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8/7/2019 Dynamics Lecture4
http://slidepdf.com/reader/full/dynamics-lecture4 1/67
ME101: Engineering Mechanics
Dynamics: Lecture 4
14th March 2011
Kinematics of Particles: Energy and Momentum Methods
8/7/2019 Dynamics Lecture4
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Introduction
• Previously, problems dealing with the motion of particles were
solved through the fundamental equation of motion,
Current chapter introduces two additional methods of analysis.
.amF
=
• Method of work and energy: directly relates force, mass,
velocity and displacement.
• Method of impulse and momentum: directly relates force,
mass, velocity, and time.
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Work of a Force
• Differential vector is the particle displacement .r d
• Work of the force is
dsF
r d F dU
=
=
•=
α cos
z y x
• Work is a scalar quantity, i.e., it has magnitude and
sign but not direction.
force.lengthו Dimensions of work are Units are( ) ( )( ) J1.356lb1ftm1N1J1 =⋅= joule
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Work of a Force
• Work of a force during a finite displacement,
( ) ∫∫
∫
==
•=→
22
2
1
cos
21
s
t
s
A
A
dsF dsF
r d F U
α
( )∫ ++=2
1
11 A
A
z y x
ss
dzF dyF dxF
• Work is represented by the area under thecurve of F t plotted against s.
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Work of a Force
• Work of a constant force in rectilinear motion,
( ) xF U ∆=→ α cos21
• Work of the force of gravity,
dyW
dzF dyF dxF dU z y x
−=
++=
( ) yW y yW
dyW U
y
y
∆−=−−=
−= ∫→
12
212
1
• Work of the weight is equal to product of weight W and vertical displacement ∆ y.
• Work of the weight is positive when ∆ y < 0,
i.e., when the weight moves down.
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Work of a Force
• Magnitude of the force exerted by a spring is
proportional to deflection,
( )lb/in.orN/mconstantspring=
=
k
kxF
• Work of the force exerted by spring,
dxkxdxF dU
x
−=−=
2221
212121
1
kxkxdxkxU x
−=−= ∫→
• Work of the force exerted by spring is positive
when x2 < x1, i.e., when the spring is returning to
its undeformed position.
• Work of the force exerted by the spring is equal to
negative of area under curve of F plotted against x,
( ) xF F U ∆+−=→ 212
121
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Work of a Force
Work of a gravitational force (assume particle M
occupies fixed position O while particle m follows path
shown),
2dr
r
MmGFdr dU −=−=
12221
2
1r
MmGr
MmGdr r
MmGU r
−=−= ∫→
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Work of a Force
Forces which do not do work (ds = 0 or cos α = 0):
• reaction at frictionless surface when body in contact
• reaction at frictionless pin supporting rotating body,
• weight of a body when its center of gravity moves
horizontally.
• reaction at a roller moving along its track, and
,
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Particle Kinetic Energy: Principle of Work & Energy
dvmvdsF
ds
dv
mvdt
ds
ds
dv
m
dt
dvmmaF
t
t t
=
==
==
• Consider a particle of mass m acted upon by force F
• Integrating from A1 to A2 ,
energykineticmvT T T U
mvmvdvvmdsF vs
t
==−=
−==
→
∫∫
221
1221
2121222
1
2
1
2
1
• The work of the force is equal to the change inkinetic energy of the particle.
F
• Units of work and kinetic energy are the same:
JmNms
mkg
s
mkg
2
22
2
1 =⋅=
=
== mvT
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Applications of the Principle of Work and Energy
• Wish to determine velocity of pendulum bob
at A2. Consider work & kinetic energy.
• Force acts normal to path and does no
work.
P
vW
Wl
T U T
10
2
2211
=+
=+ →
glv
g
22 =
• Velocity found without determining
expression for acceleration and integrating.
• All quantities are scalars and can be added
directly.
• Forces which do no work are eliminated from
the problem.
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Applications of the Principle of Work and Energy
• Principle of work and energy cannot be
applied to directly determine the acceleration
of the pendulum bob.
• Calculating the tension in the cord requires
supplementing the method of work and energy
with an application of Newton’s second law.
• As the bob passes through A2 ,
W l
gl
g
W W P
l
v
g
W W P
amF nn
32
22
=+=
=−
=∑
glv 22 =
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Power and Efficiency
• rate at which work is done.
vF
dt
r d F
dt
dU
Power
•=
•==
=
• Dimensions of power are work/time or force*velocity.
W746s
lbft550hp1or
s
mN1
s
J1(watt)W1 =
⋅=⋅==
•
inputpower
outputpower
input work
k output wor
efficiency
=
=
=η
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Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
• Determine the distance required for thework to equal the kinetic energy change.
An automobile weighing 1000 kg is
driven down a 5o incline at a speed of
72 km/h when the brakes are applied
causing a constant total breaking force
of 5000-N.
Determine the distance traveled by the
automobile as it comes to a stop.
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Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
( )( ) j000,20m/s20kg1000
sm20
s3600
1h
1km
m0001
h
km72
2
212
121
1
1
===
=
=
mvT
v
00 == T v
04145000,200
2211
=−
=+ →
x
T U T
m25.48= x
• Determine the distance required for the work to
equal the kinetic energy change.
( )( )( )
x
x xU
4145
5sinm/s81.9kg10005000 2
21
−=
°+−=→
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Sample Problem 13.2
SOLUTION:
• Apply the principle of work and
energy separately to blocks A and B.
• When the two relations are combined,
the work of the cable forces cancel.
Solve for the velocit .
Two blocks are joined by an inextensible
cable as shown. If the system is released
from rest, determine the velocity of block
A after it has moved 2 m. Assume that the
coefficient of friction between block Aand the plane is µ k = 0.25 and that the
pulley is weightless and frictionless.
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Sample Problem 13.2
SOLUTION:
• Apply the principle of work and energy separately
to blocks A and B.
( )( )
2211
2
:
N490N196225.0
N1962sm81.9kg200
T U T
W N F
W
Ak Ak A
A
=+
======
→
µ µ
( ) ( )( ) ( )( ) ( ) 2
21
2
1
kg200m2N490m2
m2m20
vF
vmF F
C
A AC
=−
=−+
( )
( ) ( )
( ) ( )( ) ( ) 2
2
1
221
2211
2
kg300m2N2940m2
m2m20
:
N2940sm81.9kg300
vF
vmW F
T U T
W
c
B Bc
B
=+−
=+−
=+
==
→
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• When the two relations are combined, the work of the
cable forces cancel. Solve for the velocity.
( ) ( )( ) ( ) 2
21 kg200m2N490m2 vF C =−
( ) ( )( ) ( ) 221 kg300m2N2940m2 vF c =+−
Sample Problem 13.2
( )( ) ( )( ) ( )
( ) 221
221
kg500J4900
kg300kg200m2N490m2N2940
v
v
=
+=−
sm43.4=v
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Sample Problem 13.3
A spring is used to stop a 60 kg package
which is sliding on a horizontal surface.
SOLUTION:
• Apply the principle of work and energy
between the initial position and the
point at which the spring is fullycompressed and the velocity is zero.
The only unknown in the relation is the
friction coefficient.e spr ng as a cons an = m
and is held by cables so that it is initially
compressed 120 mm. The package has a
velocity of 2.5 m/s in the position shown
and the maximum deflection of the spring
is 40 mm.
Determine (a) the coefficient of kinetic
friction between the package and surface
and (b) the velocity of the package as it
passes again through the position shown.
• Apply the principle of work and energyfor the rebound of the package. The
only unknown in the relation is the
velocity at the final position.
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Sample Problem 13.3
SOLUTION:
• Apply principle of work and energy between initial
position and the point at which spring is fully compressed.
( )( ) 0J5.187sm5.2kg60 22
2
121
2
11 ==== T mvT
( )
( )( )( ) ( ) k k
k f xW U
µ µ
µ
J377m640.0sm81.9kg602
21
−=−=
−=→
( )( )( ) ( )( )
( ) ( )
( )( ) J0.112m040.0N3200N2400
N3200m160.0mkN20N2400m120.0mkN20
2
1
maxmin21
21
0max
0min
−=+−=
∆+−=
==∆+====
→ xPPU
x xk PkxP
e
( ) ( ) ( ) J112J377212121 −−=+= →→→ k e f U U U µ
( ) 0J112J377-J5.187
:2211
=−
=+ →
k
T U T
µ 20.0=
k µ
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Sample Problem 13.3
• Apply the principle of work and energy for the rebound
of the package.
( )232
1232
132 kg600 vmvT T ===
( ) ( ) ( )
J36.5
J112J377323232
+=
+−=+= →→→ k e f U U U µ
( ) 232
1
3322
kg60J5.360
:
v
T U T
=+
=+ →
sm103.13 =v
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Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to
determine velocity at point 2.
• Apply Newton’s second law to find
normal force by the track at point 2.-
1 and moves without friction down thetrack shown.
Determine:
a) the force exerted by the track onthe car at point 2, and
b) the minimum safe value of the
radius of curvature at point 3.
• Apply principle of work and energy todetermine velocity at point 3.
• Apply Newton’s second law to find
minimum radius of curvature at point 3
such that a positive normal force isexerted by the track.
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Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to determine
velocity at point 2.
( )
( )2
1m120:
)m12(mgm120
2
22211
21
2
22
1
21
=+=+
=+=
==
→
→
vW
W T U T
W U mvT T
( ) ( )( ) sm34.15m81.9m242122 222
2 === vsgv
• Apply Newton’s second law to find normal force by
the track at point 2.
:nn amF =↑+ ∑
W N
W g
mmv
am N W n
5
4mg46
242
2
=
=====+− ρ
kN05.49= N
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Sample Problem 13.4
• Apply principle of work and energy to determine
velocity at point 3.
( )
( )( ) sm13.12sm81.9m1515
2
1m5.70
3
22
3
2
33311
===
=+=+ →
vgv
mvmgT U T
• Apply Newton’s second law to find minimum radius of
curva ure a po n suc a a pos ve norma orce s
exerted by the track.
:nn amF =↓+ ∑
g
g
g
v
vmmg
amW n
152
3
23
==
=
=
ρ
ρ
m15= ρ
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Sample Problem 13.5
SOLUTION:
Force exerted by the motor
cable has same direction as
the dumbwaiter velocity.Power delivered by motor is
equal to Fv D , v D = 2.5 m/s.
The dumbwaiter D and its load have acombined weight of 300 kg, while the
counterweight C weighs 400 kg.
Determine the power delivered by the
electric motor M when the dumbwaiter(a) is moving up at a constant speed of
2.5 m/s and (b) has an instantaneous
velocity of 2.5 m/s and an acceleration of
1 m/s2, both directed upwards.
• In the first case, bodies are in uniformmotion. Determine force exerted by
motor cable from conditions for static
equilibrium.
• In the second case, both bodies areaccelerating. Apply Newton’s
second law to each body to
determine the required motor cable
force.
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Sample Problem 13.5
• In the first case, bodies are in uniform motion.
Determine force exerted by motor cable from
conditions for static equilibrium.
Free-body C:
( )( )
W2450
m/s5.2N819
=
== DFvPower
y −
Free-body D:
:0=↑+ ∑ yF
N981g100g200g300g300
0g300
==−=−=
=−+
T F
T F
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Sample Problem 13.5
• In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the required
motor cable force.
↓=−=↑= 2
2
12 sm5.0sm1 DC D
aaa
Free-body C: ( )
5.040081.9400
5.04002400
−
=− T
C C y
2
==
Free-body D:
: D D y amF =↑+ ∑ ( )
( ) N138130081.93001862
1300300
==−+
=−+
F F
T F
( )( ) W3452sm5.2N3811 === DFvPower
W3450=Power
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Potential Energy
2121 yW yW U −=→
• Work of the force of gravity ,W
• Work is independent of path followed; dependsonly on the initial and final values of Wy.
=WyV g
to force of gravity.
2121 gg V V U −=→
• Units of work and potential energy are the same:
JmN =⋅==WyV g
• Choice of datum from which the elevation y is
measured is arbitrary.
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Potential Energy
• Previous expression for potential energy of a body
with respect to gravity is only valid when the
weight of the body can be assumed constant.
• For a space vehicle, the variation of the force of
gravity with distance from the center of the earth
should be considered.
• Work of a gravitational force,
1221
r
GMm
r
GMmU −=→
• Potential energy V g when the variation in the
force of gravity can not be neglected,
r
WR
r
GMmV g
2
−=−=
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Potential Energy
• Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,
222
1212
121 kxkxU −=→
• The potential energy of the body with respect
,
( ) ( )2121
221
ee
e
V V U
kxV
−=
=
→
• Note that the preceding expression for V e isvalid only if the deflection of the spring is
measured from its un-deformed position.
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Conservative Forces
• Concept of potential energy can be applied if the
work of the force is independent of the path
followed by its point of application.
( ) ( )22211121 ,,,, z y xV z y xV U −=→
Such forces are described as conservative forces.
• For any conservative force applied on a closed path,
0=• r d F
• Elementary work corresponding to displacementbetween two neighboring points,
( ) ( )
( ) z y xdV
dz zdy ydx xV z y xV dU
,,
,,,,
−=
+++−=
V z
V
y
V
x
V F
dz z
V dy
y
V dx
x
V dzF dyF dxF z y x
grad−=
∂
∂+
∂
∂+
∂
∂−=
∂
∂+∂
∂+∂
∂−=++
C i f
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Conservation of Energy
• Work of a conservative force,
2121 V V U −=→
• Concept of work and energy,
1221 T T U −=
→
• Follows that
2211 +=+ V T V T
• When a particle moves under the action of
conservative forces, the total mechanical
energy is constant.
W V T
W V T
=+
==
11
11 0
( )
W V T
V W gg
W mvT
=+
====
22
2
2
22
1
2
022
1• Friction forces are not conservative. Total
mechanical energy of a system involving
friction decreases.
• Mechanical energy is dissipated by friction
into thermal energy. Total energy is constant.
M i U d C i C l F
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Motion Under a Conservative Central Force
• When a particle moves under a conservative central
force, both the principle of conservation of angular
momentum
and the principle of conservation of energy
φ φ sinsin000
rmvmvr =
V T V T +=+ 00
may be applied.
r
m
mvr
m
mv−=− 2
2
1
0
2
02
1
• Given r , the equations may be solved for v and ϕ.
• At minimum and maximum r , ϕ = 90o. Given the
launch conditions, the equations may be solved for
r min , r max , vmin, and vmax.
S l P bl 13 6
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Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of
energy between positions 1 and 2.
• The elastic and gravitational potential
energies at 1 and 2 are evaluated from
the given information. The initial kinetic
A 1.5-kg collar slides without friction
along a circular rod in horizontal plane.
The spring attached to the collar has an
un-deformed length of 150 mm. and a
constant of K 5400 N/m.
If the collar is released from
equilibrium A, determine its velocity
(a) as it passes through B and C .
energy is zero.
• Solve for the kinetic energy and velocity
at 2.
S l P bl 13 6
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Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of energy between
positions 1 and 2.
Position 1: ( ) ( )
( ) ( )( )
J125.15
m275.0N/m400
mm275.0mm275
mm150250mmmm175
2
212
21
=
=∆=
==
−+=−∆
A
AD A
O AD
V
Lk V
L L
Position 2: ( )
( )
( )
( )( )
j125.6
.175m000N/m4
2
1)(
m175.0mm175m501mm523
mm325mm125mm003
75.0kg
2
.
22
21
2 / 122
222
21
=
=∆=
==+==∆
=+=
=
==
BD B
BD BD
BD
B B B B
k V
L
L
vvmvT
Conservation of Energy:( )
( )
222
2
/sm00.12
75.0
125.6125.15
125.675.0125.150
=
−=
+=+=+
B
B B A A
v
vT V T
sm46.3= Bv
S 13
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Sample Problem 13.7
SOLUTION:
• Since the pellet must remain in contact
with the loop, the force exerted on the
pellet must be greater than or equal to
zero. Setting the force exerted by the
loop to zero, solve for the minimum
velocity at D.
The 250 g pellet is pushed against the
spring and released from rest at A.
Neglecting friction, determine the
smallest deflection of the spring forwhich the pellet will travel around the
loop and remain in contact with the
loop at all times.
• Apply the principle of conservation of
energy between points A and D. Solve
for the spring deflection required to
produce the required velocity and
kinetic energy at D.
Sample Problem 13 7
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Sample Problem 13.7
SOLUTION:
• Setting the force exerted by the loop to zero, solve for the
minimum velocity at D.
:nn maF =↓+ ∑
( )( ) 222
2
sm905.4sm81.9m.50 ===
==
rgv
r vmmgmaW
D
Dn
• Apply the principle of conservation of energy between
.
( )0
300mN6000
1
22212
21
1
=
==+=+=
T
x xkxV V V ge
( )( )( )
( )( ) j613.0sm905.4kg25.021
j45.2m1sm81.9kg25.00
22221
2
2
2
=×==
==+=+=
D
ge
mvT
mgyV V V
j613.0 j45.230002
2211
+=+
+=+
x
V T V T
mm101m101.0==
x
Sample Problem 13 9
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Sample Problem 13.9
SOLUTION:
• For motion under a conservative central
force, the principles of conservation of
energy and conservation of angular
momentum may be applied simultaneously.
• Apply the principles to the points of
minimum and maximum altitude to
parallel to the surface of the earthwith a velocity of 36900 km/h from
an altitude of 500 km.
Determine (a) the maximum altitude
reached by the satellite, and (b) themaximum allowable error in the
direction of launching if the satellite
is to come no closer than 200 km to
the surface of the earth
determine the maximum altitude.
• Apply the principles to the orbit insertion
point and the point of minimum altitude to
determine maximum allowable orbit
insertion angle error.
Sample Problem 13 9
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Sample Problem 13.9
• Apply the principles of conservation of energy and
conservation of angular momentum to the points of minimum
and maximum altitude to determine the maximum altitude.
Conservation of energy:
12121
02021 r
GMmmv
r
GMmmvV T V T A A A A −=−+=+ ′′
Conservation of angular momentum:0
011100r
vvmvr mvr ==1r
Combining,
2001
0
1
0
02
1
202
021 2
111vr
GM
r
r
r
r
r
GM
r
r v =+
−=
−
( )( ) 23122622
60
0
sm10398m1037.6sm81.9
sm1025.10hkm36900
km6870km500km6370
×=×==
×==
=+=
gRGM
v
r
km60400m104.606
1=×=r
Sample Problem 13 9
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Sample Problem 13.9
• Apply the principles to the orbit insertion point and the point
of minimum altitude to determine maximum allowable orbit
insertion angle error.
Conservation of energy:
min
2max21
0
202100
r
GMmmvr
GMmmvV T V T A A −=−+=+
Conservation of angular momentum:0r
minmaxmaxm n
r
Combining and solving for sin ϕ 0,
°±°=
=
5.1190
9801.0sin
0
0
ϕ
φ
°±= 5.11errorallowable
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Impulsive Motion
• Force acting on a particle during a very short
time interval that is large enough to cause a
significant change in momentum is called an
impulsive force.
• When impulsive forces act on a particle,
vmt F vm
=∆+
• When a baseball is struck by a bat, contact
occurs over a short time interval but force is
large enough to change sense of ball motion.
• Non-impulsive forces are forces for whichis small and therefore, may be
neglected.t F ∆
Sample Problem 13.10
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Sample Problem 13.10
SOLUTION:
• Apply the principle of impulse and
momentum. The impulse is equal to the
product of the constant forces and thetime interval.
driven down a 5o
incline at a speed of 100 km/h when the brakes are applied,
causing a constant total braking force of
7000 N.
Determine the time required for theautomobile to come to a stop.
Sample Problem 13 10
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Sample Problem 13.10
SOLUTION:
• Apply the principle of impulse and
momentum.
2211vmvm
=+∑ →Imp
Taking components parallel to the incline,
( )
( )( ) ( ) ( ) 0N7000N5sin17660m/s7.782kg1800
m78.27h1
s3600
km1
m1000100km/h100
05sin1
=−°+
=××=
=−°+
t t
Ft t mgmv
s16.9=t
Sample Problem 13.11
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p
SOLUTION:
• Apply the principle of impulse and
momentum in terms of horizontal and
vertical component equations.
A 120 kg baseball is pitched with avelocity of 24 m/s. After the ball is hit
by the bat, it has a velocity of 36 m/s in
the direction shown. If the bat and ball
are in contact for 0.015 s, determine theaverage impulsive force exerted on the
ball during the impact.
Sample Problem 13.11
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p
SOLUTION:
• Apply the principle of impulse and momentum in
terms of horizontal and vertical component equations.
2211 vmvm
=+ →Imp
x component equation:
40cos6m/s3k 12.0s015.0m/s24k 12.0
40cos21
°=+−
°=∆+− x
F
mvt F mv
N6.412=
xF
y component equation:
( ) ( )( )N1.185
40sin6m/s3kg12.0015.0
40sin0 2
=°=
°=∆+
y
y
y
F
F
mvt F
N452F =
x
y
Sample Problem 13.12
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p
SOLUTION:
• Apply the principle of impulse and
momentum to the package-cart system
to determine the final velocity.• Apply the same principle to the package
alone to determine the impulse exerted
on it from the change in its momentum.g pac age rops rom a c u e
into a 24 kg cart with a velocity of 3m/s. Knowing that the cart is initially at
rest and can roll freely, determine (a)
the final velocity of the cart, (b) the
impulse exerted by the cart on the
package, and (c) the fraction of the
initial energy lost in the impact.
Sample Problem 13.12
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Sample Problem 13.12
SOLUTION:
• Apply the principle of impulse and momentum to the package-cart
system to determine the final velocity.
x
y
x components:
( )( ) ( ) 2
21
kg25kg1030cosm/s3kg10
030cos
v
vmmvm c p p
+=°
+=+°
m/s742.02 =v
2211 vmmvm c p p
+=+∑ →Imp
Sample Problem 13.12
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p
• Apply the same principle to the package alone to determine the impulse
exerted on it from the change in its momentum.
x
y
x components:
( )( ) ( ) 2
21
kg1030cosm/s3kg10
30cos
vt F
vmt F vm
x
p x p
=∆+°
=∆+°
sN56.18 ⋅−=∆t F x
( ) ( ) sN9.23sN51sN56.1821 ⋅=∆⋅+⋅−=∆=∑ → t F jit F
Imp
2211 p p =→
y components:
( )( ) 030sinm/s3kg10
030sin1
=∆+°−
=∆+°−
t F
t F vm
y
y p
sN15 ⋅=∆t F y
Sample Problem 13.12
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To determine the fraction of energy lost,
( )( )( ) ( )( ) J63.9sm742.0kg25kg10
J45sm3kg102
212
221
1
2
212
121
1
=+=+=
===
vmmT
vmT
c p
p
786.0J45
J9.63J45
1
21 =−
=−
T
T T
Impact
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p
• Impact: Collision between two bodies which
occurs during a small time interval and during
which the bodies exert large forces on each other.
• Line of Impact: Common normal to the surfacesin contact during impact.
• Central Impact: Impact for which the mass
centers o t e two o es e on t e ne o mpact;
otherwise, it is an eccentric impact..
rec en ra mpac
• Direct Impact: Impact for which the velocities of
the two bodies are directed along the line of
impact.
Oblique Central Impact
• Oblique Impact: Impact for which one or both of
the bodies move along a line other than the line of
impact.
Direct Central Impact
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• Bodies moving in the same straight line,
v A > v B .
• Upon impact the bodies undergo a
period of deformation, at the end of which,
they are in contact and moving at a
common velocity.
• A period of restitution follows during
which the bodies either regain their
original shape or remain permanentlydeformed.
• Wish to determine the final velocities of the
two bodies. The total momentum of the
two body system is preserved,
B B B B B B A A vmvmvmvm ′+′=+
• A second relation between the final
velocities is required.
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Oblique Central Impact
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• Final velocities are
unknown in magnitude
and direction. Four
equations are required.
• No tangential impulse component; ( ) ( ) ( ) ( ) B B A A vvvv ′=′=tangential component of momentum
for each particle is conserved.
• Normal component of total
momentum of the two particles is
conserved.
( ) ( ) ( ) ( )n B Bn A An B Bn A A vmvmvmvm ′+′=+
• Normal components of relative
velocities before and after impact
are related by the coefficient of
restitution.
( ) ( ) ( ) ( )n Bn An An B vvevv −=′−′
Oblique Central Impact
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• Block constrained to move along horizontal
surface.
• Impulses from internal forces
along the n axis and from external force
exerted by horizontal surface and directed
F F − and
ext F
along the vertical to the surface.
• Final velocity of ball unknown in direction and
magnitude and unknown final block velocity
magnitude. Three equations required.
Oblique Central Impact
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• Tangential momentum of ball is ( ) ( )t Bt B vv ′=
.
• Total horizontal momentum of block and ball is conserved.
( ) ( ) ( ) ( ) x B B A A x B B A A vmvmvmvm ′+′=+
• Normal component of relative
velocities of block and ball are related
by coefficient of restitution.
( ) ( ) ( ) ( )n Bn An An B vvevv −=′−′
• Note: Validity of last expression does not follow from previous relation for
the coefficient of restitution. A similar but separate derivation is required.
Problems Involving Energy and Momentum
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• Three methods for the analysis of kinetics problems:
- Direct application of Newton’s second law
- Method of work and energy
- Method of impulse and momentum
• Select the method best suited for the problem or part of a problem
un er cons erat on.
Sample Problem 13.14
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SOLUTION:
• Resolve ball velocity into components
normal and tangential to wall.
• Impulse exerted by the wall is normalto the wall. Component of ball
momentum tangential to wall is
conserved.
A ball is thrown against a frictionless,vertical wall. Immediately before the
ball strikes the wall, its velocity has a
magnitude v and forms angle of 30o
with the horizontal. Knowing thate = 0.90, determine the magnitude and
direction of the velocity of the ball as
it rebounds from the wall.
• Assume that the wall has infinite massso that wall velocity before and after
impact is zero. Apply coefficient of
restitution relation to find change in
normal relative velocity between wall
and ball, i.e., the normal ball velocity.
Sample Problem 13.14
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SOLUTION:• Resolve ball velocity into components parallel and
perpendicular to wall.
vvvvvv t n 500.030sin866.030cos =°==°=
• Component of ball momentum tangential to wall is conserved.
vvv t t 500.0==′
n
°=
=′
+−=′
− 7.32500.0
779.0tan926.0
500.0779.0
1vv
vvv t n λ λ
• Apply coefficient of restitution relation with zero wallvelocity.
( )
( ) vvv
vev
n
nn
779.0866.09.0
00
−=−=′
−=′−
Sample Problem 13.15
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SOLUTION:
• Resolve the ball velocities into components
normal and tangential to the contact plane.
• Tangential component of momentum for
each ball is conserved.
velocities of two identicalfrictionless balls before they strike
each other are as shown. Assuming
e = 0.9, determine the magnitude
and direction of the velocity of each
ball after the impact.
of the two ball system is conserved.
• The normal relative velocities of the
balls are related by the coefficient of
restitution.
• Solve the last two equations simultaneously
for the normal velocities of the balls after
the impact.
Sample Problem 13.15
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• Tangential component of momentum for each ball is
SOLUTION:
• Resolve the ball velocities into components normal and
tangential to the contact plane.
( ) sm79.730cos =°= An A vv ( ) sm5.430sin =°= At A vv
( ) sm660cos −=°−= Bn B vv ( ) sm39.1060sin =°= Bt Bvv
conserved.
( ) ( ) sm5.4==′ t At A vv ( ) ( ) sm39.10==′ t Bt B vv
• Total normal component of the momentum of the two
ball system is conserved.
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( ) 79.1
679.7
=′+′
′+′=−+′+′=+
n Bn A
n Bn A
n B Bn A An B Bn A A
vv
vmvmmmvmvmvmvm
Sample Problem 13.15
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• The normal relative velocities of the balls are related by the
coefficient of restitution.
( ) ( ) ( ) ( )[ ]( )[ ] 41.12679.790.0 =−−=
−=′−′n Bn An Bn A vvevv
• Solve the last two equations simultaneously for the normal
velocities of the balls after the impact.
′ =′
°=
=′
+=′
°=
=′
+−=′
−
−
6.551.7
39.10tansm58.12
39.101.7
3.4031.5
5.4tansm1.7
5.431.5
1
1
B
nt B
A
nt A
v
v
v
v
λ λ
λ λ
t
n
.−n A
.n B
Sample Problem 13.16
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SOLUTION:
• Determine orientation of impact line of
action.
• The momentum component of ball Atangential to the contact plane is
conserved.
Ball B is hanging from an inextensible
cord. An identical ball A is released
from rest when it is just touching the
cord and acquires a velocity v0
before
striking ball B. Assuming perfectly
elastic impact (e = 1) and no friction,
determine the velocity of each ball
immediately after impact.
• e o a or zon a momen um o e
two ball system is conserved.
• The relative velocities along the line of
action before and after the impact are
related by the coefficient of restitution.
• Solve the last two expressions for the
velocity of ball A along the line of action
and the velocity of ball B which is
horizontal.
Sample Problem 13.16
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SOLUTION:• Determine orientation of impact line of action.
°=
==
30
5.02
sin
θ
θ r
r
• The momentum component of ball A
tangential to the contact plane is
conserved.
( )0 030sin vmmv
vmt F vm
t A
A A
′=+°
′=∆+
• The total horizontal ( x component)
momentum of the two ball system is
conserved.
( ) ( )
( ) ( )
( ) 0
0
433.05.0
30sin30cos5.00
30sin30cos0
vvv
vvv
vmvmvm
vmvmt T vm
Bn A
Bn A
Bn At A
B A A
=′+′
′−°′−°=
′−°′−°′=′+′=∆+
( ) 05.0 vvt A =′
Sample Problem 13.16
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• The relative velocities along the line of action beforeand after the impact are related by the coefficient of
restitution.
( ) ( ) ( ) ( )
( )( ) 0
0
866.05.0030cos30sin
vvvvvv
vvevv
n A B
n A B
n Bn An An B
=′−′−°=′−°′
−=′−′
•
A along the line of action and the velocity of ball B
which is horizontal.
( ) 00 693.0520.0 vvvv Bn A =′−=′
←=′
°=°−°=
°=
==′
−=′
−
0
10
00
693.0
1.16301.46
1.465.0
52.0tan721.0
520.05.0
vv
vv
vvv
B
A
nt A
α
β
λ λ
Sample Problem 13.17
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SOLUTION:
• Apply the principle of conservation of
energy to determine the velocity of the
block at the instant of impact.
• Since the impact is perfectly plastic, the
block and pan move together at the same
velocity after im act. Determine that
A 30 kg block is dropped from a height
of 2 m onto the the 10 kg pan of a
spring scale. Assuming the impact to be
perfectly plastic, determine themaximum deflection of the pan. The
constant of the spring is k = 20 kN/m.
velocity from the requirement that the
total momentum of the block and pan isconserved.
• Apply the principle of conservation of
energy to determine the maximum
deflection of the spring.
Sample Problem 13.17
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SOLUTION:• Apply principle of conservation of energy to
determine velocity of the block at instant of impact.
( )( )( )
( ) ( )( )
030
J588281.9300
21
2211
2222
1222
12
11
+=+
===
====
A A A
A
V T V T
V vvmT
yW V T
.222
• Determine velocity after impact from requirement that
total momentum of the block and pan is conserved.
( ) ( ) ( )
( )( ) ( ) sm70.41030026.630 33
322
=+=+
+=+
vv
vmmvmvm B A B B A A
Sample Problem 13.17
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• Apply the principle of conservation of energy todetermine the maximum deflection of the spring.
( ) ( )( )
( )( )4
233
212
321
3
2
212
321
3
0
J241.01091.410200
J4427.41030
T
kx
V V V
vmmT
eg
B A
=
=××=+=
+=
=+=+=
−
Initial spring deflection due to
pan weight:
( )( )m1091.4
1020
81.910 3
33−×=
×==
k
W x B
( ) ( )( ) ( ) 2
43
213
4
24
321
34
424
10201091.4392
1020392
x x
x x x
kxhW W V V V B Aeg
×+×−−=
×+−−=
+−+=+=
−
( ) ( )m230.0
10201091.43920241.0442
4
24
3
213
4
4433
=
×+×−−=+
+=+
−
x
x x
V T V T
m1091.4m230.03
34−×−=−= x xh m225.0=h