward-going” B AT acts on disk A) Thus A AK T a I α− = && (Torque on Disk A)
( ) 2A B B BT b T b I I abαβ α φ− − = − − + = −&&&& &&&& (Torque on Disk B)
( )( ) 12A B B BT T M a b M aα φ− = + − =&&&& &&α (Force on Disk B)
Eliminate AT and BT
( )2 2 2 22 4
4 A B BK b I M a b ab
Iα= + +
&&
Integrating this equation gives :
( )2 2 2 22 4
4 A B BKt b I M a b a Ibα
= + +&
Putting Aω α= & at t gives t= o
( )2
2 2 2 2
44A
A B B
b Ktb I M a b a I
ω =+ +
o
Putting in values for AI and BI gives
2
4322
A
A B
Kt
a M Mω =
⎛ ⎞+⎜ ⎟⎝ ⎠
o
Since the angular velocity of is B2Ba
bαω β α φ= − + =&& && , we have
232 22
B A
A B
a Ktb ab M M
ω ω= =⎛ ⎞+⎜ ⎟⎝ ⎠
o
16
17
8.22 From section 8.7 (see Figure 8.7.1), the instantaneous center of rotation is the
point . If O x is the distance from the center of mass to and O2l is the distance from
the center of mass to the center of percussion O′ , then from eqn. 8.7.10 …
2 21
2 12cmIl Mlx
M M⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 12
l
6lx =
8.23 In order that no reaction occurs between the table surface and the ball, the ball must approach and recede from its collision with the cushion by rolling without slipping. Using a prime to denote velocity and rotational velocity after the collision:
ˆ
cm cmPv vm
′ = −
( )ˆ
cm cmP h aI
ω ω′ = − −
The conditions for no reaction are cm cmv aω= and cm cmv aω′ ′= .
( )ˆ ˆ
cm cmP Pa a hm I
ω ω⎡ ⎤
− = − −⎢ ⎥⎣ ⎦
a
( )1 1 a h am I= −
225
I ma=
22 7
5 5mah ama
= + = a
2da =
710
h d=
8.24 During the collision, angular momentum about the point 0 is conserved: m v l Iθ′ ′ =o
&
m v lI
θ′ ′
= o&
After the collision, energy is conserved:
21 cos cos2 2 2
l lI mg m gl mg m glθ θ θ′ ′ ′ ′− − = − −o o&
( )2 2 21 1 cos
2 2m v l mlg m
Iθ
′ ′ ⎛ ⎞l′ ′= − +⎜ ⎟⎝ ⎠
oo
17
18
2
2
3mlI m l′ ′= +
( )1
2 221 2 1 cos
2 3ml mlv g m l m l
m lθ
⎡ ⎤⎛ ⎞⎛ ⎞′ ′ ′ ′= − + +⎢ ⎥⎜ ⎟⎜ ⎟′ ′ ⎝ ⎠⎝ ⎠⎣ ⎦o o
8.25
The effect of rod BC acting on rod AB is impulse + P′ . The effect of AB on BC is P′− . 1
ˆ ˆmv P P′= + 2ˆmv P′= −
(1ˆ ˆ
2l )I P Pω ′= − 2
ˆ2lI Pω ′= −
2
12mlI =
(16 ˆ ˆP P
mlω ′= − ) 2
6 Pml
ω ′= −
1 12Blv v ω= − 2 22B
lv v ω= +
( )ˆ ˆ 6 ˆ ˆ2B
P P lv Pm ml
′+ ⎛ ⎞ P′= − −⎜ ⎟⎝ ⎠
ˆ 6 ˆ
2BP lv Pm ml′ ⎛ ⎞′= − + −⎜ ⎟
⎝ ⎠′
( )ˆ ˆ ˆ ˆ ˆ ˆ3 3P P P P P P′ ′ ′+ − − = − −
ˆ ˆ8 2P P′ =
ˆˆ4PP′ =
1
ˆ54
Pv m
= 2
ˆ
4Pvm
= −
1
ˆ92
Pml
ω = 2
ˆ32
Pml
ω = −
ˆ
BPv =m
−
-----------------------------------------------------------------------------------------------------------
18
CHAPTER 9 MOTION OF RIGID BODIES IN
THREE DIMENSIONS 9.1 (a) ( )2 2
xxI y z dm= +∫
dm dxdyρ= and 22m a ρ=
x
y
z
a
2a
( )2 2
0 00
y a x a
xx y xI y dxρ
= =
= == +∫ ∫ dy
∫
= 2
02
aa y dyρ
242
3 3xxmaI aρ= =
( ) 22 2 2
0 0
y a x a
yy y xI x z dm x dxdρ
= =
= == + =∫ ∫ ∫ y
3
0
83
a a dyρ= ∫
4 28 43 3yy
a mI ρ= =
a
From the perpendicular axis theorem:
25
3zz xx yymaI I I= + +
2
0 0
y a x a
xy yx y xI I xydm xy dxρ
= =
= == = − = −∫ ∫ ∫ dy
2
0
42
a a ydyρ= − ∫
2
4
2xy yxmaI I aρ= = − = −
0xz zx yz zyI I xzdm I= = − = = =∫ I
(b) 2cos , 5
α =1cos5
β = , cos 0γ =
From equation 9.1.10 …
2 2 24 4 1 2 12
3 5 3 5 2 5 5ma ma maI
⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
2215
ma=
(c) ( ) ( )ˆ ˆ ˆcos cos 25
i j i jωω ω α β= + =r
+
From equation 9.1.29 …
1
2 2 22 2ˆ ˆ
3 2 25 5 5 5ma ma ma maL i jω ω ω ω⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞
= ⋅ + − + − + ⋅⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣
r 243
⎤⎥⎦
( )2
ˆ ˆ26 5maL i jω
= +r
(d) From equation 9.1.32:
( )2
2 21 12 22 12 5 6 5
maT L maω ω5
ω ω= ⋅ = ⋅ + =rr
9.2 (a) ( )ˆˆ ˆ3
i j kωω = + +r
( )222 22 2
12 3xx rod yy
m a
a -a
a
-a
a
-a
zzI I ma I I= = = = =
0xyI xydm= − =∫ since, for each rod, either x or or both are 0. The same is true for the other
products of inertia. y
From equation 9.1.29:
( )22 ˆˆ ˆ3 3
L ma i j kω= ⋅ + +
r
From equation 9.1.32,
( )2 2
2 2 21 2 1 1 12 33 3 3
ma maT2ω ω ω
= + + =
(b) From equation 9.1.10, with the moments of
inertia equal to 22
3ma and the products of inertia equal to 0:
( )2
2 2 22 2cos cos cos3 3
ma 2I maα β γ= + + =
(c) For the x-axis being any axis through the center of the lamina and in the plane of the lamina, and the y-axis also in the plane of the lamina …
xx yyI I= due to the similar geometry of the mass distributions with respect to the x- and y-axes.
From the perpendicular axis theorem:
zz xx yyI I I= + 2zz xxI I=
From Table 8.3.1, 2
6zzmaI =
2
12xxmaI =
2
9.3 (a) From equation 9.2.13, 2
tan 2 xy
xx y
I
yI Iθ =
−
From Prob. 9.1, 2
3xxmaI = ,
243yy
maI = , 2
2xymaI = −
tan 2 1θ = 2 45 22θ θ= =o o.5 The 1-axis makes an angle of with the x-axis. 22.5o
(b) From symmetry, the principal axes are parallel to the sides of the lamina and perpendicular to the lamina, respectively.
9.4 (a) From symmetry, the coordinate axes are principal axes.
From Table 8.3.1:
x
y
z ωr
( ) ( )2 2 21
132 312 12mI a a ma⎡ ⎤= + =⎣ ⎦
( )22 22
10312 12mI a a ma⎡ ⎤= + =⎣ ⎦
( )22 23
5212 12mI a a ma⎡ ⎤= + =⎣ ⎦
( )1 2 3ˆ ˆ ˆ2 314
e e eωω = + +r
From equation 9.2.5,
2 2
2 2 21 13 10 4 5 92 12 14 12 14 12 14
2
a ma maT m ω ω ω⎡ ⎤= ⋅ + ⋅ + ⋅⎢ ⎥
⎣ ⎦
2 2724
ma ω=
(b) From equation 9.2.4,
2 21 2 3
13 10 2 5 3ˆ ˆ ˆ12 12 1214 14 14
L e ma e ma e ma2ω ω ω⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⋅ + ⋅ + ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
r
( )2
1 2ˆ ˆ ˆ13 20 1512 14ma
3L e e eω= + +
r
( )( ) ( )( ) ( )( )
( ) ( )1 1
2 2 2 2 2 22 2
1 13 2 20 3 15cos
1 2 3 13 20 15
LL
ωθω
+ +⎡ ⎤⋅ ⎣ ⎦= =+ + ⋅ + +
rr
( )
12
98 0.929511,116
= =
21.6θ = o
9.5 (a) Select coordinate axes such that the axis of the rod is the 3-axis, its center is at the
3
origin, and ωr lies in the 1, 3 plane.
1
2
3
ωr α
From Table 8.3.1, 2
1 2 12mlI I= = , 3 0I =
( )1 3ˆ ˆsin cose eω ω α= +r α
From equation 9.2.4,
( )2 2
1 1ˆ ˆsin 0 0 sin12 12ml mlL e e ωω α α= + + =
r
Lr
is perpendicular to the rod, and 2
sin12
mlL ω α=r
(b) Since ωr is constant, from equations. 9.3.5 … 1 0 0N = +
( )( )2
2 0 cos sin12mlN ω α ω α
⎛ ⎞= + ⎜ ⎟
⎝ ⎠
3 0 0N = +
2 2
2ˆ sin cos12
mlN e ω α α=r
Nr
is perpendicular to the rod ( direction) and to 3e Lr
( direction), and 1e2 2
sin cos12
mlN ω α α=r
9.6 From Problem 9.4 … ( )1 2 3ˆ ˆ ˆ2 314
e e eωω = + +r
21
1312
I ma= , 22
1012
I ma= , and 23
512
I ma=
From eqns. 9.3.5:
( )( ) ( ) ( )2 2 2
1 0 2 3 5 10 514 12 28
ma maN ω ω= + − = −
2
( )( ) ( ) ( )2 2 2 2
2 0 3 1 13 514 12 28
ma maN ω ω= + − = 4
( )( ) ( )2 2
3 0 1 2 10 1314 12 28
ma maN2 2ω ω
= + − = −
( )12 2 2 2
2 2 2 25 4 1 4228 28
ma maN ω ω= + + =
r
9.7 Multiplying equations 9.3.5 by 1ω , 2ω , and 3ω ,respectively … ( )1 1 1 1 2 3 3 20 I I Iω ω ω ω ω= + −&
( )2 2 2 1 2 3 1 30 I I Iω ω ω ω ω= + −&
4
( )3 3 3 1 2 3 2 10 I I Iω ω ω ω ω= + −& Adding, 1 1 1 2 2 2 3 3 30 I I Iωω ω ω ω= + +& & & ω
( )2 2 21 1 2 2 3 3
102
d I I Idt
ω ω ω⎡ ⎤= + +⎢ ⎥⎣ ⎦
From equation 9.2.5, [ ]0 rotd Tdt
=
rotT constant= Multiplying equations 9.3.5 by 1 1I ω , 2 2I ω , and 3 3I ω , respectively: ( )2
1 1 1 1 1 2 3 3 20 I I Iω ω ω ω ω= + −& I
( )22 2 2 2 1 2 3 1 30 I I Iω ω ω ω ω= + −& I
( )23 3 3 3 1 2 3 2 10 I I Iω ω ω ω ω= + −& I
Adding, 2 2 21 1 1 2 2 2 3 3 30 I I Iωω ω ω ω= + +& & & ω
( )2 2 2 2 2 21 1 2 2 3 3
102
d I I Idt
ω ω ω= + +
From equation 9.2.4, ( )20 d Ldt
=
2L constant= 9.8 From equations 9.3.5 for zero torque … ( )1 1 2 3 3 20 I I Iω ω ω= + −&
( )2 2 1 3 1 30 I I Iω ω ω= + −& From the perpendicular axis theorem, 3 1 2I I I= + 1 1 2 3 10 I Iω ω ω= +& 2 2 1 3 20 I Iω ωω= −&
Multiplying by 1
1Iω and 2
2Iω , respectively:
1 1 1 2 30 ω ω ω ω ω= +& 2 2 1 2 30 ω ω ωω ω= −&
Adding, ( )2 21 1 2 2 1 2
102
ddt
ωω ω ω ω ω= + = +& &
2 21 2 constantω ω+ =
If 1 2I I= , from the third of Euler’s equations … ( )3 3 0I ω =& 3 constantω = 9.9 (a) From symmetry … 3sI I= and 1 2I I I= =
From the perpendicular axis theorem, 3 1 2I I I= + or 2sI I= From eqn 9.5.8, ( )2 1 cosω αΩ = −
5
For , 45α = o
2ω
Ω =
For 2 1sπω
= , 12 2 1.414sT sπ
= = =Ω
1 2
3
ωr 45o
1T is the period of precession of ωr about . 3eFrom equation 9.6.12,
112 22 22
22 2
2 11 1 cos 1 11 2
sII
φ ω α ω⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞= + − = + −⎢ ⎥⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠ ⎢ ⎥⎣ ⎦ ⎣ ⎦&
52
φ ω=&
22 0.4 0.632sT sπφ
= = =&
is the period of wobble of about 2T 3e Lr
.
(b) 2 2
21 2
1712 16 16 12m a mI I I a⎛ ⎞
= = = + = ⋅⎜ ⎟⎝ ⎠
a
( )2
2 23 2
12 12sm mI I a a= = + = ⋅
a
From equation 9.5.8, 2 1 15117 172 216
ωω
⎛ ⎞⎜ ⎟
− =⎜ ⎟⎜ ⎟⎝ ⎠
Ω =
12 17 2 1.603
15T s sπ= = =Ω
From equation 9.6.12,
12 22 2
2
2 16 11 117 2
φ ω⎡ ⎤⎛ ⎞⋅ ⎛ ⎞= + −⎢ ⎥⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦&
1.5072φ ω=&
22 1 0.663s
1.5072T sπ
φ= = =&
9.10 From equation 9.6.10, tan tans
II
θ α= .
Since sI I> , θ α< and the angle between the axis of rotation ( )ωr and the
invariable line ( is )Lr
α θ− .
From your favorite table of trigonometric identities …
( ) tan tantan1 tan tan
α θα θα θ−
− =+
6
( )2
1 tantan
1 tan
s
s
IIII
αα θ
α
⎛ ⎞−⎜ ⎟
⎝ ⎠− =+
( )12
tantan
tans
s
I II I
αα θ
α− −⎡ ⎤
− = ⎢ ⎥+⎣ ⎦
9.11 α θ− is a maximum for sII
a maximum.
For 2sII= , ( ) 2
tantan2 tan
αα θα
− =+
( )( ) ( )
2 2
2 22 2 2
tan 1 2 tan 2 tantan 2 tan 2 tan 2 tan
dd
α θ α αα α α α
− −= − =
+ + +
At the maximum, ( ) 2tan0 2 tan
tand
dα θ
αα−
= = −
1tan 2 54.7α −= = o
( )( )2
11 222tan 1 41 2
2
α θ
⎛ ⎞−⎜ ⎟⎝ ⎠− ≤ =+
1 2tan 19.54
α θ − ⎛ ⎞− ≤ =⎜ ⎟⎜ ⎟
⎝ ⎠
o
9.12 (a) From Problem 9.10, for sI I> , the angle between ωr and L
r is …
( )12
tantan
tans
s
I II I
αα θ
α− −⎡ ⎤
− = ⎢ ⎥+⎣ ⎦
From Problem 9.9(a), 2sI I= and tan tan 45 1α = =o
( )
1 11 1tan tan 18.42 1 3
α θ − −− = = =+
o
(b) From Prob. 9.9(b), 2
212s
maI = ⋅ and 217
16 12maI = ⋅
1 1
1721516tan tan 17.017 492
16
α θ − −
⎡ ⎤⎛ ⎞−⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥− = = =⎢ ⎥+⎢ ⎥⎣ ⎦
o
7
9.13 From Example 9.6.2, 0.2α ′′= and 1.00327sII=
It follows from equation 9.6.10 that, for sI I> , the angle between ωr and is: Lr
1tan tans
II
α θ α α− ⎛ ⎞− = − ⎜ ⎟
⎝ ⎠
For α so small,
tanα α≈ , and 1tan tan tans s s
I I II I I
αα α− ⎛ ⎞≈ ≈⎜ ⎟
⎝ ⎠
s
s s
I III Iαα θ α α−
− ≈ − =
.00327 0.2 0.000651.00327
α θ− ≈ = arcsec
9.14 From Table 8.3.1, 2
2smaI = and
2
4maI =
3eω ω=or and 3
ˆ ˆ4
maP eω=
r
Selecting the 2-axis through the point of impact, during the collision …
2
2 3ˆ ˆ ˆ ˆ4 4
ma maNdt r p ae e e1ω ω
= × = × =∫rr r
The only non-zero component of is . During the collision
Nr
1N
2 0ω = , so integrating the first of Euler’s equations 9.3.5 …
1 1N dt I 1ω=∫ Immediately after the collision,
2
1 2
44
mama
ωω ω⎛ ⎞= =⎜ ⎟⎝ ⎠
Pr
1
2
3
0ωr
ωr
( 3ω still is equal to ω , and 2 0ω = )
After the collision, . From Prob. 9.8, with 0N =r
1 2I I= , for 0N =r
, 3 constantω = and . 2 2
1 2 constantω ω+ =
( )
12 2 21 2
3
tan constantω ω
αω
+= =
Using the values of iω immediately after the collision …
( )
12 20
tan 1ω
αω
+= =
45α = o
8
From equation 9.5.8, with ( )1 1 1 3ˆ ˆ ˆe e eω ω ω ω= + = +o
r r :
( )( )( )2 1 2 cos 45ω ωΩ = − =o
From equation 9.6.12 …
( )( )1
2 2 22 1 2 1 cos 45 5φ ω ω⎡ ⎤= + − =⎣ ⎦o&
9.15 ( )1 1 2 2 3 3ˆ ˆ ˆN c c e e eω ω ω ω= − = − + +
r r
Say the 3 axis is the symmetry axis, 3 sI I= , 1 2I I I= = For the third component of equations 9.3.5 …
3 3 0sc Iω ω− = +&
1
2
3 ωr α
Nr
3
3 s
cI
ωω
= −&
( )3
3
lns
c tI
ωω
= −o
( )3 3s
ctIeω ω
−
=o
For the first two components of equations 9.3.5 …
( )1 1 2 3 sc I I Iω ω ω ω− = + −& , and
( )2 2 1 3 sc I I Iω ω ω ω− = + −& Rearranging terms and multiplying by 1ω and 2ω respectively …
( )21 1 1 1 2 3 0s
c I II
ωω ω ωω ω+ + − =&
( )22 2 2 1 2 3 0s
c I II
ω ω ω ωω ω+ − − =&
Adding, ( )2 21 1 2 2 1 2 0c
Iωω ω ω ω ω+ + + =& &
( ) ( )2 21 22 2
1 2
1 2d cdt I
ω ωω ω
+ = −+
( )2 21 22 21 2
2ln c tI
ω ωω ω
+= −
+o
( ) ( )2
2 2 2 21 2 1 2
ctIeω ω ω ω
−+ = +
o
( ) ( )1 1
2 2 2 22 21 2 1 2
ctIeω ω ω ω+ = +
o
( ) ( )
11 12 2 2
1 2
3
tan tan sct
I Ieω ω
α αω
⎛ ⎞− −⎜ ⎟
⎝ ⎠+
= = o
9
For sI I> , 1 1 0sI I
⎛ ⎞− >⎜ ⎟
⎝ ⎠ and α decreases with time.
9.16 (a) From table 8.3.1… 2
2smaI = about the symmetry (spin) axis.
2
4cm diskaI m= about an axis through
the center of mass of the disk and ⊥ to the spin axis. From the parallel axis theorem, equation 8.3.21, …
22
4 2diskma aI m⎛ ⎞= + ⎜ ⎟
⎝ ⎠ about the pivot
point. 2
2 3rodm aI ⎛ ⎞= ⎜ ⎟
⎝ ⎠ moment of inertia of
the rod about the pivot point. Thus …
22 2
224 2 2 3 3disc rod
ma a m aI I I m ma⎡ ⎤⎛ ⎞ ⎛ ⎞= + = + + =⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
x
y
z
45o
Sr
a
a / 2
From equation 9.7.10, ( )
12 2 24 cos
2 coss sI S I S MglI
Iθ
φθ
+ −=&
12 2
21 42cos
s sI I MglS SI I I
θφθ
⎧ ⎫⎡ ⎤⎪ ⎪⎛ ⎞= ± −⎢ ⎥⎨ ⎬⎜ ⎟⎝ ⎠⎢ ⎥⎪ ⎪⎣ ⎦
⎩ ⎭
& cos
2
2
322 4
3
s
maI
maI= = , 1cos cos 45
2θ = =o
12
9 92 22 8 80
3
m amMl cm
maI a−
⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= = =
( ) ( )1
2 2 22 1
2 1
1 3 3 9 1 60900 900 4 9804 4 80 22 2
cm rpmrpm cms s
φπ
−−
⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= ± −⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎪ ⎪⎣ ⎦
⎩ ⎭
&
or 15.1rpmφ =& 939 rpm (b) From equation 9.7.12, 2 2 4sI S Mg≥ lI
10
43s
II= , 1
2
33 32 2
2 202
s
m aMl cm
maI a−
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= = =
( )2
2 1 21
4 3 4 604 98020 3 2s s
Mgl I rpmS cm cm sI I sπ
− −−
⎛ ⎞⎛ ⎞ ⎛ ⎞≥ ⋅ = ⋅ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
267S rp≥ m 9.17 (Neglecting the pencil head) From Table 8.3.1,
2
222 8s
bmmbI
⎛ ⎞⎜ ⎟⎝ ⎠= =
The moment of inertia of the pencil about an axis thru its center of mass and ⊥ to its symmetry axis from equation 8
...
.3.26 2
2 224 12 16 12
2b
a b am
⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎛ ⎞⎝ ⎠ + = +⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
cI m ⎢ ⎥=⎢ ⎥
From the parallel axis theorem, 2 2 2
2 16ca bI I m m
⎛ ⎞⎛ ⎞= + = +⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠3
a
b
a
From equation 9.7.12, 22
4
s
MglISI
≥
2 2
22 4
42 16 3
64
a b amg mS
m b
⎛ ⎞⎛ ⎞ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠≥
( )
( )( )1 1
2 2 2 22 21
22
980 2016 16 1 202 16 3 2 16 31ga b aS r
b−⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞
≥ + = +⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦
ad s⋅
118,294 2910S rad s rps−≥ ⋅ = 9.18 s rim spokes hubI I I I= + +
2
2
2rim rimmaI m a= =
3 2
3 12spokes spokesa maI m= = , assuming the spokes to be thin rods
0 , assuming its radius is small hubI =
11
From the perpendicular axis theorem, equation 8.3.14, 2sII =
From equation 9.10.14, ( )2
2s s
ImgaSI I ma
>+
12
12
22
1 62 17
12
mga gSama ma
⎡ ⎤⎢ ⎥
⎛ ⎞⎢ ⎥> = ⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠+⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
9
For rolling without slipping, v aS=
12
12 1 1
306 326 2 3.5519 19 12gav ft s ft s− −
⎡ ⎤⎛ ⎞× ×⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥> = ⋅ =⎜ ⎟ ×⎝ ⎠ ⎢ ⎥⎢ ⎥⎣ ⎦
⋅
If the spokes and hub are neglected, 2
2smaI =
12
12
22
12 3
2
mga gSama ma
⎡ ⎤⎢ ⎥
⎛ ⎞⎢ ⎥> = ⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠+⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
12
12 1 1
30322 3.65
3 3 12gav ft s ft s− −
⎡ ⎤⎛ ⎞×⎜ ⎟⎢ ⎥⎛ ⎞ ⎝ ⎠⎢ ⎥> = ⋅ = ⋅⎜ ⎟ ×⎝ ⎠ ⎢ ⎥⎢ ⎥⎣ ⎦
9.19 From equations 9.3.5 with 0N =
r …
( )1 1 3 2 2 3 0I I Iω ω ω+ − =&
( )2 2 1 3 1 3 0I I Iω ω ω+ − =&
( )3 3 2 1 1 2 0I I Iω ω ω+ − =& Differentiating the first equation with respect to t: ( )( )1 1 3 2 2 3 2 3 0I I Iω ω ω ω ω+ − + =&& & &
From the second and third equations:
( )3 12 1 3
( )2
I II
ω ωω−
=& , and 1 23 1 2
3
I II
ω ωω−
=&
12
( ) ( ) ( )1 2 3 12 21 1 3 2 1 2 1 3
3 2
0I I I I
I I II I
ω ω ω− −⎡ ⎤
+ − + =⎢ ⎥⎣ ⎦
&& ω ω
1 1 1 0Kω ω+ =&& , ( )( ) ( )( )3 2 2 1 3 2 3 12 21 2 3
1 3 1 2
I I I I I I I IK
I I I Iω ω
− − − −⎡ ⎤ ⎡= − +
⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
(a) For 3ω large and 2 0ω = , so 1 0K > 1 1 1 0Kω ω+ =&& is the harmonic oscillator equation. 1ω oscillates, but remains small. Motion is stable for initial rotation about the 3 axis if the 3 axis is the principal axis having the largest or smallest moment of inertia.
(b) For 3 0ω = and 2ω large, 1 0K < so 1 1 1 0Kω ω+ =&& is the differential equation for
exponential growth of 1ω with time: 11
k t k tAe Beω −= + 1
x y
. Motion is unstable for the initial rotation mostly about the principal axis having the median moment of inertia.
9.20 0I mxy i i i
i
= =∑ since either ix or iy is
zero for all six particles. Similarly, all the other products of inertia are zero. Therefore the coordinate axes are principle axes.
( ) ( ) ( )2 22 2 2 20 0xx i i ii
I m y z m b b c c⎡ ⎤= + = + + + − + + −⎣ ⎦∑
( )2 22xxI m b c= +
( )2 22yyI m a c= +
( )2 22zzI m a b= +
2 2
2 2
2 2
0 02 0 0
0 0
b cI m a c
a b
⎡ ⎤+⎢ ⎥= +⎢ ⎥⎢ ⎥+⎣ ⎦
t
(a,0,0)
(0,b,0)
(0,0,c)
( )
( )( )
1 2 31 12 2 2 2 2 22 2
ˆ ˆ ˆa
ae be ce ba b c a b c c
ω ωω⎡ ⎤⎢ ⎥= + + = ⎢ ⎥
+ + + + ⎢ ⎥⎣ ⎦
r
From equation 9.1.28, L Iω=r t r
( )
2 2
2 21
2 2 2 2 22
0 02 0 0
0 0
b c amL a c
a b c a b c
ω⎡ ⎤+
b⎡ ⎤
⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥+ + ⎢ ⎥+ ⎣ ⎦⎣ ⎦
r
13
( )
( )( )( )
2 2
2 21
2 2 2 2 2 2
2a b c
mL ba b c c a b
ω⎡ ⎤+⎢ ⎥⎢ ⎥= +⎢ ⎥+ + ⎢ ⎥+⎣ ⎦
ra c
From equation 9.1.32, 12
T Lω= ⋅rr
( ) [ ]( )( )( )
2 2
22 2
2 2 2
2 2
1 22
a b cmT a b c b a c
a b cc a b
ω⎡ ⎤+⎢ ⎥⎢ ⎥= +⎢ ⎥+ +⎢ ⎥+⎣ ⎦
( ) ( ) (2
2 2 2 2 2 2 2 2 22 2 2
mT a b c b a c c a ba b c
ω ⎡ ⎤= + + + +⎣ ⎦+ +)+
( )2
2 2 2 2 2 22 2 2
2mT a b a ca b c
ω= +
+ +b c+
9.21 For Problem 9.1:
2
1 1 03 21 4 02 3
50 03
I ma
⎛ ⎞−⎜ ⎟⎜ ⎟⎜ ⎟= −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
t,
21
5 0
ωω⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
r
2
1 1 03 2 21 4 0 12 35 050 0
3
maL I ωω
⎛ ⎞−⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= = − ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
⎜ ⎟⎜ ⎟⎝ ⎠
r t r
2 2
2 1 13 2 6 1
4 11 23 35 5 6 00 0
ma ma maω ω
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟= − + = = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠
⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2
5ω
( )2
11 1 2 1 0 22 2 5 6 5 0
maT L ω ωω⎛ ⎞⎜ ⎟= ⋅ = ⎜ ⎟⎜ ⎟⎝ ⎠
rr
( )2 2 2 2
2 2 060 15
ma maω ω= + + =
14
For Problem 9.4:
2
13 0 00 10 0
120 0 5
maI⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
t,
12
14 3
ωω⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
r
2
13 0 0 10 10 0 2
12 14 0 0 5 3
maL I ωω⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
⎜ ⎟= = ⎜ ⎟⎜ ⎟⎝ ⎠
r t r 21320
12 14 15
ma ω⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
( )2
131 1 1 2 3 202 2 14 12 14 15
maT L ω ωω⎛ ⎞⎜ ⎟= ⋅ = ⎜ ⎟⎜ ⎟⎝ ⎠
rr
( ) ( )2 2
2 2713 40 4524 14 24ma maω ω= + + =
9.22 Since the coordinate axes are axes of symmetry, they are principal axes and all
products of inertia are zero. From Table 8.3.1,
( ) ( ) ( )2 2 2 22 212 3xxm mI b c b c⎡ ⎤= + = +⎣ ⎦
y
x
z ωr
2a
2b
2c
( )2 2
3yymI a c= + , ( )2 2
3zzmI a b= +
2 2
2 2
2 2
0 00 0
30 0
b cmI a c
a b
⎡ ⎤+⎢ ⎥= +⎢ ⎥⎢ ⎥+⎣ ⎦
t
( )( ) ( )1 2 31 2 2 2
2 2 2 2
ˆ ˆ ˆa
ae be ce ba b ca b c c
ω ωω⎡ ⎤⎢ ⎥= + + = ⎢ ⎥+ ++ + ⎢ ⎥⎣ ⎦
r
From equation 9.1.28, L Iω=r t r
( )
2 2
2 21
2 2 2 2 22
0 00 0
3 0 0
b c amL a c
a b c a b c
ω⎡ ⎤+
b⎡ ⎤
⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥+ + ⎢ ⎥+ ⎣ ⎦⎣ ⎦
r
( )
( )( )( )
2 2
2 21
2 2 2 2 2 23
a b cmL b
a b c c a b
ω⎡ ⎤+⎢ ⎥⎢ ⎥= +⎢ ⎥+ + ⎢ ⎥+⎣ ⎦
ra c
From equation 9.1.32, 12
TT Lω= ⋅rr
15
( ) [ ]( )( )( )
2 2
22 2
2 2 2
2 2
12 3
a b cmT a b c b a c
a b cc a b
ω⎡ ⎤+⎢ ⎥⎢ ⎥= +⎢ ⎥+ +⎢ ⎥+⎣ ⎦
( ) ( ) ( ) (
22 2 2 2 2 2 2 2 2
2 2 26mT a b c b a c c
a b cω ⎡ ⎤= + + + +⎣ ⎦+ +
)a b+
( ) ( )
22 2 2 2 2 2
2 2 23mT a b a c
a b cω
= ++ +
b c+
With the origin at one corner, from the parallel axis theorem:
( ) ( ) ( )
2 22 2 2 24
3 3xx
m b c mI m b c b c+
= + + = +
( )2 243yymI a c= + , ( )2 24
3zzmI a b= +
xyI xydm xy dVρ= − = −∫ ∫
2 2 2 2 2
0 0 08
x a y b z c
xy x y zI xydxdydz a b cρ ρ
= = =
= = == − = −∫ ∫ ∫
, so ( )( )( )2 2 2 8m a b c abcρ ρ= = xyI mab= − xyI mac= − , yzI mbc= −
( )
( )
( )
2 2
2 2
2 2
43
43
43
b c ab ac
I m ab a c bc
ac bc a b
⎡ ⎤+ − −⎢ ⎥⎢ ⎥⎢ ⎥= − + −⎢ ⎥⎢ ⎥⎢ ⎥− − +⎢ ⎥⎣ ⎦
t
9.23 (See Figure 9.7.1) ( ) ( ) ( )z x y zz zz
L L L L′ ′= + + ′
( ) ( )sin cos sin sin cosz y z sL L L I I Sθ θ φ θ θ′ ′= + = +& θ 9.24 (See Figure 9.7.1) If the top precesses without nutation, it must do so at θ θ= o where ( )V θo is a minimum of ( )V θ …
( ) 0dV
dθ θ
θθ
=
=o
( ) ( )2
2
coscos
2 sinz zL L
VI
θmglθ θ
θ′−
= + (See equation 9.8.7)
16
( ) ( )2 2
3
cos cos sin cossin 0
sinz z z z zL L L L LdV mgl
d Iθ θ
θ θ θ θθ
θ θ′ ′ ′
=
− − + −= −
o
o o o oo
o
=
let cosz zL Lγ θ′= − o Then and solving for 2 2cos sin sin 0zL mglIγ θ γ θ θ′− +o o
4 =o γ
12 2
2
sin 4 cos1 12cosz
z
L mglIL
θ θγθ
′
′
⎡ ⎤⎛ ⎞⎢ ⎥= ± −⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦
o o
o
now zL ′ is large since ψ& is large and the precession rate is small, so we can expand the term in square root above and use the (-) solution since γ must be positive …
2
2
sin 2 cos1 12cosz
z
L mglIL
θ θγθ
′
′
⎡ ⎤≈ − +⎢ ⎥
⎣ ⎦o o
o
2sincosz z
z
mglIL LL
θγ θ′′
= − ≈ oo
From equations 9.7.2, 9.7.5 and 9.7.7 … ( )2 2
0 0sin cos cosz sL I I 0φ θ φ θ ψ= + +& & & θ
0
and …
0 0cos ( cos )cosz sL Iθ φ θ ψ θ′ = +& & so …
( )2 2 2sin cos cos cos coss s s sI I I I Iγ θ θ φ ψ θ φ θ ψ= + + − −o o o o& && & θo
( )2 2
2 sin sinsincosz s
mglI mglIIL I
θ θφ θψ φ θ′
= ≈ =+
o oo
o
&&&
and since 0ψ >>& , we can ignore
the φ& term in the denominator and we have …
s
mglI
φψ
≈&&
Hence, if ψ& large, 1
0θ θ
θ=
=& and 1 s
mglIθ θ
φψ=
=&&
the top will precess without nutation
at 1θ θ= o the place where ( ) minV . θ =
9.25 1 2 32ab b ⎛ ⎞+ = ⎜ ⎟
⎝ ⎠
1
2
1sin 302
bb
= =o
1 2 3ab = 2 3
ab =
2 2
2a H b2= +
2 2
2 2 2 22
23 3a aH a b a= − = − =
17
23
H a=
Thus, the coordinates of the 4 atoms are: Oxygen: ( )0,0, H
Hydrogen: 1 1, ,0 ; , ,02 2a ab b⎛ ⎞ ⎛− − −⎜ ⎟ ⎜
⎝ ⎠ ⎝⎞⎟⎠
Carbon: ( )2 ,0,0b (a) The axes 1, 2, 3 are principal axes if the products of inertia are zero.
1 10 02 2xy i i ia aI m x y b b⎛ − ⎞⎛ ⎞ ⎛ ⎞− = = + − + − + ≡⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
∑ 0
0 0 0 0 0yzI− = + + + ≡
The 1,2,3 axes are principal axes. 0 0 0 0 0xzI− = + + + ≡ (b) Find principal moments
( )2 2
2 2 21
6716 1 12 2 6xx i i ia a 2I I m y z H ⎛ ⎞ ⎛ ⎞= = + = + + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∑ a
( ) ( ) ( )2 2 2 2 2 22 1
8916 1 1 126yy i i i
21 2I I m x z H b b b= = + = + + + =∑ a
( )2 2
2 2 2 2 23 1 1
1402 2zz i i ia a 2
2 3I I m x y b b b
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = + = + + + + + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠∑ a
2
67 0 06
890 06
140 03
I a
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
(c) 3 1 2I I I< < therefore rotation about the1-axis is unstable (see discussion Sec. 9.4) ----------------------------------------------------------------------------------------------------------
18
CHAPTER 10 LAGRANGIAN MECHANICS
10.1 Solution … ( ) ( ) ( )0,x t x t tαη= +
( ) ( ) ( )0,x t x t tαη= +& & &
where ( )0, sinx t tω= and ( )0, cosx t tω ω=&
212
T mx= & 2 21 12 2
V kx m xω= = 2
so: ( ) ( )2
1
2 2 2
2
t
t
mJ x xα ω= −∫ & dt
( ) (2
1
2 22cos sin2
t
t
m t tω ω αη ω ω αη⎡ ⎤= + − +⎣ ⎦∫ & ) dt
( ) ( ) ( ) ( )2 2
1 1
22 2 2 2 2 2cos sin cos sin
2 2
t t t
t t t
m mJ t t dt m t t dt dtαα ω ω ω αω η ω ωη ω η ω η= − + − + −∫ ∫ ∫& &2
1
Examine the term linear in α :
( ) ( )2 2 2
2
11 1 1
cos sin cos sin sin 0t t
t
tt t
t t dt t t tdt tdtη ω ωη ω η ω ωη ω ωη ω− = + −∫ ∫&t
t
≡∫
(1st term vanishes at both endpoints: ( ) ( )2 1 0t tη η= = )
so ( ) ( )2 2
1 1
2 2 21 1cos 22 2
t t
t t
J m tdt mα ω ω α η ω η= + −∫ ∫ & 2 2 dt
[ ] ( )2
1
2 2 2 22 1
1 1sin 2 sin 24 2
t
t
m t t mω ω ω α η ω η= − + −∫ & dt
which is a minimum at 0α = 10.2 V mgz=
( )2 2 212
T m x y z= + +& & &
( )2 2 212
L T V m x y z mgz= − = + + −& & &
L mxx∂
=∂
&&
, L myy∂
=∂
&&
, L mzz∂
=∂
&&
d L mxdt x
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&, d L my
dt y⎛ ⎞∂
=⎜ ⎟∂⎝ ⎠&&
&, d L mz
dt z∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠
&&&
1
0L Lx y∂ ∂
= =∂ ∂
, L mgz∂
= −∂
From equations 10.4.5 … 0i i
L d Lq dt q
⎛ ⎞∂ ∂− =⎜ ⎟∂ ∂⎝ ⎠&
, 0mx =&& mx const=& , 0my =&& my const=& mz mg= −&&
10.3 Choosing generalized coordinate x as linear displacement down the inclined plane (See Figure 8.6.1), for rolling without slipping …
xa
ω =&
2
2 2 2 21 1 1 1 2 72 2 2 2 5 10
xT mx I mx ma mxa
ω ⎛ ⎞⎛ ⎞= + = + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
&& & 2&
=gx
For V at the initial position of the sphere, 0 V m sinθ= −
27 sin10
L T V mx mgx θ= − = +&
75
L mxx∂
=∂
&&
, 75
d L mxdt x
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&
sinL mgx
θ∂=
∂
7 sin5
mx mg θ=&&
5 sin7
x g θ=&&
From equations 8.6.11 - 8.6.13 … 5 sin7cmx g θ=&&
10.4 (a) For x the distance of the hanging block below the edge of the table:
2 21 12 2
2x mx mx= + =& & & x
V mx mgx= − = +&
T m and V m g= −
x
L T 2
2L mxx∂
=∂
&&
, 2d L mxdt x
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&
L mgx∂
=∂
2mx mg=&&
2gx =&&
2
(b) 2 212 2
m 2x m x m xT m′⎛ ⎞′= + = +⎜ ⎟
⎝ ⎠& & & and 2
2 2x x m ggx m g mgx xl l
′⎛ ⎞′= − − = − −⎜ ⎟⎝ ⎠
V m
2 2
2 2m mL T V m x mgx x
l′ ′⎛ ⎞= − = + + +⎜ ⎟
⎝ ⎠&
g
( )2L m m xx∂ ′= +∂
&&
, ( )2d L m m xdt x
∂⎛ ⎞ ′= +⎜ ⎟∂⎝ ⎠&&
&
L mmg xgx l
′∂= +
∂
( )2 m gm m x mg xl′
′+ = +&&
2
g ml m xxl m m
′+⎛ ⎞= ⎜ ⎟′+⎝ ⎠&&
10.5 The four masses have positions:
m1 : 1 2x x+ m2 : 1 1l x x2− + m3 : 2 2l x x3− + m4 : 2 2 3l x l x3− + −
( ) ( )1 1 2 2 1 1 2V g m x x m l x x= − + + − +⎡⎣
( ) ( )3 2 2 3 4 2 2 3 3m l x x m l x l x+ − + + − + − ⎤⎦
( ) (2 21 1 2 2 1 2
12
T m x x m x x⎡= + + − +⎣ & & & & )
)
( ) (2 23 2 3 4 2 3m x x m x x ⎤+ − + + − − ⎦& & & &
( ) ( ) ( )2 21 1 2 2 1 2 3 2 3
1 1 12 2 2
L T V m x x m x x m x x= − = + + − + + − +& & & & & &2
( ) ( ) ( ) ( )24 2 3 1 1 2 2 1 2 3 4 3 3 4
1 .2
m x x gx m m gx m m m m gx m m cons+ − − + − + + − − + − +& & t
( ) (1 1 2 2 1 21
L m x x m x xx∂
= + − − +∂
& & & &&
)
) 2+ −& &
= + ( ) (1 2 1 1 2m m x m m x
( ) (1 2 1 1 21
d L m m x m m xdt x
∂= + + −
∂&& &&
&) 2
( )1 21
L g m mx∂
= −∂
( ) ( ) ( )1 2 1 1 2 2 1 2m m x m m x g m m+ + − = −&& &&
3
( ) ( ) ( ) (1 1 2 2 1 2 3 2 3 4 2 32
L m x x m x x m x x m x xx∂
= + + − + − − + − − −∂
& & & & & & & &&
)
( )1 2 3 42
L g m m m mx∂
= + − −∂
( ) ( ) ( ) ( )1 2 1 1 2 3 4 2 4 3 3 1 2 3 4m m x m m m m x m m x g m m m m− + + + + + − = + − −&& && &&
( ) (3 2 3 4 2 33
L m x x m x xx∂
= − + − − −∂
& & & &&
)
( )3 43
L g m mx∂
= −∂
( ) ( ) ( )4 3 2 3 4 3 3 4m m x m m x g m m− + + = −&& && For , , , and 1m m= 2 4m m= 3 2m = m 4m m= :
, 1 25 3 3mx mx mg− = −&& && ( )1 235
x x g= −&& &&
1 2 33 8 2mx mx mx mg− + − =&& && &&
, 2 33mx mx mg− + =&& && ( )3 213
x x g= +&& &&
Substituting into the second equation:
2 2 29 9 1 18 25 5 3 3
x g x x g− + + − − =&& && && g
288 815 15
x g=&& , 2 11gx =&&
13 10 65 11 11
x g g⎛ ⎞= − = −⎜ ⎟⎝ ⎠
&&
31 12 43 11 11
x g g⎛ ⎞= =⎜ ⎟⎝ ⎠
&&
Accelerations:
: 1m 1 25
11x x g+ = −&& &&
: 2m 1 2711
x x g− + =&& &&
: 3m 2 33
11x x g− + =&& &&
: 4m 2 35
11x x g− − = −&& &&
10.6 See figure 10.5.3, replacing the block with a ball. The square of the speed of the ball is calculated in the same way as for the block in Example 10.5.6. 2 2 2 2 cosv x x xx θ′ ′= + +& & && The ball also rotates with angular velocity ω so …
4
2 21 1 12 2 2
T mv I Mxω= + + &2
For rolling without slipping, xa
ω′
=&
. 225
I ma=
( )2 2 21 12 cos2 5
T m x x xx mx Mxθ′ ′ ′= + + + +& & & & & &212
sinV mgx θ′= − , for V at the initial position of the ball. 0=
2 21 7 2 cos2 5
L T V m x x xx θ⎛ ⎞⎟′ ′= − = + +⎜
⎝ ⎠& & & & 21 sin
2Mx mgx θ′+ +&
7 cos5
L mx mxx
θ∂ ′= +′∂
& & , 7 cos5
d L mx mxdt x
θ∂⎛ ⎞ ′= +⎜ ⎟′∂⎝ ⎠&& &&
&
sinL mgx
θ∂=′∂
7 cos sin5
mx mx mgθ θ′ + =&& &&
( )5 sin cos7
x g xθ θ′ = −&& &&
cosL mx mx Mxx
θ∂ ′= + +∂
& &&
& , ( ) cosd L m M x mxdt x
θ∂⎛ ⎞ ′= + +⎜ ⎟∂⎝ ⎠&& &&
&
0Lx∂
=∂
( ) cos 0m M x mx θ′+ + =&& &&
( ) 25 5sin cos cos 07 7
m M x mg mxθ θ θ+ + − =&& &&
( )2
5 sin cos5 cos 7
mgxm m M
θ θθ
=− +
&&
10.7 Let x be the slant height of the particle …
2 2 2v x x 2ω= +&
tθ ω=
x ( )2 2 21 1
2 2T mv m x x 2ω= = +&
sin sinV mgx mgx tθ ω= =
( )2 21 sin2
L T V m x x mgx tω ω= − = + −&
L mxx∂
=∂
&&
, d L mxdt x
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&
2 sinL mx mg tx
ω ω∂= −
∂
2 sinmx mx mg tω ω= −&& 2 sinx x g tω ω− = −&&
5
The solution to the homogeneous equation , is 2 0x xω− =&& t tx Ae Beω ω−= + Assuming a particular solution to have the form sinpx C tω= ,
2 2sin sin sinC t C t g tω ω ω ω ω− − = −
22gCω
=
2 sin2
t t gx Ae Be tω ω ωω
−= + +
At time , 0t = x x= o and 0x =& x A B= +o
02gA Bω ωω
= − +
2
12 2
gA xω
⎛ ⎞= −⎜ ⎟⎝ ⎠
o
2
12 2
gB xω
⎛ ⎞= +⎜ ⎟⎝ ⎠
o
( ) ( )2 2
1 12 2 2 2
t t t tg gx x e e e et
ω ω ω ω
sinω ω ω− −⎡ ⎤ ⎡ ⎤= + − − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
o
From Appendix B, we use the identities for hyperbolic sine and cosine to obtain
2 2cosh sinh sin2 2
g gx x t t tω ω ωω ω
= − +o
10.8 In order that a particle continues to move in a plane in a rotating coordinate system, it is necessary that the axis of rotation be perpendicular to the plane of motion. For motion in the xy plane, kω ω=
r . ( ) ( )ˆˆ ˆ ˆ ˆv v r ix jy k ix jyω ω′ ′= + × = + + × +
r r r r& &
( ) (ˆ ˆv i x y j y x)ω ω= − + +r
& &
( )2 2 2 21 2 22 2
mT mv v x x y y y y x xω ω ω ω= ⋅ = − + + + +r r
& & & 2 2
L T V= −
( )L m x yx
ω∂= −
∂&
&, ( )d L m x y
dt xω∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠
&& &&
,
( )2L Vm y xx x
ω ω∂ ∂= + −
∂ ∂&
( ) ( )2 Vm x y m y xx
ω ω ω ∂− = + −
∂&& & &
( )22xF m x y xω ω= − −&& &
( )L m y xy
ω∂= +
∂&
&, ( )d L m y x
dt yω
⎛ ⎞∂= +⎜ ⎟∂⎝ ⎠
&& &&
,
6
( )2L Vm x yy y
ω ω∂ ∂= − + −
∂ ∂&
( ) ( )2 Vm y x m x yy
ω ω ω ∂+ = − + −
∂&& & &
( )22yF m y x yω ω= + −&& &
For comparison, from equation 5.3.2 ( 0 0A =r
and 0ω =r& )
( )2F ma m v m rω ω ω′ ′= + × + × ×r r r r r r ′r
( ) ( ) ( )ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2F m ix jy m k ix jy m k k ix jyω ω ω⎡ ⎤= + + × + + × × +⎣ ⎦r
&& && & &
( )22xF m x y xω ω= − −&& &
( )22yF m y x yω ω= + −&& &
10.9 Choosing the axis of rotation as the z axis … ( ) ( )ˆ ˆ ˆˆ ˆ ˆ ˆv v r ix jy kz k ix jy kzω ω′ ′= + × = + + + × + +
r r r r& & &
( ) ( ) ˆˆ ˆv i x y j y x kzω ω= − + + +r
& & &
( )2 2 2 2 2 21 2 22 2
mT mv v x x y y y y x x zω ω ω ω= ⋅ = − + + + + +r r
& & & & &2
L T V= −
( )L m x yx
ω∂= −
∂&
& ( )d L m x y
dt xω∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠
&& &&
,
( )2L Vm y xx x
ω ω∂ ∂= + −
∂ ∂&
( ) ( )2 Vm x y m y xx
ω ω ω ∂− = + −
∂&& & &
( )22xF m x y xω ω= − −&& &
( )L m y xy
ω∂= +
∂&
&, ( )d L m y x
dt yω
⎛ ⎞∂= +⎜ ⎟∂⎝ ⎠
&& &&
( )2L Vm x yy y
ω ω∂ ∂= − + −
∂ ∂&
( ) ( )2 Vm y x m x yy
ω ω ω ∂+ = − + −
∂&& & &
( )22yF m y x yω ω= + −&& &
L mzz∂
=∂
&&
, d L mzdt z
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&, L V
z z∂ ∂
= −∂ ∂
zVmz Fz
∂= − =
∂&&
From equation 5.3.2 ( and 0 0A =r
0ω =r& ) …
7
( )2F ma m v m rω ω ω′ ′= + × + × ×r r r r r r ′r
( ) ( ) ( )ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2 ˆF m ix jy kz m k ix jy kz m k k ix jy kzω ω ω⎡ ⎤= + + + × + + + × × + +⎣ ⎦r
&& && & &&& &
( )22xF m x y xω ω= − −&& &
( )22yF m y x yω ω= − −&& &
zF mz= && 10.10
( )2 2 212
T m r r θ= + &&
r
θ
( )21 cos2
V k r l mgr θ= − −o
( ) ( )22 2 2 cos2 2m kL T V r r r l mgrθ θ= − = + − − +o
&
L mrr∂
=∂
&&
, d L mrdt r
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&
( )2 cosL mr k r l mgr
θ θ∂= − − +
∂ o&
( )2 cosmr mr k r l mg θ= − − +o&& &
2L mr θθ∂
=∂
&& sinL mgr θ
θ∂
= −∂
( )2 sind mr mgrdt
θ θ= −&
10.11 (See Example 4.6.2) ( )2 sin 24ax θ θ= + ( )1 cos 2
4ay θ= −
at 0θ = & 0x y =
( )2 212
T m x y= +& & V mgy=
( ) ( )cos 2 1 cos 22 2a ax θθ θ θ= + = +
&& && θ
sin 22ay θ θ= &&
( ) ( )2 2
2 21 cos 2 sin 2 1 cos 28 4
ma mgaL T V θ θ θ θ⎡ ⎤= − = + + − −⎣ ⎦&
[ ] ( )2 2
1 2cos 2 1 1 cos 28 4
ma mgaθ θ θ= + + − −&
2 2
2 2cos sin2 2
ma mgaθ θ θ⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦&
8
where we used the trigonometric identies … 22cos 1 cos 2θ θ= + and 22sin 1 cos 2θ θ= − Let sins a θ= so coss aθ θ= &&
2
2 21 12 2 2 2
ms mgL s msa
= − = −&
& 2ks where mgk a
=
The equation of motion is thus
0ks sm
+ =&& or 0gs sa
+ =&& -a simple harmonic oscillator
10.12 Coordinates:
cos sinx a t bω θ= + sin cosy a t bω θ= −
sin cosx a t bω ω θ= − + && θ
cos siny a t bω ω θ= + && θ
( )2 212
L T V m x y mgy= − = + −& &
( ) (2 2 2 2 2 sin sin cos2m a b b a t mg a t b )ω θ θ ω θ ω ω⎡ ⎤+ − − −⎣ ⎦
& & θ= +
( ) ( )2 cosd L mb mba tdt
θ ω θ ω θ ωθ∂
= + − −∂
&& &&
( )cos sinL mb a t mgbθ ω θ ωθ∂
= − −∂
& θ
The equation of motion 0L d Ldtθ θ
∂ ∂− =
∂ ∂ & is
( )cos sin 0a gtb b
θ ω θ ω θ− − +&& =
Note – the equation reduces to equation of simple pendulum if 0ω → . 10.13 Coordinates:
( )cos cosx l t l tω θ ω= + +
( )sin siny l t l tω θ ω= + +
( ) ( )sin sinx l t l tω ω θ ω θ ω= − − + +&&
( ) ( )cos cosy l t l tω ω θ ω θ ω= + + +&&
9
( ) ( )22 2 2 21 1 ...2 2
L T m x y ml ω θ ω⎡ ⎤= = + = + + +⎢ ⎥⎣ ⎦&& &
( ) ( ) ( )2... sin sin cos sin cos cos cos cos sin sinml t t t t t tω θ ω ω θ ω ω θ ω θ ω ω θ+ + + −⎡ ⎤⎣ ⎦&
( ) ( )22 2 21 cos2
ml mlω θ ω ω θ ω θ⎡ ⎤= + + + +⎢ ⎥⎣ ⎦& &
0L d Ldtθ θ
∂ ∂− =
∂ ∂ &
( ) ( )2 2sin sin 0ml mlθ ωθ θ ω θ ω θ− + +&& & & =
(a) 2 sin 0θ ω θ+ =&&
(b) The bead executes simple harmonic motion ( )0θ >% about a point diametrically opposite the point of attachment.
(c) The effective length is 2" " glω
=
10.14 ( )ˆ ˆ ˆcos sinv jat l i jθ θ θ= + +
r &
( )ˆ ˆcos sinil j at lθ θ θ= + + & θ
( )2 2 2 2 2 2 2 21 cos 2 sin sin2 2
mT mv v l a t atl lθ θ θ θ θ= ⋅ = + + +r r & & θ&
21 cos2
V mg at l θ⎛ ⎞= −⎜ ⎟⎝ ⎠
( )2
2 2 2 2 2 sin cos2 2m aL T V l a t atl mg ltθ θ θ θ
⎛ ⎞= − = + + − −⎜ ⎟
⎝ ⎠& &
2 sinL ml matlθ θθ∂
= +∂
&& , 2 sin cosd L ml mal matl
dtθ θ θ
θ∂⎛ ⎞ = + +⎜ ⎟∂⎝ ⎠
&& && θ
cos sinL matl mglθ θ θθ∂
= −∂
&
2 sin cos cos sinml mal matl matl mglθ θ θ θ θ θ+ + = −&& & & θ
sin 0a gl
θ θ++ =&&
For small oscillations, sinθ θ≈
0a gl
θ θ++ =&&
2 2 lTa g
π πω
= =+o
10.15 (a) 21 12 2
T mx I 2θ= + && and 225
I ma=
10
2 21 1 22 2 5
T mx ma 2θ⎛ ⎞= + ⎜ ⎟⎝ ⎠
&&
V mgx= −
mg
T x
tθ θ= &
2 2 21 12 5
L T V mx ma mgxθ= − = + +&&
The equation of constraint is … ( , ) 0f x x aθ θ= − =
The 2 Lagrange equations with multipliers are …
0L d L fx dt x x
λ∂ ∂ ∂− + =
∂ ∂ ∂&
0L d L fdt
λθ θ θ∂ ∂ ∂
− + =∂ ∂ ∂&
L mxx∂
=∂
&&
d L mxdt x
∂=
∂&&
& L mg
x∂
=∂
fx
λ λ∂=
∂
0mg mx λ− + =&& and from the θ-equation … 22 0
5ma aθ λ− − =&&
Differentiating the equation of constraint … xa
θ =&&&& and substituting into the above …
57
x g=&& and 27
mgλ = −
(b) The generalized force that is equivalent to the tension T is …
27x
fQ mx
λ λ∂= = = −
∂g
10.16 For the velocity of a differential mass element, dm, of the spring at a distance v′r
x′ below the support …
xv vx′
′ =r r , mdm dx
x′
′=
( )2
22 2
0 0
1 1 1 12 2 2 2
m x x mT mv v dm mx x dx x
′x
′ ′⎛ ⎞′ ′= + = + ⎜ ⎟⎝ ⎠∫ ∫& &
2 21 12 2 3
mT mx x′
= +& &
( )2
0
12
mV k x l mgx gx dm
′′ ′= − − − ∫
( )2
0
12
x mk x l mgx gx dxx′
′ ′= − − − ∫ ( )21
2 2mV k x l mgx gx′
= − − −
( )221 12 3 2 2
m mL T V m x k x l m gx′ ′⎛ ⎞ ⎛ ⎞= − = + − − + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠&
11
3
L mm xx
′∂ ⎛ ⎞= +⎜ ⎟∂ ⎝ ⎠&
&,
3d L mm xdt x
′∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠&&
&,
( )2
L mk x l m gx
′∂ ⎛ ⎞= − − + +⎜ ⎟∂ ⎝ ⎠
( )3 2m mm x k x l m′ ′⎛ ⎞ ⎛ ⎞+ = − − + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠&& g
For 2
g my x l mk
′⎛ ⎞= − − +⎜ ⎟⎝ ⎠
,
03mm y ky′⎛ ⎞+ + =⎜ ⎟
⎝ ⎠&&
The block oscillates about the point 2
g mx l mk
′⎛ ⎞= + +⎜ ⎟⎝ ⎠
…
with a period …
2 32
mmT
′
kπ πω
⎛ ⎞+⎜ ⎟⎝ ⎠= =o
10.17 Note: 4 objects move – their coordinates are labeled ix : The coordinate of the
movable, massless pulley is labeled px . Two equations of constraint: ( ) ( )1 1 1, 0p pf x x x l x= − − =
( ) ( ) ( )2 2 3 2 3, , 2 0p pf x x x x x x l′= + − + =
2 2 21 1 1 1 2 2 2 3 3 3
1 1 12 2 2
L T V m x m gx mx m gx m x m gx= − = − + − + −& & & 3
0jj
ji i i
fL d Lq dt q q
λ∂∂ ∂
− + =∂ ∂ ∂∑
Thus: (1) 1 1 1 1 0m g m x λ− − + =&& (2) 2 2 2 2 0m g m x λ− − + =&& (3) 3 3 3 2 0m g m x λ− − + =&& Now – apply Lagrange’s equations to the movable pulley – note m 0p → So:
(4) 1 21 2 0
p p
f fx x
λ λ∂ ∂+ =
∂ ∂ or 1 22 0= λ − λ
Now - px can be eliminated between 1f and 2f
( )2 1 2 3 12 2 2f f x x x l l′= = + + − + = 0 (5) Thus 2 3 12 0x x x+ + =&& && &&
12
With a little algebra – we can solve (1) –(5) for the 5 unknowns 1x&& , 2x&& , 3x&& , 1λ , and 2λ
1
1 2 3
21 1 1 1
4
g
m m m
λ =⎡ ⎤⎛ ⎞
+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
2
1 2 3
1 1 1 14
g
m m m
λ =⎡ ⎤⎛ ⎞
+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
As the check, let and 2 3m m m= = 1 2m m= .
Thus, there is no acceleration and 1 2mgλ = 10.18 (See Example 5.3.3) (a) ˆrr re= ˆ ˆrr re r eθθ= + && & Constraint: ( ) 0f tθ θ ω= − =
so … θ ω=& and 0θ =&&
( )2 2 212
T m r r θ= + && L=
0L d Lr dt r∂ ∂
− =∂ ∂&
and 0L d L fdt
λθ θ θ∂ ∂ ∂
− + =∂ ∂ ∂&
2r rω=&& 22 0mrr mrθ θ λ− − +& &&& = t tr Ae Beω ω−= + ( )0r 0= apply constraint
t tr Ae Beω ωω ω −= −& ( )0r lω=& so … 0A B+ = 2mrrλ ω= & A B lω ω ω− =
2A l= 2lA =
2lB = −
thus .. ( )2t tlr e eω ω−= −
sinh
coshr l tr l t
ωω ω
= ⎫⎬= ⎭&
2 22 sinh coshm l t tλ ω ω= ω
now at so r l= t T= 11 0sinh 1T .88ω ω
−= =
(b) There are 2 ways to calculate F …
(i) See Example 5.3.3 … 2F m xω ′= &
(2tlx e eω ω−′ = − )t and ( )2
t tlx eω eω ω−′ = +&
t
22 coshF m lω ω=
(ii) fλθ∂∂
is the generalized force, in this case – a torque T, acting on the bead
13
fT rF λθ∂
= =∂
2 22 sinh coshsinh
f m l t tFr l tλ ω ω
θ ω∂
= =∂
ω
22 coshm l tω ω= 10.19 (See Example 4.6.1) The equation of constraint is ( ), 0f r r aθ = − =
( 2 2 2
2mT r r )θ= + && cosV mgr θ=
θ a
r ( )2 2 2 cos
2mL r r mgrθ θ= + −&&
0L d L fr dt r r
λ∂ ∂ ∂− + =
∂ ∂ ∂&
0L d L fdt
λθ θ θ∂ ∂ ∂
− + =∂ ∂ ∂& 1f
r∂
=∂
0fθ∂
=∂
Thus 2 cos 0mr mg mrθ θ λ− − +& && =
= =& &&
=
2sin 2 0mgr mr mrrθ θ θ− − =&& &&
Now , r r so r a= 0 2 cos 0ma mgθ θ λ− +&
2sin 0mga maθ θ− =&&
singa
θ θ=&& and ddθθ θθ
=&
&& &
so singd da
θ θ θ=∫ ∫& & θ or 2
cos2
g ga a
θ θ= − +&
hence, ( )3cos 2mgλ θ= − and when 0λ → particle falls off hemisphere at
1 2cos3
θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠
o
10.20 Let 2x mark the location of the center of curvature of the movable surface relative to a fixed origin. This point defines the position of m2. 1r marks the position of the particle of mass m1. r is the position of the particle relative to the movable center of
curvature of m2.
Kinetic energy 2 22 2 1 1
1 12 2
T m x m r= +& &
( ) ( ) 2 21 1 2 2 2 22r r x r x r x r x r⋅ = + ⋅ + = + + ⋅& & & & & & & & & &
but ˆ ˆrr re r eθ θ= + && &
( )2 2 2 2 22 2 1 2 2
1 1 ˆ ˆ22 2 rT m x m x r r x re r eθθ θ⎡ ⎤∴ = + + + + ⋅ +⎣ ⎦
& &&& & & &
14
2 2 2 2 22 2 1 2 2 2
1 1 2 sin 2 cos2 2
m x m x r r x r x rθ θ θ θ⎡ ⎤= + + + + −⎣ ⎦& && & & & & &
Potential energy sinV mgy mgr θ= − = − L T V= −
Equation of constraint ( ), 0f r r aθ = − = 1fr∂
∴ =∂
0fθ∂
=∂
2
1fx∂
=∂
Lagrange’s equations 0i i i
L d L fx dt x x
λ⎛ ⎞∂ ∂ ∂
− +⎜ ⎟∂ ∂ ∂⎝ ⎠&=
x2) 2 2 1 1 1 sin cos 0dm x m x m r rdt
θ θ θ⎡ ⎤− − − − =⎣ ⎦&&& && &
22 2 1 1 1 sin sin cos cos sin 0dm x m x m r r r r r
dtθ θ θ θ θ θ θ θ θ⎡ ⎤− − + + − + =⎣ ⎦& && & &&& && & && &
using constraint ; r a constant= = 0r r= =& &&
( 212
1 2
sin cosmx am m )θ θ θ θ= − +
+&& &&& (1)
θ) 22 2 2 22 sin sin cos sin cos cos 0r rr x r x r x r xr x r grθ θ θ θ θ θ θ θ θ θ− − − − − + + + =&& & & && && & & & & & &
using constraint ; r a constant= = 0r r= =& && )
2 sin cos 0x ga a
θ θ θ− − + =&&&& (2)
r) 22 2 2
1
cos sin sin sin 0r x x r x gmλθ θ θ θ θ θ θ− + − + + + + =& & &&& && & &
Apply constraint…
22
1
cos sin 0x a gmλθ θ θ+ + + =&&& (3)
Now, plug solution for 2x&& (1) into equation (2):
( )21
1 2
sin cos sin cos 0m gm m a
θ θ θ θ θ θ− + −+
&& && & θ =
( )1 1
2 21 sin sin cos cosm mgf fa
θ θ θ θ θ− − −&& & 0θ =
where 1
1
1 2m
mfm m
=+
Now 212
d d d d ddt d dt d dθ θ θ θθ θ θ
θ θ θ⎧ ⎫= = = = ⎨ ⎬⎩ ⎭
& & &&& & &
So – let 2x θ= &
1 1
2 21 sin 2 sin cos cos 0m mdx gf xfd a
θ θ θθ
−⎡ ⎤− − −⎣ ⎦ θ =
Let 1
21 sinmy f θ= −
( )sin2 ddx dy gy xd d a d
θθ θ θ+ =
15
Hence: ( ) (2 singd yx da
)θ= and ( )( )
( )
( )sin
00
2 sinyx
yx
gd yx da
θ θ θ
θ
θ=
=
=∫ ∫o
but at so 2 0x θ= =& 0θ = o
( ) 2 singyxa
θ θ=
( )1
22
2 sin1 sinm
gxa f
θθ θθ
= =⎡ ⎤−⎣ ⎦
&
Now we can solve for θ&& and plug 2θ& , θ&& into (3) to obtain ( )λ θ
1
22
1 si2 1 sim
d g dd a d f
θθ θ nnθ θ θ
⎡ ⎤⎧ ⎫= = ⎢ ⎥⎨ ⎬ −⎩ ⎭ ⎢ ⎥⎣ ⎦&& &
After some algebra – yields
( )( )
1
1
2
22
1 sincos
1 sinm
m
fga f
θθ θ
θ
+=
−&&
The solution for ( )λ θ is thus determined from equation (3)
1
2 2
1
sin cos cos sinmgfa m a
λθ θ θ θ θ θ θ⎡ ⎤− + +⎣ ⎦& && & = −
collecting terms:
( )1 1
2 2
1
1 cos sin cos sinm mgf fa m a
λθ θ θ θ θ θ− − + = −& &&
Plug θ&& , θ& into the above --- plus --- a lot of algebra yields …
( ) 2 1 1
1
2 2
221
2 sin sin cos 1 sinsin
1 sinm m m
m
f f fm g f
θ θ θ θλ θθ
θ
⎡ ⎤− −⎣ ⎦− = +⎡ ⎤−⎣ ⎦
where 2
2
1 2m
mfm m
=+
As a check, let so it is immoveable … then and 2m →∞2
1mf →1
0mf →and we have
( )1
3sinm gλ θ
θ− → … which checks
10.21 (a) ( )2 2 2 21 12 2
T mv m R R zφ= = + +&& &2
( )2 2 2 2
2mL T V R R z Vφ= − = + + −&& &
L mRR∂
=∂
&&
, d L mRdt R
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
& , 2L VmRR R
φ∂ ∂= −
∂ ∂&
16
2 VmR mRR
φ ∂= −
∂&&&
2RmR mR Qφ− =&&&
2L mR φφ∂
=∂
&& , 2 2d L mR mRR
dtφ φ
φ⎛ ⎞∂
= +⎜ ⎟∂⎝ ⎠&& &&
& , L Vφ φ∂ ∂
= −∂ ∂
2 2 VmR mRR Qφφ φφ
∂+ = − =
∂&& &&
L mzz∂
=∂
& , d L mzdt z
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&, L V
z z∂ ∂
= −∂ ∂
zVmz Qz
∂= − =
∂&&
For F ma=r r , using the components of ar from equation 1.12.3:
( )2RF m R Rφ= − &&& , ( )2F m R Rφ φ φ= +& &&& , zF mz= &&
From Section 10.2, since and R z are distances, and are forces. However, since RQ zQφ is an angle, Qφ is a torque. Since Fφ is coplanar with and perpendicular to , RQ RFφ φ= … and all equations agree. (b) 2 2 2 2 2 2 2sinv r r rθ φ θ= + +& &&
( )2 2 2 2 2 2sin2mL T V r r r Vθ φ θ= − = + + −& &&
L mrr∂
=∂
&&
, d L mrdt r
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&, 2 2 2sinL Vmr mr
r rθ φ θ∂ ∂
= + −∂ ∂
&
2 2 2sin rVmr mr mr Qr
θ φ θ ∂− − = − =
∂& &&&
2L mr θθ∂
=∂
&& , 2 2d L mr mrr
dtθ θ
θ∂⎛ ⎞ = +⎜ ⎟∂⎝ ⎠
&& &&& , 2 2 sin cosL Vmr φ θ θ
θ θ∂ ∂
= −∂ ∂
&
2 2 22 sin cos Vmr mrr mr Qθθ θ φ θ θθ∂
+ − = − =∂
&& & &&
2 2sinL mr φ θφ∂
=∂
&& , 2 2 2 2sin 2 sin 2 sin cosd L mr mrr mr
dtφ θ φ θ θφ θ
φ⎛ ⎞∂
= + +⎜ ⎟∂⎝ ⎠&& & & &&
& θ ,
L Vφ φ∂ ∂
= −∂ ∂
2 2 2 2sin 2 sin 2 sin cos Vmr mrr mr Qφφ θ φ θ θφ θ θφ
∂+ + = −
∂&& & & && =
is a force rQ rF . Qθ and Qφ are torques. Since φ is in the xy plane, the moment arm for φ is sinr θ ; i.e. sinQ r Fφ φθ= . Q rFθ θ= . From equation 1.12.14 …
17
( )2 2 2sin rm r r r Fφ θ θ− −& &&& =
( )22 sin cosm r r r Fθθ θ φ θ θ+ − =&& & &&
( )sin 2 sin 2 cosm r r r Fφφ θ φ θ θφ θ+ +&& & & && = The equations agree.
10.22 For a central field, ( )V V r= , so 0V Vθ φ∂ ∂
= =∂ ∂
and V Fr
∂= −
∂.
For spherical coordinates, using equation 1.12.12: 2 2 2 2 2 2sinv r r r 2φ θ θ= + +& &&
( ) ( )2 2 2 2 2 2sin2mL T V r r r V rφ θ θ= − = + + −& &&
L mrr∂
=∂
&&
, d L mrdt r
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&, 2 2 2sinL Vmr mr
r rφ θ θ∂ ∂
= + −∂ ∂
& &
2 2 2sin rmr mr mr Fφ θ θ− −& &&& =
2L mr θθ∂
=∂
&& , 2 2d L mr mrr
dtθ θ
θ∂⎛ ⎞ = +⎜ ⎟∂⎝ ⎠
&& &&& , 2 2 sin cosL mr φ θ θ
θ∂
=∂
&
2 2 22 sin comr mrr mrθ θ φ θ θ+ −&& & && s 0=
2 2sinL mr φ θφ∂
=∂
&& ,
2 2 2 2sin 2 sin 2 sin cosd L mr mrr mrdt
φ θ φ θ φθ θφ
⎛ ⎞∂= + +⎜ ⎟∂⎝ ⎠
&& & & &&& θ
0Lφ∂
=∂
2 2 2 2sin 2 sin 2 sin cosmr mrr mrφ θ φ θ φθ θ θ 0+ + =&& & & && 10.23 Since θ α= = constant, there are two degrees of freedom, r and θ . , rv r= & sinv rφ φ α= &
( )2 2 2 21 1 sin2 2
T mv m r r φ α= = + &&
cosV mgr α=
( )2 2 2 2sin cos2mL T V r r mgrφ α α= − = + −&&
L mrr∂
=∂
&&
, d L mrdt r
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠&&
&, 2 2sin cosL mr mg
rφ α α∂
= −∂
&
2 2sin cosmr mr mgφ α α= −&&&
2 2sinL mr φ αφ∂
=∂
&& , 0L
φ∂
=∂
18
( )2 sin 0d mrdt
φ α =&
Say constant 2 sinmr φ α = =& l
2
3 cosmr mgmr
α= −l
&&
212
d dr dr r rdt dr dr
= = =&
&& & & &r
( )2
23 cos
2m drd r mg dr
m rα= −
l&
2 2
2 cos2 2
mr mgr Cmr
α= − − +& l
The constant of integration C is the total energy of the particle: kinetic energy due to the component of motion in the radial direction, kinetic energy due to the component of motion in the angular direction, and the potential energy.
( )2
2 cos2
U r mgrmr
α= +l
For , , and turning points occur at 0φ ≠& 0≠l 0r =&
Then 2
20 co2
mgr Cmr
α= − − +l s
( )2
3 2cos 02
mg r Crm
α − + =l
The above equation is quadratic in r ( 2 r 4∝l ) and has two roots.
10.24 Note that the relation obtained in Problem 10.23, ( )2
3 2cos 02lmg r Crm
α − + = ,
defines the turning points. For the particle to remain on a single horizontal circle, there must be two roots with r . Thus ( divided into the above expression leaves a term that is linear in r.
r= o )2r r− o
( )
( )( ) ( ) ( )
( ) ( )
2 2 3 2 2
3 2 2
2 2
2 2
2 2 2
22 2
2
2
2 2 2 2
2 2 2 2
r r cr rr r r Cr m
r r r rr
r C r r r
r C r r r C r r r C
r r C r m r r C
+ −
− + − +
− +
− −
− − − + −
⎡ ⎤− − + − −⎣ ⎦
o
o o
o o
o o
o o o o o
o o o o o
l
l
For the remainder to vanish, both terms must equal zero. 2 24 2r r C r− − =o o o 0
32
C r= o
19
2
3 22 02
r Crm− + =o o
l
2
22 2r
mr= o
o
l
And with 2 sinmr φ α= o o&l
12
2
1sinmr
φα
⎛ ⎞= ⎜ ⎟⎝ ⎠
o
o
&
From Prob. 10.23, 2
3 cosmr mgmr
α= −l
&&
For small oscillations about r , o r r ε= +o . r ε= &&&&
3
3 3 3
1 1 1 31 1r r r r r
ε ε−
⎛ ⎞ ⎛= + ≈ −⎜ ⎟ ⎜
⎝ ⎠ ⎝o o o o
⎞⎟⎠
2
3
31 cm mmr r
ε osgε α⎛ ⎞
= − −⎜ ⎟⎝ ⎠o o
l&&
2 2
4 3
3 cosm mmr mr
gε ε α+ = −o o
l l&&
2 4
2
2 223 sin
m rT π ππω φ
= = =o
o&l
13α
10.25 ( ) ( )2 2 2 212 2 x y z
mL mv qv A x y z q xA yA zA= + ⋅ = + + + + +rr
& & & && &
xL mx qAx∂
= +∂
&&
, xdAd L mx qdt x dt
∂⎛ ⎞ = +⎜ ⎟∂⎝ ⎠&&
&
Using the hint, x x xdA A A Ax y zdt x y z
∂ ∂ ∂= + +
∂ ∂ ∂& & & x
yx zAAL Aq x y zx x x x
∂⎛ ⎞∂∂ ∂= + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠
& & &
yx x x x AA A A A zAmx q x y z q x y zx y z x x x
∂⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂+ + + = + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
&& & & & && &
y x x zA A A Amx q y zx y z x
⎡ ∂ ⎤⎛ ⎞∂ ∂ ∂⎛ ⎞= − − −⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠⎣ ⎦&& & &
( ) ( )z y
q y A z A⎡ ⎤= ∇× − ∇×⎢ ⎥⎣ ⎦
r rr r& &
( )x
q v A⎡ ⎤= × ∇×⎣ ⎦rrr
20
( )x
mx q v B= ×rr
&&
Due to the cyclic nature of the Cartesian coordinates, i.e., ˆˆ ˆi j k× = …
( )y
my q v B= ×rr
&&
( )z
mz q v B= ×rr
&&
Altogether, ( )mr q v B= ×rr r&&
10.26 V mgz= ( )2 2 212
T m x y z= + +& & &
xTp mxx
∂= =∂
&&
, xpxm
=& similarly, ypy
m=& and zpz
m=&
( )2 2 212 x y zH T V p p p mgz
m= + = + + +
x
x
pH xp m∂
= =∂
&
0 xH px
∂= = −
∂& , xp constant=
( ) 0xdp mx mxdt
= =& & &&= similarly, yp constant= , or 0my =&&
y
y
pH yp m∂
= =∂
&
z
z
H p zp m∂
= =∂
&
zH mg pz
∂= = −
∂& , ( )z
dp mz mz mgdt
= = = −& & &&
These agree with the differential equations for projectile motion in Section 4.3. 10.27 (a) Simple pendulum …
cosV mgl θ= − 2 212
T ml θ= &
2Tp mlθ θθ∂
= =∂
&& , 2
pml
θθ =&
2
2 cos2
pH T V mglmlθ θ= + = −
2
pHp ml
θ
θ
θ∂= =
∂&
sinH mgl pθθθ
∂= =
∂&−
21
(b) Atwood’s machine …
( )1 2V m gx m g l x= − − − 2
2 21 2 2
1 1 12 2 2
xT m x m x Ia
= + +&
& & [Includes the pulley]
1 2 2
T Ip m m xx a
∂ ⎛ ⎞= = + +⎜ ⎟∂ ⎝ ⎠& ,
1 2 2
pxIm m
a
=⎛ ⎞+ +⎜ ⎟⎝ ⎠
&
( )2
1 2 2
1 2 22
pH T V m m gx m glIm m
a
= + = − − −⎛ ⎞+ +⎜ ⎟⎝ ⎠
1 2 2
H p xIp m m
a
∂= =
∂ ⎛ ⎞+ +⎜ ⎟⎝ ⎠
&
( )1 2H m m g px
∂= − − = −
∂& , ( )1 2p m m g= −&
(c) Particle sliding down a smooth inclined plane …
sinV mgx θ= − 212
T mx= &
Tp mxx
∂= =∂
&&
, pxm
=&
2
sin2pH T V mgxm
θ= + = −
H p xp m
∂= =
∂&
sinH mg px
θ∂= − = −
∂& , sinp mg θ=&
10.28 (a)
r
F
( , ) ( , )i i iL T q q T q t= −& …note, potential energy is time dependent.
VFr
∂= −
∂ so tkV Fdr e
rβ−= = −∫
( )2 2 21 12 2
T mv v m r r θ= ⋅ = +r r &&
22
rLp mrr∂
= =∂
&&
rprm
=&
2Lp mrθ θθ∂
= =∂
&& 2
pmr
θθ =&
22
22 2tr
i i rpp kH p q L p r p e
m mr rβθ
θθ−= − = + − − −∑ && & &
Substituting for and r& θ& …
22
22 2tr pp kH e
m mr rβθ −= + −
(b) H = T + V … which is time-dependent (c) E is not conserved 10.29 Locate center of coordinate system at C.M. The potential is independent of the center of mass coordinates. Therefore, they are ignorable.
(r1,θ1)
(r2,θ2)
L T V= −
( ) ( ) ( )22 2 2 2 2 21 1 1 1 2 2 2 2 1 2
1 1 12 2 2
m r r m r r k r r lθ θ= + + + − + −& && &
where l is the length of the relaxed spring and k is the spring constant.
1 11
rLp mrr∂
= =∂
&&
11
1
rprm
=&
21 1 1
1
Lp mrθ θθ∂
= =∂
&& 1
1 21 1
pθ =& θ
m r… and similarly for m2
i iH p q L= −∑ & &
2 22 2
21 21 21 22 2
1 1 1 2 2 2
1 ( )2 2 2 2 2
r rp pp pH km m r m m r
θ θ= + + + − + −r r l
Equations of motion … First, θ1 , θ2 are ignorable coordinates, so 1 2,p pθ θ are each conserved.
( )1 1 21
rH
1 1p k r r l m rr
∂= − = + − = −
∂& &&
( )2 1 22
rH
2 2p k r r l m rr
∂= − = + − = −
∂& &&
1 2r rp p=& & 0rpΔ =& rpΔ = constant. The radial momenta are equal and opposite.
23
10.30 2 21 12 2
L mx kx= −&
2 2 2
1 1 1
2 21 102 2
t t t
t t tLdt Ldt mx kx dtδ δ δ ⎛ ⎞= = = −⎜ ⎟
⎝ ⎠∫ ∫ ∫ &
( )2
1
0t
tmx x kx x dtδ δ= −∫ & &
dx xdt
δ δ=&
( ) ( )2 2 2
1 1 1
t t t
t t t
dmx xdt mx x dt mxd xdt
δ δ δ= =∫ ∫ ∫& & & &
Integrating by parts:
( )2 22
11 1
t tttt t
mx xdt mx x x d mxδ δ δ= −∫ ∫& & & &
0xδ = at and
1t 2t
( ) ( )dd mx mx dt mx dtdt
= =& & &&
2 2
1 1
t t
t tmx x dx x mx dtδ δ= −∫ ∫& & &&
( )2
1
0t
tmx x kx x dtδ δ= − −∫ &&
0mx kx+ =&&
10.31 (a) 2
20 21 vL m c V
c= − − −
Let 2
2
1
1 vc
γ =−
0 xL m x px
γ∂= ≡
∂&
& This is the generalized momentum for part (b).
Thus, Lagrange’s equations for the x-component are …
0xd L L d Vpdt x x dt x
∂ ∂ ∂− = − =
∂ ∂ ∂&
… and so on for the y and z components. (b) i i
i
H v p= −∑ L
but … 0
ii
pvmγ
=
So … 2 2 2
02
0
i
i
p c m cH Vm cγ γ
= +∑ +
22 2
02
0
m cp cH Vm cγ γ
= + +
24
( )2 2 2 402
0
1H p c m cm cγ
= + V+
(c) Now, if then we have … 2
0T m cγ=
2 2
2 2 2 4 2 40 0 2 4
0
1 p cp c m c m cm c
⎛ ⎞+ = +⎜ ⎟
⎝ ⎠
( )2 2 2 2
2 4 2 4 2 2 200 02 4
0
1 1m v cm c m c v cm c
γ γ⎛ ⎞
= + = +⎜ ⎟⎝ ⎠
2 2
2 4 2 4 2 2 40 02 2 2 2
111 1
v cm c m c m cv c v c
γ⎛ ⎞ ⎛ ⎞
= + = =⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠0
Thus … 20H m c V T Vγ= + = +
(d) 2 2
2002 2 2
11 ...1 2
m c vT m cv c c
⎛ ⎞= ≈ + +⎜ ⎟− ⎝ ⎠
2 20 0
12
T m v m c≈ +
------------------------------------------------------------------------------------------------------------
25
1
CHAPTER 11 DYNAMICS OF OSCILLATING SYSTEMS
11.1
(a) 2
2( )2k kV x x
x= +
2
2
kV kxx
′ = − At equilibrium, 0V ′ =
1 3x k=
2
3
2kV kx
′′ = +
1 3 2 3x k
V k k k=
′′ = + = > 0 Stable (b) ( ) bxV x kxe−= bx bxV ke bkxe− −′ = − At equilibrium 0bx bxke bkxe− −− =
1xb
=
2bx bx bxV bke bkxe b kxe− −′′ = − − + −
1 1 11
2 0x b
V bke bke bke− − −=
′′ = − + = − < Unstable
(c) ( )4 2 2( )V x k x b x= −
( )3 24 2V k x b x′′ = −
At equilibrium ( )3 24 2k x b x 0− =
0, 2x b= ± ( )2 212 2V k x b′′ = −
20
2x
V kb=
′′ = − < 0 Unstable
( )2 2 22
6 2 4x b
V k b b kb=±
′′ = − = 0> Stable
(d) for case (a) 2 3km
ω = 2 223 3mT sk
π ππω
= = =
for case (c) at 2x b= ± 2
2 4kbm
ω = 2
2 24
mT skb
π π πω
= = =
2
11.2 ( )2 2( , ) 2 4V x y k x y bx by= + − −
2 ( ) 2 ( 2 )V Vk x b k y bx y
∂ ∂= − = −
∂ ∂
at equilibrium 2x b and y b= =
2 2 2
2 22 0V V Vk kx x y y
∂ ∂ ∂= =
∂ ∂ ∂ ∂2=
k
11 12 21 222 0 0 2k k k k k= > = = =
11 12 2
21 22
4 0k k
kk k
= − > 0
The equilibrium is stable. 11.3
21( )2
V x kx= −
( )( ) dV x dxF x kx mx mxdx dx
= − = = =&&& &
kx dx mx dx= & &
0 0
x v
x
kxdx mx dx=∫ ∫ & & ( )2
2 202 2
k vx x m− =
( )1 22 20
dxv k m x xdt
= = −
( )0
1 22 200
;x t
x
dx dt where k mx x
α α= =−
∫ ∫
( )2 20 0ln lnx x x x tα+ − − =
2 20 0
tx x x x eα+ − =
2 2 2 20 0 02t t 2x x x e xx e xα α− = − +
0 0 cosh2
t te ex x xα α
tα−+
= =
11.4 Let the length of the unstretched, elastic cord bd. Then
e
y
m
2l
2 22d l y= +
( )21 22
V k d l mg= − − y
3
( )2 2 2 2 24 4 8 42kV l y l l y l m= + − + + − gy
( )2 2 2 22 2 2V k l y l l y mg= + − + − y
The first term, , is an additive constant to the potential energy, so with appropriate adjustment of the zero reference point …
24kl
( )2 2 2( ) 2 2V y k y l l y mgy= − + −
( ) 1 22 2( ) 2 2 2dV y k y yl l y mgdy
−⎡ ⎤= − + −⎢ ⎥⎣ ⎦
At equilibrium, the above expression is zero, so …
2 2
44 0klyky mgl y
− −+
=
2 2
44 klyky mgl y
− =+
2 2 2
2 2 2 22 2
1616 8 k l yk y kmgy m gl y
− + =+
( )2 4 3 2 2 2 2 2 2 2 2 2 2 216 8 16 16 8 0k y kmgy k l m g k l y kl mgy l m g− + + − − + =
4 2 2 2
3 24 4 2 4 2 2 2 0
2 16 2 16y mg m g mg m gy y yl kl k l kl k l
− + − +2
=
letting yul
= and 4mga
kl=
4 3 2 2 22 2u au a u au a− + − + = 0 11.5
( )1 2V mg h h= +
h2
h1
a
b θ
θφ d
1 cosh b θ=
( )2 sinh d θ ϕ= +
( )2 sin cos cos sinh d θ ϕ θ= + ϕ
cos sinb ad dθϕ ϕ= =
2 sin cosh b aθ θ ϕ= +
4
( )cos sinV mg a b bθ θ θ= + +⎡ ⎤⎣ ⎦
( )( )sin sin cosV mg a b b bθ θ θ θ′= + − + +⎡ ⎤⎣ ⎦
[ ]sin cosV mg a bθ θ θ′= − +
[ ]cos sin cosV mg b b aθ θ θ θ′′= − −
( )0V mg b a′′ = − Equilibrium … Stable Unstable a b< a b> 11.6
2 4
cos 1 ...2! 4!θ θθ = − + −
3 5
sin ...3! 5!θ θθ θ= − + −
( )2 4 3 5
1 ... ...2! 4! 3! 5!
V mg a b bθ θ θ θθ θ⎡ ⎤⎛ ⎞ ⎛ ⎞
= + − + − + − + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦
2 43 ...2 24
a b a bV mg a b θ θ− −⎡ ⎤= + − + −⎢ ⎥⎣ ⎦
For a = b, 42 ...12aV mg a θ⎡ ⎤= − +⎢ ⎥⎣ ⎦
3 ...3
mgaV θ′=− + terms in higher order of θ 2 ...V mgaθ′′=− +
2 ...V mgaθ′′′=− + 2 0V mga′′′′=− < ∴ Equilibrium is unstable
11.7
The center of mass (CM) of the hemisphere is 38
a from the flat side (see Equation 8.1.8).
The height of CM) above the point of contact between the two hemispheres is designated by h2 in the figure. h1 is the height of the point of contact above the ground.
h2
h1 b θ
θφ a
5
( ) (1 23cos cos cos8
V mg h h mg b a a )θ θ θ⎡ ⎤= + = + − +⎢ ⎥⎣ ⎦ϕ and a bϕ θ=
( ) ( )1 23cos cos8
bV mg h h mg a b aaθθ θ⎡ ⎤⎛ ⎞= + = + − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
( ) 3sin sin8a a b a bV mg a b
a aθ θ
⎡ ⎤+ ⎛ +⎛ ⎞ ⎛ ⎞′= − + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦
⎞⎟ Equilibrium occurs at 0oθ =
( )23cos cos
8a a b a bV mg a b
a aθ θ
⎡ ⎤+ ⎛ + ⎞⎛ ⎞ ⎛ ⎞′′= − + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦
( ) ( )(2
03 3 58 8a a b mgV mg a b a b b a
a a⎡ ⎤+⎛ ⎞′′= − + + = + −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦)
a
0 0 3 5V for b′′ > >
Therefore, the equilibrium is stable for 35ba<
11.8 From Problem 11.4, we have
( )1
2 22 2 2 2 2
2( ) 2 2 2 2 1 yV y k y l l y mgy k y l mgyl
⎛ ⎞⎛ ⎞⎜ ⎟= − + − = − + −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟
⎝ ⎠
Expanding the square root for small yl
…
12 22
2 2
1 11 1 ...2 8
y y yl l l
⎛ ⎞ 4
4+ = + − +⎜ ⎟⎝ ⎠
42 2 2
2( ) 2 24yV y k y l y mgyl
⎡ ⎤≈ − − + −⎢ ⎥
⎣ ⎦
32
2kV yl
′ ≈ −mg
at equilibrium,
12 3
02
mglV yk
⎛ ⎞′= ⇒ =⎜ ⎟
⎝ ⎠
22
6kV yl
′′=
( )1
2 3
2 212 3 33
22
6 62 2y mgl k
k mgl mgV kl k l=
⎛ ⎞ ⎛ ⎞′′ = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
6
1 2 13 3 3
6 62 2
V k g g km m l l m
ω′′ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
16
11.9
From Problem 11.5, ( )0V mg b a′′ = −
0VM
ω′′
= where 212
T Mθ= &
The vector r locates the CM of the block (see figure in 11.5).
( )2 21 12 2 CMT m r Iβ θ= +& & (The CM motion is
essentially horizontal. Also, r a b const≈ + = and d a const≈ = )
( )sin sin cosr b dβ θ ϕ θ= − +
( ) ( )cos cos sinr b dβ β θ θ ϕ θ ϕ θ= + + +& & &&
θ
β θ+φ
b
d r
For small β, θ and for 2πϕ ≈ , we obtain
( )r b aβ θ ϕ θ≈ + +& & &&
( )sin tan 2b aθ π ϕ≈ − or 2
b a πθ ϕ⎛ ⎞≈ −⎜ ⎟⎝ ⎠
b aθ ϕ≈ −& & So r aβ θ≈& & and substituting into the kinetic energy expression, we get
( )22 2 2 2 21 1 1 1 522 2 6 2 3
T ma m a maθ θ θ⎡ ⎤ ⎡= + =⎢ ⎥ ⎢⎣ ⎦ ⎣& & ⎤
⎥⎦&
Thus, 253
M ma= and ( )025 3
g b aVM a
ω′′ −
= =
( )( )02 2 5 3T a g b aπ πω
= = −
11.10
From Problem 11.7, ( )(0 3 58mg )b b a
a′′= + −V a
( )( )0
0
22 43 5
m aTg a b b aV
π π= =+ −′′
7
11.11
2 21 12 2cmT mv Iϕ= + &
θ
φ b a ( )cmv b a θ= − &
225
I ma=
The relationship between the angles, andθ ϕ (see Figure), can be determined from the condition that there is no slipping as the ball rolls in the hemisphere, so the length of roll measured along the ball, (a )θ ϕ+ , must equal the length of roll measured along the hemisphere, bθ … so we have …
( ) ( )b a and b aθ θ ϕ θ θ ϕ= + ∴ = +& & &
and … b aa
ϕ θ−⎛ ⎞=⎜ ⎟⎝ ⎠
&&
( ) ( ) ( )2 2
2 22 22
1 1 2 1 12 2 5 2
b aT m b a ma m b a
aθ 22
5θ θ
− ⎛ ⎞= − + = − +⎜ ⎟⎝ ⎠
&& &
( )cosV mg b a θ=− −
( )sin @ 0oV mg b a equilibriumθ θ′= − =
( ) ( )0cosV mg b a V mg b aθ ′′′′= − ∴ = −
( )( ) ( )
0
2
57 75
mg b aV gM b am b a
ω′′ −
= = =−−
( )0
72 25b a
Tg
π πω
−= =
11.12 The potential energy of the satellite shaped like a “thin rod” is (See Example 11.2.2, Figure 11.2.1) …
eM dmV Gr
= −∫ but 2mdm dxa
= where 2a is the length of the rod.
2 2a ae ea a
GM GM mm dV dxr a a r− −
= − =−∫ ∫x
8
( )1
2 2 20 02 cosr r x r x φ= + +
( ) ( )1 1
2 2 2 20 1 cosr r x ε φ= + + where 0
2 20
2xrr x
ε =+
( )
( )
12
12 2 2
0
1 cos2
aea
dxGM mVa r x
ε φ −
−
+=−
+∫
For ( )1
2 2 2 20 0 0 0
0
2, , 1 cxr x r x r and r rr
osε ε φ>> + ≈ ≈ ≈ +
Thus, for small ε , the expression for the potential energy, V, can be approximated …
22
0 0 0
1 2 3 21 cos cos2 2 8
aea
GM m x xV dar r r
φ φ−
⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟≈− − +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
∫ x
2 3
20 0
3cos 22 02 2
eGM aV aar r
φ⎡ ⎤≈− + +⎢ ⎥
⎣ ⎦3
2 2
20 0
cos12
eGM aVr r
φ⎡ ⎤≈− +⎢ ⎥
⎣ ⎦
22
2 30 0 0
2cos sin sin 22 2
e eGM GM aaVr r r
φ φ φ⎛ ⎞
′≈ =⎜ ⎟⎝ ⎠
Equilibrium @ 0φ =
2
30
cos 2eGM aVr
φ′′≈ and 2
0 30
eGM aVr
′ ≈
213
M I ma= =
03
0
3 eV GM r
ω′′
= =M
30
02 2
3 e
rTGM
π πω
= =
11.13 The amplitude of the symmetric component is A1 and the amplitude of the anti-symmetric component is A2 (See Equations 11.3.19a through 11.3.20b).
9
[ ]221 1 2
1 (0) (0)4
A x x= +
[ ] 01 1 2
1 (0) (0)2 2
AA x x= + =
[ ]222 2 1
1 (0) (0)4
A x x= −
[ ] 02 2 1
1 (0) (0)2 2
AA x x= − = 1 2A A∴ =
From Equation 11.3.18 the solution for x1 is …
( )01 1( ) cos cos
2A
2x t t tω ω= + (The phase 2δ is 180o, which insures that 1 0(0)x A= )
( ) ( )0 1 2 1 21( ) 2cos cos
2 2 2A t t
x tω ω ω ω+ −⎛ ⎞
= ⎜ ⎟⎝ ⎠
Letting … 1 2 1
2 2and 2ω ω ωω + −
= Δ =ω
( )1 0( ) cos cosx t A t tω= Δ From Equation 11.3.18, the solution for x2 is …
( )02 1( ) cos cos
2A
2x t t tω ω= −
( ) ( )0 1 2 1 22 ( ) 2sin sin
2 2 2A t t
x tω ω ω ω+ −⎛ ⎞
= ⎜ ⎟⎝ ⎠
( )2 0( ) sin sinx t A t tω= Δ 11.14 At time t = 0 and short times thereafter … cos 1tΔ ≈ and sin 0tΔ ≈ . Thus, …
1 0 2cos 0x A t and xω≈ ≈ . This situation occurs again when 2t πΔ = .
1 12 2
2 1 1 22 2
k k km m
ω ω ⎡ ⎤′− +⎛ ⎞ ⎛ ⎞⎢ ⎥Δ = = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
1 12 21 21 1
2k km k
⎡ ⎤′⎛ ⎞ ⎛ ⎞⎢ ⎥Δ = + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
for , k k′<<
122 1 21 1 1 ... 1
2k kk k
kk
′ ′ ′⎛ ⎞ ⎛ ⎞+ − ≈ + + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
10
121
2k km k
′⎛ ⎞ ⎛ ⎞Δ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
12
1
2 2 22 m k kTk k k
π ππω
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟′ ′Δ ⎝ ⎠ ⎝ ⎠2
12kT Tk
⎛ ⎞= ⎜ ⎟′⎝ ⎠
11.15
( )2 2 21 2
12
T ml θ θ= +& &
( ) ( )1 21 cos 1 cos kV mglr
θ θ= − + − −⎡ ⎤⎣ ⎦
0 2sin sinr r l l 1θ θ= + −
( )1
1 21 0 2
cossinsin sinklV mgl
r l l 1
θθθ θ θ∂
= −∂ + −
( ) ( )
2 221 1
1 2 321 0 2 1 0 2
sin 2 coscossin sin sin sin
kl klV mglr l l r l l
θ θθθ
1θ θ θ∂
= − −∂ + − + − θ
1θ 2θ
r
l
2 2
11 2 31 001 2
2V kk mgr
θ θθ
= =
∂= =∂
ll −
( )
221 2
32 1 0 2
2 cos cossin sin
klVr l l
θ θθ θ
1θ θ∂
=∂ ∂ + −
2 2
12 31 2 2 1 00 01 2 1 2
2V Vkr
θ θ θ θθ θ θ θ
= = = =
∂ ∂= =∂ ∂ ∂ ∂
2kl=
( )2
2 22 0 2
cossinsin sinklV mgl
r l l 1
θθθ θ θ∂
= +∂ + −
( ) ( )
2 222 2
2 2 322 0 2 1 0 2
sin 2 coscossin sin sin sinkl klV mgl
r l l r l lθ θθ
θ1θ θ θ
∂= − −
∂ + − + − θ
2 2
22 2 32 001 2
2V kk mr
θ θθ
= =
∂= =∂
lgl −
Thus, from Equation 11.3.37a or b, we have
11
2 211 1 12 1 2 22 2
1 22
V k k kθ θ θ θ⎡ ⎤= + +⎣ ⎦
But from Equation 11.3.9, for the coupled oscillator, we have
( )2 2 21 1 1 2 2
1 1 122 2 2
V kx k x x x x kx′= + − + + 21
The forms of the potential energy function are similar with …
11 22 12k k k k and k k′ ′= + = = − In the case here …
2
30
2klk mgl and kr
′= = −
11.16
2 21 1 2 2
1 12 2
T m x m x= +& &
( )22 21 1 2 1 2 2
1 1 12 2 2
V k x k x x k x′= + − +
L T V= −
1 11
d L m xdt x
∂=
∂&&
& ( )1 1 2 1
1
L k x k x xx∂ ′= − + −∂
( )1 1 1 1 2 1 0m x k x k x x′+ − − =&&
2 22
d L m xdt x
∂=
∂&&
& ( )2 2 2 1
2
L k x k x xx∂ ′= − − −∂
( )2 2 2 2 2 1m x k x k x x′+ + −&&
21 1
22 2
0m k k k
k m k kω
ω′ ′− + + −
=′ ′− − + +
( ) ( ) ( )( )4 2
1 2 2 1 1 2 1 2 0m m m k k m k k k k k k kω ω′ ′ ′ ′− + + + + + + −⎡ ⎤⎣ ⎦2′ =
( ) ( ) ( )4 21 2 2 1 1 2 1 2 1 2 0m m m k k m k k k k k k kω ω′ ′− + + + + +⎡ ⎤⎣ ⎦ ′ =
( ) ( ) ( ) ( ) ( )22 1 1 2 2 1 1 2 1 2 1 2 1 22
1 2
40
2
m k k m k k m k k m k k m m k k k k k
m mω
′ ′ ′ ′ ′+ + + ± + + + − + +⎡ ⎤ ⎡⎣ ⎦ ⎣= =⎤⎦
2 For 1 2 1 2, 2 , , 2 ,m m m m k k k k k k′= = = = =
12
( ) ( ) ( ) ( )( )( )
2 2 2 22
2
2 3 4 6 4 4 2 2 6
2 2
m k m k mk mk m k k
mω
+ ± + − +=
2 10 64 4
k km m
ω = ±
0 0 02 kand wherem
ω ω ω ω= =
11.17
x1
x2
m
2m
k
k
Note: As discussed in Section 3.2, the effect of any constant external force on a harmonic oscillator is to shift the equilibrium position. x1 and x2 are the positions of the harmonic oscillator masses away from their respective “shifted” equilibrium positions.
( )2 21 2
1 1 22 2
T mx m x= +& &
( )221 2
1 12 2
V kx k x x= + − 1
L T V= −
( )1 1 21 1
,d L Lmx kx k x xdt x x
∂ ∂= =− +
∂ ∂&&
& 1−
1 1 22 0mx kx kx+ − =&&
( )2 22 2
2 ,d L Lmx k x xdt x x
∂ ∂= =−
∂ ∂&&
& 1−
2 2 12 0mx kx kx+ − =&& The secular equation (11.4.12) is thus
2
2
20
2m k k
k m kω
ω− + −
=− − +
2 4 2 2 22 5 2m mk k kω ω− + + 0=
The eigenfrequencies are thus … 2 5 17
4km
ω ± ⎛ ⎞= ⎜ ⎟⎝ ⎠
The homogeneous equations (Equations 11.4.10) for the two components of the jth eigenvector are …
21
22
20
2j
j
am k kak m k
ωω
⎛ ⎞⎛ ⎞− + −=⎜ ⎟⎜ ⎟
− − +⎝ ⎠⎝ ⎠
For the first eigenvector (the anti-symmetric mode, j = 1) …
13
Inserting 21
5 174
km
ω + ⎛ ⎞= ⎜ ⎟⎝ ⎠
into the first of the two homogeneous equations yields
11 215 17 2
4k k a ka
⎡ ⎤+− + =⎢ ⎥⎣ ⎦
21 113 17
4a a−
=
Letting a11 = 1, then a21 = -0.281 (Thus, in the anti-symmetric normal mode, the amplitude of the vibration of the second mass is 0.281 that of the first mass and 180o out of phase with it.) For the second eigenvector (the symmetric mode, j = 2) …
Inserting 22
5 174
km
ω − ⎛ ⎞= ⎜ ⎟⎝ ⎠
into the first of the two homogeneous equations yields
12 225 17 2
4k k a ka
⎡ ⎤−− + =⎢ ⎥⎣ ⎦
22 123 17
4a a+
=
Letting a12 = 1, then a22 = 1.781 (Thus, in the symmetric normal mode, the amplitude of the vibration of the second mass is 1.781 that of the first mass and in phase with it.) The two eigenvectors (Equation 11.4.13 and see accompanying Table) are …
( ) (111 1 1 1
21
1cos cos
0.281a
Q ta )1tω δ ω⎛ ⎞ ⎛ ⎞
= − =⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠δ−
)2t
( ) (122 2 2 2
22
1cos cos
1.781a
Q ta
ω δ ω⎛ ⎞ ⎛ ⎞
= − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
δ−
11.18
( )22 21 1
1 12 2
T ml m l l2θ θ φ= + +& & &
( )1 1 2cos cos cosV mgl mg l lθ θ φ=− − + For small angular displacements …
( )2 222 2
1 1 2 1 22 1 12 2 2m mL T V l l l mgl mgl
2θ φθ θ φ
⎛ ⎞ ⎛= − ≈ + + + − + −⎜ ⎟ ⎜
⎝ ⎠ ⎝& & & ⎞
⎟⎠
( )21 1 1 2 , 2d L Lml ml l l mgl
dt 1θ θ φθ θ∂ ∂
= + + = −∂ ∂
&& && &&& θ
0
1 22 2l l gθ φ θ+ + =&& &&
14
( )2 1 2 2,d L Lml l l mgldt
θ φ φφ φ∂ ∂
= + = −∂ ∂
&& &&&
1 2 0l l gθ φ φ+ + =&& && The secular equation (Equation 11.4.12) is …
2 21 2
2 21 2
2 20
l g ll l gω ωω ω
− + −=
− − +
( )4 2 21 2 1 2 1 22 2 2l l g l l g l lω ω− + + − =4 0ω
Solving for the eigenfrequencies 2ω …
( ) ( ) ( )22 2
1 2 1 2 1 22 2 21 2 1 2
1 2 1 2
2 4 82
g l l g l l l l g g l l l ll l l l
ω+ ± + −
= = + + +
The homogeneous equations (Equations 11.4.10) for the two components of the jth eigenvector are …
2 211 2
2 221 2
2 20j
j
al g lal l g
ω ωω ω
⎛ ⎞⎛ ⎞− + −=⎜ ⎟⎜ ⎟
− − +⎝ ⎠⎝ ⎠
Inserting the larger eigenfrequency (the (+) solution for 2ω above) into the upper homogeneous equation yields the solution for the components of the 1st eigenvector … ( )2 2
1 1 11 2 1 212 2l g a l aω ω− + − = 0 2 2 2 2
1 2 1 2 1 2 1 211 21
2 1
2 1l l l l l l l l
ga gal l
⎡ ⎤+ + + + + +⎢ ⎥− =⎢ ⎥⎣ ⎦
2 22 1 1 2
21 112
l l l la a
l− − +
= for the higher frequency, anti-symmetric mode.
Inserting the smaller eigenfrequency (the (-) solution for 2ω ) into the upper homogeneous equation yields the solution for the components of the 2nd eigenvector … ( )2 2
1 2 12 2 2 222 2l g a l aω ω− + − = 0 2 2
2 1 1 222 12
2
l l l la a
l− + +
= for the lower frequency, symmetric mode.
Again, we let a11 = 1 and a21 = 1, since only ratios of the components of a given eigen vector can be determined. The two eigenvectors are thus (Equation 11.4.12 and accompanying table)
(2 21 12 1 1 2
2
1
cosQ tl l l ll
)1ω δ⎛ ⎞⎜ ⎟
= −− − +⎜ ⎟⎜ ⎟⎝ ⎠
… anti-symmetric
15
( )2 22 22 1 1 2
2
1
cosQ tl l l ll
2ω δ⎛ ⎞⎜ ⎟
= −− + +⎜ ⎟⎜ ⎟⎝ ⎠
… symmetric
As a check, set and compare with the solution for Example 11.3.1. 1 2l l l= = 11.19
( )2 2 21 2 3
12
T m x x x= + +& & &
x1x2 x3 ( ) ( )2 22 2
1 2 1 3 2 312
V k x x x x x x⎡ ⎤= + − + − +⎣ ⎦
L T V= −
0i i
d L Ldt x x
∂ ∂− =
∂ ∂&
( )1 1 2 1 0mx kx k x x+ − − =&&
1 1 22 0mx kx kx+ − =&& ( ) ( )2 2 1 3 2 0mx k x x k x x+ − − − =&&
1 2 2 32 0kx mx kx kx− + + − =&& ( )3 3 2 3 0mx k x x kx+ − + =&&
2 3 32 0kx mx kx− + + =&& The secular equation (Equation 11.4.12) is …
2
2
2
2 02 0
0 2
m k kk m k k
k m k
ωω
ω
− + −− − + −
− − +=
02 0
( ) ( )32 2 22 2 2m k k m kω ω− + − − + =
( )22 22 0, 2 2m k or m k kω ω− + = − + − =
2 20
2 2km
ω ω= =
2 2 2m kω− + = ± k
( ) ( )2 202 2 2 2k
mω ω= ± = ±
From Equation 11.5.17 (N = 1 and n = 3)
1 02 sin8πω ω=
16
Because 1 cossin2 2θ θ−=
1 0 01 cos 42 2
2πω ω ω−
= = 1 2 2−
1 0 2 2ω ω= −
2 0 02 22 sin 2 28 2π
0ω ω ω= = = ω
3 0 03 1 cos32 sin 28 2
4π πω ω ω −= =
3 0 02 1 2 2 2ω ω ω= + = + 2 11.20
Generalized coordinates: X, s a( )θ=
θ a
m
M X
(x,y)
x
y
and
sin (1 cos )x X a y aθ θ= + = −
cos sinx X a y aθ θ θ= + =& &&& & θ
( )2 21 12 2
T MX m x y= + +& & & 2
( ) (2 221 1 cos sin2 2
MX m X a aθ θ θ θ )⎡ ⎤= + + +⎢ ⎥⎣ ⎦& && &
( )1 cosV mgy mga θ= = −
For small oscillations, in terms of the generalized coordinates X and s …
( )22 21 1 12 2 2
T MX m X s V mga
≈ + + ≈& & & s
( )22 21 1 12 2 2
L T V MX m X s mgsa
= − ≈ + + −& & &
Lagrange’s equations of motion yield …
( )0 0gX s s M m X msa
+ + = + + =&& &&&& &&
Assuming … i t i tX Ae s Beω ω= =
we obtain the matrix equation …
17
( )
2 2
2 2 0Ag aBM m m
ω ωω ω
⎛ ⎞− ⎛ ⎞=⎜ ⎟⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠
Setting the determinant of the above matrix equal to zero yields… 2
1 0ω = , ( )22
g m MM
ω+
=
The mode corresponding to 1ω is … 0
00 0
g a AB
−⎛ ⎞⎛ ⎞=⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ implies that 0A B= =
Thus, mode 1 exhibits no oscillation! It is a pure translation with … 0θ = and 1 2X A t A= +
The mode corresponding to 2ω is …
( )2 22 2 0A g a Bω ω+ − =
or… 2
22
2
B m MA g a m
ωω
+= = −
−
Setting 1A = , we have for the 2nd mode … 2i tX e ω= and 2i tm Ms a e
mωθ +
= = −
This mode corresponds to an oscillation about the CM where … ( ) (m M X ms ma )θ+ = − = − The normal mode vectors are …
1 21 0
A t AQ
+⎛= ⎜⎝ ⎠
⎞⎟ and 2
2
1i tQ e ω
m Mm
⎛ ⎞⎜ ⎟= +⎜ ⎟−⎜ ⎟⎝ ⎠
11.21 (a) We can solve for the normal modes using Equation 11.4.9 … ( )2 0ω− =K M a K and M are the potential energy and kinetic energy matrices respectively. a is a two-component vector whose elements are the amplitudes of the coordinates q. The kinetic energy in Example 11.3.2, assuming small displacements from equilibrium, are …
( )( )22 21 1 22 2
T mX m X r Xrθ θ≈ + + +& && & &
( ) ( )221 1 12 22 2 2
T mX m Xr m rθ θ≈ + +& && &
or, in matrix form
θ r
M
m
X
12
T = %qMq where Xrθ⎛ ⎞
=q ⎜ ⎟⎝ ⎠
18
Thus, 2 11 1
m⎛ ⎞= ⎜ ⎟
⎝ ⎠M
The potential energy is …
( )2212 2
gV kX m rr
θ≈ +&
( ) ( )2 22 20
1 2 12 22 2 2
k mgV mX m r m X rm kr
θ ω θ⎡ ⎤ 2⎡ ⎤≈ + = +⎢ ⎥ ⎣ ⎦⎣ ⎦& &
where 20 2
k kM m m
ω = =+
Thus The above matrix equation is thus … 20
2 00 1
mω ⎛ ⎞= ⎜
⎝ ⎠K ⎟
12 20
2
2 0 2 10
0 1 1 1a
m ma
ω ω⎡ ⎤ ⎛ ⎞⎛ ⎞ ⎛ ⎞
− =⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
( )2 2 20 1
2 2 220
20
aa
ω ω ω
ω ω ω
⎛ ⎞− − ⎛ ⎞=⎜ ⎟⎜ ⎟⎜ ⎟− − ⎝ ⎠⎝ ⎠
From the top equation in the matrix equation above, we get … ( )2 2 2
0 1 22 0a aω ω ω− − =
( )2 202
21
2aa
ω ω
ω
−=
The amplitudes for the two normal modes can be found by setting 2 21 2or 2ω ω ω= .
In each case we set , which we are free to do since only ratios of the amplitudes can be determined.
1 1a =
For ( )2 21 2 2 2
0ω ω= = − ω we obtain …
( ) ( )( )
2 20 1
2 21
2 1 2 222
2 2a
ω ω
ω
⎡ ⎤− −− ⎣ ⎦= =−
=
Thus, for mode 1, the lower frequency, symmetric mode …
11
1
2i te ω⎛ ⎞
= ⎜ ⎟⎝ ⎠
q
For ( )2 22 2 2 2
0ω ω= = + ω we obtain …
( ) ( )( )
2 20 2
2 21
2 1 2 222
2 2a
ω ω
ω
⎡ ⎤− +− ⎣ ⎦= =+
= −
Thus, for mode 2, the higher frequency, anti-symmetric mode …
22
1
2i te ω⎛ ⎞
= ⎜ ⎟−⎝ ⎠
q
19
(b) In this case, 1m or M mm M
<< >>+
. We also assume that the spring is “slack”,
i.e., k gM m r
<<+
, an assumption not stated in the problem (which needs to be rectified in
the next edition, I suppose)
Also, , so we let 2mg kr≠ 2 20 0
k k and gM m M r
ω = ≈ Ω+
=
The kinetic energy is
( )( )22 21 1 22 2
T MX m X r Xrθ θ≈ + + +& && & &
( ) ( )( ) ( ) ( )2 22 21 1 1 1 12 22 2 2 2 2
T M m X m r Xr MX m Xr m rθ θ θ≈ + + + ≈ + +& & && & & & θ&
and the M-matrix is M mm m
⎛ ⎞≈ ⎜ ⎟⎝ ⎠
M
The potential energy is
( )22 20 0
1 1 1 12 2 2 2
gV kX m r M mr
θ ω≈ + = + Ω& 2
The K-matrix is thus 2
02
0
00M
mω⎛ ⎞
≈ ⎜ ⎟Ω⎝ ⎠
K
To solve for 2ω , we set 2 0ω− =K M ( )
( )
2 2 20
2 2 20
0M m
m m
ω ω ω
ω ω
− −=
− Ω −
which yields … 2 2 4 2 2 2 2 4 2
0 0 0 0 0mM mω ω ω ω ω ω⎡ ⎤Ω + − Ω − − =⎣ ⎦
( ) ( )4 2 2 2 2 20 0 0 0 0M m M Mω ω ω ω− − +Ω + Ω =
Neglecting m with respect to M and simplifying yields … ( )4 2 2 2 2 2
0 0 0 0 0ω ω ω ω− +Ω + Ω =
Solving for 2ω …
( ) ( )22 2 22 20 0 0 00 02
4
2 2
ω ωωω
+Ω − Ω+Ω= ±
2
( ) ( )2 2 2 20 0 0 02
2 2ω ω
ω+Ω −Ω
= ± Thus, we get …
20
2 2 21 0 2andω ω ω= = 2
0Ω Now, we solve for the amplitudes a of the normal mode vectors … ( )2 0ω− =K M a
( )( )
2 2 20 1
2 2 220
0M m a
am m
ω ω ω
ω ω
⎛ ⎞− − ⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟− Ω − ⎝ ⎠⎝ ⎠
Using the first of the above matrix equations for 2 21
20ω ω ω= = gives …
2 10 1a and a= = Using the second equation for 2 2
2ω ω 20= = Ω gives …
1 20 1a and a= = Thus, the normal modes are approximately …
0 01 2
1 00 1
i t i te and eω Ω⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Q Q
Note that we could have guessed this almost immediately. The above assumption is tantamount to omitting the cross elements 2mω− . This completely eliminates the small coupling between the two oscillators, which reduces the matrix to the purely diagonal terms …
2ω−K M
⎟( )
( )
2 20
2 20
0
0
M
m
ω ω
ω
⎛ ⎞−⎜⎜ ⎟Ω −⎝ ⎠
, which leads directly to the above solution.
11.22
θ1
θ2θ3
k
K
k
2m
m
m
We “scale” the force constants and masses to 1 unit, namely,
1 1m and k= = . Let k K 1′ and l= = such that 3l = circumference
So: ( )2 2 21 2 3
1 22
T θ θ θ= + +& & &
( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 21 2 2 1 1 3 3 1 2 3 3 2
1 1 1 1 1 12 2 2 2 2 2
V K 2Kθ θ θ θ θ θ θ θ θ θ θ θ= − + − + − + − + − + −
Collecting terms …
( ) ( )2 2 21 2 3 1 2 1 3
1 4 2 1 2 1 4 4 42
V K K 2 3Kθ θ θ θ θ θ θ θ θ⎡ ⎤= + + + + − − −⎣ ⎦
K - Matrix M - Matrix
21
4 -2 -2= -2 2+2K -2K
-2 -2K 2+2K
⎛ ⎞⎜⎜⎜ ⎟⎝ ⎠
K ⎟⎟
2 0 00 1 00 0 1
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
M
The ttansformation matrix A that diagonalizes these matrices is made up of the three eigenvectors Qi whose amplitudes are ai … we guess that … 1. Uniform rotation
1 2 3 1
111
orθ θ θ⎛ ⎞⎜ ⎟= = = ⎜ ⎟⎜ ⎟⎝ ⎠
a
2. Anti-symmetric oscillation of 2&3, while 1 remains fixed
1 3 2 1
00; 1
1orθ θ θ
⎛ ⎞⎜ ⎟= = − = ⎜ ⎟⎜ ⎟−⎝ ⎠
a
3. Anti-symmetric oscillation of 1&2 together with respect to 3
1 3 2 1
11; 1
1orθ θ θ
⎛ ⎞⎜ ⎟= = = −⎜ ⎟⎜ ⎟−⎝ ⎠
a
Thus, and⎛ ⎞ ⎛⎜ ⎟ ⎜⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎝
%A A =1 0 1 1 1 1
= 1 1 -1 0 1 -11 -1 -1 1 -1 -1
⎞⎟⎟⎟⎠
⎞⎟⎟⎟⎠
We can now diagonalize Kand M …
diag
4 -2 -2-2 2+2K -2K-2 -2K 2+ 2K
⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟ =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
%AKA K1 1 1 1 0 1
= 0 1 -1 1 1 -11 -1 -1 1 -1 -1
0 0 0 4 0 00 4 8 0 2 00 16 0 0 4
diag diagK Likewise⎛ ⎞ ⎛⎜ ⎟ ⎜= + =⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎝
K M
The eigenfrequencies are 2
i diag diag iω ⎡ ⎤= ⎣ ⎦K M
2 2 22 2 30; 2 4 ; 4Kω ω ω= = + =
The general solution can be generated from the following table …
22
1 2
1 1 3
2 1 2
3 1 2
0
Q Q Q
a aa a aa a a
θθθ
−− −
3
3
3
where ( )1 1 1 1
11 cos1
a tω δ⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠
Q , ( )2 2 2 2
01 cos1
a tω δ⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟−⎝ ⎠
Q , ( )3 3 3 3
11 cos1
a tω δ⎛ ⎞⎜ ⎟= − −⎜ ⎟⎜ ⎟−⎝ ⎠
Q
Thus, ( ) ( )1 1 2 2 3 3 3cos cosa t a tθ ω δ ω δ= − + −
( ) ( ) ( )1 1 2 2 2 2 2 3 3 3cos cos cosa t a t a tθ ω δ ω δ ω δ= − + − − −
( ) ( ) ( )3 1 2 2 2 2 2 3 3 3cos cos cosa t a t a tθ ω δ ω δ ω δ= − − − − −
Initial conditions are: 1 0 2 3 1 2 310 ; 0; 0oθ θ θ θ θ θ θ= = = = = = =& & &
The conditions generate 6 equations with 6 unknowns and solving gives …
0 01 1cos cos
2 2t t3
θ θθ ω ω= +
0 02 1cos cos
2 2t t3
θ θθ ω ω= −
0 02 1cos cos
2 2t t3
θ θθ ω ω= −
11.23 See Ex. 11.4.1, page 497-498 The amplitudes of the eigenvectors are …
1 2 3
1 1 11 0 21 1 1
m M⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠
a a a
K - Matrix M - MatrixK -K 0
= -K 2K -K0 -K K
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
K0 0
0 00 0
mM
m
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
M
1 1 1 0 1 1 11 0 1 2 1 0 21 2 1 0 1 1 1
diag
K KK K K m M
m M K K
−⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= = − − − −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟− − −⎝ ⎠⎝ ⎠⎝ ⎠
%K AKA
2 2
0 0 00 2 00 2 8 8
diag KK K m M K m M
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟+⎝ ⎠
K
23
1 1 1 0 0 1 1 11 0 1 0 0 1 0 21 2 1 0 0 1 1 1
diag
mM m M
m M m
⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= = − −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠⎝ ⎠
%M AMA
2
2 0 00 2 00 0 2 4
diag
m Mm
m m M
+⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟+⎝ ⎠
M
So, we have … 2
1 0ω = , 22
22
K Km m
ω = = , and …
( )( )
2 22 22
3 2
1 4 42 8 82 4 1 2
K m M m MK K m M K m Mm m M m m M
ω+ ++ +
= =+ +
( )( ) ( )
22
3
1 21 2
1 2m MK K m M
m m M mω
+= = +
+
11.24
x3
y3
m x1
y1
m
x2
y2
M
α α a a
Select coordinates (xi , yi) as shown, then …
( ) ( )2 2 2 2 2 21 1 3 3 2 2
1 12 2
T m x y x y M x y= + + + + +& & & & & &
The potential energy depends only on the compression (or stretching) of the two springs connecting each m to M (hydrogen to sulfur). Let 1aδ and 2aδ be incremental changes in the distances or and or . We have … 1a (1 2)HS → 2a (3 2)HS →
( ) ( ) ( )1 1 2 2 1 1 2 2 11sin cos2
a x x y y x x y yδ α α= − + − = − + −
24
( ) ( ) ( )2 2 3 2 3 2 3 2 31sin cos2
a x x y y x x y yδ α α= − + − = − + −
( ) ( )2 21 2
12
V k a aδ δ⎡ ⎤= +⎣ ⎦
We can reduce the degrees of freedom from 6 to 3 by ignoring the two translational
odes and the rotational mode. Thus we consider only vibrational modes. The mcoordinates must obey the following constraints … No center of mass motion:
( ) ( )1 3 2 1 30 0m y y My and m x x Mx+ + = + + = 2
No angular momentum about any point. We choose that point to be the sulfur atom (M)… my a my a mx a mx a3 1 3 1sin sin sin sin 0α α α α− − − =
e introduce three generalized coordinates Q, q1 and q2 that should be close to what we uess would be normal modes …
d q2 using the above 3 equations of constraint …
3 1 3 1 0y y x x− − − = Wg 1 3 1 1 3 2 1 3; ;Q x x q x x q y y= + = − = + Solve for xi , yi in terms of Q, q1 an
( ) ( ) 1 1 2 1 3 3 12 21 1; ; ;mx Q q x Q x x x Q q= + =− = − = −
M
( ) ( )1 2 2 2 31 1; ;2 2
my q Q y q y Q qM
= − =− = + 2
in terms of these generalized coordinates, is … Thus, the kinetic energy, 2 2 21 1 1mT m Q mq m qμ⎛ ⎞= + + +&
where
1 212 4 4M M⎜ ⎟
⎝ ⎠& &
2M mμ = + is the mass of the H2 olecule
S m
The potential energy is … 2 2
2 2 21 22
1 1 1 11 28 4
m k q k q qM Mμ μ⎛ ⎞1
2 8V k Q kq
M= + + −
Note, that in the Lagrangian
+⎜ ⎟⎝ ⎠
L T V= − , the only cross term is one involving . erefore, Q is a normal mode with eigenfrequency given by the ratio …
1 2q qth
2 1Q Q Qk mK Mm M
ω ⎛ ⎞= = ⎜ ⎟⎝ ⎠+
onstructing the residual 2x2 K and matrices involving only q1 and q2 terms, which we ill call Kq and Mq gives …
C Mw
2 2
1 1 01 14
Mand
02M Mμ
μ μ−⎛ ⎞ ⎛ ⎞
= =Mμ⎜ ⎟
⎝ ⎠⎜ ⎟−⎝ ⎠
Kq Mq
25
We have omitted the factors k and m, which we’ll repla e final solution, remembering that the eigenfrequencies that we find as a solution to
ce in th2- =ωKq Mq 0 will
be multiples of k / m.
( )21 21
42
M-
M M
ω μω
μ
− −=
−Kq Mq 2 0
2Mμ μ ω=
−
which reduces to … ( )2 22 2 1 Mω ω μ⎡ ⎤− + =⎣ ⎦ 0 which has the non-trivial solution …
( )2 1 1 kω μ= +1,2 2M
m in which, we have put back the factor . These two modes are
“degenerate.”
k / m
( )2- =ωKq Mq qPlugging 2ω back into the matrix equation 0 gives …
3q or x x y y= − − + = + be or example, with 3
1 2 1 3 1qwhich can satisfied in a variety of ways, f 1 3x x an 1d y y= = − , etc
Q-mode: ant-symmetric about y-axis: 3
Pictorially, the three normal modes are …
1 3 1x x and y y= = −
2 1Qk mω ⎛ ⎞m M
= +⎝ ⎠
Breathing mode: symmetric about y-axis:
3
⎜ ⎟
1 3 1x x and y y= − =
( )21 1k
2M
m
Stretching mode: symmetric about y-axis:
3
ω μ= +
1 3 1x x and y y= − =
( )22 1k
2M
m 11.25
Plug each of these functions into the wave equation and it is satisfied!
ω μ= +
(a)
26
(b) ( ) ( )1 2
i t i k k xi t ikxQ q q e e e eω ωω +Δ − +Δ−= + = + ( ) ( ) ( ) ( ) ( ) ( )2 22 2 i t k x i t k xi t i k k xe e e eω ωω ω − Δ − Δ + Δ − Δ⎡ ⎤ ⎡ ⎤+Δ − +Δ ⎣ ⎦ ⎣ ⎦⎡ ⎤= +⎣ ⎦
The real part of the above is … ( ) ( )2cos cos
2 2Q
2t k x kt k x
ω ωωΔ − Δ⎡ ⎤ ⎡ Δ Δ ⎤⎛ ⎞ ⎛ ⎞+ − +⎜ ⎟ ⎜ ⎟= ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
⎣ ⎦( ) ( ) ( )2cos cos
2t k x
t kxω
ωΔ − Δ⎡ ⎤
≈ −⎢ ⎥⎣ ⎦
(c) The group speed is …
gudt
= (the phadx se of the amplitude remains the same)
gukωΔ
=Δ
11.26
27
From Equation 11.5.17 ...
32sin sin
10 10,N 4
1 1 1 1
2 3 4sin sin10 101.28, 2.62, 3.08
sin sin sin sin10 10 10 10
Nπ π π πωω ω ω
π π π πω ω ω ω= = = = =
1.27
is the tension in the cord
= =
1 F k= Δl
1 is its stretched length l ld
n+ Δ
=+
From the equation following Equation 11.5.7… FKd
=
From Equation 11.6.4c …
( )2v l l lm
= + Δ Δ k
( )1
122
transkv l lm
⎛ ⎞= + Δ⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠lΔ for transverse waves
For longitudinal vibrations, we use Y, the tension in the cord per unit stretched length
( ) ( )Y k l ll l l
= = + ΔΔ + Δ
k lΔ
YKd
= (Equation 11.6.8)
( )2 l l ndd + Δ (Eq2 kkv
m n m= = uation 11.6.9a)
( )1nd n d l l≈ + = + Δ
( )12
longkv lm
⎛ ⎞= +⎜ ⎟⎝ ⎠
lΔ
11.28
28
From Equation 11.6.7b …
12Fv
μ⎛ ⎞
= ⎜ ⎟ ⎝ ⎠
As in problem 11.27, F k l= Δ 12
transk lv ⎛ ⎞Δ
= ⎜ ⎟ μ⎝ ⎠
From Equation 11.6.9b … 12Yv
μ⎛ ⎞
= ⎜ ⎟ and from problem ( )Y k l l= + Δ so … ( )12
long
k l lv
μ+ Δ⎛ ⎞
= ⎜ ⎟⎝ ⎠
⎝ ⎠
11.27 …
11.29
neral solution to the wave equation (Equation 11.6.10) that yields a standing wave f any arbitrary shape can be obtained as a linear combination of standing sine waves of a
The geoform given by Equation 11.6.14, i.e.,
( )1
2( , ) sin cos sinn n n nn n
xy x t A t B t πω ω∞
=
⎛ ⎞= +∑ λ⎜ ⎟
⎝ ⎠
where 2n
nTπω =
since th oe speed f a wave is … 12
0n n n Fv λ ω λ2nT π μ⎝ ⎠
we have …
⎛ ⎞= = = ⎜ ⎟
12
0n
F 2
n
πωμ λ⎝ ⎠
But the wavelength of the standing wave is constrained by the fixed endpoints of the string, i.e. …
⎛ ⎞= ⎜ ⎟
2n
ln
λ = , so …
12
0 nn
Flπω
μ⎛
= ⎜ ⎟⎝ ⎠
Now, at t = 0, the wave starts from rest in the configuration specified, so…
⎞
( ,0) sinny x Bl
n xπ=∑
From the discussion of Fourier analysis in Appendix G or in Section 3.9, the Fourier coefficients are given by …
29
0
2 ( ,0)sinl
nn xB y x dx
l lπ
= ∫
Since the string starts from rest, we have …
0
( , ) sin 0n nAt l
ω= ≡∂ ∑
t
y x t n xπ
=
Therefore, 0
onfiguration is shown in the Figure P11.29. Thus …
∂
nA =
The initial c2( ,0) 0
2y x x x
l= < < a l
( )2( ,0)2
a ly x l x x ll
= − < <
We can now determine Bn …
( )2
0 2
2 2 2sinna n x a sin
l l
l
n xB x dxl l l l
π π= + l x dx
l
⎡ ⎤−⎢ ⎥
⎢ ⎥⎣ ⎦∫
Using integration tables, we obtain the result …
∫
( )1
22 21 sin 1,3,5,...
2nB nn π8n a nπ−⎡ ⎤= −
Thus, the general solution is …
=⎢ ⎥⎣ ⎦
0 2,4,6,...nB n= =
2( , ) cos sin c9
y x tl lπ
= −8 1 3 3 1 5 5os sin cos sin ...
25a vt x vt x vt x
l l l lπ π π π π π⎛ ⎞+ −⎜ ⎟
⎝ ⎠
None of the harmonics which have a node at the midpoint have been stimulated … only the odd harmonics have been excited.
logy with the generation of the traveling sine wave of Equation 11.6.14 from quation 11.6.13, we get …
11.30 By anaE
( ) ( ) ( ) ( )2
3 34 1( , ) sin sin sin sin9
x vt x vt x vt x vtay x tl l
π π π ππ
⎡ + − + −⎛ ⎞ ⎛ ⎞= +⎢⎜
⎢⎝ ⎠ l l− +⎟ ⎜ ⎟
⎝ ⎠⎣
( ) ( )5 51 sin sin ...25
x vt x vtl l
π π ⎤+ −⎛+ +
⎞⎥⎜ ⎟⎥⎝ ⎠ ⎦
- --------------- ------------------------------------------------------------------------------------