DYNAMICS Ferdinand P. Beer Plane Motion of Rigid Bodies: E. - - PowerPoint PPT Presentation

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Seventh Edition Ferdinand P. Beer

• E. Russell Johnston, Jr.

Lecture Notes:

• J. Walt Oler

Texas Tech University CHAPTER

• Plane Motion of Rigid Bodies:

Energy and Momentum Methods

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 2

Contents

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 Sample Problem 17.1 Sample Problem 17.2 Sample Problem 17.3 Sample Problem 17.4 Sample Problem 17.5 Principle of Impulse and Momentum Systems of Rigid Bodies Conservation of Angular Momentum

Sample Problem 17.6 Sample Problem 17.7 Sample Problem 17.8 Eccentric Impact Sample Problem 17.9 Sample Problem 17.10 Sample Problem 17.11

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 3

Introduction

• Method of work and energy and the method of impulse and

momentum will be used to analyze the plane motion of rigid bodies and systems of rigid bodies.

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

problems involving displacements and velocities.

2 2 1 1

T U T = +

→

• Principle of impulse and momentum is appropriate for problems

involving velocities and time.

( ) ( )2

1 2 1

2 1 2 1

O t t O O t t

H dt M H L dt F L

• =

+ = +

• Problems involving eccentric impact are solved by supplementing the

principle of impulse and momentum with the application of the coefficient of restitution.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 4

Principle of Work and Energy for a Rigid Body

• Method of work and energy is well adapted to

problems involving velocities and displacements. Main advantage is that the work and kinetic energy are scalar quantities.

• Assume that the rigid body is made of a large number
• f particles.

2 2 1 1

T U T = +

→

=

2 1, T

T =

→2 1

U initial and final total kinetic energy of particles forming body total work of internal and external forces acting on particles of body.

• Internal forces between particles A and B are equal

and opposite.

• Therefore, the net work of internal forces is zero.
• In general, small displacements of the particles A and

B are not equal but the components of the displacements along AB are equal.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 5

Work of Forces Acting on a Rigid Body

• Work of a force during a displacement of its point of

application,

( )

• =

⋅ =

→

2 1 2 1

cos

2 1 s s A A

ds F r d F U α

• Consider the net work of two forces

forming a couple of moment during a displacement of their points of application. F F

• −

and M

• θ

θ d M d Fr ds F r d F r d F r d F dU = = = ⋅ + ⋅ − ⋅ =

2 2 1 1

• (

)

1 2 2 1

2 1

θ θ θ

θ θ

− = =

→

M d M U if M is constant.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 6

Work of Forces Acting on a Rigid Body

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 the point of contact of a body rolling without

sliding on a fixed surface.

( )

= = = dt v F ds F dU

c C

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 7

Kinetic Energy of a Rigid Body in Plane Motion

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

2 2 1 2 2 1 2 2 2 1 2 2 1 2 2 1 2 2 1

)

• (
• ω

ω I v m m r v m v m v m T

i i i i

+ = ′ + =

• ′

+ =

• Kinetic energy of a rigid body can be separated into:
• the kinetic energy associated with the motion of the

mass center G and

• the kinetic energy associated with the rotation of the

body about G.

• Consider a rigid body rotating about a fixed axis

through O. ( )

2 2 1 2 2 2 1 2 2 1 2 2 1

)

• (
• ω

ω ω

O i i i i i i

I m r r m v m T =

• +
• +
• =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 8

Systems of Rigid Bodies

• For problems involving systems consisting of several rigid bodies, the principle
• f work and energy can be applied to each body.
• We may also apply the principle of work and energy to the entire system,

2 2 1 1

T U T = +

→

= arithmetic sum of the kinetic energies of all bodies forming the system = work of all forces acting on the various bodies, whether these forces are internal or external to the system as a whole.

2 1,T

T

2 1→

U

• 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 the internal forces is zero
• work on the system reduces to the work of the external forces

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 9

Conservation of Energy

• Expressing the work of conservative forces as a

change in potential energy, the principle of work and energy becomes

2 2 1 1

V T V T + = +

• =

− = + = + θ ω θ ω sin 3 sin 2 1 3 2 1

2 2 2 2 1 1

l g mgl ml V T V T ,

1 1

= = V T

( )

( )

2 2 2 2 12 1 2 1 2 2 1 2 1 2 2 2 1 2 2 2 1 2

3 2 1 ω ω ω ω ml ml l m I v m T = + = + =

θ θ sin sin

2 1 2 1 2

mgl Wl V − = − =

• Consider the slender rod of mass m.
• mass m
• released with zero velocity
• determine ω at θ

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 10

Power

• Power = rate at which work is done
• For a body acted upon by force and moving with velocity ,

F

• v
• v

F dt dU

• ⋅

= = Power

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

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

• M
• ω

θ M dt d M dt dU = = = Power

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 11

Sample Problem 17.1

For the drum and flywheel, The bearing friction is equivalent to a couple

• f

At the instant shown, the block is moving downward at 1.8 m/s. . m kg 15

2

⋅ = I m. N 80 ⋅ Determine the velocity of the block after it has moved 1.2 m downward. SOLUTION:

• Consider the system of the flywheel

and block. The work done by the internal forces exerted by the cable cancels.

• Apply the principle of work and

kinetic energy to develop an expression for the final velocity.

• Note that the velocity of the block

and the angular velocity of the drum and flywheel are related by ω r v =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 12

Sample Problem 17.1

SOLUTION:

• Consider the system of the flywheel and block. The work done by

the internal forces exerted by the cable cancels.

• Note that the velocity of the block and the angular velocity of the

drum and flywheel are related by

0.4 s rad 5 . 4 m 0.4 s m 8 . 1

2 2 2 1 1

v r v r v r v = = = = = = ω ω ω

• Apply the principle of work and kinetic energy to develop an

expression for the final velocity. J 330 ) s rad 5 . 4 ( ) m kg 15 ( 2 1 s) / m (1.8 kg) 110 ( 2 1

2 2 2 2 1 2 1 2 1 2 1 1

= ⋅ + = + = ω I mv T

2 2 2 2 2 2 2 2 2 1 2 2 2 1 2

9 . 101 4 . ) 15 ( 2 1 ) 110 ( 2 1 v v v I v m T =

• +

= + = ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 13

Sample Problem 17.1

J 330

2 1 2 1 2 1 2 1 1

= + = ω I mv T

2 2 2 2 2 1 2 2 2 1 2

9 . 101 v I v m T = + = ω

• Note that the block displacement and pulley

rotation are related by

rad . 3 m 4 . m 2 . 1

2 2

= = = r s θ

• Principle of work and energy:

s m 7 . 3 .9 01 1 J 8 . 1054 J 300

2 2 2 2 2 1 1

= = + = +

→

v v T U T s m 7 . 3

2 =

v

( ) ( ) ( )( ) ( )( )

J 8 . 1054 rad . 3 m N 80 N 2 . 1 N 1079

1 2 1 2 2 1

= ⋅ − = − − − =

→

θ θ M s s W U

Then,

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 14

Sample Problem 17.2

mm 80 kg 3 mm 200 kg 10 = = = =

B B A A

k m k m

The system is at rest when a moment of is applied to gear B. Neglecting friction, a) determine the number of revolutions of gear B before its angular velocity reaches 600 rpm, and b) tangential force exerted by gear B on gear A. m N 6 ⋅ = M SOLUTION:

• Consider a system consisting of the two
• gears. Noting that the gear rotational speeds

are related, evaluate the final kinetic energy

• f the system.
• Apply the principle of work and energy.

Calculate the number of revolutions required for the work of the applied moment to equal the final kinetic energy of the system.

• Apply the principle of work and energy to a

system consisting of gear A. With the final kinetic energy and number of revolutions known, calculate the moment and tangential force required for the indicated work.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 15

Sample Problem 17.2

SOLUTION:

• Consider a system consisting of the two gears. Noting that

the gear rotational speeds are related, evaluate the final kinetic energy of the system.

( )( )

s rad 1 . 25 250 . 100 . 8 . 62 s rad 8 . 62 min s 60 rev rad 2 rpm 600 = = = = =

A B B A B

r r ω ω π ω

( )( ) ( )( )

2 2 2 2 2 2

m kg 0192 . m 080 . kg 3 m kg 400 . m 200 . kg 10 ⋅ = = = ⋅ = = =

B B B A A A

k m I k m I

( )( ) ( )( )

J 9 . 163 8 . 62 0192 . 1 . 25 400 .

2 2 1 2 2 1 2 2 1 2 2 1 2

= + = + =

B B A A

I I T ω ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 16

Sample Problem 17.2

• Apply the principle of work and energy. Calculate the

number of revolutions required for the work.

( )

rad 32 . 27 163.9J J 6

2 2 1 1

= = + = +

→ B B

T U T θ θ rev 35 . 4 2 32 . 27 = = π θB

• Apply the principle of work and energy to a system

consisting of gear A. Calculate the moment and tangential force required for the indicated work.

( )( )

J . 126 1 . 25 400 .

2 2 1 2 2 1 2

= = =

A A

I T ω

( )

m N 52 . 11 J . 26 1 rad 10.93

2 2 1 1

⋅ = = = + = +

→

F r M M T U T

A A A

rad 93 . 10 250 . 100 . 32 . 27 = = =

A B B A

r r θ θ N 2 . 46 250 . 52 . 11 = = F

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 17

Sample Problem 17.3

A sphere, cylinder, and hoop, each having the same mass and radius, are released from rest on an incline. Determine the velocity

• f each body after it has rolled through a

distance corresponding to a change of elevation h. SOLUTION:

• The work done by the 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, the distribution of the total kinetic energy between the linear and rotational components will be different as well.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 18

Sample Problem 17.3

SOLUTION:

• The work done by the 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. r v = ω With

2 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1 2

v r I m r v I v m I v m T

• +

=

• +

= + = ω

2 2 2 2 2 2 1 2 2 1 1

1 2 2 mr I gh r I m Wh v v r I m Wh T U T + = + =

• +

= + = +

→

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 19

Sample Problem 17.3

2 2

1 2 mr I gh v + = gh v mr I Hoop gh v mr I Cylinder gh v mr I Sphere 2 707 . : 2 816 . : 2 845 . :

2 2 2 1 2 5 2

= = = = = =

• Because each of the bodies has a different centroidal

moment of inertia, the distribution of the total kinetic energy between the linear and rotational components will be different as well.

• The velocity of the body is independent of its mass and

radius. NOTE:

• For a frictionless block sliding through the same

distance, gh v 2 , = = ω

• The velocity of the body does depend on

2 2 2

r k mr I =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 20

Sample Problem 17.4

A 15 kg slender rod pivots about the point

• O. The other end is pressed against a

spring (k = 324 kN/m) until the spring is compressed one inch and the rod is in a horizontal position. If the rod is released from this position, determine its angular velocity and the reaction at the pivot as the rod passes through a vertical position. SOLUTION:

• The weight and spring forces are
• conservative. The principle of work and

energy can be expressed as

2 2 1 1

V T V T + = +

• Evaluate the initial and final potential

energy.

• Express the final kinetic energy in terms of

the final angular velocity of the rod.

• Based on the free-body-diagram equation,

solve for the reactions at the pivot.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 21

Sample Problem 17.4

SOLUTION:

• The weight and spring forces are conservative. The

principle of work and energy can be expressed as

2 2 1 1

V T V T + = +

• Evaluate the initial and final potential energy.

( )( )

J 25 . 101 J 25 . 101 m 025 . m kN 324

2 2 1 2 1 2 1 1

= = = + = + = kx V V V

e g

( )( )

J 2 . 66 m 0.45 N 147

2

= = + = + = Wh V V V

e g

• Express the final kinetic energy in terms of the angular

velocity of the rod.

( )( )

2 2 2 12 1

m kg 81 . 2 m 5 . 1 kg 15 12 1 ⋅ = = = ml I

( ) ( ) ( )

2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2

92 . 2 81 . 2 45 . ) 15 ( 2 1 ω ω ω ω ω ω = + = + = + = I r m I v m T

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 22

Sample Problem 17.4

s rad 46 . 3

2 =

ω 2 . 66 2.92 J 25 . 101

2 2 2 2 1 1

+ = + + = + ω V T V T

From the principle of work and energy,

• Based on the free-body-diagram equation, solve for the

reactions at the pivot.

( )( )

α ω r a r a

t n

= = = =

2 2 2 2

s m 39 . 5 s rad 46 . 3 m 45 .

α r a a

t n

= =

• 2

s m 39 . 5

( )

• =

eff O O

M M

( )r

r m I α α + = = α

( )

• =

eff x x

F F

( )

α r m Rx = =

x

R

( )

• =

eff y y

F F 15 . 66 = R

• (

)

N 15 . 66 s m 39 . 5 s m 15 N 147 N 147

2 2

= − = − = −

y n y

R ma R

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 23

Sample Problem 17.5

Each of the two slender rods has a mass

• f 6 kg. The system is released from rest

with β = 60o. Determine a) the angular velocity of rod AB when β = 20o, and b) the velocity of the point D at the same instant. SOLUTION:

• Consider a system consisting of the two rods.

With the conservative weight force,

2 2 1 1

V T V T + = +

• Express the final kinetic energy of the system

in terms of the angular velocities of the rods.

• Evaluate the initial and final potential energy.
• Solve the energy equation for the angular

velocity, then evaluate the velocity of the point D.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 24

Sample Problem 17.5

• Evaluate the initial and final potential energy.

( )( )

J 26 . 38 m 325 . N 86 . 58 2 2

1 1

= = = Wy V

( )( )

J 10 . 15 m 1283 . N 86 . 58 2 2

2 2

= = = Wy V SOLUTION:

• Consider a system consisting of the two rods. With the

conservative weight force,

2 2 1 1

V T V T + = +

( )(

)

N 86 . 58 s m 81 . 9 kg 6

2

= = = mg W

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 25

Sample Problem 17.5

Since is perpendicular to AB and is horizontal, the instantaneous center of rotation for rod BD is C. m 75 . = BC

( )

m 513 . 20 sin m 75 . 2 = ° = CD and applying the law of cosines to CDE, EC = 0.522 m

B

v

• D

v

• Express the final kinetic energy of the system in terms of

the angular velocities of the rods.

( )ω

m 375 . =

AB

v

• (

) ( )

AB B

BC AB v ω ω = = ω ω =

BD

• Consider the velocity of point B

( )ω

m 522 . =

BD

v

• (

)( )

2 2 12 1 2 12 1

m kg 281 . m 75 . kg 6 ⋅ = = = = ml I I

BD AB

For the final kinetic energy,

( )( ) ( ) ( )( ) ( )

2 2 2 1 2 12 1 2 2 1 2 12 1 2 2 1 2 12 1 2 2 1 2 12 1 2

520 . 1 281 . 522 . 6 281 . 375 . 6 ω ω ω ω ω ω ω = + + + = + + + =

BD BD BD AB AB AB

I v m I v m T

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 26

Sample Problem 17.5

s rad 3.90 J 10 . 15 1.520 J 26 . 38

2 2 2 1 1

= + = + + = + ω ω V T V T

• Solve the energy equation for the angular velocity, then

evaluate the velocity of the point D. s rad 90 . 3 =

AB

ω

• (

) ( )( )

s m 00 . 2 s rad 90 . 3 m 513 . = = = ω CD vD s m 00 . 2 =

D

v

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 27

Principle of Impulse and Momentum

• Method of impulse and momentum:
• well suited to the solution of problems involving time and velocity
• the only practicable method for problems involving impulsive motion

and impact. Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 28

Principle of Impulse and Momentum

v m m v L

i i

• =

=

• The momenta of the particles of a system may be reduced to a vector attached

to the mass center equal to their sum,

i i i G

m v r H

• ×

′ = and a couple equal to the sum of their moments about the mass center, ω I H G =

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

respect to the reference plane,

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 29

Principle of Impulse and Momentum

• 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 free-body-diagram equation,

• Leads to three equations of motion:
• summing and equating momenta and impulses in the x and y directions
• summing and equating the moments of the momenta and impulses with

respect to any given point

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 30

Principle of Impulse and Momentum

• Noncentroidal rotation:
• The angular momentum about O

( ) ( )

( )ω

ω ω ω ω

2

r m I r r m I r v m I IO + = + = + =

• Equating the moments of the momenta and

impulses about O,

2 1

2 1

ω ω

O t t O O

I dt M I = +

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 31

Systems of Rigid Bodies

• Motion of several rigid bodies can be analyzed by applying the

principle of impulse and momentum to each body separately.

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

convenient to apply the principle of impulse and momentum to the system as a whole.

• For each moving part of the system, the 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.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 32

Conservation of Angular Momentum

• When the sum of the angular impulses pass through O, the linear

momentum may not be conserved, yet the angular momentum about O is conserved,

( ) ( )2

1

H H =

• Two additional equations may be written by summing x and y

components of momenta and may be used to determine two unknown linear impulses, such as the impulses of the 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. The total linear momentum and angular momentum about any point are conserved,

( ) ( )2

1

H H =

2 1

L L

• =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 33

Sample Problem 17.6

The system is at rest when a moment of is applied to gear B. Neglecting friction, a) determine the time required for gear B to reach an angular velocity of 600 rpm, and b) the tangential force exerted by gear B on gear A. m N 6 ⋅ = M

mm 80 kg 3 mm 200 kg 10 = = = =

B B A A

k m k m

SOLUTION:

• Considering each gear separately, apply the

method of impulse and momentum.

• Solve the angular momentum equations for

the two gears simultaneously for the unknown time and tangential force.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 34

Sample Problem 17.6

SOLUTION:

• Considering each gear separately, apply the method of impulse and

momentum.

( ) ( ) ( )( )

s N 2 . 40 s rad 1 . 25 m kg 400 . m 250 .

2

⋅ = ⋅ = − = − Ft Ft I Ftr

A A A

ω moments about A: moments about B:

( ) ( ) ( )

( )(

)

s rad 8 . 62 m kg 0192 . m 100 . m N 6

2 2

⋅ = − ⋅ = − + Ft t I Ftr Mt

B B B

ω

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

the unknown time and tangential force. N 46.2 s 871 . = = F t

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 35

Sample Problem 17.7

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) the time t2 at which the sphere will start rolling without sliding and b) the linear and angular velocities

• f the sphere at time t2.

.

k

µ

1

v SOLUTION:

• Apply principle of impulse and momentum to

find variation of linear and angular velocities with time.

• Relate the linear and angular velocities when

the sphere stops sliding by noting that the velocity of the point of contact is zero at that instant.

• Substitute for the linear and angular velocities

and solve for the time at which sliding stops.

• Evaluate the linear and angular velocities at

that instant.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 36

Sample Problem 17.7

SOLUTION:

• Apply principle of impulse and momentum to

find variation of linear and angular velocities with time. = − Wt Nt y components: x components:

2 1 2 1

v m mgt v m v m Ft v m

k

= − = − µ gt v v

k

µ − = 1

2

mg W N = = moments about G:

( )

( )

2 2 5 2 2

ω µ ω mr tr mg I Ftr

k

= = t r g

k

µ ω 2 5

2 =

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.

• =

− = t r g r gt v r v

k k

µ µ ω 2 5

1 2 2

• Substitute for the linear and angular velocities

and solve for the time at which sliding stops. g v t

k

µ

1

7 2 =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 37

Sample Problem 17.7

x components: gt v v

k

µ − =

1 2

y components: mg W N = = moments about G: t r g

k

µ ω 2 5

2 =

Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2

• =

− = t r g r gt v r v

k k

µ µ ω 2 5

1 2 2

g v t

k

µ

1

7 2 =

• Evaluate the linear and angular velocities at

that instant.

• −

= g v g v v

k k

µ µ

1 1 2

7 2

• =

g v r g

k k

µ µ ω

1 2

7 2 2 5

1 2

7 5 v v = r v1

2

7 5 = ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 38

Sample Problem 17.8

Two solid spheres (radius = 75 mm, W = 1 kg) are mounted on a spinning horizontal rod ( ω = 6 rad/sec) as shown. The balls are held together by a string which is suddenly

• cut. Determine a) angular velocity of the

rod after the balls have moved to A’ and B’, and b) the energy lost due to the plastic impact of the spheres and stops. , m kg 0.3

2

⋅ =

R

I SOLUTION:

• Observing that none of the external forces

produce a moment about the y axis, the angular momentum is conserved.

• Equate the initial and final angular
• momenta. Solve for the final angular

velocity.

• The energy lost due to the plastic impact is

equal to the change in kinetic energy of the system.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 39

Sample Problem 17.8

Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2

( )

[ ]

( )

[ ]

2 2 2 2 2 1 1 1 1 1

2 2 ω ω ω ω ω ω

R S s R S s

I I r r m I I r r m + + = + +

SOLUTION:

• Observing that none of the

external forces produce a moment about the y axis, the angular momentum is conserved.

• Equate the initial and final

angular momenta. Solve for the final angular velocity.

( )( )

2 3 2 5 2 2 5 2

m kg 10 25 . 2 m 075 . kg 1 ⋅ × = = =

−

ma IS

( )( ) ( )( )

3 2 2 2 3 2 2 1

10 625 . 390 625 . 1 10 625 . 15 125 . 1

− −

× = = × = = r m r m

S S

R S s R S s

I I r m I I r m + + + + =

2 2 2 1 1 2

ω ω

s rad 6

1 =

ω

2

m kg 3 . ⋅ =

R

I

s rad 08 . 2

2 =

ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 40

Sample Problem 17.8

• The energy lost due to the plastic

impact is equal to the change in kinetic energy of the system.

2 3

m kg 10 25 . 2 ⋅ × =

− S

I

2 3 2 1

m kg 10 625 . 15 ⋅ × =

−

r mS

s rad 6

1 =

ω

2

m kg 31 . ⋅ =

R

I

s rad 08 . 2

2 =

ω

2 3 2 2

m kg 10 625 . 390 ⋅ × =

−

r mS

( ) ( )

2 2 2 1 2 2 1 2 2 1 2 2 1

2 2 2 ω ω ω

R S S R S S

I I r m I I v m T + + = + + =

( )( ) ( )( )

04 . 6 86 . 1 J 86 . 1 85 . 1 08575 . J 04 . 6 6 33575 .

1 2 2 2 1 2 2 2 1 1

− = − = = = = = T T T T T

J 18 . 4 − = ∆T

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 41

Eccentric Impact

Period of deformation Period of restitution

• =

dt R Impulse

• =

dt P Impulse

• (

) ( )n

B n A

u u

• =
• Principle of impulse and momentum is supplemented by

( ) ( ) ( ) ( )n

B n A n A n B

v v v v dt P dt R n restitutio

• f

t coefficien e − ′ − ′ = = =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 42

Sample Problem 17.9

A 20 g bullet is fired into the side of a 10 kg square panel which is initially at rest. Determine a) the angular velocity of the panel immediately after the bullet becomes embedded and b) the 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 the principle of impulse and momentum.

• The final angular velocity is found from

the moments of the momenta and impulses about A.

• The reaction at A is found from the

horizontal and vertical momenta and impulses.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 43

Sample Problem 17.9

SOLUTION:

• Consider a system consisting of

the bullet and panel. Apply the principle of impulse and momentum.

• The final angular velocity is

found from the moments of the momenta and impulses about A. moments about A:

( ) ( )

2 2

m 0.225 m 0.35 ω

P P B B

I v m v m + = +

( )

2 2

m 0.225 ω = v

( )( )

2 2 2 6 1

m kg 3375 m 45 . kg 10 6 1 ⋅ = = = b m I

P P

( )( )( ) ( )( )( )

2 2

3375 . 225 . 225 . kg 10 35 . 450 02 . ω ω + =

( )

s m 839 . 225 . s rad 67 . 4

2 2 2

= = = ω ω v

s rad 67 . 4

2 =

ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 44

Sample Problem 17.9

( )

s m 839 . 225 . s rad 67 . 4

2 2 2

= = = ω ω v

• The reactions at A are found

from the horizontal and vertical momenta and impulses. x components:

( )( ) ( ) ( )( )

839 . 10 0006 . 450 02 .

2

= + = ∆ +

x p x B B

A v m t A v m N 1017 − =

x

A N 1017 =

x

A

y components: = + t Ay∆ =

y

A

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 45

Sample Problem 17.10

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 the principle of impulse and

momentum.

• The moments about A of the momenta and

impulses provide a relation between the final angular velocity of the rod and velocity of the sphere.

• The definition of the coefficient of

restitution provides a second relationship between the final angular velocity of the rod and velocity of the sphere.

• Solve the two relations simultaneously for

the angular velocity of the rod and velocity

• f the sphere.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 46

Sample Problem 17.10

SOLUTION:

• Consider the sphere and rod as a

single system. Apply the principle

• f impulse and momentum.
• The moments about A of the

momenta and impulses provide a relation between the final angular velocity of the rod and velocity of the rod. moments about A:

( ) ( ) ( )

ω′ + ′ + ′ = I v m v m v m

R R s s s s

m 6 . m 2 . 1 m 2 . 1

( ) ( )( )

2 2 12 1 2 12 1

m kg 96 . m 2 . 1 kg 8 m 6 . ⋅ = = = ′ = ′ = ′ mL I r vR ω ω

( )( )( ) ( ) ( ) ( )( ) ( )

( )ω

ω ′ ⋅ + ′ + ′ =

2

m kg 96 . m 6 . m 6 . kg 8 m 2 . 1 kg 2 m 2 . 1 s m 5 kg 2

s

v ω′ + ′ = 84 . 3 4 . 2 12

s

v

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 47

Sample Problem 17.10

Moments about A: ω′ + ′ = 84 . 3 4 . 2 12

s

v

• The definition of the coefficient of

restitution provides a second relationship between the final angular velocity of the rod and velocity of the sphere.

( )

( ) ( )

s m 5 8 . m 2 . 1 = ′ − ′ − = ′ − ′

s s B s B

v v v e v v ω Relative velocities:

• Solve the two relations

simultaneously for the angular velocity of the rod and velocity of the sphere. Solving, s m 143 . − = ′

s

v s m 143 . = ′

s

v rad/s 21 . 3 = ′ ω rad/s 21 . 3 = ′ ω

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 48

Sample Problem 17.11

A square package of mass m moves down conveyor belt A with constant velocity. At the 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:

• Apply the principle of impulse and

momentum to relate the velocity of the package on conveyor belt A before the impact at B to the angular velocity about B after impact.

• Apply the principle of conservation of

energy to determine the minimum initial angular velocity such that the mass center

• f the package will reach a position

directly above B.

• Relate the required angular velocity to the

velocity of conveyor belt A.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 49

Sample Problem 17.11

SOLUTION:

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

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

( )(

)

( )(

)

2 2 2 2 2 1 1

ω I a v m a v m + = +

( )

2 6 1 2 2 2 2

a m I a v = = ω

( )(

)

( )( ) ( )

2 2 6 1 2 2 2 2 2 2 1 1

ω ω a m a a m a v m + = +

2 3 4 1

ω a v =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 17 - 50

Sample Problem 17.11

• Apply the principle of conservation of energy to determine the

minimum initial angular velocity such that the mass center of the package will reach a position directly above B.

3 3 2 2

V T V T + = +

2 2

Wh V =

( ) ( )

2 2 2 3 1 2 2 2 6 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2

ω ω ω ω ma ma a m I mv T = + = + =

3 3

Wh V =

3 =

T (solving for the minimum ω2)

( ) ( )

( )

a a GB h 612 . 60 sin 15 45 sin

2 2 2

= ° = ° + ° = a a h 707 .

2 2 3

= =

( ) ( )

a g a a a g h h ma W Wh Wh ma 285 . 612 . 707 . 3 3

2 2 3 2 2 2 3 2 2 2 2 3 1

= − = − = + = + ω ω a g a a v 285 .

3 4 2 3 4 1

= = ω ga v 712 .

1 =