chapter 17 - islamic university of gazasite.iugaza.edu.ps/mhaiba/files/2012/01/ch17-plane...rigid...

Chapter 17Plane Motion of Rigid

Bodies: Energy and

Momentum Methods

2 - 1

5/12/2012 9:28 AM

Dr. Mohammad Abuhaiba, PE

17 - 2

Introduction

Principle of Work and Energy for a Rigid Body

Work of Forces Acting on a Rigid Body

Kinetic Energy of a Rigid Body in Plane Motion

Systems of Rigid Bodies

Conservation of Energy

Power

Principle of Impulse and Momentum


Conservation of Angular Momentum

Eccentric Impact

Chapter Outline

5/12/2012 9:28 AM


Introduction

17 - 3

• Principle of work and energy is well suited to the solution of

problems involving displacements and velocities.

2211 TUT

• Principle of impulse and momentum is appropriate for problems

involving velocities and time.

2121

2

1

2

1

O

t

tOO

t

t

HdtMHLdtFL

• Problems involving eccentric impact are solved by supplementing

the principle of impulse and momentum with the application of

coefficient of restitution.

5/12/2012 9:28 AM


Principle of

Work and

Energy for a

Rigid Body

17 - 4

• Method of work and energy is well adapted to

problems involving velocities &

displacements.

• Work and kinetic energy are scalar quantities.

• Assume that rigid body is made of a large

number of particles.

2211 TUT

21, TT

21U

initial & final total kinetic energy of

particles forming body

total work of internal & external

forces acting on particles of body.

• Internal forces between particles A and B are

equal and opposite.

• Therefore, net work of internal forces is zero.

• In general, small displacements of particles A

and B are not equal but components of

displacements along AB are equal.

5/12/2012 9:28 AM


Work of

Forces Acting

on a Rigid

Body

17 - 5

• Work of a force during a displacement of

its point of application,

2

1

2

1

cos21

s

s

A

A

dsFrdFU

• Consider net work of two forces

forming a couple of moment during a

displacement of their points of application.

FF

and

M

dM

dFrdsF

rdFrdFrdFdU

2

211

12

21

2

1

M

dMU

if M is constant.

5/12/2012 9:28 AM


Work of Forces Acting on a Rigid

Body

17 - 6

Forces acting on rigid bodies which do no work:

• Forces applied to fixed points:

reactions at a frictionless pin when the supported body

rotates about the pin.

• Forces acting in a direction perpendicular to the

displacement of their point of application:

reaction at a frictionless surface to a body moving along

the surface

weight of a body when its center of gravity moves

horizontally

• Friction force at point of contact of a body rolling without

sliding on a fixed surface.

0 dtvFdsFdU cC

5/12/2012 9:28 AM


17 - 7

• Consider a rigid body of mass m in plane motion.

2

212

21

22

212

21

2

212

21

Δ

Δ

Ivm

mrvm

vmvmT

ii

ii

• Kinetic energy of a rigid body can be separated

into:

kinetic energy associated with motion of

mass center G and

kinetic energy associated with rotation of

body about G.• Consider a rigid body rotating about a fixed

axis through O.

2

21

22

212

212

21 ΔΔΔ

O

iiiiii

I

mrrmvmT

5/12/2012 9:28 AM


Kinetic Energy of a Rigid Body in Plane

Motion


17 - 8

• For problems involving systems consisting of several rigid bodies,

principle of work and energy can be applied to each body.

• We may also apply principle of work and energy to the entire system,

2211 TUT = arithmetic sum of kinetic energies of all

bodies forming the system

= work of all forces acting on various

bodies, whether these forces are internal

or external to the system as a whole.

21,TT

21U

• For problems involving pin connected members, blocks and pulleys

connected by inextensible cords, and meshed gears,

internal forces occur in pairs of equal and opposite forces

points of application of each pair move through equal distances

net work of internal forces is zero

work on the system reduces to work of the external forces

5/12/2012 9:28 AM


Conservation of Energy

17 - 9

• Expressing work of conservative forces as a

change in potential energy, principle of work

and energy becomes

2211 VTVT

sin3

sin2

1

32

10 2

22211

l

g

mglml

VTVT

0,0 11 VT

22

22121

212

21

21

222

1222

12

32

1

mlmllm

IvmT

sinsin21

21

2 mglWlV

• Consider the slender rod of mass m.

• mass m

• released with zero velocity

• determine at

5/12/2012 9:28 AM


Power

17 - 10

• Power = rate at which work is done

• For a body acted upon by force and moving with velocity ,F

v

vFdt

dU Power

• For a rigid body rotating with an angular velocity and acted

upon by a couple of moment parallel to axis of rotation,

M

Mdt

dM

dt

dUPower

5/12/2012 9:28 AM


Sample Problem 17.1

17 - 11

For the drum and flywheel,

The bearing friction is equivalent to a

couple of At the instant shown,

the block is moving downward at 6 ft/s.

Determine velocity of block after it has

moved 4 ft downward.

.sftlb5.10 2I

ft.lb60

SOLUTION:

• Consider the system of

flywheel and block. The

work done by internal forces

exerted by the cable cancels.

• Apply principle of work and

kinetic energy to develop an

expression for final velocity.

• Note that velocity of block

and angular velocity of drum

and flywheel are related by

rv

5/12/2012 9:28 AM


Sample Problem 17.1

17 - 12

SOLUTION:

• Consider the system of flywheel and block. The work

done by internal forces exerted by the cable cancels.

• Note that velocity of block and angular velocity of drum

and flywheel are related by

1.25srad80.4

ft1.25

sft6 222

11

v

r

v

r

vrv

• Apply principle of work and kinetic energy to develop

an expression for the final velocity.

lbft255

srad80.4sftlb5.102

1sft6

sft32.2

lb240

2

1 22

2

212

1212

11

ImvT

2

2

2

22

2

2

2212

221

2 09.725.1

5.102

1

2.32

240

2

1v

vvIvmT

5/12/2012 9:28 AM


Sample Problem 17.117 - 13

lbft255212

1212

11 ImvT

22

222

1222

12 09.7 vIvmT

• Note that block displacement and pulley rotation

are related by

rad20.3ft25.1

ft422

r

s

• Principle of work and energy:

sft01.12

7.09lbft768lbft255

2

22

2211

v

v

TUT

sft01.122 v

lbft768

rad20.3ftlb60ft4lb240

121221

MssWU

Then,

5/12/2012 9:28 AM


Sample Problem 17.2

17 - 14

mm80kg3

mm200kg10

BB

AA

km

km

The system is at rest when a

moment of is applied to

gear B. Neglecting friction,

determine

a. number of revolutions of gear Bbefore its angular velocity

reaches 600 rpm

b. tangential force exerted by gear

B on gear A.

mN6 M

SOLUTION:

• Consider a system consisting of two

gears. Noting that gear rotational

speeds are related, evaluate final

kinetic energy of the system.

• Apply principle of work and energy.

Calculate number of revolutions

required for work of applied moment

to equal final kinetic energy of

system.

• Apply principle of work and energy

to system consisting of gear A. With

final kinetic energy and number of

revolutions known, calculate

moment and tangential force

required for the indicated work.

5/12/2012 9:28 AM


Sample

Problem 17.2

17 - 15

SOLUTION:

• Consider a system consisting of two gears.

Noting that gear rotational speeds are related,

evaluate final kinetic energy of the system.

srad1.25250.0

100.08.62

srad8.62mins60

revrad2rpm600

A

BBA

B

r

r

222

222

mkg0192.0m080.0kg3

mkg400.0m200.0kg10

BBB

AAA

kmI

kmI

J9.163

8.620192.01.25400.02

212

21

2

212

21

2

BBAA IIT

5/12/2012 9:28 AM


Sample Problem

17.2

17 - 16

• Apply principle of work and energy. Calculate

number of revolutions required for the work.

rad32.27

163.9JJ60

2211

B

B

TUT

rev35.4

2

32.27

B

• Apply principle of work and energy to a system

consisting of gear A. Calculate moment and

tangential force required for the indicated work.

J0.1261.25400.02

212

21

2 AAIT

mN52.11

J0.261rad10.930

2211

FrM

M

TUT

AA

A

rad93.10250.0

100.032.27

A

BBA

r

r

N2.46250.0

52.11F

5/12/2012 9:28 AM


Sample Problem

17.3

17 - 17

A sphere, cylinder, and hoop, each

having the same mass and radius, are

released from rest on an incline.

Determine the velocity of each body

after it has rolled through a distance

corresponding to a change of elevation

h.

SOLUTION:

• Work done by weight of the bodies

is the same. From the principle of

work and energy, it follows that

each body will have the same

kinetic energy after the change of

elevation.

• Because each of the bodies has a

different centroidal moment of

inertia, distribution of total kinetic

energy between linear and

rotational components will be

different as well.

5/12/2012 9:28 AM


Sample

Problem 17.3

17 - 18

SOLUTION:

• Work done by weight of the bodies is the

same. From principle of work and energy, it

follows that each body will have the same

kinetic energy after change of elevation.

r

vWith

2

221

2

212

212

212

21

2

vr

Im

r

vIvmIvmT

22

2

2

221

2211

1

22

0

mrI

gh

rIm

Whv

vr

ImWh

TUT

5/12/2012 9:28 AM


Sample

Problem 17.3

17 - 19

2

2

1

2

mrI

ghv

ghvmrIHoop

ghvmrICylinder

ghvmrISphere

2707.0:

2816.0:

2845.0:

2

2

21

2

52

• Because each of the bodies has a different

centroidal moment of inertia, distribution of

total kinetic energy between linear and

rotational components will be different as

well.

• The velocity of body is independent of its

mass and radius.

NOTE:

• For a frictionless block sliding through the

same distance, ghv 2,0

• The velocity of body does depend on

2

2

2 rk

mrI

5/12/2012 9:28 AM


Sample Problem

17.4

17 - 20

A 30-lb slender rod pivots about

point O. The other end is pressed

against a spring (k = 1800 lb/in)

until spring is compressed one inch

and rod is in a horizontal position.

If rod is released from this position,

determine its angular velocity and

reaction at the pivot as the rod

passes through a vertical position.

SOLUTION:

• The weight and spring forces are

conservative. Principle of work

and energy can be expressed as

2211 VTVT

• Evaluate initial and final potential

energy.

• Express final kinetic energy in

terms of final angular velocity of

the rod.

• Based on the FBD equation, solve

for reactions at the pivot.

5/12/2012 9:28 AM


Sample

Problem 17.4

17 - 21

SOLUTION:

• The weight and spring forces are conservative.

Principle of work and energy can be expressed as

2211 VTVT

• Evaluate initial and final potential energy.

lbft75lbin900

in.1in.lb180002

212

121

1

kxVVV eg

lbft45

ft1.5lb3002

WhVVV eg

• Express final kinetic energy in terms of angular

velocity of the rod.

2

2

2

2

121

sftlb941.1

ft5sft32.2

lb30

12

1

mlI

22

222

122

222

1222

1222

1222

12

019.2941.15.12.32

30

2

1

IrmIvmT

5/12/2012 9:28 AM


Sample

Problem 17.4

17 - 22

srad86.32 lbft452.019lbft750 22

2211

VTVT

From the principle of work and energy,

• Based on the FBD equation, solve for reactions at the

pivot.

ra

ra

t

n

2

22

2

sft3.22

srad86.3ft5.1

ra

a

t

n

2sft3.22

effOO MM rrmI 0 0

effxx FF rmRx 0xR

effyy FF

lb22.9

sft3.22sft32.2

lb30

lb30

2

2

y

ny

R

maR

22.9R

5/12/2012 9:28 AM


Sample Problem

17.5

17 - 23

Each of the two slender rods has a

mass of 6 kg. The system is

released from rest with b = 60o.

Determine

a. angular velocity of rod AB

when b = 20o

b. velocity of point D at the

same instant.

SOLUTION:

• Consider a system consisting of two

rods. With conservative weight force,

2211 VTVT

• Express final kinetic energy of system

in terms of angular velocities of rods.

• Evaluate initial and final potential

energy.

• Solve energy equation for angular

velocity, then evaluate velocity of

point D.

5/12/2012 9:28 AM


Sample

Problem 17.5

17 - 24

• Evaluate initial and final potential energy.

J26.38

m325.0N86.5822 11

WyV

J10.15

m1283.0N86.5822 22

WyV

SOLUTION:

• Consider a system consisting of two rods.

With the conservative weight force,

2211 VTVT

N86.58

sm81.9kg6 2

mgW

5/12/2012 9:28 AM


Sample

Problem 17.5

17 - 25

Since is perpendicular to AB and is

horizontal, instantaneous center of rotation for rod

BD is C.

m75.0BC m513.020sinm75.02 CD

apply the law of cosines to CDE, EC = 0.522 m

Bv

Dv

• Express final kinetic energy of the system in terms

of angular velocities of the rods.

m375.0ABv

ABB BCABv BD

Consider velocity of point B

m522.0BDv

2

2

1212

121

mkg281.0

m75.0kg6

mlII BDAB

For the final kinetic energy,

22

21

2

1212

212

121

2

212

1212

212

121

2

520.1281.0

522.06281.0375.06

BDBDBDABABAB IvmIvmT

5/12/2012 9:28 AM


Sample

Problem 17.5

17 - 26

srad3.90

J10.151.520J26.380 2

2211

VTVT

• Solve energy equation for angular velocity,

then evaluate velocity of point D.

srad90.3AB

sm00.2

srad90.3m513.0

CDvD

sm00.2Dv

5/12/2012 9:28 AM


Home Work Assignment 17-1

3, 11, 18, 25, 32, 40, 47

Due Sunday 13/5/2011

5/12/2012 9:28 AM


2 - 27


17 - 28

• Method of impulse and momentum:

well suited to solution of problems involving time and velocity

only practicable method for problems involving impulsive motion

and impact.

Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2

5/12/2012 9:28 AM



17 - 29

vmmvL ii

Δ

• Momenta of particles of a system may be reduced to a vector attached to

the mass center equal to their sum,

iiiG mvrH Δ

and a couple equal to sum of their moments about the mass center,

IHG

• For plane motion of a rigid slab or of a rigid body symmetrical with

respect to the reference plane,

5/12/2012 9:28 AM



17 - 30

• Principle of impulse and momentum for the plane motion of a rigid slab

or of a rigid body symmetrical with respect to the reference plane

expressed as a FBD equation,

• Leads to three equations of motion:

summing and equating momenta and impulses in x and y directions

summing and equating moments of momenta and impulses with

respect to any given point

5/12/2012 9:28 AM



17 - 31

• Noncentroidal rotation:

- The angular momentum about O

2rmI

rrmI

rvmIIO

- Equating moments of momenta and

impulses about O,

21

2

1

O

t

tOO IdtMI

5/12/2012 9:28 AM



17 - 32

• Motion of several rigid bodies can be analyzed by applying

principle of impulse and momentum to each body separately.

• For problems involving no more than three unknowns, it may

be convenient to apply principle of impulse and momentum to

the system as a whole.

• For each moving part of the system, diagrams of momenta

should include a momentum vector and/or a momentum couple.

• Internal forces occur in equal and opposite pairs of vectors and

do not generate nonzero net impulses.

5/12/2012 9:28 AM


Conservation of Angular Momentum

17 - 33

• When sum of angular impulses pass through O, linear

momentum may not be conserved, yet angular momentum about

O is conserved,

2010 HH

• Two additional equations may be written by summing x and ycomponents of momenta and may be used to determine two

unknown linear impulses, such as impulses of reaction

components at a fixed point.

• When no external force acts on a rigid body or a system of rigid

bodies, the system of momenta at t1 is equipollent to the system

at t2. Total linear momentum and angular momentum about any

point are conserved,

2010 HH

21 LL

5/12/2012 9:28 AM


Sample Problem 17.6

17 - 34

The system is at rest when a

moment of is applied

to gear B. Neglecting friction,

a. determine time required for

gear B to reach an angular

velocity of 600 rpm

b. tangential force exerted by gear

B on gear A.

mN6 M

mm80kg3

mm200kg10

BB

AA

km

km

SOLUTION:

• Considering each gear separately,

apply method of impulse and

momentum.

• Solve angular momentum

equations for the two gears

simultaneously for the unknown

time and tangential force.

5/12/2012 9:28 AM


Sample Problem 17.6

17 - 35

SOLUTION:

• Considering each gear separately, apply method of impulse and momentum.

sN2.40

srad1.25mkg400.0m250.0

02

Ft

Ft

IFtr AAA

moments about A:

moments about B:

srad8.62mkg0192.0

m100.0mN6

0

2

2

Ftt

IFtrMt BBB

• Solve angular momentum equations for two gears simultaneously for the unknown

time and tangential force.

N 46.2s 871.0 Ft

5/12/2012 9:28 AM


Sample Problem 17.7

17 - 36

Uniform sphere of mass m and

radius r is projected along a rough

horizontal surface with a linear

velocity and no angular

velocity. The coefficient of

kinetic friction is Determine

a. time t2 at which the sphere

will start rolling without

sliding

b. linear and angular velocities

of the sphere at time t2.

.k

1v

SOLUTION:

• Apply principle of impulse and

momentum to find variation of linear

and angular velocities with time.

• Relate linear and angular velocities

when the sphere stops sliding by

noting that velocity of the point of

contact is zero at that instant.

• Substitute for linear and angular

velocities and solve for the time at

which sliding stops.

• Evaluate linear and angular

velocities at that instant.

5/12/2012 9:28 AM


Sample Problem

17.7

17 - 37

SOLUTION:


momentum to find variation of linear

and angular velocities with time.

0WtNt

y components:

x components:

21

21

vmmgtvm

vmFtvm

k

gtvv k 12

mgWN

moments about G:

22

52

2

mrtrmg

IFtr

k

tr

gk2

52

Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2• Relate linear and angular velocities

when sphere stops sliding by noting

that velocity of point of contact is

zero at that instant.

tr

grgtv

rv

kk

2

51

22

• Substitute for linear and angular

velocities and solve for time at which

sliding stops.

g

vt

k1

7

2

5/12/2012 9:28 AM


Sample Problem

17.7

17 - 38

x components: gtvv k 12

y components: mgWN

moments about G: tr

gk2

52


tr

grgtv

rv

kk

2

51

22

g

vt

k1

7

2

• Evaluate linear and angular

velocities at that instant.

g

vgvv

kk

1

127

2

g

v

r

g

k

k

1

27

2

2

5

127

5vv

r

v12

7

5

5/12/2012 9:28 AM


Sample Problem 17.8

17 - 39

Two solid spheres (radius = 3 in.,

W = 2 lb) are

mounted on a spinning horizontal

rod ( = 6 rad/sec) as shown. The

balls are held together by a string

which is suddenly cut. Determine

a. angular velocity of rod after the

balls have moved to A’ and B’

b. energy lost due to the plastic

impact of the spheres and stops.

,sftlb 0.25 2RI

SOLUTION:

• Observing that none of the external

forces produce a moment about the

y axis, the angular momentum is

conserved.

• Equate initial and final angular

momenta. Solve for final angular

velocity.

• The energy lost due to plastic

impact is equal to change in kinetic

energy of the system.

5/12/2012 9:28 AM


Sample

Problem

17.8

17 - 40


22222

11111

2

2

RSs

RSs

IIrrm

IIrrm

SOLUTION:

• Observing that none of

external forces produce

a moment about the y

axis, angular

momentum is

conserved.• Equate initial and final

angular momenta.

Solve for final angular

velocity.

22

2522

52 sftlb00155.0ft

12

2

sft32.2

lb2

maIS

2696.012

25

2.32

20108.0

12

5

2.32

22

2

2

2

2

1

rmrm SS

RSs

RSs

IIrm

IIrm

22

21

12

srad61 2sftlb 25.0 RI

srad08.22

5/12/2012 9:28 AM


Sample

Problem

17.8

17 - 41

• The energy lost due to

plastic impact is equal to

change in kinetic energy of

the system.2sftlb00155.0 SI

221 sftlb0108.0 rmS

srad61

2sftlb 25.0 RI

srad08.22

222 sftlb2696.0 rmS

22

212

212

212

21 222 RSSRSS IIrmIIvmT

95.471.1

lbft71.108.2792.0

lbft95.46275.0

12

2

21

2

2

21

1

TTΔT

T

T

lbft 24.3 T

5/12/2012 9:28 AM



54, 60, 69, 76, 83, 91

Due Tuesday 15/5/2011

5/12/2012 9:28 AM


2 - 42

Eccentric Impact17 - 43

Period of deformation Period of restitution

dtRImpulse

dtPImpulse

nBnA uu

• Principle of impulse and momentum is supplemented by

nBnA

nAnB

vv

vv

dtP

dtRnrestitutiooftcoefficiene

5/12/2012 9:28 AM


Sample Problem 17.9

17 - 44

A 0.05-lb bullet is fired into the side

of a 20-lb square panel which is

initially at rest. Determine

a. angular velocity of the panel

immediately after the bullet

becomes embedded

b. impulsive reaction at A,

assuming that the bullet

becomes embedded in 0.0006 s.

SOLUTION:

• Consider a system consisting of

the bullet and panel. Apply

principle of impulse and

momentum.

• Final angular velocity is found

from moments of momenta and

impulses about A.

• The reaction at A is found from

horizontal and vertical momenta

and impulses.

5/12/2012 9:28 AM


Sample Problem 17.9

17 - 45

SOLUTION:

• Consider a system

consisting of the bullet

and panel. Apply

principle of impulse and

momentum.

• Final angular velocity is

found from moments of

momenta and impulses

about A.

moments about

A: 2129

21214 ft0ft PPBB Ivmvm

2129

2 ft v 22

2

61 sftlb2329.0

12

18

2.32

20

6

1

bmI PP

2129

2129

1214 2329.0

2.32

201500

2.32

05.0

sft50.3

srad67.4

2129

2

2

v

srad67.42

5/12/2012 9:28 AM


Sample Problem 17.9

17 - 46

sft50.3srad67.4 2129

22 v

• Reactions at A are found from horizontal and vertical momenta and impulses.

x components:

50.32.32

200006.01500

2.32

05.0

2

x

pxBB

A

vmtAvm

lb259xA lb259xA

y components: 00 tAy 0yA

5/12/2012 9:28 AM


Sample Problem 17.10

17 - 47

A 2-kg sphere with an initial

velocity of 5 m/s strikes the lower

end of an 8-kg rod AB. The rod is

hinged at A and initially at rest.

The coefficient of restitution

between the rod and sphere is 0.8.

Determine the angular velocity of

the rod and the velocity of the

sphere immediately after impact.

SOLUTION:

• Consider the sphere and rod as a

single system. Apply principle of

impulse and momentum.

• Moments about A of momenta and

impulses provide a relation

between final angular velocity of

the rod and velocity of the sphere.

• The definition of the coefficient of

restitution provides a 2nd

relationship between final angular

velocity of rod and velocity of the

sphere.

• Solve two relations simultaneously

for angular velocity of the rod and

velocity of the sphere.

5/12/2012 9:28 AM



17 - 48

SOLUTION:

• Consider the sphere and

rod as a single system.

Apply principle of impulse

and momentum.

• Moments about A of

momenta and impulses

provide a relation between

final angular velocity of

the rod and velocity of the

rod.

moments about A:

Ivmvmvm RRssss m6.0m2.1m2.1

22

1212

121 mkg96.0m2.1kg8

m6.0

mLI

rvR

2mkg96.0

m6.0m6.0kg8m2.1kg2m2.1sm5kg2 sv

84.34.212 sv

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Sample Problem

17.10

17 - 49

Moments about A: 84.34.212 sv

• The definition of the

coefficient of restitution

provides a 2nd

relationship between final

angular velocity of the

rod and velocity of the

sphere.

sm58.0m2.1

s

sBsB

v

vvevv

Relative velocities:

• Solve two relations

simultaneously for the

angular velocity of the

rod and velocity of the

sphere.

Solving,

sm143.0sv sm143.0sv

rad/s21.3 rad/s21.3

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Sample Problem

17.11

17 - 50

A square package of mass m moves

down conveyor belt A with constant

velocity. At end of the conveyor,

the corner of the package strikes a

rigid support at B. The impact is

perfectly plastic. Derive an

expression for the minimum

velocity of conveyor belt A for

which the package will rotate about

B and reach conveyor belt C.

SOLUTION:


momentum to relate velocity of

package on conveyor belt A before

impact at B to angular velocity

about B after impact.

• Apply principle of conservation of

energy to determine minimum

initial angular velocity such that

the mass center of package will

reach a position directly above B.

• Relate required angular velocity to

velocity of conveyor belt A.

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17 - 51

SOLUTION:

• Apply principle of impulse and momentum to relate velocity of package on

conveyor belt A before impact at B to angular velocity about B after impact.

Moments about B:

222

221

1 0 Iavmavm 2

61

222

2 amIav

22

61

22

222

21

1 0 amaamavm

234

1 av

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Sample Problem 17.1117 - 52

• Apply principle of conservation of energy to determine

minimum initial angular velocity such that mass center

of package will reach a position directly above B.

3322 VTVT

22 WhV

22

2

312

22

61

21

2

222

21

222

1222

12

mamaam

ImvT

33 WhV

03 T (solving for the minimum 2)

aa

GBh

612.060sin

1545sin

22

2

aah 707.022

3

agaaa

ghh

ma

W

WhWhma

285.0612.0707.033

0

2232

22

3222

2

31

agaav 285.034

234

1 gav 712.01

5/12/2012 9:28 AM



96, 104, 111, 121, 130

Due Saturday 19/5/2011

5/12/2012 9:28 AM


2 - 53

chapter 17 - islamic university of gazasite.iugaza.edu.ps/mhaiba/files/2012/01/ch17-plane...rigid...

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