jump to first page chapter 4 kinetics of a particle xx f(x)f(x) x ff max min
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Chapter 4 Kinetics of a Particle
0" and 0d
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0" and 0d
d when occurs, maximum Local
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d sloped aldifferenti
tangentd
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Integration: the reverse of differentiation
xox+x
xx
f(x)
x)(
)(
)()(lim
)(
)()( lim
: from curve under the area thecalculate To
0
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constant a is C
Ced e ed
de
Clnd 1
1
d
lnd
1- n ,C1n
dx n d
d
Ccotdcsc csc- dx
cotd
Ctand sec sec d
tand
cosd sin ,sin - dx
cosd
1nn1-n
n
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4
2F
1F RF
zzz
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i
FFF
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maFmaFmaF
amFi
21
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where
and ,
Newton’s 2nd law
constant ,0 ,0 ,0 vaamFi
iNewton’s 1st law
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6
Work done,cos FdsrdFdU
zFyFxF zyx ddd
rdds
where
rdFU
Total work done
Example 1 What is the work done by a force on a article: (a)in circular motion? (b)horizontal motion? (c)from A to B?
F
rd
v
F
A
B
hg
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7
rd
F
Kinetic energy K.E.
Work done by an external force
K.E.)in (change ..
2
1
2
1
d2
2
) d(
d d
d d
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2121
2
EK
mvmv
vmvvm
vvmrdtvdm
ramrFW
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9
Dissipative force (e.g. friction): work done from one point to another point depends on the path.
f
A
Bpath 1
path 2
e.dissipativ is force themeanssign ""
W Smg -
rdmgW
Smg -
)d toopposite is ( rdmg
dW
)2path (BA
2
)2path (BA
1
)1path (BA
2path
1path
1path
rf
rf
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10
P.E. rF
dF
EKdF
dFdF
dFdF
dFdF
energy, potential of change theas defined isIt only, points end and initial offunction a is dc Therefore,
same. theis rc
0.. force, veconservati aunder path closed a of completionAfter 0 r
rr
rr
rr
pathany
pathany
c
2path reversed c1path c
2path c1path c
2path c1path c
0
0
path close
cF
Non-dissipative force (conservative force): work done from one point to another point is independent on the path.
A
Bpath 1
path 2
P.E. between two points is equal to the work done by an external force against the field of a conservative force for bringing the particle from the starting point to the end point, with the external force = .
,d)c(- pathany rF
c- F
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11
Example 2 (gravitational potential)
r
mMG
r
mMG r
r
mMGrV
V
rrrr
mMGV
rrr
mMGV
rVP.E.
rr
mMGF
rr
2
r
2
r
2
d)(
. 0such that ,at set ispoint reference The
ˆd]ˆ[
d]ˆ[
)(
;ˆ2
R
Mr m
r̂
X=0 X
X
Example 3 Find V of a spring. Ans. kx2/2
Example 4 Potential energy of a mass m, positioned at h from the ground. Ans. mgh
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12
.Δd K.ErF
In general, the two types of forces coexist:
P.E.K.E.rfrF-K.E.rf
K.E.rfF( K.E.rF
ΔΔdd)c(Δd
Δd)c
Δd
If there is no dissipative force, K.E. + P.E. = 0, i.e. conservation of mechanical energy.
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13
ExampleThe rod is released at rest from = 0, find :(a) velocity of m when the rod arrives at the horizontal position.(b) the max velocity of m.(c) the max. value of .
2m
r
m
r
(a) At = 45o, v = 0.865 (gr)1/2
P.E.=-2mgr sin+mg(r-r cos )K.E. = (2m+m)v2/2(P.E.+ K.E.) = 03mv2/2 – mgr(2 sin + cos -1) = 0v = [2gr(2 sin + cos -1)/3]1/2
2/1
max
o2
)(908.0
63.4 θor 2 θ tan 0θsinθcos23
2/
dθ
d (b)
grv
grv
omaxmaxmax
maxmax2
maxmax2
maxmax
9.126θor0θ ,6.0or18.02.0θcos 03θcos2θcos5
θcos1θcos12 01θ cos θsin 2
0 when valuemaximum has θ (c)
v
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14
xx
Vd
d
d
x
VFx
d
d
BA
rd
From definition of potential energy: dV(x) = -Fdx
From the concept of differential dV =
mgh
mghF
mghhV
kxx
kxF
kxxV
rmMGrr
mMGr
mMG
rr
rVF
r
mMGrV
d
d)( with ground on the reference with force nalGravitatio
d
2/d2/)( with spring a of force Restoring
/)1
(d
d)(
d
d
d
)(d
)( with force nalGravitatio
Examples
2
2
2
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15
vmG With defined as the linear momentum
Gdt
Gdvm
dt
dvmF
)(
Linear momentum
When (i) the total (external) force is zero, or (ii) the collision time t1 t2 is extremely short.
1. In a motion, linear momentum can be conserved,
0dor 0,2
1
t
ttF G
2. Define impulse = change in linear momentum:
1221 GGGddtF
F
time
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momentum)linear ofion (conservat 0 ,- Since
)(
ofmoment linear in change
ofmoment linear in change
0 force External
2
1
2
1
2
1
GFF
dtFFGGG
dtFGB
dtFGA
BA
t
tBABA
t
tBB
t
tAA
A B
RF aF
Collision between systems A and B.
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Angular Momentum
Take moment about O
Angular momentum about O is :
oo HmrvGrvrmH ˆ sin
mr
v
v
rv
O
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Torque = Moment of force about O is defined as :
)d
d to(analogous
d
d
d
)d(
d
d
o
o
t
GF
t
Ht
vmrt
vmr
amrFrM