dynamics lecture4

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ME101: Engineering Mechanics

Dynamics: Lecture 4

14th March 2011

Kinematics of Particles: Energy and Momentum Methods



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.



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.length×• Dimensions of work are Units are( ) ( )( ) J1.356lb1ftm1N1J1 =⋅= joule



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.



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.



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



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

−=−= ∫→



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

,



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



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.



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 =



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



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.



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

−=

°+−=→



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.



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

=+−

=+−

=+

==

→



• 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



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.



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



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



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.



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



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



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.



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



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



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.



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

−=−=



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.



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



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



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



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



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



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.


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



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



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



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



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



Impulsive Motion



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




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




SOLUTION:


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




p

SOLUTION:


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.




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




p

SOLUTION:


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.





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




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




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



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



• 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.



• 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 −=′−′




• 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.




• 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



• 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.




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.




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

−=−=′

−=′−




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.




• 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




• 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




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.




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 =′




• 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




SOLUTION:


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.


energy to determine the maximum

deflection of the spring.




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




• 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

dynamics lecture4

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