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This lecture note is taken and modified from Engineering Mechanics -Dynamics, R.C.Hibbeler, Prentice Hall whichcopyright belongs to Pearson Education South Asia Pte Ltd.
Trimester 2 2014/2015
Chapter 4Kinematics of a Particle:
Impulse and Momentum
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Chapter Objectives
Principle of linear impulse and momentum for a particle
Conservation of linear momentum for particles
Analyze the mechanics of impact
Concept of angular impulse and momentum
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Chapter Outline
1. Principle of Linear Impulse and Momentum
2. Principle of Linear Impulse and Momentum for a System of
Particles
3. Conservation of Linear Impulse for a System of Particles4. Impact
5. Angular Momentum
6. Relation between Momentum of a Force and Angular
Momentum7. Principle of Angular Impulse and Momentum
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15.1 Principle of Linear Impulse and
Momentum
Equation of motion for a particle of mass m isF= ma= mdv/dt
Rearranging the terms and integrating between the
limits v= v1at t = t1and v= v2at t = t2
Referred to as the
principle of linear impulseand momentum
12
2
1
2
1
2
1
vvFvF mmdtdmdtt
t
v
v
t
t
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15.1 Principle of Linear Impulse and
Momentum
L inear Momentum Vectors of the form L = mvis called
l inear momentum
Magnitude mv has unit of mass-velocity, kg.m/s
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15.1 Principle of Linear Impulse and
Momentum
L inear Impulse is called linear impulse, and measure the
effect of a force during the time the force acts
The impulse acts in the same direction as the force,
magnitude has unit of force-time, N.s
dtFI
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15.1 Principle of Linear Impulse and
Momentum
Principle of L inear Impulse and Momentum The equation is rewritten in the form
Principle of l inear impulse and momentumin itsx, y,
zcomponents is
212
1
vFv mdtmt
t
21
21
21
)()(
)()(
)()(
2
1
2
1
2
1
z
t
t
zz
y
t
t yy
x
t
t xx
vmdtFvm
vmdtFvm
vmdtFvm
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15.1 Principle of Linear Impulse and
Momentum
Procedures for AnalysisFree-Body Diagram
Establish thex, y, zinertial frame of reference and
draw FBD
Establish direction and sense of initial and final
velocities
Assume the sense of vector components in the
direction of the positive inertial coordinates
Draw the impulse and momentum diagrams for the
particle
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15.1 Principle of Linear Impulse and
Momentum
Procedures for AnalysisPrinciple of Impulse and Momentum
Apply the principle of linear impulse andmomentum,
If the motion occurs in thex-yplane, itcan resolve the vector components of F
Every force acting on the particles FBDcan create an impulse
The impulse is equal to the area under theforce-time curve
21 21
mvdtFmvt
t
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The 100-kg stone is originally at rest on the smoothhorizontally surface. If a towing force of 200 N, acting at
an angle of 45, is applied to the stone for 10 s, determinethe final velocity and the normal force which the surface
exerts on the stone during the time interval.
Example 15.1
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SolutionFree-Body Diagram
Since all forces acting are constant, the impulses are
I= Fc(t2t1)
Principle of Impulse and Momentum
Resolving the vectors along thex, y, zaxes,
Example 15.1
m/s1.14)100(45cos)10(2000
)()(
22
21
2
1
vv
vmdtFvm xt
t xx
N840045sin)10(200)10(981)10(0
)()( 212
1
CC
y
t
t yy
NN
vmdtFvm
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Block A and B have a mass of 3 kg and 5 kg respectively.If the system is released from rest, determine the velocity
of block B in 6 s.
Example 15.3
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SolutionFree-Body Diagram
Since weight of each block is constant,
the cord tensions will also be constant.
Since mass of pulleyDis neglected,
the cord tension is TA= 2TB.
Example 15.3
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SolutionPrinciple of Impulse and Momentum
BlockA:
BlockB:
Example 15.3
1))(3()6)(81.9(3)6(20
)()(
2
21
2
1
AB
A
t
t yA
vT
vmdtFvm
2))(5()6()6)(81.9(50
)()(
2
21
2
1
BB
B
t
t yB
vT
vmdtFvm
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SolutionKinematics
We have
Taking time derivative yields
When B moves downward A moves upward.
Sub this result into Eq. 1 and solving Eqs. 1 and 2 yields
(vB)2= 35.8 m/s and TB= 19.2 N
Example 15.3
lss BA 2
BA vv 2
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15.2 Principle of Linear Impulse and Momentum for
a System of Particles
Principle of linear impulse and momentum for asystem of moving particles is
Internal forces fiacting between particles do not
appear with this summation,
dt
dm i
ii
vF
21
2
1 ii
t
t iii mdtm vFv
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15.3 Conservation of Linear Momentum
for a System of Particles
When the sum of the external impulsesacting on asystem of particles iszero, the equation is
This is called conservation of linear momentum
The total momentum for a system of particles remains
constant during the time period t1to t2
21 iiii mm vv
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15.3 Conservation of Linear Momentum
for a System of Particles
Internal impulsesfor the system will always cancelout as they occur in equal but opposite collinear pairs
The forces causing negligible impulses are called non-
impulsive forces
Forces that are very large and act for a very short
period of time are called impulsive forces
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15.3 Conservation of Linear Momentum
for a System of Particles
Procedure for AnalysisFree-Body Diagram
Establish thex, y, zinertial frame of reference and
draw the FBD
Apply conservation of linear momentum in a given
direction
Establish the direction and sense of the particles
initial and final velocities.
Draw the impulse and momentum diagrams for each
particle of the system
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15.3 Conservation of Linear Momentum
for a System of Particles
Procedure for AnalysisMomentum Equations
Apply the principle of linear impulse and momentum
or the conservation of linear momentum
Determine the internal impulse F.dt acting on onlyone particle of a system
Average impulsive forceFavgcan be determined from
Favg= F dt/t.
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The 15-Mg boxcar A is coasting at 1.5 m/s on thehorizontal track when it encounters a 12-Mg tank B
coasting at 0.75 m/s toward it. If the cars meet and couple
together, determine
(a) the speed of both cars just after the coupling, and (b)
the average force between them if the coupling takes place
in 0.8 s.
Example 15.4
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SolutionPart (a)
Free-Body Diagram.
Consider bothcars as a single system.
Momentum is conserved in thexdirection since the
coupling force Fis internalto the system.
Example 15.4
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SolutionConservation of L inear Momentum
Part (b)
Average (impulsive) coupling forceFavg, can be
determined by applying the principle of linear momentum
to eitherone of the cars
Example 15.4
smv
v
vmmvmvm BABBAA
/5.0
)27000()75.0)(12000()5.1)(15000(
)()()(
2
2
211( )
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SolutionPart (b)
Conservation of M omentum
Example 15.4
kNF
F
vmdtFvm
avg
avg
AAA
8.18
)5.0)(15000()8.0()5.1)(15000(
)( 21
( )
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The bumper carsA andB each have a mass of 150 kg andare coasting with the velocities shown before they freely
collide head on. If no energy is lost during the collision,
determine their velocities after collision.
Example 15.6
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SolutionFree-Body Diagram
The cars will be considered as a single system.
The free-body diagram is shown.
Conservation of Momentum
Example 15.6
11
150)150(2150)3)(150(
)()()(
22
22
2211
BA
BA
BBAABBAA
vv
vv
vmvmvmvm
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SolutionConservation of Energy
Sub Eq. (1) into (2), we get
Since refer before the collision, hence
and
Example 15.6
213
002
2
2
2
2
22
12
22
12
12
12
12
1
2211
BA
BBAABBAA
vvvmvmvmvm
VTVT
m/s2orm/s2andm/s3 22 BB vv
27
m/s22 Bv
m/s32
B
v m/s22 Av
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An 800-kg rigid pile P is driven into the ground using a300-kk hammer H. the hammer falls from rest at a height
y0= 0.5 m and strikes the top of the pile. Determine the
impulse which the hammer imparts on the pile if the pile
is surrounded entirely by loose sand so that after striking,the hammer does not rebound off the pile.
Example 15.7
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SolutionConservation of Energy
Velocity can be determined using the conservation of
energy equation applied to the hammer.
Example 15.7
m/s13.3)(
0))(300(21)5.0)(81.9(3000
)(2
1)(
2
1
1
2
1
1
2
10
2
0
1100
H
H
HHHHHH
v
v
yWvmyWvm
VTVT
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SolutionFree-Body Diagram
Weight of the hammer and pile and the resistance force Fs
of the sand are all non-impulsive,
Momentum is conserved in the vertical direction during
this short time.
Example 15.7
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SolutionConservation of Momentum
Since the hammer does not rebound off the pile just after
the collision, then (vH)2= (vP)2= v2
Example 15.7
m/s8542.0
8003000)13.3)(300(
)()()(
2
22
2211
v
vv
vmvmvmvm pHppHH
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SolutionPrinciple of Impulse and Momentum
The impulse which the pile imparts to the hammer can
now be determined since v2is known.
Example 15.7
sN683
)854.0)(300()13.3)(300(
)()( 212
1
dtR
dtR
vmdtFvm Ht
t yHH
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15.4 Impact
Impact occurs when two bodies collide and causeimpulsive forces to be exerted between them
1. Central impactoccurs when the direction of motion of
the mass centers of the two colliding particles is along
the line of impact
2. Oblique impactoccurs when one or both of the
particles is at an angle with the line of impact
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15.4 Impact
Central Impact When (vA)1> (vB)1, collision will occur
During collision, particles undergo aperiod of
deformable
Only at maximum deformationwill both particles have
common velocity as their relative motion is zero
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15.4 Impact
Central Impact Afterward aperiod of restitution occurs, particles will
return to their original shape or remain permanently
deformed
Equal but opposite restitution impulseRdtpushesthe particle apart from one another
After separation the particles will have the final
momentum, where (vB)2> (vA)2
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15.4 Impact
Central Impact Ratio of the restitution impulse to the deformation
impulse is called the coefficient of restitution
Can be expressed in terms of the particles initial andfinal velocities,
11
22
)()(
)()(
BA
AB
vv
vve
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15.4 Impact
Coefficient of Restitution ehas a value between zero and one
Elastic Impact (e = 1) Deformation impulse (Pdt) is equal and opposite to
the restitution impulse (Rdt)
Plastic (inelastic) Impact (e = 0)
After collision both particles couple or stick together
and move with a common velocity
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15.4 Impact
Procedure for Analysis (Central Impact) The conservation of momentum applies to the system
of particles, mv1= mv2
Coefficient of restitution relates the relative velocities
of the particles along the line of impact, just before
and just after collision
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15.4 Impact
Oblique Impact When oblique impact occurs, particles move away
from each other with velocities having unknown
directions and unknown magnitudes
Provided the initial velocities are known, four
unknown are present in the problem
Unknown are (vA)2, (vB)2, 2and 2, orxandy
components of the final velocities
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15.4 Impact
Oblique Impact Wheny axis is established within the plane of contact
and thexaxis along the line of impact, the impulsive
forces of deformation and restitution act only in the x
direction
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15.4 Impact
Procedure for Analysis Momentum of the system is conserved along the line
of impact, x axis so that m(vx)1= m(vx)2 The coefficient of restitution, e, relates the relative-
velocity components of the particles along the line ofimpact(xaxis).
Momentum of particleAis conserved along theyaxis,perpendicular to the line of impact, since no impulse
acts on particleAin this direction. Momentum of particleBis conserved along theyaxis,
perpendicular to the line of impact, since no impulseacts on particleBin this direction.
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The bagA, having a mass of 6 kg is released from rest atthe position = 0. After falling to = 90, is strikes an18 kg boxB. If the coefficient of restitution between the
bag and the box is e= 0.5, determine the velocities of the
bag and box just after impact and the loss of energyduring collision.
Example 15.9
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SolutionConservation of Energy
At = 0, we have
Conservation of Momentum
After impact,AandBassume travel to the left
Example 15.9
smv
v
VTVT
A
A
/43.4)(
0))(6(21181.960
1
2
1
1100
2222
2211
)(343.4)()(6))(18()43.4)(6(0
)()()()(
BAAB
AABBAABB
vvvv
vmvmvmvm
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SolutionConservation of Resti tution
Realizing that for separation to occur after collision (vB)2
> (vA)2,
Solving the two equations,
Example 15.9
215.2)()()()(
)()( 22
11
22
BA
BA
AB vvvv
vve
m/s66.1)(
m/s554.0m/s554.0)(
2
2
B
A
v
v
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SolutionLoss of Energy
Applying the principle of work and energy to the bag and
box just before and after collision,
Example 15.9
J
U
TTU
15.33
)33.4)(6(
2
1)544.0)(6(
2
1)66.1)(18(
2
1 22221
1221
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Two smooth disksAandB, having mass of 1 kg and 2 kgrespectively, collide with the velocities shown. If the
coefficient of restitution for the disks is e= 0.75,
determine thexandycomponents of the final velocity of
each disk just after collision.
Example 15.11
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SolutionResolving each of the initial velocities into x and y
components, we have
Example 15.9
smv
smv
smvsmv
By
Bx
Ay
Ax
/707.045sin1)(
/707.045cos1)(
/50.130sin3)(/60.230cos3)(
1
1
1
1
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SolutionConservation of x Momentum
We have
Conservation of (x) Resti tution
We have
Example 15.9
18.1)(2)()(2)(1)707.0(2)60.2(1
)()()()(
2222
2211
BxAxBxAx
BxBAxABxBAxA
vvvv
vmvmvmvm
48.2)()()07.0(60.2
)()(75.0
)()(
)()(
2222
11
22
AxBxAxBx
BxAx
AxBx
vvvv
vv
vve
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SolutionSolving the two simultaneous equations,
Conservation of y Momentum
The momentum of each diskis conservedin the y
direction (plane of contact),
Example 15.9
smv
smsmv
Bx
Ax
/22.1)(
/26.1/26.1)(
2
2
smsmvvmvm
smvvmvm
ByByBByB
AyAyAAyA
/707.0/707.0)()()(
/5.1)()()(
221
221
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15.5 Angular Momentum
Angular momentum, HO, of a particle about point Oisdefined as the moment of the particles linearmomentum about O
Scalar Formulation
If a particle is moving along a curve, the angular
momentum can be determine by using a scalar
formulation
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15.5 Angular Momentum
Scalar Formulation The magnitudeof HOis
dis the moment arm or perpendicular distance from O
to the line of action of mv.
Common for (HO)zis kg.m2/s
The directionof HOis defined by the right-hand rule.
))(()( mvdH zO
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15.5 Angular Momentum
Vector Formulation If the particle is moving along a space curve, the
vector cross product is used to determine the angular
momentumabout O
We have
Angular momentum is determined by evaluating the
determinant
vrH mO
zyx
zyxO
mvmvmv
rrr
kji
H
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15.6 Relation Between Moment of a Force and
Angular Momentum
The moments about Point Oof all forces acting isrelated by
The moments of the forces
about point Ocan be obtained by
Derivation of rx mvcan be written as
Since therefore
vF m
vrFrM mO
vrvrvrH mmmdt
dO )(
0)( rrvr mm
OO HM 53
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15.6 Relation Between Moment of a Force and
Angular Momentum
The equation states that resultant momentum aboutpoint O of all the forces acting on the particle is equal
to the time rate of change of the particles angular
momentum about point O
The result is also similar to
System of Particles
Forces acting on arbitrary ith particle of the system
consist of a resultant external forceFi, and a resultantinternal forcefi
LF
Oiiiii )()()( HfrFr 54
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15.6 Relation Between Moment of a Force and
Angular Momentum
System of Particles Similar equations can be written for each of the other
particles of the system
The second term is sero since the internal forces occur
in equal but opposite collinear pairs, and hence the
moment of each pair about point Ois zero
Oiiiii )()()( HfrFr
OO HM 55
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The box has a mass m and is traveling down the smoothcircular ramp such that when it is at the angle it is a
speed v. Determine its angular momentum about point O
at this instant and the rate of increase in its speed, i.e., at.
Example 15.12
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SolutionSince vis tangent to the path, the angular momentum is
From the FBD, the weight W = mgcontributes a momentabout O
Since rand mare constant,
Example 15.12
rmvHO
)()sin(; rmvdt
drmgHM OO
sinsin gdt
dv
dt
dvrmmgr
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15.7 Relation Between Moment of a Force and
Angular Momentum
Principle of Angular Impulse and Momentum We have MOdt= dHOand integrated, assuming at
time t = t1, HO= (HO)1and time t = t2, HO= (HO)2
This equation is referred to as theprinciple of angular
impulse and momentum
21
12
)()(
)()(
2
1
2
1
O
t
t OO
OO
t
t
O
dt
dt
HMH
HHM
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15.7 Relation Between Moment of a Force and
Angular Momentum
Principle of Angular Impulse and Momentum Since the moment of a force about point Ois MO= r
x F, the angular impulse may be expressed in vector
form as
The principle of angular impulse and momentum for a
system of particles may be written as
2
1
2
1
)(tt
t
t O dtdt FrMAngular impulse
21 )()(2
1O
t
tOO dt HMH
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15.7 Relation Between Moment of a Force and
Angular Momentum
Vector Formulation Using impulse and momentum principles, it is
possible to write which define the particles motion,
212
1
vFv mdtmt
t
21 )()(2
1
O
t
t OO dt HMH
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15.7 Relation Between Moment of a Force and
Angular Momentum
Scalar Formulation If the particle is confined to move in the x-y plane,
three independent equations may be written to express
the motion,
21 )()(2
1x
t
t xx vmdtFvm
21 )()(2
1O
t
t OO HdtMH
21 )()(2
1
y
t
t yy vmdtFvm
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15.7 Relation Between Moment of a Force and
Angular Momentum
Conservation of Angular Momentum When the angular impulse acting on a particle are all
zero during the time t1to t2, it may be written as
It is known as the conservation of angular momentum.
If no external impulse is applied to the particle, both
linear and angular momentum is conserved.
21 )()( OO HH
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5.7 e at o etwee o e t o a o ce a d
Angular Momentum
Conservation of Angular Momentum In some cases, the particles angular momentum will
be conserved and linear momentum may not.
This occurs when the particle is subjected onlyto a
central force.
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Angular Momentum
Procedure for AnalysisFree-Body Diagram
Draw the particles FBD where angular momentum isconserved.
The direction and sense of the particles initial andfinal velocities is established.
Momentum Equations
Apply the principle of angular impulseand momentum,
2121 )()(or)()(2
1OOO
t
t OO dt HHHMH
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The 0.4 kg ballBis attached to a cord which passesthrough a hole atAin a smooth table. When the ball is r1
= 0.5 m from the hole, it is rotating around in a circle such
that its speed is v1= 1.2 m/s. By applying a force Fthe
cord is pulled downward through the hole with a constantspeed vc= 2 m/s.
Example 15.14
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Determine (a) the speed of the ball at the instant it is r2=0.2 m from the hole, and (b) the amount of work done by
Fin shortening the radial distance from r1to r2.
Example 15.14
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SolutionPart (a) Free-Body Diagram
As the ball the cord force Fon the ball passes through the
zaxis.
Weight and NBare parallel and the conservation ofangular momentum applies about thezaxis.
Conservation of Angular Momentum
Example 15.14
m/s3)4.0)(2.0()2.1)(4.0)(5.0( 22
2211
21
vv
vmrvmr BB
HH
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SolutionConservation of Angular Momentum
The speed of the ball is thus
Part (b)
The only force that does work on the ball is F.
Example 15.14
m/s606.3)2()0.3( 222 v
JUU
TUT
FF 313.2)606.3)(4.0(2
1)2.1)(4.0(
2
1 22
2211
68
8/10/2019 Chapter 4 Linear Impulse and Momentum
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