solutionsto supplementary problems - springer · d.gross•w.hauger j.schr¨oder •w.a.wall...
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D. Gross • W. HaugerJ. Schroder • W.A. WallS. Govindjee
Engineering Mechanics 3Dynamics
Solutions toSupplementary
Problems
The numbers of the problems and the figures correspondto the numbers in the textbook Gross et al., EngineeringMechanics 3, Dynamics, 2nd Edition, Springer 2013
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1Chapter 1
Motion of a Point Mass
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2 1 Motion of a Point Mass
E1.17 Example 1.17 A point P moves on a given path from A to B
(Fig. 1.42). Its velocity v decreases linearly with arc-length s from
a value v0 at A to zero at B.
How long does it take P to reach point B?
l s
v
s
B
P
A
s = l
v
v0
Fig. 1.42
Solution First we write down the velocity v as a linear function
of the arc-length s:
v(s) = v0(1− s
l
).
Separation of variables
ds
dt= v → ds
v0(1− s
l
) = dt
and indeterminate integration lead to∫
ds
1− sl
= v0
∫dt → s = l
[1− C exp
(−v0t
l
)].
The integration constant C is determined from the initial condi-
tion:
s(0) = 0 : 0 = 1− C → s = l
[1− exp
(−v0t
l
)].
Point B is reached when s = l. This yields the corresponding time
tB :
s(tB) = l → tB →∞ .
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1 Motion of a Point Mass 3
E1.18Example 1.18 A radar screen tracks a rocket which rises vertically
with a constant acceleration
a (Fig. 1.43). The rocket is
launched at t = 0.
Determine the angular ve-
locity ϕ and the angular acce-
leration ϕ of the radar screen.
Calculate the maximum an-
gular velocity ϕ and the cor-
responding angle ϕ.
ϕ
lFig. 1.43
Solution Since the acceleration
a is constant, the velocity v
and the position x of the rocket
are
v = at+ v(0) ,
x = at2/2 + v(0)t+ x(0) .
xϕ
Applying the initial conditions yields
v(0) = 0 → v = at,
x(0) = 0 → x =at2
2.
The angle ϕ of the radar screen follows as
tanϕ =x
l→ ϕ(t) = arctan
(at2
2l
).
Differentiation leads to the angular velocity
ϕ(t) =at
l/
[1 +
(at2
2l
)2]
and the angular acceleration
ϕ(t) =
(a
l− 3a3t4
4l3
)/
[1 +
(at2
2l
)2]2.
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4 1 Motion of a Point Mass
The time t∗ of the maximum angular velocity ϕmax is obtained
from
ϕ(t∗) = 0 → t∗ =
(4l2
3a2
)1/4
.
Thus,
ϕmax = ϕ(t∗) → ϕmax =
√3√3a
8l
and
ϕ(t∗) = arctan(1/√3) → ϕ(t∗) = 30 .
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1 Motion of a Point Mass 5
E1.19Example 1.19 Two point masses P1 and P2 start at point A with
zero initial velocities and travel on a circular path. P1 moves with
a uniform tangential acceleration at1 and P2 moves with a given
uniform angular velocity ω2.
a) What value must at1 have
in order for the two masses
to meet at point B?
b) What is the angular veloci-
ty of P1 at B?
c) What are the normal acce-
lerations of the two masses
at B?
r
P2
P1
A B
Fig. 1.44
Solution The velocity and position of P1 follow from at1 = const =
v with the initial conditions s01 = 0 and v01 = 0 as
v1 = at1t , s1 =1
2at1t
2 .
Similarly, for point P2 we obtain from at2 = 0 with s02 = 0 and
v02 = rω2:
v2 = rω2 , s2 = rω2t .
a) Both points meet at time tB at point B. Thus,
πr =1
2at1t
2B , πr = rω2tB .
This leads to
tB =π
ω2→ at1 =
2πr
t2B=
2rω22
π.
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6 1 Motion of a Point Mass
b) The tangential acceleration at1 and the time tB are now known.
Hence, we can calculate the angular velocity of P1 at B:
v1(tB) = at1tB = rω1(tB)
→ ω1(tB) =at1tBr
=2rω2
2π
πrω2= 2ω2 .
c) The normal accelerations at B follow from an = rω2:
an1 = rω21(tB) = 4rω2
2 , an2 = rω22 .
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1 Motion of a Point Mass 7
E1.20Example 1.20 A child of massm jumps up and down on a trampo-
line in a periodic manner. The child’s jumping velocity (upwards
upon leaving the trampoline) is v0 and during the contact time
∆t the contact force F (t) has a triangular form.
Find the necessary contact force amplitude F0 and the jum-
ping period T0.
Fig. 1.45
F
t∆t T0
zF0
Solution The child is subjected
to its constant weight W = mg
and, during the contact with the
trampoline, to the contact for-
ce F (t). While the child is in the
air, the equation of motion is gi-
ven by
↑: mz = −mg.
F
mg
F
t20 tt1
T0
F0
Integration between t = t0 = 0 (end of a contact) and t = t1(beginning of a new contact) yields
mv1 −mv0 = −mgt1,
where v1 = v(t1) and v0 = v(0). Recall that the velocity at the
beginning of a vertical motion and the velocity at the end of a
free fall of a body are equal in magnitude: v1 = −v0 (note the
different signs). Therefore, we obtain
2v0 = gt1 → t1 = 2v0g.
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8 1 Motion of a Point Mass
The jumping period T0 follows with ∆t = t2 − t1 as
T0 = t1 +∆t → T0 =2v0g
+∆t .
We now apply the Impulse Law between time t0 and time t2 (see
the figure):
↑ : mv2 −mv0 = −mgT0 +1
2F0(t2 − t1) .
In order to have a periodic process, the velocities v2 and v0 have
to coincide: v2 = v0. With this condition the Impulse Law yields
−mgT0 +1
2F0∆t = 0 → F0 =
2T0∆t
mg = 2( 2v0g∆t
+ 1)mg .
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1 Motion of a Point Mass 9
E1.21Example 1.21 A car is travelling in a circular arc with radius R
and velocity v0 when it starts to brake.
If the tangential deceleration is
at(v) = −(a0+κv), where a0 and κ are
given constants, find the time to brake
tB, the stopping distance sB, and the
normal acceleration an during the bra-
king.
v0
R
s
Fig. 1.46
Solution The acceleration at is given as a function of the velocity:
at(v) = v = −(a0 + κv) .
We separate the variables and integrate (initial condition: v(0) = v0):
t(v) = −∫ v
v0
dv
a0 + κv=
1
κlna0 + κv0a0 + κv
.
The time tB when the car comes to a stop follows from the con-
dition v = 0:
tB = t(v = 0) =1
κln(1 +
κv0a0
).
Now we determine the inverse function of t(v):
eκt =a0 + κv0a0 + κv
→ v(t) =a0κ
[(1 +
κv0a0
)e−κt − 1
].
Integration with the initial condition s(0) = 0 leads to the position
s(t) =
∫ t
0
v(t)dt =a0κ2
[(1 +
κv0a0
)(1− e−κt
)− κt
].
The stopping distance sB is obtained for t = tB :
sB = s(tB) =a0κ2
[(1 +
κv0a0
)(1− 1
1 + κv0a0
)− ln
(1 +
κv0a0
)]
=a0κ2
[κv0a0− ln
(1 +
κv0a0
)].
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10 1 Motion of a Point Mass
The normal acceleration an(t) during the breaking is found as
an =v2
R=
a20Rκ2
(1 +
κv0a0
)e−κt − 1
2
.
As a check on the correctness of the result we calculate an for
t = tB and obtain an = 0.
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1 Motion of a Point Mass 11
E1.22Example 1.22 A point P moves on
a parabola y = b(x/a)2 from A to
B. Its position as a function of the
time is given by the angle ϕ(t) =
arctanω0 t (see Fig. 1.47).
Determine the magnitude of ve-
locity v(t) of point P . How much
time elapses until P reaches point
B? Calculate its velocity at B.
r
B
P
a xA
y
b
ϕ
Fig. 1.47
Solution The parabola y = b(x/a)2 and the angle ϕ(t) = arctanω0t
are given. The angle ϕ can also be expressed as ϕ = arctan y/x.
Thus, y/x = ω0t. Solving for x and y yields the position of
point P :
x(t) =a2
bω0t , y(t) =
a2
b(ω0t)
2 .
To obtain the velocity of P , we differentiate:
x(t) =a2
bω0 , y(t) = 2
a2
bω20t ,
v =√x2 + y2 → v(t) =
a2
bω0
√1 + 4ω2
0t2 .
Point B is reached at time tB when x = a:
x(tB) = a → tB =b
aω0.
Thus, the velocity at B is obtained as
vB = v(tB) → vB =a2
bω0
√
1 + 4b2
a2.
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12 1 Motion of a Point Mass
E1.23 Example 1.23 A rod with length l rota-
tes about support A with angular po-
sition given by ϕ(t) = κ t2. A body
G slides along the rod with position
r(t) = l(1− κ t2).a) Find the magnitude of velocity and
acceleration of G when ϕ = 45.
b) At what angle ϕ does G hit the sup-
port?
Given: l = 2 m, κ = 0.2 s−2.
A
l
B
r G
ϕ(t)
Fig. 1.48
Solution a) First we calculate the time t = t1 that it takes the
rod to reach the position ϕ = ϕ1 = 45:
ϕ1 = π/4 = κt21 → t1 =√π/(4κ) = 1.98 s .
Then we determine the derivatives of the given functions r(t) and
ϕ(t):
r = l(1− κt2) , r = −2κlt , r = −2κl ,
ϕ = κt2 , ϕ = 2κt , ϕ = 2κ .
With t = t1 we obtain the velocity
vr = r = −2κlt1 = −1.58 m/s ,
vϕ = rϕ = l(1− κt21)2κt1 = 0.34 m/s ,
v =√v2r + v2ϕ = 1.62 m/s
v vrπ
4
vϕ
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1 Motion of a Point Mass 13
and the acceleration
ar = r − rϕ2 = −2κl− l(1− κt21)4κ2t21= −1.07 m/s2 ,
aϕ = rϕ+ 2rϕ = l(1− κt21)2κ− 2 · 2κlt12κt1= −2.34 m/s
2,
a =√a2r + a2ϕ = 2.57 m/s2 .
ar aϕ
a
π
4
b) Body G reaches the support (r = 0) at time tE :
r = 0 = l(1− κt2E) → tE =√1/κ = 2.24 s .
This yields the corresponding angle ϕE :
ϕE = κt2E = 1 (= 57.3) .
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14 1 Motion of a Point Mass
E1.24 Example 1.24 A mouse sits in a tower (with radius R) at point A
and a cat sits at the center 0.
If the mouse runs at a constant
velocity vM along the tower wall and
the cat chases it in an Archimedian
spiral r(ϕ) = Rϕ/π, what must the
cat’s constant velocity vC be in order
to catch the mouse just as the mou-
se reaches its escape hole H? At what
time does it catch the mouse?
H
R
0r
A
ϕ
Fig. 1.49
Solution We determine the components of the velocity of the cat
from the given equation r(ϕ) = Rϕ/π of the Archimedian spiral:
vr = r =dr
dϕ
dϕ
dt=R
πϕ and vϕ = rϕ =
R
πϕϕ .
Thus, the constant velocity vC can be written as
vC =√v2r + v2ϕ =
R
πϕ√1 + ϕ2 =
R
π
dϕ
dt
√1 + ϕ2 .
Separation of variables and integration lead to
vC
t∫
0
dt = vCt =R
π
ϕ∫
0
√1 + ϕ2 dϕ .
With the integral∫ √
1 + x2dx =1
2
[x√1 + x2 + arsinh x
]
this results in
vCt =R
2π
(ϕ√1 + ϕ2 + arcsinh ϕ
).
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1 Motion of a Point Mass 15
The time T when both the cat and the mouse reach the hole
follows from the constant velocity vM of the mouse along the wall:
πR = vMT → T =πR
vM.
Introduction of T and ϕ(T ) = π into the expression for vCt yields
the velocity of the cat:
vC =vM2π2
(π√1 + π2 + arcsinh π
)= 0.62 vM .
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16 1 Motion of a Point Mass
E1.25 Example 1.25 A soccer player kicks the ball (mass m) so that
it leaves the ground at an angle α0 with an initial velocity v0(Fig. 1.50). The air exerts a drag force Fd = k v on the ball; it
acts in the direction opposite to the velocity.
Determine the velocity v(t) of the ball. Calculate the horizon-
tal component vH of v when the ball reaches the teammate at a
distance l.
x
z
l
α0
v0
Fig. 1.50
Solution The equations of motion are
mx = −Fd cosα , mz = mg + Fd sinα .
We introduce the kinematic relations
x =dx
dt, z =
dz
dt
x
z
vFd
mg
α
and the drag force Fd = kv. Noting that the components of the
velocity are given by
x = v cosα , z = −v sinα
we obtain
mdx
dt= −kx , m
dz
dt= mg − kz .
Separation of variables gives
dx
x= − k
mdt ,
dz
z − mg
k
= − kmdt
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1 Motion of a Point Mass 17
and integration yields
lnx
C1= − k
mt , ln
z − mg
kC2
= − kmt .
The constants of integration C1 and C2 can be determined from
the initial conditions. For t = 0 the left-hand sides of the equations
above have to be zero. Thus, the arguments of the ln-functions
have to be equal to one (numerator and denominator have to be
equal):
C1 = x(0) = v0 cosα0 , C2 = z(0)− mg
k= −v0 sinα0 −
mg
k.
This leads to the components of the velocity:
x(t) = v0 cosα0e−kt/m , z(t) =
mg
k−(mgk
+ v0 sinα0
)e−kt/m .
We now integrate the x-component to obtain the horizontal posi-
tion of the ball:
x(t) = −mkv0 cosα0e
−kt/m + C .
Exploiting the initial condition
x(0) = 0 → C =m
kv0 cosα0
yields
x(t) =m
kv0 cosα0
(1− e−kt/m
).
The time t∗ when the ball reaches the teammate at the distance
l is found to be
x(t∗) = l → t∗ =m
kln
mv0 cosα0
mv0 cosα0 − kl.
This leads to the corresponding horizontal component of the ve-
locity:
x(t∗) = v0 cosα0 −kl
m.
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18 1 Motion of a Point Mass
E1.26 Example 1.26 A body with mass m
starts at a height h at time t = 0
with an initial horizontal velocity v0.
If the wind resistance can be
approximated by a horizontal force
H = c0mx2, at what time tB and
location xB does it hit the ground?
z
m
x
h
B
v0
Fig. 1.51
Solution The equations of
motion are given by
→ : mx = −mc0x2 , ↑ : mz = −mg .
We integrate and apply the
z
x
mg
H
initial conditions x(0) = 0, z(0) = h, x(0) = v0, z(0) = 0 to obtain
∫ x
v0
d ˙x˙x2
= −c0∫ t
0
dt → x =1
1v0 + c0t
,
∫ x
0dx =
∫ t
0
11v0 + c0t
dt → x =1
c0ln(1 + c0v0t) ,
z = −gt , z = −g2t2 + h .
The time tB when the body hits the ground follows from zB = 0:
tB =
√2h
g.
Thus, the location of point B is obtained as
xB = x(t = tB) =1
c0ln
(1 + c0v0
√2h
g
).
In the case of a vanishing wind resistance (c0 = 0) this result
reduces to
xB = limc0→0
1
c0ln
(1 + c0v0
√2h
g
)= v0
√2h
g.
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1 Motion of a Point Mass 19
E1.27Example 1.27 A mass m slides on a
rotating frictionless and massless
rod S such that it is pressed against
a rough circular wall (coefficient of
friction µ).
If the mass starts in contact with
the wall at a velocity v0, how many
rotations will it take for its velocity
to drop to v0/10?
g
rS
m
µ
Fig. 1.52
Solution If we express the accelera-
tion vector in terms of the Serret-
Frenet frame, then the equations of
motion are written as
տ : mat = −R , ւ : man = N .R
at
an
N
Note that the weight of the mass does not influence the motion
since the motion takes place in a horizontal plane. If we use the ki-
nematic relations at = v, an = v2/r, and the friction law R = µN
we can write the first equation of motion in the form
mv = −µm v2
r.
Separation of variables and integration yield∫ v
v0
dv
v2= −
∫ t
0
µ
rdt → v(t) =
v0
1 +µv0r t
.
We integrate again to obtain the distance traveled:∫ s
0
ds = v0
∫ t
0
dt
1 + µv0r t→ s(t) =
r
µln (1 +
µv0r
t) .
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20 1 Motion of a Point Mass
Now we calculate the time t1 that it takes for the velocity to drop
to v0/10:
v010
=v0
1 +µv0r t1
→ t1 =9r
µv0.
The corresponding value s(t1) is found as
s1 = s(t1) =r
µln (1 +
µv0r
t1) =r
µln 10 .
This yields the corresponding number of rotations:
n =s12πr
=ln 10
2πµ.
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1 Motion of a Point Mass 21
E1.28Example 1.28 A car (mass m) is traveling with constant velocity
v along a banked circular curve (radius r, angle of slope α), see
Fig. 1.53. The coefficient of static friction µ0 between the tires of
the car and the surface of the road is given.
Fig. 1.53
r
m
α
µ0
Determine the region of the allowable velocities so that sliding
(down or up the slope) does not take place.
Solution We model the car as a point mass and express the ac-
celeration vector in terms of the Serret-Frenet frame. Then the
equation of motion in the direction of the normal vector is (see
the free-body diagram)
man = N sinα+H cosα .
v2/gr
µ0 = 0.3
π/60 π/3
2.0
1.5
1.0
0.5
αN
mg
H
M r
α
vr
M
eten
en
g
Since the car does not move in the vertical direction we can apply
the equilibrium condition
↑: 0 = N cosα−H sinα−mg .
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22 1 Motion of a Point Mass
We now solve these two equations for the normal force N and the
force of static friction H :
H = man cosα−mg sinα ,N = man sinα+mg cosα .
The car does not slide if the condition of static friction
|H | ≤ µ0N
is satisfied. With an = v2/r this leads to the allowable region of
the velocity:
tanα− µ0
1 + µ0 tanα≤ v2
gr≤ tanα+ µ0
1− µ0 tanα.
This is displayed for µ = 0.3 as a function of the angle α in the
figure.
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1 Motion of a Point Mass 23
E1.29Example 1.29 A car (mass m) has a velocity v0 at the beginning of
a curve (Fig. 1.54). Then it slows down with constant tangential
acceleration at = −a0. The coefficient of static friction between
the road and the tires is µ0.
Calculate the velocity v
of the car as a function of
the arc-length s. What is the
necessary radius of curvature
ρ(s) of the road so that the
car does not slide?
s
v0
m
Mρ(s)
Fig. 1.54
Solution We model the car as a point mass. Its velocity v can
be determined from the given constant tangential acceleration atthrough integration (initial condition v(0) = v0):
v = at = −a0 → v(t) = v0 − a0t .
The distance traveled follows from
s = v → s(t) = s0 + v0t− a0t2/2 .
With the initial condition s(0) = s0 = 0 we obtain
s(t) = v0t− a0t2/2 .
Elimination of the time t yields the velocity as a function of the
arc-length:
v(s) =√v20 − 2a0s .
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24 1 Motion of a Point Mass
In order to determine the necessary radius of curvature ρ(s) we
write down the equations of motion:
mat = Ht → Ht = −ma0 ,
man = Hn → Hn = mv2
ρ,
0 = N −mg → N = mg .
The static friction forceH = (H2t +H
2n)
1/2 and the normal force N
have to satisfy the condition of static friction to avoid sliding of
the car:
|H | ≤ µ0N → a20 +v4
ρ2≤ µ2
0g2 .
Solving for ρ yields
ρ(s) ≥ v20 − 2a0s√µ20g
2 − a20.
This condition can not be satisfied if a0 > µ0g.
M
mg
Hn
Ht
et
Nen
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1 Motion of a Point Mass 25
E1.30Example 1.30 A bowling ball (mass m) moves with constant ve-
locity v0 on the frictionless return of a bowling alley. It is lifted
on a circular path (radius r) to the height 2r at the end of the
return. The upper part of the circular path has a frictionless guide
of length r ϕG (Fig. 1.55).
Given the angle ϕG, determine the velocity v0 such that the
bowling ball reaches the upper level.
Fig. 1.55
g
v0 mr
ϕG
Solution The velocity v0 has to be large enough so that the bow-
ling ball reaches the upper level (height 2r) with a velocity v ≥ 0.
It is convenient to use the Conservation of Energy Law to deter-
mine the relation between v0 and v. With T0 = mv20/2, V0 = 0,
T1 = mv2/2 and V1 = 2mgr we obtain
T0 + V0 = T1 + V1 → mv20/2 + 0 = mv2/2 + 2mgr .
This yields a first requirement on v0:
v20 = v2 + 4gr → v20 ≥ 4gr .
In addition, the velocity v0 has to be large enough so that the
normal force between the bowling ball and the lower part of the
circular path does not become zero before the ball reaches the
guide of length rϕG. The necessary velocity v(ϕG) follows from
the equation of motion in the normal direction (note that ϕ is
measured from the upper level):
ց: mv2(ϕ)
r= N +mg cosϕ .
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26 1 Motion of a Point Mass
The condition N ≥ 0 for ϕ = ϕG leads to
v2(ϕG) ≥ gr cosϕG .
In order to calculate the corresponding velocity v0 we use the Con-
servation of Energy Law between the lower level and the beginning
of the guide:
T0 + V0 = T2 + V2
→ mv20/2 + 0 = mv2(ϕG)/2 +mgr(1 + cosϕG) .
Thus, we obtain the second requirement
v20 ≥ (2 + 3 cosϕG)gr .
Both requirements are plotted in the figure, which shows that
the minimum velocity v0 is obtained as
v20 =
(2 + 3 cosϕG)gr for ϕG < ϕ∗
G = arccos2/3 ,
4gr for ϕG > ϕ∗
G .
2+3 cos ϕG
8
6
4
2
0.2π0 0.6π π
ϕG
v20/gr
0.4π 0.8π
ϕ∗
G
N
mg
ϕ
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1 Motion of a Point Mass 27
E1.31Example 1.31 A point mass m is
subjected to a central force
F = mk2r, where k is a constant
and r is the distance of the mass
from the origin 0. At time t = 0
the mass is located at P0 and has
velocity components vx = v0 and
vy = 0.
Find the trajectory of the
mass.
α
y
F
m
x
r0
0
P0
r
v0
Fig. 1.56
Solution The equations of motion in the directions x and y are
→: mx = −F cosα ,
↑ : my = −F sinα .
With F = mk2r, x = r cosα and y = r sinα we obtain
x+ k2x = 0 , y + k2y = 0 .
Both equations are equivalent to the differential equation that de-
scribes undamped free vibrations of a point mass (see Chapter 5).
The solutions are
x = A cos kt+B sin kt , y = C cos kt+D sin kt .
The constants of integration are calculated from the initial condi-
tions:
x(0) = 0 → A = 0 , y(0) = r0 → C = r0 ,
x(0) = v0 → B =v0k, y(0) = 0 → D = 0 .
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28 1 Motion of a Point Mass
We now have a parametric representation of the curve describing
the motion:
x =v0k
sin kt , y = r0 cos kt .
To eliminate the time, these equations are squared and then ad-
ded. Thus, we finally obtain
(x
v0/k
)2
+
(y
r0
)2
= 1 .
The point mass moves along an ellipse which reduces to a circle
for k = v0/r0.
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1 Motion of a Point Mass 29
E1.32Example 1.32 A centrifuge with radius r rotates with constant an-
gular velocity ω0. A massm is to be placed at rest in the centrifuge
and accelerated within a time t1 to the angular velocity ω0.
What will be the needed (con-
stant) moment M acting on the
mass? What is the power P of this
moment?ω0
mr
Fig. 1.57
Solution The centrifuge rotates with a
constant angular velocity. Thus, the
driving torque M needed to accelera-
te the point mass is equal to the mo-
ment of the friction force R which acts
between the mass and the cylinder:
M = rR .
R
N
Nat
R
The equation of motion in the tangential direction for the mass is
↑ : mat = R ,
where at = rω. Thus,
mrω =M
r→ ω =
M
mr2.
Integration with the initial condition ω(0) = 0 yields
ω =M
r2mt .
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30 1 Motion of a Point Mass
The condition ω(t1) = ω0 leads to the required moment:
M =r2mω0
t1.
The corresponding power is
P = M · ω0 =Mω0 =r2mω2
0
t1.
Note that the moment M has to be reduced to zero after time t1.
Otherwise the mass would be further accelerated.
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1 Motion of a Point Mass 31
E1.33Example 1.33 A skier (mass m) has the velocity vA = v0 at point
A of the cross country course (Fig. 1.58). Although he tries hard
not to lose velocity skiing uphill, he reaches point B with only a
velocity vB = 2 v0/5. Skiing downhill between point B and the
finish C he again gains speed and reaches C with vC = 4 v0.
Between B and C assume that a constant friction force acts due
to the soft snow in this region; the drag force from the air on the
skier can be neglected.
Calculate the work done by the skier on the path from A to B
(here the friction force is negligible). Determine the coefficient of
kinetic friction between B and C.
finish
10h
B
3h
C
A
h
Fig. 1.58
Solution In order to calculate the work done by the skier on the
path from A to B we use the work-energy theorem:
TB + VB = TA + VA + U
→ mv2B/2 +mgh = mv2A/2 + 0 + U
→ U = m(gh− 21v20/50) .
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32 1 Motion of a Point Mass
Now we apply the work-energy theorem between points B and C,
where the work of the friction force is given by −RlBC :
TC + VC = TB + VB + U
→ mv2C/2 + 0 = mv2B/2 + 3mgh−RlBC .
With the normal force
N = mg cosα → N = mg10h
lBC
and the law of kinetic friction
R = µN → R = µmg10h
lBC
mg
R
Nα
we obtain
µ =3
10− v2C − v2B
20gh→ µ ≈ 3
10− 4v20
5gh.
The result is valid only for µ ≥ 0.
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1 Motion of a Point Mass 33
E1.34Example 1.34 A circular disk (radius R)
rotates with the constant angular veloci-
ty Ω. A point P moves along a straight
guide; its distance from the center of
the disk is given by ξ = R sinω t whe-
re ω = const (Fig. 1.59).
Determine the velocity and the acce-
leration of P .
P
R ξ
Ω
Fig. 1.59
Solution We use polar coordinates r, ϕ to
solve the problem. From Ω = const we find
ϕ = Ω = const → ϕ = Ωt , ϕ = 0 .
The position vector is written as
r = ξ er → r = R sinωt er .
eϕ ϕ
Prer
Differentiation yields the velocity vector
v = ξ er + ξϕ eϕ → v = Rω cosωt er +RΩ sinωt eϕ
and the acceleration vector
a = (ξ − ξϕ2) er + (ξϕ+ 2ξϕ) eϕ
→ a = −R(ω2 +Ω2) sinωt er + 2RωΩcosωt eϕ .
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34 1 Motion of a Point Mass
The paths of point P are displayed for several values of Ω/ω in
the following figures.
Ω/ω = 0.5Ω/ω = 0.25
Ω/ω = 2.0Ω/ω = 1.0
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1 Motion of a Point Mass 35
E1.35Example 1.35 A chain with length l and
mass m hangs over the edge of a friction-
less table by an amount e.
If the chain starts with zero initial
velocity, find the position of the end of
the chain as a function of time.
e
Fig. 1.60
Solution All the links of the chain have the same displacement,
velocity and acceleration. The corner only produces a change of
direction. We therefore consider the chain to be a single mass with
an applied force that depends on the length x of the overhanging
part. Thus, with a = x, the equation of
motion is
ma = mx
lg → x− g
lx = 0 .
This differential equation of second order
with constant coefficients has the soluti-
on
mx
lg
x
x(t) = A cosh
√g
lt+B sinh
√g
lt .
We calculate the integration constants from the initial conditions:
x(0) = 0 → B = 0
x(0) = e → A = e
→ x(t) = e cosh
√g
lt .
This solution is valid only for x ≤ l.We may also solve the problem by
making an imaginary section cut at
the corner. Then the equations of
motion for each part of the chain are
d
dt
(ml− xl
x)= S ,
d
dt
(mlxx)=m
lxg − S .
If we eliminate S we again obtain the
differential equation
x− g
lx = 0 .
x
S
S
mx− ll
mx
l
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36 1 Motion of a Point Mass
E1.36 Example 1.36 A rotating source of
light throws a beam of light onto a
screen (Fig. 1.61). Point P on the
screen should move with constant
velocity v0.
Determine the required angu-
lar acceleration ϕ(ϕ) of the source
of light. Sketch ϕ(ϕ) and ϕ(ϕ).
x
v0
r0
ϕ
P
Fig. 1.61
Solution The position of point P is given by
x = r0 tanϕ .
We write down the inverse function
ϕ = arctanx
r0.
Differentiation with x = v = v0 = const yields
ϕ =1
1 +(xr0
)2x
r0=
v0r0r20 + x2
=v0
r0(1 + tan2 ϕ)=v0r0
cos2 ϕ ,
ϕ =v0r0
2 cosϕ(− sinϕ)ϕ = −2(v0r0
)2sinϕ cos3 ϕ .
v0r
ϕ
ϕ−π2
π
2
ϕ
ϕ−π2
π
2
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1 Motion of a Point Mass 37
Note: the maximum value of ϕ, located at ϕ = ±30, is obtainedas
|ϕmax| =3
8
√3(v0r0
)2.
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38 1 Motion of a Point Mass
E1.37 Example 1.37 A car travels with velocity v0 = 100 km/h. At time
t = 0, the driver fully applies the brakes. At that moment the car
starts sliding on a rough road (coefficient of kinetic friction µ).
Use the simplest model for the car (point mass) and calculate
the time t∗ and the distance x∗ until the car comes to a stop
a) on a dry road (µ = 0.8),
b) on a wet road (µ = 0.35).
Solution The equation of motion in the horizontal direction
← : ma = mx = −R ,
the equilibrium condition in the
vertical direction
↑ : 0 = N −mg
x
R
N
mg
and the law of friction
R = µN
lead to
a = −µg .
Integration with v(t=0) = v0 and x(t=0) = 0 yields
v(t) = v0 − µgt , x(t) = v0t−1
2µgt2 .
The time t∗ follows from the condition v = 0:
t∗ =v0µg
.
This leads to
x∗ = x(t∗) =v20µg− v20
2µg=
v202µg
.
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1 Motion of a Point Mass 39
a) On a dry road (µ = 0.8) we obtain
t∗ =100 · 1000
3600 · 0.8 · 9.81 = 3.55 s ,
x∗ =
(100
3.6
)2 1
2 · 0.8 · 9.81 = 49 m .
b) On a wet road (µ = 0.35) we are led to
t∗ =100 · 1000
3600 · 0.35 · 9.81 = 8.1 s ,
x∗ =(1003.6
)2 1
2 · 0.35 · 9.81 = 112 m .
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40 1 Motion of a Point Mass
E1.38 Example 1.38 In order to determine the
coefficient of restitution e experimental-
ly, a ball is dropped from height h0 onto
a horizontal rigid surface (Fig. 1.62). Af-
ter the ball has hit the surface 7 times, it
reaches only 20% of the original height
h0.
Calculate the coefficient of restituti-
on.
h0
h7=0.2h0
Fig. 1.62
Solution With the positive direction of
the velocity taken as upwards, the velo-
city of the ball immediately before the
first impact is given by
v = −√2gh0 .
The definition
e = −vv
h1h0
h2
h7
of the coefficient of restitution in terms of the velocities yields the
velocity v immediately after the impact:
v = −ev = e√2gh0 .
From the Conservation of Energy Law we obtain the height which
the body will reach:
1
2mv2 = mgh1 → h1 =
v2
2g= e2h0 .
Similarly, we obtain
hi = e2hi−1
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1 Motion of a Point Mass 41
for the subsequent impacts. Thus,
h7 = e2h6 = e4h5 = . . . = e14h0
and
e =
(h7h0
)1/14
= 0.21/14 = 0.891 .
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2Chapter 2
Dynamics of Systems of PointMasses
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44 2 Dynamics of Systems of Point Masses
E2.10 Example 2.10 Two vehicles (masses m1 and m2, velocities v1 and
v2) crash head-on, see Fig. 2.20. After a plastic impact the vehicles
are entangled and slide with locked wheels a distance s to the right.
The coefficient of kinetic
friction between the wheels
and the road is µ.
Calculate v1 if v2 and s
are known.
v1 v2
m1 m2
Fig. 2.20
Solution The total momentum of the system remains unchanged
during the collision. We assume that positive velocities are direc-
ted to the right (note the direction of v2). We then obtain the
velocity v of the vehicles after the impact:
m1v1 −m2v2 = (m1 +m2)v → v =m1v1 −m2v2m1 +m2
.
In order to relate the distance s to the velocity v we apply the
work-energy theorem
T1 − T0 = U .
We insert the kinetic energies
T0 = (m1 +m2)v2/2 , T1 = 0 ,
the work of the friction force
U = −Rs
and Coulomb’s friction law
R = µN → R = µ(m1 +m2)g .
We then obtain
−1
2(m1 +m2)v
2 = −µ(m1 +m2)gs
→ v1 =m2
m1v2 +
(1 +
m2
m1
)√2µgs .
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2 Dynamics of Systems of Point Masses 45
E2.11Example 2.11 A block (mass m2) rests on a horizontal platform
(mass m1) which is also initially at rest (Fig. 2.21). A constant
force F accelerates the
platform (wheels rolling
without friction) which
causes the block to slide
on the rough surface of
the platform (coefficient of
kinetic friction µ).
m2F
m1
l
µ
Fig. 2.21
Determine the time t∗ that it takes the block to fall off the
platform.
Solution We separate the
block and the platform and
introduce the coordinates
x1 and x2 as shown in the
free-body diagram. Then
the equations of motion for
the two bodies are
① → : m1x1 = F −R ,
② → : m2x2 = R .
With the friction law R =
µN = µm2g, we obtain
N
1
m1g
F
BA
x2
x1
m2g
R
R
2
x1 =F − µm2g
m1, x2 = µg .
Now, we integrate twice. Using the initial conditions x1(0) =
x2(0) = 0 and x1(0) = x2(0) = 0 we obtain
x1 =F − µm2g
m1t , x2 = µgt ,
x1 =F − µm2g
m1
t2
2, x2 = µg
t2
2.
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46 2 Dynamics of Systems of Point Masses
The block falls off the platform if
x1 − x2 = l .
This yields
F − µm2g
m1
(t∗)2
2− µg (t
∗)2
2= l → t∗ =
√2lm1
F − µg(m1 +m2).
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2 Dynamics of Systems of Point Masses 47
E2.12Example 2.12 A railroad wagon (mass m1) has a velocity v1(Fig. 2.22). It collides with a wagon (mass m2) which is initi-
ally at rest. Both wagons roll without friction after the collision.
The second wagon is connected via a spring (spring constant k)
with a block (mass m3) that lies on a rough surface (coefficient of
static friction µ0).
Assume the impact to be plastic and determine the maximum
value of v1 so that the block stays at rest.
v1
m3
km1 m2 µ0
Fig. 2.22
Solution The total momentum of the system remains unchanged
during the impact. This yields the velocity v of the wagons after
the collision:
m1v1 = (m1 +m2)v → v =m1
m1 +m2v1 .
xm1 m2
m3 F
H
m3g
N
We assume that the block stays at rest. Then we can write down
the equilibrium conditions
→: H = F , ↑: N = m3g .
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48 2 Dynamics of Systems of Point Masses
The force of static friction H and the normal force N have to
satisfy the condition of static friction:
H ≤ µ0N → F ≤ µ0m3g .
With
F = kx → x ≤ µ0m3g
k
we obtain the maximum compression of the spring:
xmax =µ0m3g
k.
Now we apply the Conservation of Energy Law to determine v1max:
1
2(m1 +m2)v
2 =1
2kx2max → v1max =
µ0m3g
m1
√m1 +m2
k.
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2 Dynamics of Systems of Point Masses 49
E2.13Example 2.13 A point mass m1 strikes a point mass m2 which is
suspended from a string (length l, negli-
gible mass) as shown in Fig. 2.23. The
maximum force S∗ that the string can
sustain is given.
Assume an elastic impact and de-
termine the velocity v0 that causes the
string to break.
l
m1 m2
v0
Fig. 2.23
Solution First we formulate the con-
servation of linear momentum
m1v0 = m1v1 +m2v2
and of energy (elastic impact!):
m1v20/2 = m1v
21/2 +m2v
22/2 .
From these equations we can calcu-
late the velocity v2 of mass m2 im-
mediately after the impact:
S
m2g
ϕ
v2 =2m1
m1 +m2v0 .
Now, we write down the equation of motion
տ: m2an = S −m2g cosϕ .
With an = v22/l we obtain the force in the string:
S = m2v22/l +m2g cosϕ .
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50 2 Dynamics of Systems of Point Masses
The maximum force Smax in the string is found for ϕ = 0:
Smax = m2(v22/l + g) .
The string breaks if the maximum force Smax is larger than the
allowable force S∗:
Smax > S∗ .
This yields
v0 >m1 +m2
2m1
√l(S∗/m2 − g) .
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2 Dynamics of Systems of Point Masses 51
E2.14Example 2.14 A ball ① (mass m1) hits a second ball ② (mass
m2, velocity v2 = 0) with a velocity v1 as shown in Fig. 2.24.
Assume that the impact
is partially elastic (coeffi-
cient of restitution e) and
all surfaces are smooth.
Given: r2 = 3 r1, m2 =
4m1.
Determine the veloci-
ties of the balls after the
collision.
1v1
m2
m1
r22
r1
Fig. 2.24
Solution We introduce the auxiliary angle α. Since the surfaces
are smooth, the linear impulse F acts in the direction of the line
of impact and the Impulse Laws are given by
① → : m1(v1 − v1) = −F cosα ,
② ր : m2v2 = F .
Note that the weights of
the balls can be neglected
during impact and that
ball ① moves only horizon-
tally.
With the hypothesis
e = −v1x − v2xv1x − v2x
and
1α
N
F
F2
v1, v1α
αr2
r1
x
v2
r2−r1
v1x = v1 cosα , v2x = 0 , v1x = v1 cosα , v2x = v2
we obtain
v1 = v1
1− e m2
m1cos2 α
1 +m2
m1cos2 α
, v2 = v1(1 + e) cosα
1 +m2
m1cos2 α
.
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52 2 Dynamics of Systems of Point Masses
Introduction of m2 = 4m1 and
sinα =r2 − r1r1 + r2
=1
2→ cosα =
√1− sin2 α =
√3
2
yields
v1 =1− 3e
4v1 , v2 =
√3
8(1 + e) v1 .
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2 Dynamics of Systems of Point Masses 53
E2.15Example 2.15 A hunter (mass m1) sits in a boat (mass m2 =
2m1) which can move in the water without resistance. The boat
is initially at rest.
a) Determine the velocity vB1of the boat after the hunter fires a
bullet (mass m3 = m1/1000) with a velocity v0 = 500 m/s.
b) Find the direction of the velocity of the boat after a second
shot is fired at an angle of 45 with respect to the first one.
Solution a) We introduce a coordinate system and assume that
the bullet is fired in the direction of the negative x-axis. Since the
linear momentum before the firing of the bullet is zero, the total
linear momentum of the system after the firing (velocities positive
to the right) also has to be zero:
→ : (m1 +m2 −m3)vB1−m3v0 = 0 .
This yields the velocity vB1of the boat:
vB1=
m3
m1 +m2 −m3v0 =
1
1000(1 + 2− 1
1000
) 500 =500
2999
≈ 0.167m
s.
The algebraic sign shows that the boat moves in the direction of
the positive x-axis.
b) Now, we have to formu-
late the Impulse Laws in
the x- and y-direction:45
m3
v0
wB2
m1 +m2α
vB2
y
x
→ : (m1 +m2 −m3)vB1= (m1 +m2 − 2m3)vB2
−m3v0 cos 45 ,
↑ : 0 = (m1 +m2 − 2m3)wB2−m3v0 sin 45
.
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54 2 Dynamics of Systems of Point Masses
They lead to
vB2=
m1 +m2 −m3
m1 +m2 − 2m3vB1
+m3
m1 +m2 − 2m3v0 cos 45
≈ 0.285m
s,
wB2=
m3
m1 +m2 − 2m3v0 sin 45
≈ 0.118m
s,
→ tanα =wB2
vB2
= 0.414 → α = 22.5 .
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2 Dynamics of Systems of Point Masses 55
E2.16Example 2.16 A car ② goes into a skid on a wet road and co-
mes to a stop sideways across the road as shown in Fig. 2.25.
In spite of having fully applied the brakes
a distance s1 from car ②, a second car
① (sliding with the coefficient of kinetic
friction µ) collides with car ②. This cau-
ses car ② to slide an additional distance
s2. Assume a partially elastic central im-
pact. Given: m1 = 2m2, µ = 1/3, e =
0.2, s1 = 50m, s2 = 10m.
Determine the velocity v0 of car ① be-
fore the brakes were applied.
s1
2m2
v01m1
s2
Fig. 2.25
Solution The velocity v1 of car ① immediately before impact fol-
lows from the work-energy theorem:
1
2m1v
21 −
1
2m1v
20 = −(µm1g)s1 → v21 = v20 − 2µ gs1 .
The velocity v2 of car ② immediately after impact can be deter-
mined from the conservation of linear momentum (note v2 = 0)
m1v1 = m1v1 +m2v2
and the hypothesis
e =v2 − v1v1
.
We obtain
v2 =1 + e
1 +m2
m1
v1 .
Now we apply the work-energy theorem to the sliding of car ②:
−1
2m2v
22 = −(µm2g)s2 → v22 = 2µ gs2 .
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56 2 Dynamics of Systems of Point Masses
If we eliminate v1 and v2 we obtain
v20 =
(1 +m2
m1
1 + e
)2
2µ gs2 + 2µ gs1
=(1.51.2
)2· 23· 9.81 · 10 + 2
3· 9.81 · 50 = 429.19m2/s2
or
v0 = 20.7m/s → v0 = 74.6 km/h .
Note: the velocity of car ① at impact is v1 = 36.4 km/h.
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2 Dynamics of Systems of Point Masses 57
E2.17Example 2.17 Two cars (point masses m1 and m2) collide at an
intersection with velocities
v1 and v2 at an angle α
(Fig. 2.26). Assume a per-
fectly plastic collision.
Determine the magni-
tude and the direction of
the velocity immediately
after the impact. Calcula-
te the loss of energy during
the collision.
βm1
m2
α
v2
v1
Fig. 2.26
Solution Both cars have the
same velocity v after the
collision (plastic impact).
We write down the Impul-
se Laws in the x- and y-
directions: m2
α
v2
v1m1
y
x
m1+m2
β
v
→ : m1v1 +m2v2 cosα= (m1 +m2)v cosβ ,
↑ : m2v2 sinα = (m1 +m2)v sinβ .
If we square and then add these equations we obtain
v =1
m1 +m2
√(m1v1)2 + 2m1m2v1v2 cosα+ (m2v2)2 ,
whereas a division leads to
tanβ =m2v2 sinα
m1v1 +m2v2 cosα.
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58 2 Dynamics of Systems of Point Masses
The loss of mechanical energy during impact is given by the dif-
ference ∆T of the kinetic energies before and after impact:
∆T =m1v
21
2+m2v
22
2− (m1 +m2)v
2
2
=m1m2
2(m1 +m2)(v21 + v22 − 2v1v2 cosα).
Note that the loss of energy is a maximum if the cars collide head-
on (α = π). The energy ∆T causes the deformation of the cars.
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2 Dynamics of Systems of Point Masses 59
E2.18Example 2.18 A bullet (mass m)
has a velocity v0 (Fig. 2.27). An
explosion causes the bullet to
break into two parts (point mas-
ses m1 and m2). The directions
α1 and α2 of the two parts and
the velocity v1 immediately after
the explosion are given.
Calculate m1 and v2. Deter-
mine the trajectory of the center
of mass of the two parts.
v2m2
m
m1
v1
α2
y
xv0
α1 =π
2
Fig. 2.27
Solution Since no external forces are acting on the system, its
linear momentum does not change (linear momentum before the
explosion = linear momentum after the explosion). Component-
wise, the principle of conservation of linear momentum gives
→ : mv0 = m2v2 cosα2 ,
↑ : 0 = m1v1 −m2v2 sinα2 .
Thus, we have two equations for the two unknowns m1 and v2.
Solving with m = m1 +m2 yields
m1 = mv0v1
tanα2 , v2 =v0
cosα2 − v0v1
sinα2.
We measure the time t from the moment of the explosion. Then,
with |yi| = vit sinαi, the location of the center of mass is given by
yc =1
m(m1y1 +m2y2)
=1
m
mv0v1
tanα2 v1t+(m−m v0
v1tanα2
) (−v0 sinα2)
cosα2 − v0v1
sinα2t
= 0 .
The center of mass of the system continues to move on the original
straight line y = 0 (law of motion for the center of mass).
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60 2 Dynamics of Systems of Point Masses
E2.19 Example 2.19 A ball (point mass m1) is attached to a cable. It is
released from rest at a height h1 (Fig. 2.28). After falling to the
vertical position at A it collides with
a second ball (point mass m2 = 2m1)
which is also initially at rest. The co-
efficient of restitution is e = 0.8.
Determine the height h2 which the
first ball can reach after the collision
and the velocity of the second ball im-
mediately after impact.
m2
h1
m1
A
Fig. 2.28
Solution The velocities of the massesm1 andm2 before the impact
are
v1 =√2gh1 , v2 = 0 .
The Impulse Laws
① → : m1(v1 − v1) = −F ,
② → : m2v2 = +F ,
2v2v1, v1
1
Fm1 m2F
and the hypothesis
e = −v1 − v2v1
yield the velocities immediately after the impact:
v1 = v1m1 − em2
m1 +m2= v1
1− 1.6
1 + 2= −0.2
√2gh1 ,
v2 = v1(1 + e)m1
m1 +m2= v1 1.8
1
1 + 2= 0.6
√2gh1 .
The height h2 follows from
the conservation of energy
1
2m1v
21 = m1gh2
as
h2 =v212g
= 0.04 h1 .
h2m2v1 m1 v2
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2 Dynamics of Systems of Point Masses 61
E2.20Example 2.20 The rigid rod (negli-
gible mass) in Fig. 2.29 carries two
point masses. It is struck by an im-
pulsive force F at a distance a from
the support A.
Determine the angular veloci-
ty of the rod immediately after the
impact and the impulsive reaction
at A. Calculate a so that the reac-
tion force at A is zero.m
m
a
Fl
l
A
Fig. 2.29
Solution We introduce a coordinate
system. The y-component of the ve-
locity of the center of mass is zero
after the impact. Since the impulsi-
ve force F also has no y-component,
the impulsive reaction at A has on-
ly an x-component, i.e. Ay = 0. We
apply the principle of linear impulse
and momentum
y
3l/2
C
Ax
Ay
Fx
→: 2m(vc − vc) = Ax − F
and the principle of angular impulse and momentum (compare
Chapter 3)
y
C : 2
(l
2
)2
m(ω − ω) = Ax3
2l − F
(3
2l − a
).
Inserting the kinematic relation
vc = −3lω/2
and the velocities
vc = 0 , ω = 0
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62 2 Dynamics of Systems of Point Masses
before the impact we obtain
ω =F a
5l2m, Ax =
(1− 3a
5l
)F .
The reaction force at A is zero (Ax = 0) if the distance a is chosen
as
a = 5l/3 .
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2 Dynamics of Systems of Point Masses 63
E2.21Example 2.21 A wagon (weight W1 = m1g) hits a buffer (spring
constant k) with velocity v. The
collision causes a block (weight
W2 = m2g) to slide on the
rough surface (coefficient of sta-
tic friction µ0) of the wagon
(Fig. 2.30).
Determine a lower bound for
the wagon’s speed before the
collision.
k
µ0
W1
v
W2
Fig. 2.30
Solution We separate the wagon and the block. The orientation
of the friction force H is assumed arbitrarily in the free-body
diagram. Both bodies have the same acceleration (x1= x2= x) as
long as the block does not sli-
de. Therefore, the equations of
motion are
① ← : m1x = −H − F ,
② ← : m2x = H .
m2g
x
H
N
m1g
1
2
F
H
With F = k x we obtain
H = − m2
m1 +m2k x .
The maximum friction force Hmax is obtained at maximal com-
pression xmax of the spring which follows from the Conservation
of Energy Law:
1
2(m1 +m2)v
2 =1
2k x2max → xmax =
√m1 +m2
kv .
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64 2 Dynamics of Systems of Point Masses
The block is on the verge of slipping if the condition of “limiting
friction”
|Hmax| = µ0N = µ0m2g → k
m1 +m2xmax = µ0g
is satisfied. This yields the velocity that is necessary to cause
slippage of the block:
v = µ0g
√m1 +m2
k.
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2 Dynamics of Systems of Point Masses 65
E2.22Example 2.22 The system shown in Fig. 2.31 consists of two blocks
(massesm1 and m2), a spring (spring constant k), a massless rope
and two massless pulleys. Block 1 lies on a rough surface (coeffi-
cient of kinetic friction µ).
At the beginning of the
motion (t = 0), the spring
is unstretched and the po-
sition of block 1 is given
by x1 = 0.
Determine the velocity
x1 as a function of the po-
sition x1.
x2
x1
m2
µ
g
k m1
Fig. 2.31
Solution Since we want to find the velocity as a function of the
position, we use the work-energy theorem
T1 − T0 = U .
In moving from the initial position to an arbitrary position, the
gravitational force, the force in the spring and the friction force
perform work:
UW = m2gx2 , Uk = −1
2kx21 ,
UR = −Rx1 = −µNx1 = −µm1gx1 .
The kinetic energies are given by
T1 =1
2m1x
21 +
1
2m2x
22 , T0 = 0 .
With the kinematic relation
x2 = 2x1
the work-energy theorem yields
1
2m1x
21 +
1
2m24x
21 = m2gx2 −
1
2kx21 − µm1gx1
→ x1 =
√2
5(2− µ)gx1 −
1
5
k
mx21 .
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3Chapter 3
Dynamics of Rigid Bodies
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68 3 Dynamics of Rigid Bodies
E3.24 Example 3.24 Point A of the rod
in Fig. 3.47 moves with constant
velocity vA to the left.
Determine the velocity and
the acceleration of point B of the
rod (point of contact with the
step) as a function of the angle
ϕ. Find the path y(x) of the in-
stantaneous center of rotation.
x
eϕ
y
ϕ
A
B
vA
er
h
Fig. 3.47
Solution We use the coordinate system with the basis vectors erand eϕ as shown in the figure. Then, the velocity of point A is
given by
vA = vA cosϕer − vA sinϕeϕ .
The velocity vB of point B
points in the direction of the
rod (the rod does not lift off the
step). Thus,
vB = vBer .
With the kinematic relation
eϕ
aϕAB
ϕ
aB
A
er
vB
vAB
arAB
vA
vB = vA + vAB where vAB = aϕeϕ
and with
a =h
sinϕ,
we obtain
vBer = vA cosϕer − vA sinϕeϕ +hϕ
sinϕeϕ
→ vB = vA cosϕ and ϕ =vAh
sin2 ϕ .
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3 Dynamics of Rigid Bodies 69
In order to determine the acceleration of point B we use the rela-
tion
aB = aA + arAB + a
ϕAB ,
where
aA = 0 , arAB = −aϕ2
er , aϕAB = aϕeϕ .
The angular acceleration of the rod is found by differentiating its
angular velocity ϕ:
ϕ =vAh(sin2 ϕ)˙ → ϕ = 2
v2Ah2
sin3 ϕ cosϕ .
Hence, we obtain
aB =v2Ah
sin2 ϕ(− sinϕer + 2 cosϕeϕ) .
The instantaneous center of rotation Π is given by the point of
intersection of two straight lines which are perpendicular to two
velocities. In the present example, the directions of the velocities
vA and vB are known. The-
refore, the instantaneous cen-
ter of rotation can easily be
constructed (see the figure).
Point Π has the coordinates
x = h cotϕ , y = h+ x cotϕ .
Elimination of the angle ϕ
yields the path of Π:
y =x2
h+ h .
y/h
path of Π
Π1.5
2.0
1.0
0.5
00.5-1.0 0 0.5 1.0
x/h
ϕ
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70 3 Dynamics of Rigid Bodies
E3.25 Example 3.25 The wheel of a
crank drive rotates with
constant angular velocity ω
(Fig. 3.48).
Determine the velocity and
the acceleration of the pi-
ston P .
l
Pψ
rω ϕ
Fig. 3.48
Solution If we introduce the
auxiliary angle ψ, then the
position xP of the piston is
given by
xP = r cosϕ+ l cosψ .
Now we differentia-
te and obtain (note
ϕ = ω = const)
xP = −rω sinϕ− lψ sinψ ,
xP = −rω2 cosϕ− lψ sinψ − lψ2 cosψ .
P
lψ
x
y
rϕ = ωt
The quantities sinψ, cosψ, ψ and ψ are as yet unknown. They
follow from the condition that the piston can move only in the
horizontal direction. Therefore, its vertical displacement is zero:
yP = 0 = r sinϕ− l sinψ .
Differentiation yields
yP = 0 = rω cosϕ− lψ cosψ ,
yP = 0 = −rω2 sinϕ− lψ cosψ + lψ2 sinψ .
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3 Dynamics of Rigid Bodies 71
Thus,
sinψ =r
lsinϕ , cosψ =
√1−
(rlsinϕ
)2,
ψ = ωr
l
cosϕ
cosψ, ψ = −ω2 r
l
sinϕ
cosψ+ ψ2 sinψ
cosψ.
Hence, we finally obtain
xP = −rωsinϕ+
r
l
sinϕ cosϕ
cosψ
,
xP = −rω2
cosϕ− r
l
[sin2 ϕ
cosψ− cos2 ϕ
cos3 ψ
].
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72 3 Dynamics of Rigid Bodies
E3.26 Example 3.26 LinkMA of the mechanism in Fig. 3.49 rotates with
angular velocity ϕ(t).
Determine the velocities of points B and C, the angular velo-
city ω and the angular acceleration ω of the angled member ABC
at the instant shown.
A C
MB
z x
y
r
l
l
α
ϕ
Fig. 3.49
Solution With the given x, y, z-coordinate system, the velocities
of points A and B at the instant shown can be written in the form
vA = rϕ
− sinϕ
cosϕ
0
, vB = vB
− cosα
sinα
0
,
where vB is as yet unknown (note that the direction of vB is
known). The angular velocity of the angled member ABC is re-
presented by
ω =
0
0
ω
.
We now use the kinematic relation
vB = vA + ω × rAB
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3 Dynamics of Rigid Bodies 73
which, with rAB = (0,−l, 0)T , in coordinates reads
vB
− cosα
sinα
0
= rϕ
− sinϕ
cosϕ
0
+
ωl
0
0
.
Solving yields
vB = rϕcosϕ
sinα, ω =
rϕ
l
(sinϕ− cosϕ
tanα
).
In an analogous way we obtain the velocity of point C:
vC = vA + ω × rAC → vC = rϕ
− sinϕ
sinϕ+ cosϕ− cosϕ
tanα0
.
In order to determine the angular acceleration ω = (0, 0, ω)T of
the angled member, we first write down the relation
aB = aA + ω × rAB + ω × (ω × rAB) ,
which in coordinates reads
aB
− cosα
sinα
0
= rϕ
− sinϕ
cosϕ
0
− rϕ
2
cosϕ
sinϕ
0
+
ωl
0
0
+
0
ω2l
0
.
Solving for aB and ω yields
aB =1
sinα(rϕ cosϕ− rϕ2 sinϕ+ lω2) ,
ω =1
l
[rϕ sinϕ+ rϕ2 cosϕ− 1
tanα(rϕ cosϕ− rϕ2 sinϕ+ lω2)
].
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74 3 Dynamics of Rigid Bodies
E3.27 Example 3.27 Wheel ① rolls in
a gear mechanism without slip
along circle ②. The mechanism
is driven with a constant angu-
lar velocity Ω (Fig. 3.50).
Determine the magnitudes
of the velocity and the accele-
ration of point P on the wheel.
1
RΩ
ϕ
P
2
r
Fig. 3.50
Solution The motion of P is composed of the rotation of B about
A with
vB = RΩ (vB ⊥ AB) , aB = RΩ2 (aB ‖ AB)
and the rotation of P about B. In order to determine the angular
velocity ω of wheel ① we consider two positions of the wheel. The
figure shows that the relation
(R+ r)α = r(β + α)
→ Rα = rβ
for the arclengths holds. Diffe-
rentiating and using α = Ω and
β = ω yields
RΩ = rω → ω = ΩR
r.
R
A
r
C
B′
β
C ′
αB
α
Note that Ω is positive counterclockwise, whereas ω is positive
clockwise.
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3 Dynamics of Rigid Bodies 75
We can now construct the velocity
diagram and the acceleration dia-
gram. Since the velocity diagram
is represented by an isosceles tri-
angle, we obtain
vP = 2ΩR sinϕ
2.
vP
ϕ = Rω
vPB = rω
vB = RΩ
Finally, the law of cosines yields
a2P = (Ω2R)2 +
(Ω2R2
r
)2
−2Ω2RΩ2R2
rcos(π − ϕ) ,
aP = Ω2R
√
1 +
(R
r
)2
+ 2R
rcosϕ .
anPB = rω2ϕ
aB = RΩ2
aP
=Ω2R2
r
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76 3 Dynamics of Rigid Bodies
E3.28 Example 3.28 A disk 1 (mass m, radius r1) rests in a frictionless
support (ω0 = 0). A second disk
2 (mass m, radius r2) rotates
with the angular velocity ω2. It
is placed on disk 1 as shown in
Fig. 3.51. Due to friction, both
disks eventually rotate with the
same angular velocity ω.
Determine ω. Calculate the
change ∆T of the kinetic energy.
1
ω2
2
r2
r1
mm
Fig. 3.51
Solution Since there is no moment of external forces acting on the
system, the angular momentum is conserved during the process:
Θ2ω2 = (Θ1 +Θ2)ω .
With the mass moments of inertia
Θ1 = mr21/2 , Θ2 = mr22/2
we obtain the resulting angular velocity ω:
ω =r22
r21 + r22ω2 .
The change ∆T of the kinetic energy follows as
∆T =1
2(Θ1 +Θ2)ω
2 − 1
2Θ2ω
22 → ∆T = −1
4mω2
2
r21r22
r21 + r22.
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3 Dynamics of Rigid Bodies 77
E3.29Example 3.29 The door (mass m, moment
of inertia ΘA) of a car is open (Fig. 3.52).
Its center of mass C has a distance b from
the frictionless hinges. The car starts to
move with the constant acceleration a0.
Determine the angular velocity of the
door when it slams shut.
AC
a0
b
m
Fig. 3.52
Solution First, we write down
the acceleration of the center
of mass C:
aC = aA + arAC + a
ϕAC ,
where the magnitudes of the
individual terms are given by
At
Cϕ
AnA
aA = a0 , arAC = bϕ2 , aϕAC = bϕ .
We then formulate the principle of linear momentum (in the t-
direction, see the free-body diagram)
ւ: m(bϕ− a0 cosϕ) = At
and the principle of angular momentum (about the center of mass
C):
y
C : ΘC ϕ = −Atb .
The mass moment of inertia ΘC follows from the parallel-axis
theorem:
ΘC = ΘA −mb2 .
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78 3 Dynamics of Rigid Bodies
Eliminating the force At yields
ϕ =ma0b
ΘAcosϕ .
The angular velocity of the door can be obtained through integra-
tion:
1
2ϕ2(ϕ) =
ma0b
ΘAsinϕ .
Thus, for ϕ = π/2 (closed door) we find
ϕ(π/2) =
√2ma0b
ΘA.
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3 Dynamics of Rigid Bodies 79
E3.30Example 3.30 A child (mass m) runs along the rim of a circular
platform (mass M , radius r) starting from point A (Fig. 3.53).
The platform is initially at rest; its support is frictionless.
Determine the angle
of rotation of the plat-
form when the child ar-
rives again at point A.
0m
r
M
A
Fig. 3.53
Solution We apply the principle of angular momentum: the time
rate of change of the angular momentum is equal to the moment of
the applied forces. Since there are no external applied forces acting
in the present example, the angular momentum is constant:
dL(0)
dt= 0 → L(0) = const
with
L(0) = Θ0ω +mr2(ω + ωrel) .
Here,
Θ0 =Mr2/2
is the mass moment of inertia of the platform, ω is the angular
velocity of the platform and ωrel is the angular velocity of the child
relative to the platform. Integration of the principle of angular
momentum with the initial condition
L(0)(0) = 0 → L(0) ≡ 0
yields∫ t
0
L(0)dt = 0 → Θ0ϕ+mr2(ϕ+ ϕrel) = 0 .
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80 3 Dynamics of Rigid Bodies
The child arrives again at
point A when the relati-
ve angle attains the value
ϕrel = 2π. Solving for ϕ
yields
ϕ = − 2π
M
2m+ 1
.
The result is displayed in
a diagram.
ϕ
2π
π
0 2 4 6 8 10
M/m
ϕrel= 2π
|ϕ|ϕchild =ϕ+ϕrel
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3 Dynamics of Rigid Bodies 81
E3.31Example 3.31 A homogeneous
triangular plate of weight W =
mg is suspended from three
strings with negligible mass.
Determine the acceleration of
the plate and the forces in the
strings just after string 3 is cut.
l3
21
2l
αα
W
Fig. 3.54
Solution The motion of the plate is a translation after string 3 is
cut. Therefore, all the points of the plate have the same accele-
ration (and the same velocity). We choose point A which rotates
about a fixed point to represent the motion of the plate. Its acce-
leration can be written in the Serret-Frenet frame as
a = aA = vet +v2
ρAen .
Its velocity is zero at the mo-
ment of the cut:
v = 0 → a = vet .
The principles of linear and an-
gular momentum are then given
by (see the free-body diagram)
etC
mg
A
2
3l
en
S2
α
1
3l
S1
ւ : mv = mg sinα ,
տ : 0 = S1 + S2 −mg cosα ,y
C : 0 =2
3lS1 cosα−
l
3S1 sinα−
l
3S2 sinα−
4
3S2 cosα .
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82 3 Dynamics of Rigid Bodies
Solving these equations yields
v = g sinα,
S1 =mg
6(sinα+ 4 cosα), S2 =
mg
6(− sinα+ 2 cosα) .
Note that S2 = 0 for sinα = 2 cosα (i.e., for tanα = 2 →α = 63.4). In this case the line of action of S1 passes through the
center of mass C.
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3 Dynamics of Rigid Bodies 83
E3.32Example 3.32 A sphere (mass
m1, radius r) and a cylindrical
wheel (mass m2, radius r) are
connected by two bars (mass
of each bar m3/2, length l).
They roll down a rough incli-
ned plane (with angle α) wi-
thout slipping (Fig. 3.55).
Find the acceleration of the
bars.α
r
r
l
m1
m3
m2
xϕ
Fig. 3.55
Solution The only external forces that act on the system are the
weights of the individual parts. Therefore, conservation of energy
T + V = const
leads to the solution. The kinetic energy is given by
T = (m1 +m2 +m3)x2/2 + (Θ1 +Θ2)ϕ
2/2
and the potential energy is
V = −(m1 +m2 +m3)gx sinα .
With the kinematic relation (rolling without slipping)
x = rϕ
and the mass moments of inertia
Θ1 = 2m1r2/5 , Θ2 = m2r
2/2
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84 3 Dynamics of Rigid Bodies
we obtain
(7m1/10 + 3m2/4 +m3/2)x2 − (m1 +m2 +m3)gx sinα = const .
Differentiation finally yields the acceleration of the bars:
2(7m1/10 + 3m2/4 +m3/2)xx− (m1 +m2 +m3)gx sinα = 0
→ a = x =(m1 +m2 +m3) sinα
7m1/5 + 3m2/2 +m3g .
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3 Dynamics of Rigid Bodies 85
E3.33Example 3.33 The cylindrical
shaft shown in Fig. 3.56 has
an inhomogeneous mass den-
sity given by ρ = ρ0(1 + αr).
Find the moments of iner-
tia Θx and Θy.
xRr
y
y z
l
Fig. 3.56
Solution We determine the
mass moment of inertia
Θx from
Θx =
∫(y2 + z2)dm =
∫r2dm .
r
y
dr
R
z
With
dm = ρ 2πr dr dx = 2πρ0(r + αr2)dr dx
we obtain
Θx = 2πρ0l∫0
R∫0
(r3 + αr4)dr dx = 2πρ0l(R4
4+ α
R5
5
)
=π
2ρ0lR
4(1 +
4
5αR).
The mass moment of inertia Θy = Θz (symmetry) follows from
Θy =
∫(z2 + x2)dm
with
dm = ρr dϕ dr dx
= ρ0(r + αr2) dϕ dr dx ,
z = r sinϕ .
r
rdϕz
yϕ
dϕ
dr
r sinϕ
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86 3 Dynamics of Rigid Bodies
Using the symmetry we obtain
Θy = 4ρ0
l∫
0
R∫
0
π/2∫
0
(r2 sin2 ϕ+ x2)(r + αr2)dϕ dr dx
= πρ0
l∫
0
R∫
0
(r2 + 2x2)(r + αr2)dr dx
=πρ0R2l
[R2
4+l2
3+ α
(R3
5+
2
9Rl2
)].
Note: For α = 0 the results reduce with m = ρ0πR2l to the mass
moments of inertia of a homogeneous cylindrical shaft (note the
term due to the parallel-axis theorem):
Θx =1
2mR2 , Θy =
1
12m(3R2 + 4l2
).
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3 Dynamics of Rigid Bodies 87
E3.34Example 3.34 Determine the
moment of inertia Θa of a ho-
mogeneous torus with a circular
cross section and mass m.
c
Ra
a
Fig. 3.57
Solution The mass moment of inertia Θa is determined from
Θa =
∫r2dm .
We can determine the geo-
metrical relations
r = R+ c sinϕ ,
dm = ρ 2πr2c cosϕdr
from the figure. With
dr = c cosϕdϕ
2c cosϕR c
r
ϕ
c sinϕ
a
a dr
we obtain
dm = 4πρc2(R + c sinϕ) cos2 ϕdϕ .
Thus,
Θa = 4πρc2+π/2∫
−π/2
(R+ c sinϕ)3 cos2 ϕdϕ
= 4πρc2+π/2∫
−π/2
[R3 cos2 ϕ + 3R2c sinϕ cos2 ϕ
+ 3Rc2 sin2 ϕ cos2 ϕ+ c3 sin3 ϕ cos2 ϕ]dϕ
= 4πρc2(π
2R3 +
3π
8Rc2
)= 2π2ρc2R
(R2 +
3
4c2).
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88 3 Dynamics of Rigid Bodies
Note that the integrals of the odd functions over the even interval
are zero. Using m = 2π2ρc2R we finally get
Θa = m(R2 +
3
4c2).
This result reduces to Θa = mR2 in the case of a thin ring
(c≪ R).
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3 Dynamics of Rigid Bodies 89
E3.35Example 3.35 A rope drum on a
rough surface is set into motion
by pulling the rope with a con-
stant force F0.
Determine the acceleration of
point C assuming that the drum
rolls (no slipping). What coeffi-
cient of static friction µ0 is ne-
cessary to ensure rolling? µ0
m,Θc
F0
Cr1
r2
α
Fig. 3.58
Solution The free-body diagram shows the forces that act on the
drum. We apply the principles of linear and angular momentum:
→ : mac = F0 cosα−H ,
↑ : 0 = N −mg + F0 sinα ,
y
C : Θcω = r1H − r2F0 .
In addition we have the kinema-
tic relation
vc = r1ω → vc = ac = r1ω
F0
α
r1r2vc
HN
ω
Cmg
between the velocity and the angular velocity (rolling without
slip). We solve these four equations to obtain the acceleration and
the forces:
ac =F0
m
cosα− r2r1
1 +Θcr21m
,
H = F0
Θc cosα
r21m+r2r1
1 +Θcr21m
, N = mg − F0 sinα .
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90 3 Dynamics of Rigid Bodies
In order to ensure rolling, the condition of static friction H ≤ µ0N
has to be satisfied. This leads to
µ0 ≥
Θc cosα
r21m+r2r1(
mg
F0− sinα
)(1 +
Θcr21m
) .
Note: For cosα > r2/r1 we have ac > 0 (motion to the right),
whereas for cosα < r2/r1 we obtain ac < 0 (motion to the left).
In the case of F0 sinα > mg the drum lifts off the ground.
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3 Dynamics of Rigid Bodies 91
E3.36Example 3.36 A homogeneous beam (massM , length l) is initially
in vertical position ① (Fig. 3.59). A small disturbance causes the
beam to rotate about the frictionless
support A (initial velocity equal to ze-
ro). In position ② it strikes a small
sphere (mass m, radius r ≪ l). As-
sume the impact to be elastic (e = 1).
Determine the angular velocities of
the beam immediately before and af-
ter the impact and the velocity of the
sphere after the impact.
l
m
M①
②
A
Fig. 3.59
Solution The angular velocity ω immediately before the impact
follows from the conservation of energy (ΘA =Ml2/3):
Mgl/2 = ΘAω2/2−Mgl/2 → ω =
√6g/l .
Since the impact is assumed to be elastic, the conservation of the
angular momentum
ΘAω = ΘAω +mlv
as well as the conservation of energy
ΘAω2/2 = ΘAω
2/2 +mv2/2
hold. Solving these two equations yields
ω =M − 3m
M + 3mω , v =
2lM
M + 3mω .
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92 3 Dynamics of Rigid Bodies
E3.37 Example 3.37 A thin half-cylin-
drical shell of weight W = mg
rolls without sliding on a flat sur-
face (Fig. 3.60).
Determine the angular velocity
as a function of ϕ when the initial
condition ϕ(ϕ = 0) = 0 is given.
R
ϕ
W
Fig. 3.60
Solution The position of the center of mass and the mass moment
of inertia are given by
c =1
Rπ
∫ π
0
(R sinα)Rdα =2R
π,
Θc= Θ0 −mc2 = mR2 − 4
π2mR2
= mR2(1− 4
π2) .
Rdα
R sinα0
CR
y
xyccC
Rϕ c cosϕ
cα
ϕ
Thus,
xc = Rϕ+ c cosϕ = R(ϕ+2
πcosϕ) ,
yc = c sinϕ =2R
πsinϕ ,
xc = Rϕ(1− 2
πsinϕ) ,
yc =2R
πϕ cosϕ .
Since we want to find the angular velocity as a function of the
angle we apply the conservation of energy:
T + V = T0 + V0 .
With
T0 = 0 , V0 = 0 ,
T =1
2m(x2c + y2c) +
1
2Θcϕ
2 =1
2mR2ϕ2
(1− 4
πsinϕ+
4
π2sin2 ϕ
+4
π2cos2 ϕ
)+
1
2mR2ϕ2
(1− 4
π2
)= mR2ϕ2
(1− 2
πsinϕ
),
V = −mgyc = −2
πmgR sinϕGro
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3 Dynamics of Rigid Bodies 93
we obtain
mR2ϕ2(1− 2
πsinϕ)− 2
πmgr sinϕ = 0
→ ϕ(ϕ) =
√2g sinϕ
R(π − 2 sinϕ).
Note that the angular velocity attains its maximum for ϕ = 90
(lowest position of the center of mass):
ϕmax =√2g/R(π − 2) .
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94 3 Dynamics of Rigid Bodies
E3.38 Example 3.38 An elevator con-
sists of a cabin (weightW = mg)
which is connected through a ro-
pe (of negligible mass) with a ro-
pe drum and a band brake (coef-
ficient of dynamic friction µ).
Determine the necessary bra-
king force F such that a cabin
travelling downwards with velo-
city v0 stops after a distance h.
0r1
r2µ
l
v0W
FΘ0
Fig. 3.61
Solution First we determine the
brake momentum MB that acts
on the drum. To this end we se-
parate the drum and the lever.
Equilibrium at the lever
y
A : 0 = −2r2S2 + lF
→ S2 =Fl
2r2
and the formula S1 = S2e−µπ
for belt friction (see Volume 1,
Chapter 9) lead to
MB = r2S2 − r2S1 = (1− e−µπ)Fl
2.
r2
F
S1
S1
S2
S2
A
l
2r2
W
ω0
Now we apply the work-energy theorem
T1 − T0 = U .
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3 Dynamics of Rigid Bodies 95
With the kinematic relations (constraints imposed by the rope)
v0 = r1ω0 , h = r1ϕh
the kinetic energies and the work are given by
T1 = 0 , T0 =1
2mv20 +
1
2Θ0ω
20 =
v202
(m+
Θ0
r21
),
U = mgh−∫ ϕh
0
MBdϕ = mgh− (1− e−µπ)Flh
2r1.
Thus, we obtain
−v20
2
(m+
Θ0
r21
)= mgh− (1− e−µπ)
Flh
2r1.
Solving for the force F yields
F =r1v
20m
lh(1− e−µπ)
(1 +
Θ0
mr21+
2gh
v20
).
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96 3 Dynamics of Rigid Bodies
E3.39 Example 3.39 A homogeneous beam (mass m, length l) rotates
about frictionless support A (Fig. 3.62) until it hits support B.
The motion starts with zero
initial velocity in the vertical
position. The coefficient of re-
stitution e is given.
Calculate the impulsive
forces at A and B. Determi-
ne the distance a so that the
impulsive force at A vanishes.
Calculate the change of the
kinetic energy.
m
l
A
a
B
Fig. 3.62
Solution First we calculate the angular velocity of the beam im-
mediately before the impact with the aid of the conservation of
energy (ΘA = ml2/3):
mgl/2 = ΘAϕ2/2 → ϕ2 = 3g/l .
C
a
AH
AV Bx
y ϕ
We then apply the principles of impulse and momentum:
→: m(¯xC − xC) = AH ,
↑: m(¯yC − yC) = AV + B ,
y
A : ΘA( ¯ϕ− ϕ) = −Ba .
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3 Dynamics of Rigid Bodies 97
Here, the velocity of the center of mass and the angular velocity
before the impact are given by
xC = 0 , yC = −√3gl/2 , ϕ =
√3g/l .
In order to obtain the velocities immediately after the impact we
use the hypothesis
e = −¯yByB
→ ¯ϕ = −eϕ ,
which leads to
¯xC = 0 , ¯yC = e√3gl/2 , ¯ϕ = −e
√3g/l .
Solving for the impulsive reactions yields
AH = 0 , AV =(1 + e)(3a− 2l)m
6a
√3gl ,
B =(1 + e)ml
3a
√3gl .
The impulsive force at A vanishes if
AV = 0 → a = 2l/3 .
The change of the kinetic energy is the difference ∆T = T1 − T0of the energies before and after impact:
∆T =1
2ΘA ˙ϕ2 − 1
2ΘAϕ
2 → ∆T = −(1− e2)mgl/2 .
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98 3 Dynamics of Rigid Bodies
E3.40 Example 3.40 A homogeneous cir-
cular disk of weight W = mg is sus-
pended from a pin-supported bar (of
negligible mass). Initially the disk
rotates with the angular velocity ω0.
a) Determine the amplitude of os-
cillation of the pendulum, if the
bar suddenly prevents the disk
from rotating.
b) Calculate the energy loss ∆E.
lm
ω0
r
ϕ
Fig. 3.63
Solution a) First we determine the angular velocity ω of the bar
immediately after the rotation of the disk is prevented (note that
the bar is still in the vertical position). Since there are no external
moments acting with respect to A,
the angular momentum is conser-
ved. With
ΘB =1
2mr2 , ΘA =
1
2mr2 +ml2 =
1
2m(r2 + 2l2)
we obtain
ΘBω0 = ΘAω → ω =ΘB
ΘAω0 =
r2
r2 + 2l2ω0 .
The maximum value ϕ1 of the angle
ϕ (= amplitude of the oscillation)
can be calculated from the conser-
vation of energy:
T1+V1=T0+V0 .
If we choose the potential energy V0(initial position) to be equal to zero,
we can write
T1 = 0 ,
V1 = mgl(1− cosϕ1) ,
T0 =1
2ΘAω
2 =mr2ω2
0
4
r2
r2 + 2l2,
V0 = 0 .
ω
Diskblocked
B
A
ϕ1
l(1− cosϕ1)
position 0
position 1
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3 Dynamics of Rigid Bodies 99
Then, conservation of energy leads to
cosϕ1 = 1− r2ω20
4gl
r2
r2 + 2l2.
b) The loss of energy ∆E is given by the difference ∆T of the
kinetic energies before and after the blocking:
∆E =ΘBω
20
2− ΘAω
2
2=mr2ω2
0
4− m
4(r2 + 2l2)
r4ω20
(r2 + 2l2)2
=mr2ω2
0
2
l2
r2 + 2l2.
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100 3 Dynamics of Rigid Bodies
E3.41 Example 3.41 A homogeneous an-
gled bar of mass m is attached to a
shaft with negligible mass. The ro-
tation of the system is driven by the
moment M0.
Determine the angular accelera-
tion and the support reactions.
M0
m
2l
2l
2l
l
ξη
Fig. 3.64
Solution The following moments
and products of inertia with respect
to the body-fixed coordinate system
ξ, η, ζ are needed:
Θζ =2
3m
(2l)2
3+m
3(2l)2 =
20
9ml2 ,
Θξζ = −m3
2ll
2= −1
3ml2 ,
Θηζ = 0 .
ω, ω
AηAξ
Bξ Bη
M0
ξc
Cη
ξ
ζ
With the moments
Mξ = 2lBη − 2lAη , Mη = 2lAξ − 2lBξ , Mζ =M0
the principle of angular momentum in component form yields
y
ξ : 2l(Bη −Aη) = −ωml2
3→ Bη −Aη = −mlω
6,
y
η : 2l(Aξ −Bξ) = −ω2ml2
3→ Aξ −Bξ = −
mlω2
6,
y
ζ : M0 =20
9ml2ω → ω =
9M0
20ml2.
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3 Dynamics of Rigid Bodies 101
In order to be able to calculate the support reactions, we now
have to formulate the principle of linear momentum. The center
of mass moves on a circle. With the distance ξc = 4l/3 from the
axis of rotation, we obtain the components of its acceleration as
acξ = −ξcω2 and acη = ξcω. Thus,
−mξc ω2 = Aξ +Bξ , m ξc ω = Aη +Bη
which leads to
Aξ = −3
4mlω2 , Aη =
27
80
M0
l,
Bξ = − 7
12mlω2 , Bη =
21
80
M0
l.
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102 3 Dynamics of Rigid Bodies
E3.42 Example 3.42 A shaft (principal moments of inertia Θ1, Θ2, Θ3)
rotates with constant angular velocity ω0 about its longitudinal
axis. This axis undergoes a rotation
α(t) about the z-axis of the fixed in
space system x, y, z.
Calculate the moment which is
exerted by the bearings on the shaft
for
a) uniform rotation α = Ωt,
b) harmonic rotation α = α0 sinΩt.
2
3, zx
1
ω0
y
α(t)
Fig. 3.65
Solution We solve the problem with the aid of Euler’s equations
Θ1ω1 − (Θ2 −Θ3)ω2ω3 =M1 ,
Θ2ω2 − (Θ3 −Θ1)ω3ω1 =M2 ,
Θ3ω3 − (Θ1 −Θ2)ω1ω2 =M3 .
With
ω1 = ω0 , ω1 = 0 , ω2 = ω2 = 0 , ω3 = α , ω3 = α
we obtain
M1 = 0 , M2 = −(Θ3 −Θ1)ω0α , M3 = Θ3α
where α(t) represents an arbitrary rotation¡.
a) In the special case of a uniform rotation α = Ωt we get
α = Ω , α = 0 .
Thus,
M1 =M3 = 0 , M2 = (Θ1 −Θ3)ω0Ω .
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3 Dynamics of Rigid Bodies 103
b) In the case of a harmonic rotation α = α0 sinΩt we find
α = α0ΩcosΩt , α = −α0Ω2 sinΩt
and
M1 = 0 , M2 = (Θ1 −Θ3)ω0Ωα0 cosΩt ,
M3 = −Θ3Ω2α0 sinΩt .
Note: Only a moment about the 2-axis acts in case a). It is caused
by two opposite support reactions of equal magnitude (= couple)
in the 3- and z-directions, respectively.
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104 3 Dynamics of Rigid Bodies
E3.43 Example 3.43 A pin-supported rigid beam
(mass m, length l) is initially at rest. At
time t0 = 0 it starts to rotate due to an
applied constant moment M0.
Determine the stress resultants (inter-
nal forces and moments) as functions of x
for t > t0. Neglect gravitational effects.
ϕA
l
x
M0
Fig. 3.66
Solution First we determine the angular acceleration and the an-
gular velocity of the beam. The principle of angular momentum
with respect to A yields
x
A : ΘA ϕ =M0 → ϕ =M0
ΘA.
Integration with the initial condition ϕ(0) = 0 gives
ϕ(t) =M0
ΘAt .
Now we cut the beam at an arbitrary position and introduce the
bending moment M and the shear force V into the free-body dia-
gram (the normal force N will be considered later). The principle
of angular momentum (with respect to the center of mass C) and
the principle of linear momentum (in the y-direction) yield
x
C : ΘC ϕ = −M(x)− V (x)l − x2
,
տ : m yC = V (x) .
The acceleration yC follows from the ki-
nematics (circular motion):
yC = rC ϕ =l + x
2ϕ .
MV
xy
C1
2(l − x)
m, ΘC
rC = 1
2(l + x)
Thus, with m = (1 − x
l)m and ΘA =
ml2
3we obtain the shear
force:
V (x) = ml + x
2ϕ = m
l + x
2
M0
ΘA=
3
2
M0
l
[1−
(xl
)2 ].
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3 Dynamics of Rigid Bodies 105
Introduction of the moment of inertia ΘC =1
12m (l − x)2 leads
to the bending moment:
M(x) = −V l − x2− ΘC ϕ
= −3
4M0
(1− x
l
)2 (1 +
x
l
)− ml2
12
(1− x
l
)3 M0
ΘA
= −M0
(1− x
l
)2 (1 +
1
2
x
l
).
The normal force can be determined from the equation of motion
in the x-direction:
ր : m xC = −N(x)
where xC = −rC ϕ2 is the centripetal
acceleration. This leads to
m
xy
N C
rC = 1
2(l + x)
N(x) = m rC ϕ2
= m (1− x
l)x+ l
2
(M0
ΘAt
)2
=9
2
M20 t
2
ml3
[1−
(xl
)2 ].
Note that the normal force increases with time t in contrast to
the bending moment and the shear force.
The stress resultant diagrams are presented below.
M0+3
2
M0
l
+9
2
M20 t
2
ml3
M(x)V (x)
N(x)
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106 3 Dynamics of Rigid Bodies
E3.44 Example 3.44 The angled member
(weightW = mg) in Fig. 3.67 con-
sists of two homogeneous bars.
Derive the equation of motion
for the member’s center of mass. 2l
l
g
A
Fig. 3.67
Solution First, we locate the center
of mass of the angled member. With
the coordinate system as shown in
the figure we obtain
xc =
m
3
l
2m
=l
6,
yc =
m
32l+
2m
3l
m=
4
3l .
Thus, the distance of the center of
mass from point A is given by
x
y
1
3m
a
C
xc
yc
A
mg
ϕa
C
a sinϕ
2
3m
a =√y2c + x2c =
√65
6l .
The member rotates about a fixed axis that passes through A. We
introduce the angle ϕ and apply the principle of angular momen-
tum:
ΘA ϕ =MA .
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3 Dynamics of Rigid Bodies 107
We measure ϕ from the position of equilibrium (C is vertically
below A). With the mass moment of inertia
ΘA =m
3
l2
12+
[(2l)2 +
(l
2
)2]
+2
3m
(2l)2
3=
7
3ml2
we obtain
x
A :7
3ml2ϕ = −mga sinϕ → ϕ+
√65
14
g
lsinϕ = 0 .
Note: In the case of small oscillations (ϕ ≪ 1, sinϕ ≈ ϕ) the
equation of motion reduces to the differential equation
ϕ+
√65
14
g
lϕ = 0
for harmonic vibrations.
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108 3 Dynamics of Rigid Bodies
E3.45 Example 3.45 A bowling ball (massm) is placed on a rough surface
(coefficient of kinetic friction
µ = 0.3) with velocity v0 = 5 m/s
(Fig. 3.68). Initially, the ball does
not rotate.
What is the position xr of the
ball when it stops sliding? Calculate
the corresponding velocity vr.
v0
µ
Fig. 3.68
Solution When the ball is placed on the rough surface it slides.
RN
ω
x
mg
rC
The friction force R is opposed to the direction of the motion.
Thus, the equations of motion are given by
→ : mx = −R ,
↑ : 0 = N −mg → N = mg ,
y
C : Θcω = rR .
With Θc = 2mr2/5 and the law of friction R = µN = µmg, the
first and the third equation lead to (initial conditions v(t = 0) =
v0, x(t = 0) = 0, ω(t = 0) = 0)
v = x = v0 − µgt , x = v0t−1
2µg t2 , ω =
5µg
2rt .
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3 Dynamics of Rigid Bodies 109
The ball rolls without sliding if the velocity of its center of mass
is given by
v = rω .
This condition leads to the corresponding time tr:
v0 − µgtr =5
2µ g tr → tr =
2v07µg
= 0.49 s .
Thus,
xr = x(tr) =12v2049µg
= 2.08 m ,
vr = v(tr) =5
7v0 = 3.57
m
s.
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110 3 Dynamics of Rigid Bodies
E3.46 Example 3.46 The double pendulum in
Fig. 3.69 consists of two identical ho-
mogeneous bars (each mass m, length
l). It is struck by a linear impulse F at
point D.
Determine the distance d of point D
from the lower end of the pendulum so
that the angular velocity ω2 of the lower
bar is zero immediately after the im-
pact. Calculate the impulsive forces at
A and B.
A
B
l
ld
FD
g
Fig. 3.69
Solution We separate the two bars and
draw the free-body diagram. Note that
there is no linear impulse in the vertical
direction. The bars are at rest before the
impact. The principles of linear and an-
gular impulse and momentum are given
by
①x
A : ΘA ω1 = l B ,
② → : mv2 = F − B ,x
C2 : ΘC2ω2 =
l
2B −
(d− l
2
)F
where
F
B
B
A
ω2
v2
l ①
②
1
2l
d− 1
2lC2
ω1
ΘA =1
3ml2 , ΘC2
=1
12ml2 .
We desire the motion of bar ② to be a translation. Therefore,
ω2 = 0 , v2 = l ω1 .
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3 Dynamics of Rigid Bodies 111
This leads to
ω1 =3
4
F
ml, B =
F
1 +ml2
ΘA
=F
4, d =
l
2
2 +ml2
ΘA
1 +ml2
ΘA
=5
8l .
The impulsive force at A follows from the principle of linear im-
pulse for bar ① with the kinematic relation v1 =1
2l ω1:
→ : mv1 = A+ B → A =F
8.
1
v1C1
B
ω1
A
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4Chapter 4
Principles of Mechanics
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114 4 Principles of Mechanics
E4.8 Example 4.8 A homogeneous disk (mass m, radius r) rolls without
slipping on a rough surface (Fig. 4.9). Its center of mass C is
connected with the wall by a spring (spring constant k).
Derive the equation of motion
using
a) Newton’s 2nd Law,
b) dynamic equilibrium conditions.
ϕ
k
mA
r
C
Fig. 4.9
Solution a) The free-body diagram shows the forces that act on
the disk.
x
FC
H
N
W ϕ
We use the coordinates x and ϕ. Then the principles of linear
and angular momentum yield
←: mxc = −F −H ,
↑: 0 = N −W → N =W ,
x
C : ΘCϕ = rH where ΘC = mr2/2 .
In addition, we have the kinematic relation
xc = rϕ → xc = rϕ ,
and the relation
F = kxc → F = krϕ
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4 Principles of Mechanics 115
for the force F in the spring. Solving these equations for the angle
ϕ yields
ϕ+ ω2ϕ = 0 where ω2 =2k
3m.
b) If we apply the dynamic equilibrium conditions, the inertial
force mxc and the pseudo moment ΘCϕ (both acting in the ne-
gative coordinate directions) have to be drawn into the free-body
diagram.
ΘCϕ
FC
H
N
B
mxcW
Dynamic moment equilibrium about point B then yields
y
B : ΘC ϕ+ rmxc + rF = 0 .
If we introduce the kinematic relation (see a)), the force F = krϕ
and ΘC = mr2/2 we again obtain
ϕ+ ω2ϕ = 0 , ω2 =2c
3m.
Note that we may choose point B to be the reference point for
the moment equilibrium. This is advantageous since then the lever
arm of the unknown force of static friction H is zero. The mass
moment of inertia must be ΘC (not ΘB!).
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116 4 Principles of Mechanics
E4.9 Example 4.9 A cylinder (mass
m, radius r) rolls without slip-
ping on a circular path (radius
R); see Fig. 4.10.
Derive the equation of moti-
on using dynamic equilibrium
conditions.
ϕ
Rr
m
Fig. 4.10
Solution We isolate the cylinder and introduce the coordinates
ϕ and ψ (angle of rotation of the cylinder). With the tangential
acceleration at = (R − r)ϕ (in the positive ϕ-direction) of the
center of mass C and the normal acceleration an = (R − r)ϕ2
(directed towards the center of the circular path), the inertial
forcesmat (opposite to at) andman (opposite to an) can be drawn
on the free-body diagram.
m(R − r)ϕ
m(R − r)ϕ2ϕ
ϕ
C
ψ
B H
mg N
ΘCψ
The pseudo moment ΘCψ acts in the negative ψ-direction. Mo-
ment equilibrium about point B yields the equation of motion:
y
B : ΘC ψ +m(R − r)ϕ+mgr sinϕ = 0 where ΘC = mr2/2 .
The system has one degree of freedom. Therefore a relation exists
between the two coordinates:
vC = (R− r)ϕ = rψ → ψ = (R/r − 1)ϕ .
This leads to
ϕ+2g
3(R− r) sinϕ = 0 .
Note that the moment equilibrium may be established with re-
spect to point B. This is advantageous since the lever arms of
the unknown force of static friction H and of the inertial force
m(R − r)ϕ2 are zero. The mass moment of inertia, however, has
to be taken with respect to the center of mass C.
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4 Principles of Mechanics 117
E4.10Example 4.10 Two blocks of weights
W1 = m1g and W2 = m2g are
suspended by a pin-supported ro-
pe drum (moment of inertia ΘA) as
shown in Fig. 4.11.
Determine the angular accelerati-
on of the drum and the force in ro-
pe ① using dynamic equilibrium con-
ditions. Neglect the mass of the ro-
pes.
ΘA
m2
1 2
m1
r1A
r2
ϕ
Fig. 4.11
Solution We first introduce the co-
ordinates x1 and x2 describing the
motion of the blocks. The inerti-
al forces −mixi point in the ne-
gative xi-directions (see the free-
body diagram). In addition, we ha-
ve to consider the pseudo moment
−ΘAϕ which acts in the negative ϕ-
direction. Moment equilibrium about
point A then yields
y
A : −r1m1(g + x1) + r2m2(g − x2)−ΘAϕ = 0 .
m2x2
m1g
m1x1
m2g
ϕ
A
x2
x1
ΘAϕ
Using the kinematic relations
x1 = r1ϕ → x1 = r1ϕ ,
x2 = r2ϕ → x2 = r2ϕ
we obtain the angular acceleration of the drum:
ϕ =r2m2 − r1m1
r21m1 + r22m2 +ΘAg .
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118 4 Principles of Mechanics
In order to determine the force in rope ① we cut the rope. Force
equilibrium (see the free-body diagram) yields
↑ : S1 −m1g −m1x1 = 0
m1g
S1
x1
m1x1
or
S1 = m1(g + r1ϕ) = m1gr2(r1 + r2)m2 +ΘA
r21m1 + r22m2 +ΘA.
Note: For r2m2 > r1m1 the drum rotates clockwise, for r2m2 <
r1m1 it rotates counterclockwise. In the special case r2m2 = r1m1
the system is in static equilibrium (ϕ = 0).
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4 Principles of Mechanics 119
E4.11Example 4.11 An angled arm (mass
m) rotates with constant angular ve-
locity Ω about point 0 (Fig. 4.12).
Calculate the bending moment,
shear force and normal force as func-
tions of position using dynamic equi-
librium conditions.
b
a
0
Ω
Fig. 4.12
Solution We introduce two coordinate systems xi, zi. Then, we
make a cut at the arbitrary position x1. The acceleration of the
mass element dm at the position s (distance from the left end of
the arm) is given by an = rΩ2 (pointing towards point 0; note
that at = 0). Therefore, this element is subjected to the inertial
force dmrΩ2 (see the free-body diagram).
ϕ Ω
s dsr
V
M
Ω
N
V (x2 = 0)
x2
z2M(x2 = 0)
N(x2 = 0)
ds
s
N(x1 = b)
M
NV
dmrΩ2 dm(a− s)Ω2
V (x1 = b)
M(x1 = b)
z1
x1
With
dm =m
a+ bds = µds ,
where µ = m/(a + b) is the mass per unit length, and with the
geometrical relations
cosϕ = (b − s)/r , sinϕ = a/r
we can determine the stress resultants through integration (note
that M ′ = V ).
Normal force (0 ≤ x1 ≤ b):
N(x1) =
∫rΩ2 cosϕdm =
∫ x1
0
Ω2(b − s)µds
= µΩ2[bs− s2/2]|x1
0 → N(x1) = µΩ2(bx1 − x21/2) .
Shear force (0 ≤ x1 ≤ b):
V (x1) =
∫rΩ2 sinϕdm =
∫ x1
0
Ω2aµds → V (x1) = µΩ2ax1 .
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120 4 Principles of Mechanics
Bending moment (0 ≤ x1 ≤ b):
M(x1) =
∫ x1
0
V (s)ds → M(x1) = µΩ2ax21/2 .
The matching conditions at the corner (x1 = b, x2 = 0) are given
by
N0 = N(x2 = 0) = V (x1 = b) = µΩ2ab ,
V0 = V (x2 = 0) = −N(x1 = b) = −µΩ2b2/2 ,
M0 =M(x2 = 0) =M(x1 = b) = µΩ2ab2/2 .
Now we make a cut at the position x2. The mass element dm at
the position s is subjected to the inertial force dm(a− s)Ω2. This
leads to the following stress resultants:
Normal force, shear force, bending moment (0 ≤ x2 ≤ a):
N(x2) = N0 +∫ x2
0 µΩ2(a− s)ds
→ N(x2) = µΩ2(ab+ ax2 − x22/2) ,
V (x2) = V0 → V (x2) = −µΩ2b2/2 ,
M(x2) =M0 + x2V0 → M(x2) = µΩ2b2(a− x2)/2 .
+
+
+
−
+ +N0 V0M0
N0| V0 | VN M
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4 Principles of Mechanics 121
E4.12Example 4.12 A wheel (weight W1 = m1g, moment of inertia ΘA)
on an inclined plane is connected to a
block (weight W2 = m2g) by a rope
which is guided over an ideal pulley
(Fig. 4.13). The wheel rolls on the pla-
ne without slipping.
Determine the acceleration of the
block applying d’Alembert’s principle.
Neglect the masses of the rope and the
pulley.
A
m1,ΘA
r
α
m2
Fig. 4.13
Solution Since the constraint forces (force in the rope, static
friction force) need not be determined, it is advantageous to apply
d’Alembert’s principle. The motion
is described by the coordinates xiand ϕ. The inertial forces mixi and
the pseudo moment ΘAϕ (acting in
the directions opposite to the chosen
positive coordinate directions) are
shown in the figure. D’Alembert’s
principle (principle of virtual work)
requires that the virtual work of all
forces vanishes:
δU + δUI = 0
ΘAϕ
α
m1g
ϕ
Ax2
m2g
x1
m2x2
m1x1
→ −m1x1δx1 −m1g sinαδx1 −ΘAϕδϕ+m2gδx2 −m2x2δx2 = 0 .
With the kinematic relations
x1 = x2 = rϕ = x →
δx1 = δx2 = rδϕ = δx
x1 = x2 = rϕ = x
we obtain[−m1x−m1g sinα−
ΘA
r2x+m2g −m2x
]δx = 0 .
Since δx 6= 0, the expression in the brackets must vanish. Thus,
x = x2 = gm2 −m1 sinα
m1 +m2 +ΘA
r2
.
Note that x < 0 for m1 sinα > m2. In this case, the wheel rolls
down the inclined plane.
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122 4 Principles of Mechanics
E4.13 Example 4.13 Two drums are connected by a rope and carry blocks
of weights m1 g and m2 g
(Fig. 4.14). Drum ① is dri-
ven by the moment M0.
Determine the accele-
ration of block ② using
d’Alembert’s principle.
Neglect the mass of the
ropes. m1
m2
ΘA ΘB
r2
B1 r1
M0
r2
A
2
Fig. 4.14
Solution We introduce the
inertial forces mixi and
the pseudo moments ΘAϕ1,
ΘBϕ2. They act in the
directions opposite to the
chosen positive coordina-
te directions. D’Alembert’s
principle requires
δU + δUI = 0 ,m1g
m1x1
B
M0
ϕ1
ΘAϕ1
x1
x2
m2g
m2x2
A
ΘBϕ2ϕ2
which leads to
−m1(g + x1)δx1 +m2(g − x2)δx2
+M0δϕ1 −ΘAϕ1δϕ1 −ΘBϕ2δϕ2 = 0 .
With the kinematic relations
x1 = r1ϕ1
x2 = r2ϕ2
ϕ1 = ϕ2
→ϕ1 = ϕ2 =
x2r2
, x1 =r1r2x2 ,
δϕ1 = δϕ2 =δx2r2
, δx1 =r1r2δx2
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4 Principles of Mechanics 123
we obtain
−m1
(g +
r1r2x2)r1r2
+m2(g − x2)
+M0
r2− ΘA
r22x2 −
ΘB
r22x2
δx2 = 0 .
Since δx2 6= 0, the expression in the curly brackets must vanish.
Thus, the acceleration of block ② is
x2 = g
1− m1r1m2r2
+M0
r2m2g
1 +m1
m2
(r1r2
)2+
ΘA
m2r22+
ΘB
m2r22
.
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124 4 Principles of Mechanics
E4.14 Example 4.14 The system shown in Fig. 4.15 consists of a block
(massM), a homogeneous disk (massm, radius r) and two springs
(spring constant k). The block moves on a frictionless surface; the
disk rolls without slipping
on the block. A force F (t)
acts on the block.
Derive the equations of mo-
tion using Lagrange’s for-
malism.
kM
m
rk
no friction
F (t)
Fig. 4.15
Solution The system has
two degrees of freedom. We
choose the displacement x
of the block and the angle
of rotation ϕ of the disk to
F (t)
ϕC
x
be the generalized coordinates.
The two springs are unstressed for x = 0 and ϕ = 0. The kinetic
energy and the potential energy of the springs, respectively, are
T =Mx2/2 + ΘCϕ2/2 +mv2C/2 ,
V = kx2/2 + k(rϕ)2/2 .
With ΘC = mr2/2 and the kinematic relation
vC = x− ϕ
we obtain the Lagrangian
L = T − V
→ L =Mx2/2 +mr2ϕ2/4 +m(x− rϕ)2/2− kx2/2− kr2ϕ2/2 .
Since the force F (t) is not given from a potential, we have to apply
the Lagrangian equations in the form
d
dt
(∂L
∂x
)− ∂L
∂x= Qx ,
d
dt
(∂L
∂ϕ
)− ∂L
∂ϕ= Qϕ .
The generalized forces Qx and Qϕ follow from the virtual work
δU of the force F (t):
δU = Qxδx+Qϕδϕ = F (t)δx → Qx = F (t) , Qϕ = 0 .
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4 Principles of Mechanics 125
To set up the equations of motion, the following derivatives
have to be calculated:
∂L
∂x=Mx+m(x− rϕ) ,
d
dt
(∂L
∂x
)=Mx+m(x− rϕ) ,
∂L
∂ϕ= mr2ϕ/2−mr(x − rϕ) ,
d
dt
(∂L
∂ϕ
)= mr2ϕ/2−mr(x − rϕ) ,
∂L
∂x= −kx , ∂L
∂ϕ= −kr2ϕ .
Thus, we obtain
(M +m)x−mrϕ+ kx = F (t) ,
−mx+3
2mrϕ + krϕ = 0 .
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126 4 Principles of Mechanics
E4.15 Example 4.15 Fig. 4.16 shows two blocks of mass m1 and m2
which can glide on a friction-
less surface. They are cou-
pled by springs (stiffnesses
k1, k2, k3).
Derive the equations of
motion using the Lagrange
formalism.
k1m1 m2
k3 k2
Fig. 4.16
Solution The system is conservative; it has two degrees of freedom.
We introduce the two coordinates x1 and x2 which describe the
positions of the two blocks.
They are measured from the
equilibrium positions of the
blocks. The kinetic and the
potential energy, respective-
ly, are given by
x1 x2
k1m1 m2
k3 k2
T =1
2m1x
21 +
1
2m2x
22 ,
V =1
2k1x
21 +
1
2k2x
22 +
1
2k3(x2 − x1)2 .
Thus, the Lagrangian of the system is
L = T − V =1
2m1x
21 +
1
2m2x
22 −
1
2k1x
21 −
1
2k2x
22 −
1
2k3(x2 − x1)2 .
To set up the Lagrange equations
d
dt
( ∂L∂x1
)− ∂L
∂x1= 0 ,
d
dt
( ∂L∂x2
)− ∂L
∂x2= 0
the following derivatives must be calculated:
∂L
∂x1= m1x1 ,
d
dt
( ∂L∂x1
)= m1x1 ,
∂L
∂x1= −k1x1 + k3(x2 − x1) ,
∂L
∂x2= m2x2 ,
d
dt
( ∂L∂x2
)= m2x2 ,
∂L
∂x2= −k2x2 − k3(x2 − x1) .
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4 Principles of Mechanics 127
Hence, we obtain
m1x1 + k1x1 − k3(x2 − x1) = 0
→ m1x1 + (k1 + k3)x1 − k3x2 = 0 ,
m2x2 + k2x2 + k3(x2 − x1) = 0
→ m2x2 + (k2 + k3)x2 − k3x1 = 0 .
Note: The two coupled differential equations describe the coupled
free vibrations of the two blocks. In the special case of k3 = 0 the
system is decoupled and we obtain two independent equations of
motion for two systems, each with one degree of freedom.
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128 4 Principles of Mechanics
E4.16 Example 4.16 Two simple pen-
dulums (each mass m, length
l) are connected by a spring
(spring constant k, unstretched
length b) as shown in Fig. 4.17.
Derive the equations of mo-
tion using the Lagrange forma-
lism.
g
l
m
mk
l
Fig. 4.17
Solution The system has two
degrees of freedom. We choose
the generalized coordinates ϕ1
and ϕ2 as shown in the figure.
With the kinetic and the poten-
tial energy, respectively,
2l sin ϕ2
0
ϕ1
ϕ2
m
mk
T = ml2(ϕ1 − ϕ2)2/2 +ml2(ϕ1 + ϕ2)
2/2
→ T = ml2(ϕ21 + ϕ2
2) ,
V = −mgl cos(ϕ1 − ϕ2)−mgl cos(ϕ1 + ϕ2) + k(2l sinϕ2 − b)2/2
→ V = −2mgl cosϕ1 cosϕ2 + k(2l sinϕ2 − b)2/2
(zero level: point 0 and unstressed spring) the Lagrangian becomes
L = T − V
→ L = ml2(ϕ21 + ϕ2
2) + 2mgl cosϕ1 cosϕ2 − k(2l sinϕ2 − b)2/2 .
To set up the Lagrange equations
d
dt
(∂L
∂ϕ1
)− ∂L
∂ϕ1= 0 ,
d
dt
(∂L
∂ϕ2
)− ∂L
∂ϕ2= 0
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4 Principles of Mechanics 129
the following derivatives are needed:
∂L
∂ϕ1= 2ml2ϕ1 ,
d
dt
(∂L
∂ϕ1
)= 2ml2ϕ1 ,
∂L
∂ϕ1= −2mgl sinϕ1 cosϕ2 ,
∂L
∂ϕ2= 2ml2ϕ2 ,
d
dt
(∂L
∂ϕ2
)= 2ml2ϕ2 ,
∂L
∂ϕ2= −2mgl cosϕ1 sinϕ2 − k(2l sinϕ2 − b)2l cosϕ2 .
This leads to the equations of motion:
lϕ1 + g sinϕ1 cosϕ2 = 0 ,
mlϕ2 +mg cosϕ1 sinϕ2 + k(2l sinϕ2 − b) cosϕ2 = 0 .
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130 4 Principles of Mechanics
E4.17 Example 4.17 A disk (weight m2g,
moment of inertia Θ2) glides along
a frictionless homogeneous bar of
weight m1g (Fig. 4.18).
Find the equations of motion
using the Lagrange formalism.
x
ϕ m1
l
m2,Θ2
Fig. 4.18
Solution The system is conservative;
it has two degrees of freedom. Its po-
sition is uniquely determined by the
distance x of point C1 from pin 0 and
by the angle ϕ (generalized coordina-
tes). With the kinetic and the poten-
tial energy, respectively,
0l
2cosϕ
x
x cosϕ
ϕC2
C1
T =1
2
(m1l2
3
)ϕ2 +
12m2[(xϕ)
2 + x2] +1
2Θ2ϕ
22
=1
2
[m1l2
3+m2x
2 + Θ2
]ϕ2 +
1
2m2x
2 ,
V = −m1gl
2cosϕ−m2gx cosϕ = −
[m1
l
2+m2x
]g cosϕ
(zero level at 0) and the Lagrangian L = T − V , we can write
down the Lagrange equations
d
dt
(∂L∂ϕ
)− ∂L
∂ϕ= 0 ,
d
dt
(∂L∂x
)− ∂L
∂x= 0 .
To this end, the following derivatives are needed:
∂L
∂ϕ=(m1l
2
3+m2x
2 +Θ2
)ϕ ,
∂L
∂ϕ= −
(m1
l
2+m2x
)g sinϕ ,
d
dt
(∂L∂ϕ
)=(m1l
2
3+m2x
2 +Θ2
)ϕ+ 2m2xxϕ ,
∂L
∂x= m2x ,
d
dt
(∂L∂x
)= m2x ,
∂L
∂x= m2xϕ
2 +m2g cosϕ .
This leads to the coupled equations of motion:
(m1l2
3+m2x
2 +Θ2
)ϕ+ 2m2xxϕ+
(m1
l
2+m2x
)g sinϕ = 0 ,
m2x−m2xϕ2 −m2g cosϕ = 0 → x− xϕ2 − g cosϕ = 0 .
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4 Principles of Mechanics 131
E4.18Example 4.18 A thin half-cylindri-
cal shell of weight W = mg rolls
without sliding on a flat surface
(Fig. 4.19).
Derive the equation of motion
using the Lagrange formalism.
mr
Fig. 4.19
Solution The system is conservative;
it has one degree of freedom. We in-
troduce the coordinate ϕ as shown
in the figure. With the distance a =
2r/π of the center of mass C from
the center M of the shell we have
M
ϕa
x
y
C
rϕ
ΘC = ΘM − a2m = r2m− a2m= (1− 4/π2)mr2 ,
xc = rϕ− a sinϕ → xc = rϕ − aϕ cosϕ ,
yc = a cosϕ → yc = −aϕ sinϕ .
Thus, the potential energy V , the kinetic energy T , the Lagrangian
and the pertinent derivatives are
V = mga(1− cosϕ) =2
πmgr(1− cosϕ) ,
T =1
2m(x2c + y2c) +
1
2ΘC ϕ
2 =1
2mr2ϕ2
[(1− 2
πcosϕ
)2
+( 2πsinϕ
)2+(1− 4
π2
)]= mr2ϕ2
(1− 2
πcosϕ
),
L = T − V = mr[rϕ2
(1− 2
πcosϕ
)− 2
πg(1− cosϕ)
],
∂L
∂ϕ= mr
[2rϕ
(1− 2
πcosϕ
)],
d
dt
(∂L∂ϕ
)= mr
[2rϕ
(1− 2
πcosϕ
)+
4
πrϕ2 sinϕ
],
∂L
∂ϕ= mr
[ 2πrϕ2 sinϕ− 2
πg sinϕ
].
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132 4 Principles of Mechanics
Introduction into
d
dt
(∂L∂ϕ
)− ∂L
∂ϕ= 0
yields the equation of motion:
ϕ(π − 2 cosϕ) + ϕ2 sinϕ+g
rsinϕ = 0 .
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4 Principles of Mechanics 133
E4.19Example 4.19 A block (mass m1)
can move horizontally on a smooth
surface (Fig. 4.20). A simple pendu-
lum (mass m2) is connected to the
block by a pin.
Find the equations of motion
using the Lagrange formalism.
m1
l
m2
g
Fig. 4.20
Solution The system is conserva-
tive; it has two degrees of free-
dom. We use the generalized coor-
dinates x and ϕ as shown in the
figure. With the zero-level of the
potential of the force m2g chosen
at the height of the mass m1, we
have
l cosϕϕ
m2
x
m1x
l sinϕ
y
V = −m2gl cosϕ ,
T =1
2m1x
2 +1
2m2[(x+ lϕ cosϕ)2 + (lϕ sinϕ)2] ,
L = T − V =1
2(m1 +m2)x
2 +m2lxϕ cosϕ+1
2m2l
2ϕ2 +m2gl cosϕ .
Introduction of the derivatives
∂L
∂ϕ= m2lx cosϕ+m2l
2ϕ ,∂L
∂ϕ= −m2lxϕ sinϕ−m2gl sinϕ ,
d
dt
(∂L∂ϕ
)= m2lx cosϕ−m2lxϕ sinϕ+m2l
2ϕ ,
∂L
∂x= (m1 +m2)x +m2lϕ cosϕ ,
∂L
∂x= 0 ,
d
dt
(∂L∂x
)= (m1 +m2)x+m2lϕ cosϕ−m2lϕ
2 sinϕ
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134 4 Principles of Mechanics
into the Lagrange equations
d
dt
(∂L∂ϕ
)− ∂L
∂ϕ= 0 ,
d
dt
(∂L∂x
)− ∂L
∂x= 0
yields the coupled equations of motion
x cosϕ+ lϕ+ g sinϕ = 0 ,
(m1 +m2)x+m2lϕ cosϕ−m2lϕ2 sinϕ = 0 .
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5Chapter 5
Vibrations
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136 5 Vibrations
E5.11 Example 5.11 The system in Fig 5.30
consists of three bars and a beam
(with negligible masses) and a block
(mass m).
Determine the circular frequency
of the free vertical vibrations. l l
l
EA
EA
m
EA
EI
Fig. 5.30
Solution The system consisting of
the truss and the beam is equiva-
lent to a system consisting of two
springs in parallel (both springs
undergo the same elongation when
the block is displaced). To deter-
mine the spring constant kB of the
beam, we subject the beam to the
force FB = 1 which acts at the
location of the block. This force
produces the deflection (see Engi-
neering Mechanics 2: Mechanics of
Materials, Table 4.3)
FB=1
wB
③
wT
FT =1
②
①
wB =1 · (2l)348EI
.
Thus, the spring constant is given by
kB =1
wB=
48 EI
(2l)3=
6 EI
l3.
In order to find the spring constant kT of the truss, we apply the
force FT = 1 at bar ①. This force causes the displacement (see
Engineering Mechanics 2: Mechanics of Materials, Section 6.3)
wT =∑ S2
i liEA
.
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With
S1 = 1 , S2 = S3 = −1
2
√2 , l1 = l , l2 = l3 =
√2 l
we obtain
wT =1
EA
[12 · l + 2 ·
(√22
)2√2 l]= (1 +
√2)
l
EA
→ kT =1
wT=
EA
(1 +√2)l
.
Now, we replace the two springs in parallel by an equivalent single
spring. Its spring constant k∗ is given by
k∗ = kB + kT =6 EI
l3+
EA
(1 +√2)l
.
Thus, the eigenfrequency is
ω =
√k∗
m=
1
l
√1
ml
(6 EI +
EAl2
1 +√2
).
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E5.12 Example 5.12 The system in Fig. 5.31 consists
of a homogeneous drum (mass M , radius r), a
block (mass m), a spring (spring constant k)
and a string (with negligible mass). The sup-
port of the drum is frictionless. Assume that
there is no slip between the string and the
drum.
Determine the natural frequency of the sys-
tem.k
Mr
m
Fig. 5.31
Solution We first draw the free-body dia-
gram. The position of the block is given by
the coordinate x, measured from the posi-
tion with an unstrechted spring. Next, we
write down the equations of motion for the
block
↓: mx = mg − S1
and for the drum
y
B : Mr2ϕ/2 = S1r − S2r .
In addition, we need the kinematic relation
x = rϕ
and the equation
B
S2 S1
ϕ
S1S2
mg
x
S2 = kx
for the restoring force in the spring. Solving yields the differential
equation for harmonic oscillations:
x+2k
M + 2mx =
2mg
M + 2m.
Thus, the natural frequency is
ω =
√2k
M + 2m.G
ross
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E5.13Example 5.13 Two drums rotate in opposite directions as shown in
Fig. 5.32. They support a homogeneous board of weight W (mass
m). The coefficient of kinetic
friction between the drums
and the board is µ.
Show that the board un-
dergoes a harmonic vibrati-
on and determine the natural
frequency.
x
a
Ω Ω
W
µ
Fig. 5.32
Solution We separate the various parts of the system. The free-
body diagram shows the forces acting if the board is displaced by
an amount x (the friction forces act on the drums in the opposite
directions of the rotations).
W
R1
R2 R1
R2
a/2
a/2
N1N2
x
Thus, the equation of motion in the x-direction of the board is
given by
mx = R2 −R1 .
The normal forces follow from force equilibrium in the vertical
direction and from moment equilibrium:
N1 =W
a
2+ x
a, N2 =W
a
2− xa
.
With the law of kinetic friction R = µN , we obtain
R2 −R1 = µ(N2 −N1) = −µmg2x
a.
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The equation of motion
mx = −µmg 2x
a
thus leads to the differential equation for harmonic vibrations:
x+ 2µg
ax = 0 .
The natural frequency is given by
ω =
√2µ
g
a.
Note that the natural frequency does not depend on the angular
velocity Ω of the drums.
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E5.14Example 5.14 A homogeneous bar (weight W = mg, length l) is
submerged in a viscous fluid and undergoes vibrations about point
A (Fig. 5.33). The drag force Fd acting
on every point of the bar is proportio-
nal to the local velocity (proportiona-
lity factor β).
a) Derive the equation of motion.
Assume small amplitudes and neglect
the buoyancy. b) Calculate the value
β = β∗ for critical damping.
ϕ
A
l
Fig. 5.33
Solution a) We consider an arbitrary
element of length dx of the bar. It is
subjected to the drag force
dFd = β v(x) dx = βxϕ dx .
We restrict ourselves to small amplitu-
des (sinϕ ≈ ϕ). In this case the prin-
ciple of angular momentum yields
ϕ
A
dx
x
mg
l
2sinϕ≈ lϕ
2
dFd
x
A : ΘAϕ = −mg l2ϕ−
l∫
0
βx2ϕdx .
Evaluation of the integral and introduction of ΘA = ml2/3 lead
to the equation of motion
ϕ+βl
mϕ+
3
2
g
lϕ = 0 → ϕ+ 2ξϕ+ ω2ϕ = 0
where
ξ =βl
2m, ω2 =
3g
2l.
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142 5 Vibrations
b) Critical damping is characterized by
ξ = ω or ζ = 1 .
Thus,
β∗l
2m=
√3
2
g
l→ β∗ =
m
l
√6g
l.
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E5.15Example 5.15 The pendulum of a clock consists of a homogeneous
rod (mass m, length l) and a homoge-
neous disk (mass M , radius r) whose
center is located at a distance a from
point A (Fig. 5.34). Assume small am-
plitudes and determine the natural fre-
quency of the corresponding oscillati-
ons. Choose m = M and r ≪ a and
calculate the ratio a/l which yields the
maximum eigenfrequency.
A
M
m
2r
la
Fig. 5.34
Solution We introduce the angle ϕ
as shown and apply the principle
of angular momentum:
x
A : ΘAϕ = −mgl/2 sinϕ−Mga sinϕ
where
ϕ
Mg
mg
A
ΘA = ml2/3 +M(r2 + 2a2)/2 .
If we assume small amplitudes (sinϕ ≈ ϕ), we can linearize the
equation of motion:
ϕ+(ml + 2Ma)g
2ΘAϕ = 0 .
Hence, the eigenfrequency is obtained as
ω =
√(ml + 2Ma)g
2ΘA.
In the special case of m = M and r ≪ a the natural frequency
simplifies to
ω =
√3l+ 6a
2l2 + 6a2g .
The ratio a/l which yields the maximum eigenfrequency is found
by setting the derivative dω/da equal to zero:
dω
da= 0 → a/l =
1
2
(√7
3− 1
).G
ross
, Hau
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144 5 Vibrations
E5.16 Example 5.16 The system in Fig 5.35
consists of a homogeneous pulley
(mass M , radius r), a block (mass m)
and a spring (spring constant k).
Determine the equation of motion
for the block and its solution for the
initial conditions x(0) = 0, v(0) = v0.
Neglect the mass of the string and
any lateral motion.
M
A
m
r
k
Fig. 5.35
Solution We separate the pulley and
the block and measure the displace-
ments x of the block and xA of point A
from the position of equilibrium. With
this choice, we do not have to consider
the weights Mg and mg in the free-
body diagram. Thus, the equations of
motion are
① ↓ : mx = −S1 ,
② ↓ : MxA = S1 + S2 − kxA ,
x
A : ΘAϕ = rS1 − rS2 .
If we use the kinematic relations (Π=:
instantaneous center of rotation, see
the figure)
x
kxA2
xAΠ
ϕ
1
S2S1
x
xA
xA =x
2, 2rϕ = x → xA =
x
2, ϕ =
x
2r
and ΘA =Mr2/2 we can solve the equations of motion for x and
obtain
x+k
4m+3
2M
x = 0 → ω =
√√√√k
4m+3
2M
.
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The general solution of this differential equation is given by
x(t) = A cosωt+B sinωt .
The initial conditions
x(0) = 0 → A = 0 , x(0) = v0 → B =v0ω
lead to
x(t) =v0ω
sinωt .
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E5.17 Example 5.17 A wheel (mass m, radius r) rolls without slipping
on a circular path (Fig. 5.36). The mass of the rod (length l) can
be neglected; the joints are
frictionless.
Derive the equation of
motion and determine the
natural frequency of small
oscillations. ϕ
g
m
r
l
A
Fig. 5.36
Solution We apply conserva-
tion of energy
T + V = const
to derive the equation of mo-
tion. The kinetic energy of
the rolling wheel is given by
(see the figure)
Cψ
vc
ϕ
T = mv2c/2 + ΘCψ2/2 ,
the potential energy is (zero-level at the height of point A)
V = −mgl cosϕ .
With the mass moment of inertia ΘC = mr2/2 and the kinematic
relations
vc = lϕ , vc = rψ → lϕ = rψ
the kinetic energy can be written as
T = 3ml2ϕ2/4 .
Introduction into the expression for conservation of energy gives
3lϕ2/4− g cosϕ = const .
Differentiation yields
3
2lϕϕ+ gϕ sinϕ = 0 → ϕ+
2g
3lsinϕ = 0 .G
ross
, Hau
ger,
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5 Vibrations 147
If we restrict ourselves to small amplitudes (sinϕ ≈ ϕ) this equa-tion reduces to
ϕ+2g
3lϕ = 0 .
Thus, the natural frequency is obtained as
ω =
√2g
3l.
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E5.18 Example 5.18 The simple pendulum in Fig. 5.37 is attached to a
spring (spring constant k) and a dashpot (damping coefficient d).
a) Determine the maximum value of
the damping coefficient d so that
the system undergoes vibrations.
Assume small amplitudes.
b) Find the damping ratio ζ so that
the amplitude is reduced to 1/10
of its initial value after 10 full cy-
cles. Calculate the corresponding
period Td.
a
ad
m
k
Fig. 5.37
Solution a) The equation of motion
follows from the principle of angular
momentum (rotation about point A;
small amplitudes: sinϕ ≈ ϕ, cosϕ ≈ 1):
x
A : ΘAϕ = −Fda− Fk2a−mg2aϕ .
With mg
Fk
Fd
ϕ
A
ΘA = m (2a)2 , Fd = da ϕ , Fk = k2aϕ
we obtain
ϕ+d
4mϕ+
( km
+g
2a
)ϕ = 0 → ϕ+ 2ξϕ+ ω2ϕ = 0 ,
where
ξ =d
8m, ω2 =
k
m+
g
2a.
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In order to have oscillations, the system must be underdamped:
ζ < 1 → ξ < ω → d
8m<
√k
m+
g
2a
→ d < 8
√km+
gm2
2a.
b) The necessary damping ratio follows with xn+10 = xn/10 from
the logarithmic decrement:
102πζ√1− ζ2
= lnxn
xn+10= ln 10
→ ζ =
√√√√1
( 20π
ln 10
)2+ 1
= 0.037 .
This leads to the period
Td =2π
ω√1− ζ2
≈ 2π
ω= 2π
√2am
2ak + gm.
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E5.19 Example 5.19 The structure in Fig. 5.38 consists of an elastic beam
(flexural rigidity EI, axial rigidity EA→∞, negligible mass) and
three rigid bars (with negligible masses). The block (mass m) is
suspended from a spring (spring constant k).
Determine the eigenfrequency of the vertical oscillations of
the block.
k
a
a a a
mFig. 5.38
Solution We reduce the structure to an
equivalent system of a spring (spring
constant k∗, see the figure) and a mass.
To this end we first replace the beam
and the bars by a spring with the spring
constant kB. We can determine kB if we
subject the beam to a force F which acts
at the free end. This force produces the
deflection
m
k∗
w =Fa3
EI
(see Engineering Mechanics 2: Mechanics of Materials, Example
6.22). The spring constant is obtained from
kB =F
w→ kB =
EI
a3.
The displacement of the block is the sum of the elongations of the
springs with spring constants k and kB. Therefore, the beam/bars
and the given spring act as springs in series. Thus, the spring
constant k∗ of the equivalent system follows from
1
k∗=
1
kB+
1
k→ k∗ =
kEI
ka3 + EI.
This yields the eigenfrequency
ω =
√k∗
m→ ω =
√kEI
(ka3 + EI)m.
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E5.20Example 5.20 A rod (length l, with negligible mass) is elastically
supported at point A (Fig. 5.39). The rotational spring (spring
constant kT ) is unstretched for ϕ = 0. The rod carries a point
mass m at its free end.
Derive the equation of motion.
Determine the spring constant so
that ϕ = π/6 is an equilibrium posi-
tion. Calculate the natural frequen-
cy of small oscillations about this
equilibrium position.
g
m
kT
l
ϕ
A
Fig. 5.39
Solution To derive the equation of
motion we apply the principle of an-
gular momentum:
y
A : ΘAϕ = mgl cosϕ−MA .
With the mass moment of inertia
mg
MA
ϕ
ΘA = ml2 and the restoring moment MA = kTϕ we obtain
ϕ =g
lcosϕ− kT
ml2ϕ .
Since ϕ = ϕ0 = π/6 is a position of equilibrium, the conditi-
on ϕ(π/6) = 0 leads to the required spring constant (note that
cos(π/6) =√3/2):
√3
2
g
l− π
6
kTml2
= 0 → kT =3√3
πmgl .
Now we consider small oscillations about this position of equili-
brium. We assume that
ϕ = ϕ0 + ψ with |ψ| ≪ 1 .
Introduction into the equation of motion yields
ψ =g
lcos(ϕ0 + ψ)− kT
ml2(ϕ0 + ψ) .
We use the trigonometric relation
cos(ϕ0 + ψ) = cosϕ0 cosψ − sinϕ0 sinψ
→ cos(ϕ0 + ψ) =
√3
2cosψ − 1
2sinψG
ross
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and linearize (cosψ ≈ 1, sinψ ≈ ψ) to obtain
ψ = −1
2
g
lψ − kT
ml2ψ +
√3
2
g
l− π
6
kTml2
→ ψ +
(1
2+
3√3
π
)g
lψ = 0 .
Thus, the natural frequency is given by
ω =
√√√√(1
2+
3√3
π
)g
l.
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E5.21Example 5.21 A single story frame
consists of two rigid columns (with
negligible masses), a rigid beam of
mass m and a spring-dashpot sys-
tem as shown in Fig. 5.40. The
ground is forced to vibration by an
earthquake; the acceleration uE =
b0cosΩt is known from measure-
ments.
Determine the maximum ampli-
tude of the steady state vibrations.
d
uE
m
k
45
Fig. 5.40
Assume that the system is underdamped and that the vibrations
have small amplitudes.
Solution We assume that the
amplitudes of the vibrations
are small. Then the elongati-
on of the diagonal is obtained
as
∆ =
√2
2(x− uE) .
The elongation produces the
force
uE
F
xm
x− uE
∆
45
F = k∆+ d ∆
in the spring-dashpot system (see the figure). The equation of
motion of the beam is given by
→: mx = −F√2
2
→ mx+d
2(x− uE) +
k
2(x− uE) = 0 .
Thus, the relative displacement y = x− uE is described by
my +d
2y +
k
2y = mb0 cosΩt
→ 1
ω2y +
2ζ
ωy + y = y0 cosΩt ,
where
ω2 =k
2m, ζ =
d
2
√1
2km, y0 =
2mb0k
.
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The maximum amplitude A is obtained for η = Ω/ω ≈ 1 (reso-
nance!). In the case of small damping (ζ ≪ 1) we obtain
A = y0Vmax ≈y02ζ
= 2√2b0d
√m3
k.
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E5.22Example 5.22 The undamped system in Fig. 5.41 consists of a
block (mass m = 4 kg) and a spring (spring constant k = 1
N/m). The block is subjected to a force F (t). The initial conditions
x(0) = x0 = 1 m, x(0) = 0 and the response
x(t) = x0
[cos
t
2t0+ 20
(1− cos
t− T2t0
)〈t− T 〉0
]
to the excitation are given. Here,
t0 = 1 s, T = 5 s and 〈t− T 〉0 = 0
for t < T and 〈t − T 〉0 = 1 for
t > T .
Calculate the force F (t).
F (t)
x(t)
mk
Fig. 5.41
Solution First we calculate the force F for t < T . Then 〈t−T 〉0 = 0
and the response is given by
x(t) = x0 cost
2t0.
The unknown force follows from the equation of motion:
F = mx+ kx .
With
x = − x02t0
sint
2t0and x = − x0
4t20cos
t
2t0
we obtain
F (t) =
(−mx0
4t20+ kx0
)cos
t
2t0.
Inserting the given numerical values of the parameters leads to
F (t) ≡ 0 for t < T .
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Now we consider the case t > T . Then 〈t − T 〉0 = 1 and the
response follows as
x(t) = x0
[cos
t
2t0+ 20
(1− cos
t− T2t0
)]
→ x =x02t0
[− sin
t
2t0+ 20 sin
t− T2t0
]
→ x =x04t20
[− cos
t
2t0+ 20 cos
t− T2t0
].
Introduction into the equation of motion yields
F (t) =
(−mx0
4t20+ kx0
)cos
t
2t0
+
(20mx04t20
− 20kx0
)cos
t− T2t0
+ 20kx0
→ F (t) ≡ 20 N for t > T .
105
20
10
30
t[s]
F (t) [N]
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E5.23Example 5.23 A simplified model of a car (mass m) is given by
a spring-mass system (Fig. 5.42). The car drives with constant
velocity v0 over an uneven surface in the form of a sine function
(amplitude U0, wavelength L).
a) Derive the equation of moti-
on and determine the forcing
frequency Ω.
b) Find the amplitude of the ver-
tical vibrations as a function
of the velocity v0.
c) Calculate the critical velocity
vc (resonance!).
x
L
cU0
v0
m
Fig. 5.42
Solution a) We denote the verti-
cal displacement of the car by x,
the uneven surface is described
by u. Then Newton’s law reads
↑ : mx = −k (x − u) .
With the position of the car, s =
v0t, in the horizontal direction
we obtain
LU0
u
s
m
x
u
u = U0 cos2πs
L= U0 cos
2πv0t
L= U0 cosΩt .
Thus,
mx+ k x = k U0 cosΩt with Ω =2πv0L
.
b) We assume the solution of the equation of motion to be of
the form of the right-hand side: x = x0 cosΩt. This leads to the
amplitude of the steady state vibrations:
x0 =U0
1− Ω2
ω2
=U0
1− 4π2v20L2
m
k
where ω2 = k/m.
c) Resonance occurs for Ω = ω:
4π2v2cL2
m
k= 1 → vc =
L
2π
√k
m.
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E5.24 Example 5.24 A homogeneous wheel (mass m) is attached to a
spring (spring constant k). The
wheel rolls without slipping on
a rough surface which moves
according to the function u =
u0cosΩt (Fig. 5.43).
a) Determine the amplitude of
the steady state vibrations.
x
k C
m
r
uµ0
Fig. 5.43
b) Calculate the coefficient µ0 of static friction which is necessary
to prevent slipping.
Solution The equations of mo-
tion for the wheel are given by
↑ : 0 = N −mg , (a)
→ : mx = −k x+H , (b)
y
C : ΘC ϕ = −r H . (c)
With the kinematic relation
x
ϕr
H
kx
C
mg
N
x = u+ rϕ → x = u+ rϕ = −u0 Ω2 cosΩt+ rϕ
and ΘC = mr2/2 we obtain from (b) and (c) the differential equa-
tion for forced vibrations:
x+2
3
k
mx = −1
3u0Ω
2 cosΩt .
a) We assume the solution to be of the form of the right-hand
side: x = x0 cosΩt. This leads to the amplitude of the steady state
vibrations:
|x0| =u0
3
∣∣∣∣2
3
k
mΩ2− 1
∣∣∣∣.
b) The condition |H |max ≤ µ0N for static friction and (a), (b)
yield the required coefficient of static friction:
µ0 ≥|H |max
N=
∣∣∣∣x+k
mx
∣∣∣∣max
g=
∣∣∣∣−x0Ω2 +k
mx0
∣∣∣∣g
→ µ0 ≥u0Ω
2
3g
∣∣∣∣∣∣∣
k
mΩ2− 1
2
3
k
mΩ2− 1
∣∣∣∣∣∣∣.
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E5.25Example 5.25 A small homogeneous disk (mass m, radius r) is at-
tached to a large homogeneous disk (mass M , radius R) as shown
in Fig. 5.44. The torsion spring
(spring constant kT ) is unstret-
ched in the position shown.
Determine the eigenfrequency
of the oscillations. Assume small
amplitudes.
M AR
a
m
kT
r
g
Fig. 5.44
Solution We apply the principle
of angular momentum to derive
the equation of motion:
x
A : ΘAϕ = −kTϕ−mga sinϕ .
In the case of small amplitudes
(sinϕ ≈ ϕ) this equation reduces
to
mg
kTϕ
A
ϕ
ϕ+ ω2ϕ = 0 with ω2 =kT +mga
ΘA.
Inserting the mass moment of inertia
ΘA =MR2
2+
[mr2
2+ma2
]
we can write the eigenfrequency in the form
ω =
√√√√√kT +mga
1
2MR2 +m(
r2
2+ a2)
.
Note that the problem can also be solved with the aid of the
conservation of energy (we choose V = 0 for ϕ = 0):
T + V = const → 1
2ΘAϕ
2 +1
2kTϕ
2 +mga(1− cosϕ) = const.
Differentiation yields
ΘAϕϕ+ kTϕϕ+mga sinϕ ϕ = 0 .
With sinϕ ≈ ϕ and ϕ 6= 0 we obtain the same result as above.Gro
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E5.26 Example 5.26 The systems ➀
and ➁ in Fig. 5.45 consist of
two beams (negligible masses,
flexural rigidity EI), a spring
(spring constant k) and a box
(mass m).
Determine the spring con-
stants k∗ of the equivalent
springs for the two systems.
a2a
EIEIk
m
EIEI
m
k
1
2
Fig. 5.45
Solution We reduce both sys-
tems to the equivalent sim-
ple systems of a mass and a
spring.
1l , EI
w
In system ①, the three “springs” are attached to the mass. The-
refore, they undergo the same deflection when the box is displaced:
they act as springs in parallel. The equivalent spring constant k∗
is the sum of the individual spring constants:
k∗ =∑
ki .
We obtain the spring constants kL and kR of the right and the
left beam, respectively, if we subject the cantilevers to a force 1
at their free ends. The corresponding deflections are
w =1 · l33EI
(see Engineering Mechanics 2: Mechanics of Materials, Section
4.5) which leads to
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kL =1
wL=
3EI
(2a)3, kR =
1
wR=
3EI
a3.
Thus,
k∗ = kL + kR + k =27
8
EI
a3+ k =
27EI + 8ka3
8a3.
m m m
kR k
kk
k∗kL
Now we consider system ②. Here, the two beams act as springs
in parallel with equivalent spring constant k. This then acts in
series with given spring (spring constant k). Hence,
k = kL + kR =27
8
EI
a3,
1
k∗=
1
k+
1
k=
8a3
27EI+
1
k
→ k∗ =27EIk
27EI + 8ka3=
27EI
8a3 + 27EI
k
.
Note that the stiffness of system ② is smaller than the one of
system ①. Therefore it vibrates with a smaller frequency.
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E5.27 Example 5.27 A homogeneous wheel (mass m, moment of inertia
ΘC , radius r) rolls without slipping on a rough beam (mass M).
The beam moves without
friction on roller supports
(Fig. 5.46).
Determine the natural fre-
quency of the system.
k C
m,ΘC
r
M
Fig. 5.46
Solution We separate the wheel
and the beam and introduce the
coordinates x1, x2 and ϕ (see
the figure). The coordinates are
measured from the position of
equilibrium. Then the equati-
ons of motion are
① → : mx1 = −kx1 −H ,
y
C : ΘC ϕ = r H ,
② → : M x2 = H .
If we use the kinematic relation
1
2 BA
x1
x2
H
N
ϕr
H
mg
Ckx1
x1 = x2 + rϕ → x1 = x2 + rϕ → x1 = x2 + rϕ
and solve for x1 we obtain
x1 +k
m+M
1 +Mr2/ΘC
x1 = 0 .
Thus, the natural frequency is given by
ω =
√√√√√k
m+M
1 +Mr2/ΘC
.
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6Chapter 6
Non-Inertial Reference Frames
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E6.5 Example 6.5 Point A of the simple
pendulum (mass m, length l) in
Fig. 6.8 moves with a constant ac-
celeration a0 to the right.
Derive the equation of motion.
g
a0A
lϕ
m
Fig. 6.8
Solution We introduce the ξ, η-co-
ordinate system as shown in the fi-
gure. It is a translating coordinate
system with point A as the origin.
The equation of motion in the mo-
ving system is
mar = F + Ff .
The (real) force F acting at the
mass is given by
F = −S sinϕeξ + (S cosϕ−mg)eη
and the fictitious force Ff is
W =mg
eη
S
η
Aeξ
ϕξ
m
Ff = −maf = −ma0eξ .
Note that the Coriolis force is zero since ω = 0.
The components of the relative acceleration ar follow from the
coordinates of the point mass in the moving system through dif-
ferentiation:
ξ = l sinϕ, η = − l cosϕ,ξ = lϕ cosϕ, η = lϕ sinϕ,
ξ = lϕ cosϕ− lϕ2 sinϕ, η = lϕ sinϕ+ lϕ2 cosϕ .
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This yields the relative acceleration
ar = ξ eξ + η eη = (lϕ cosϕ− lϕ2 sinϕ)eξ
+(lϕ sinϕ+ lϕ2 cosϕ)eη .
Introduction into the equation of motion leads to the components
of the equation of motion in the direction of the axes ξ and η:
m (lϕ cosϕ− lϕ2 sinϕ) = −S sinϕ−ma0 ,
m (lϕ sinϕ+ lϕ2 cosϕ) = S cosϕ−mg.
These are two equations for the unknowns ϕ and S. Solving for ϕ
yields the equation of motion
lϕ+ g sinϕ+ a0 cosϕ = 0 .
Note that the position ϕ0 = − arctana0/g is obtained for ϕ = 0.
The pendulum oscillates about this position for ϕ 6= 0.
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166 6 Non-Inertial Reference Frames
E6.6 Example 6.6 The two disks in
Fig. 6.9 rotate with constant
angular velocities Ω and ω
about their respective axes.
Determine the absolute ac-
celeration of point P at the in-
stant shown.
ϕ
Ω
Pr
ω
a
z
xy
Fig. 6.9
Solution We describe the mo-
tion of point P in a coordina-
te system x, y, z which is fixed
to the large disk. The absolute
acceleration of P is
aP = af + ar + ac ,
where the acceleration of the
reference frame and the relati-
ve acceleration are given by
ϕPr
z
y
x
af =
0
−(a+ r cosϕ)Ω2
0
, ar =
0
−rω2 cosϕ
−rω2 sinϕ
.
We also write the angular velocity of the reference frame and the
relative velocity as column vectors:
Ω =
0
0
Ω
, vr =
0
−rω sinϕ
rω cosϕ
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and we calculate the Coriolis acceleration:
ac = 2Ω× vr → ac =
2rωΩ sinϕ
0
0
.
Combining yields
aP =
2rωΩ sinϕ
−(a+ r cosϕ)Ω2 − rω2 cosϕ
−rω2 sinϕ
.
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E6.7 Example 6.7 A horizontal circular platform (radius r) rotates with
constant angular velocity Ω (Fig. 6.10). A block (massm) is locked
in a frictionless slot at a distance a
from the center of the platform. At
time t = 0 the block is released.
Determine the velocity vr of the
block relative to the platform when
it reaches the rim of the platform.
r a
m
Ω
Fig. 6.10
Solution We describe the motion of
the block in a coordinate system x, y
which is fixed to the platform. The
absolute acceleration of the block is
given by
a
y x
aB = af + ar + ac.
Here, the acceleration of the reference frame, the relative accele-
ration and the Coriolis acceleration are
af =
[−xΩ2
0
], ar =
[x
0
], ac =
[0
2Ωx
].
Thus, the absolute acceleration becomes
aB =
[x− Ω2x
2Ωx
].
The equation of motion for the block is
maB = F ,
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where the force F (which is exerted from the slot on the block) is
F =
[0
Fy
].
We now write down the x-component of the equation of motion:
m(x − Ω2x) = 0 → x− Ω2x = 0 .
The general solution of this differential equation is given by
x(t) = A coshΩt+B sinhΩt .
With the initial conditions
x(0) = a → A = a ,
x(0) = 0 → B = 0
we obtain
x(t) = a coshΩt .
When the block reaches the rim of the platform, the condition
x(tR) = r → coshΩtR = r/a
is satisfied. Thus the relative velocity is (note that cosh2 x −sinh2 x = 1)
x(tR) = aΩ sinhΩtR → x(tR) = Ω√r2 − a2 .
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E6.8 Example 6.8 A simple pendulum is
attached to point 0 of a circular disk
(Fig. 6.11). The disk rotates with
a constant angular velocity Ω; the
pendulum oscillates in the horizon-
tal plane.
Determine the circular frequency
of the oscillations. Assume small
amplitudes and neglect the weight
of the mass.
l mr
0
Ω
Fig. 6.11
Solution We introduce a rotating ξ, η, ζ-coordinate system. Then
Ω = Ωeζ , a0 = r0 = −rΩ2eξ ,
Ω = 0 , r0P = l cosϕ eξ + l sinϕ eη .
The relative velocity can be ex-
pressed by the relative angular ve-
locity ϕ∗ (∗: time derivative rela-
tive to the moving frame):
vr
r0P
ϕr0
P
ξ
Ωη
vr = lϕ∗ → vr = −lϕ∗ sinϕ eξ + lϕ∗ cosϕ eη .
Thus, the fictitious forces F f and F c are
F f = − ma0 −mΩ× (Ω× r0P ) = m(rΩ2 + lΩ2 cosϕ)eξ
+ mΩ2l sinϕ eη ,
F c = − 2mΩ× vr = 2mΩlϕ∗(eξ cosϕ+ eη sinϕ) .
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With the tangential relative acceleration art = lϕ∗∗ the equa-
tion of motion in the tangential direction is obtained as
տ: mlϕ∗∗ = m(lΩ2 + 2lϕ∗Ω) sinϕ cosϕ
− m[rΩ2 + (lΩ2 + 2lϕ∗Ω) cosϕ] sinϕ
= − mrΩ2 sinϕ .
We assume small amplitudes
(sinϕ ≈ ϕ). This yields
ϕ∗∗ +rΩ2
lϕ = 0 .
ξ
m
art
ϕS
mΩ2l sinϕ
2mlϕ∗Ω sinϕ
η2mlϕ∗Ωcosϕ
m(rΩ2+lΩ2 cosϕ)
Hence, the circular frequency of the oscillations is
ω =√r/l Ω .
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E6.9 Example 6.9 A drum rotates with
angular velocity ω about point B
(Fig. 6.12). Pin C is fixed to the
drum; it moves in the slot of link
AD.
Determine the angular velocity
ωAD of link AD and the velocity
vr of the pin relative to the link at
the instant shown. ω
A
C
B
D
4 l
3 l
Fig. 6.12
Solution We use the rotating co-
ordinate system x, y as shown in
the figure. The (absolute) velocity
of pin C is given bya
β
A
C
y
3 ℓ
x
4 ℓ
vC = 3lω
[cosβ
− sinβ
].
With the geometrical relations
a =√16l2 + 9l2 = 5l , sinβ = 3/5 , cosβ = 4/5
we can write
vC =3lω
5
[4
−3
].
The velocity of the reference frame at point C and the velocity of
pin C relative to the moving frame are
vf = βl
[0
5
], vr = vr
[1
0
].
With
vC = vf + vr → 3lω
5
[4
−3
]= βl
[0
5
]+ vr
[1
0
],
we finally obtain
ωAD = β = −9ω
25, vr =
12
5ωl
[1
0
].G
ross
, Hau
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E6.10Example 6.10 A point P moves along a circular path (radius r)
on a platform with a constant relative velocity vr (Fig. 6.13). The
platform rotates with a con-
stant angular velocity ω about
point A. The eccentricity e is gi-
ven.
Determine the relative, fixed
frame-, Coriolis, and absolute
accelerations of P .
0
vr
A e
ω rP
Fig. 6.13
Solution We introduce the coor-
dinate system ξ, η, ζ as shown in
the figure. Its origin is located at
the center 0 of the platform; it
rotates with the platform. Thus
point P undergoes a circular
motion relative to this system.
With the magnitude ar = v2r/r
of the relative acceleration and
its direction (from P to 0) we
can write
ξϕ
η
0
vr
A e
ω r0P
P
ar = −v2r
r(eξ cosϕ+ eη sinϕ) .
With
ω = ωeζ , ω = 0 , r0P = eξr cosϕ+ eηr sinϕ ,
vr = vr(−eξ sinϕ+ eη cosϕ) , r0 = a0 = −e ω2eξ
we obtain
af = a0 + ω × (ω × r0P )
= −e ω2eξ + rω2[eζ × (eζ × eξ cosϕ) + eζ × (eζ × eη sinϕ)]
= −(e+ r cosϕ)ω2eξ − rω2 sinϕeη ,
ac = 2ω × vr = 2ωvr[eζ × (−eξ sinϕ) + eζ × eη cosϕ]
= −2ωvr(eξ cosϕ+ eη sinϕ) .Gro
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Thus, the absolute acceleration is found as
a = af + ar + ac
= −[eω2 + (rω2 +v2rr
+ 2ωvr) cosϕ]eξ − [rω2 +v2rr
+ 2ωvr] sinϕ eη
= −[eω2 + r(ω +vrr)2 cosϕ]eξ − r(ω +
vrr)2 sinϕ eη .
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E6.11Example 6.11 A circular ring (ra-
dius r) rotates with constant an-
gular velocity Ω about the x-
axis (Fig. 6.14). A point mass m
moves without friction inside the
ring.
Derive the equations of moti-
on and determine the equilibrium
positions of the point mass relati-
ve to the ring.
Ω
g
m
x
yz
r
ϕ
Fig. 6.14
Solution We describe the motion of the point mass in the x, y, z-
coordinate system (see the figure) which rotates with the ring.
The absolute acceleration a of the point mass is given by
a = af + ar + ac ,
where the acceleration of the reference frame and the relative ac-
celeration are
af =
0
−rΩ2 sinϕ
0
, ar =
−rϕ2 cosϕ− rϕ sinϕ
−rϕ2 sinϕ+ rϕ cosϕ
0
.
With the angular velocity of the ring and the relative velocity
Ω =
Ω
0
0
, vr =
−rϕ sinϕ
rϕ cosϕ
0
we can calculate the Coriolis acceleration
ac = 2Ω× vr → ac =
0
0
2rΩϕ cosϕ
.
The equation of motion is given by
ma = W +N ,Gro
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where
W =
−mg0
0
is the weight of the point mass and
N =
Nx
Ny
Nz
with Ny/Nx = tanϕ
is the force exerted from the ring on the point mass. Now, we can
write down the components of the equation of motion:
−m(rϕ2 cosϕ+ rϕ sinϕ) = −mg +Nx ,
−m(rϕ2 sinϕ− rϕ cosϕ+ rΩ2 sinϕ) = Nx tanϕ ,
2mrΩϕ cosϕ = Nz .
Positions of equilibrium relative to the ring are characterized by
ϕ = 0 , ϕ = 0 →(rΩ2 +
g
cosϕ
)sinϕ = 0 .
This yields
ϕ1 = 0 , ϕ2 = π , ϕ3,4 = π ± arccosg
rΩ2.
The positions ϕ3,4 exist only for Ω2 > g/r.
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E6.12Example 6.12 A point P moves
on a square plate along a cir-
cular path (radius r) with a
constant relative velocity vr.
The plate moves horizontally
with the constant acceleration
a0 (Fig. 6.15).
Determine the magnitude of
the absolute acceleration of P .
a0
vrP
r
Fig. 6.15
Solution The components of the
relative velocity in the moving
reference frame ξ, η are
vrξ = ξ∗ = −vr sinϕ ,
vrη = η∗ = vr cosϕξ0
y
x
fixed
η
ϕ
r
vrP
(∗: time derivative relative to the moving frame). Differentiation
with respect to the moving frame leads to the components of the
relative acceleration (note: rϕ∗ = vr ):
arξ = ξ∗∗ = −vrϕ∗ cosϕ = −v2r
rcosϕ ,
arη = η∗∗ = −vrϕ∗ sinϕ = −v2r
rsinϕ .
Since the reference frame undergoes a translation, the absolute
acceleration is given by
ax = a0 + arξ = a0 −v2rr
cosϕ ,
ay = arη = −v2r
rsinϕ .
It has the magnitude
a =√a2x + a2y =
√a20 +
v4rr2− 2a0
v2rr
cosϕ .
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E6.13 Example 6.13 A crane starts to move from rest with a constant
acceleration b0 along a straight track. At the same time, the jib
begins to rotate with constant angu-
lar velocity ω, and the trolley on the
jib begins to move towards point 0
with constant relative acceleration bc(Fig. 6.16). The initial positions of the
jib and of the trolley are given by ϕ0
and s0.
Determine the absolute velocity and
the absolute acceleration of the trolley
as functions of the time t.
0
ϕ
s
b0
bc
b0
ω
bc
Fig. 6.16
Solution We use the fixed coordinate system x, y, z, where the
x-axis coincides with the track. In ad-
dition, we introduce the rotating coor-
dinate system ξ, η, ζ, where the ξ-axis
rotates with the jib. Then, the gene-
ral equations for the absolute velocity
and the absolute acceleration are
r0P
ζ
ηy
xz
0
P
ω
ξ
ϕ
v = v0 + ω × r0P + vr ,
a = a0 + ω × r0P + ω × (ω × r0P ) + 2ω × vr + ar .
We measure the time t from the beginning of the motion. Then,
making use of the given accelerations, angular velocity and initial
conditions, we obtain
a0 = b0ex → v0 = b0t ex ,
ω = ωeζ , ω = 0 ,
ar = −bceξ → vr = −bct eξ → r0P = (− 12bct
2 + s0)eξ ,
and
ω × r0P = ω(− 12bct
2 + s0)eη , 2ω × vr = −2ω bct eη ,ω × (ω × r0P ) = −ω2(− 1
2bct2 + s0)eξ .
Gro
ss, H
auge
r, Sc
hröd
er, W
all,
Gov
idje
e
Engi
neer
ing
Mec
hani
cs 3
, Dyn
amics
Sprin
ger 2
013
6 Non-Inertial Reference Frames 179
With the relation ex = eξ cosϕ−eη sinϕ, where ϕ = ϕ0 +ω t, we
finally obtain
v = [b0t cosϕ− bct]eξ + [−b0t sinϕ+ ω(− 12 bct
2 + s0)]eη ,
a = [b0 cosϕ− ω2(− 12bct
2 + s0)− bc]eξ − [b0 sinϕ+ 2ω bct]eη .
Gro
ss, H
auge
r, Sc
hröd
er, W
all,
Gov
idje
e
Engi
neer
ing
Mec
hani
cs 3
, Dyn
amics
Sprin
ger 2
013