Introduction to Dynamics (N. Zabaras)

Plane Motion of Rigid Bodies: Energy and Momentum

MethodsProf. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: nzabaras@gmail.com

URL: http://www.zabaras.com/

February 26, 2016

1

Introduction to Dynamics (N. Zabaras)

Introduction

17 - 2

• 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.

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 the coefficient of restitution.

Closely following the Vector Mechanics for Engineers, Beer and Johnston (Chapter 17), and

Engineering Mechanics: Dynamics Hibbeler (Chapter 18)

Introduction to Dynamics (N. Zabaras)

2

2

/

1

2

1(1)

2

( ) ( ) ( )

[( ) ] [( ) ]

i i

im

i P i P

P x P y

P x P y

T d mv

T d mv

v v ω x y

v y v ωx

v v v

i j k i j

i j

2 2 2

2 2 2 2 2 2

2 2 2

2 2 2

2 2

[( ) ] [( ) ]

( ) 2( ) ( ) 2( )

2( ) 2( ) (2)

(2) (1)

1 1( ) ( )

2 2

1 1( ) ( )

2 2

(

i i i P x P y

P x P x P y P y

P P x P y

P P x P ym m m m

P P x P y p

G

v v y v x

v v y ω y v v x ω x

v v y v x ω r

T dm v v ω y dm v ω xdm ω r dm

mv v ωym v ωxm I ω

P mass c

v v

) 0enter y x

2 21 1

2 2G GT mv I ω

Kinetic Energy

3

Introduction to Dynamics (N. Zabaras)

21

2GT mv 2 21 1

2 2G GT mv I ω

2 21 1

2 2

Translational Rotational

Kinetic energy Kinetic energy

G GT mv I ω

ωrv GG 2 2 21 1

2 2G GT m r ω I ω

2 21( )

2G GI m r ω

21

2OI ω

Translation Rotation General

Kinetic Energy for Rigid Bodies

4

Introduction to Dynamics (N. Zabaras)

mB=6 kg, mD=10 kg, mC=12 kg

No slipping, Total K.E = ?Block (Translation)

2 21 1(6 )(0.8 / ) 1.92 J

2 2B B BT m kg m s

Disk (Rotation)2 2 21 1 1

( )2 2 2

D A D D D DT I m r

2 21 1[ (10 )(0.1 ) ](8 / ) 1.60 J

2 2DT kg m rad s

Cylinder ( Translation & Rotation)2 2 2 2 21 1 1 1

( )2 2 2 2

C G G C G C C CT m I m m r

2 2 21 1 1(12 )(0.4 / ) [ (12 )(0.1 ) ](4 / ) 1.44 J, where 4 0.1 0.4 /

2 2 2C G C CT kg m s kg m rad s v r m s

Total B D CT T T T 1.92 1.6 1.44 4.96 JTotalT

0.88 /

0.1

BD

D

rad sr

0.84 /

2 0.2

BC

C

rad sr

Example

5

Introduction to Dynamics (N. Zabaras)

1 1 2 2T U T

2 2 2 2 2 2

1 1 2 1 2 1 2 2

1 1 1 1 1. ( ) ( ) ( )

2 2 2 2 2G G G Gm I F S W y k s s M m I

S rr

Translation

n

Rotation

Principle of Work and Energy

6

Introduction to Dynamics (N. Zabaras)

Principle of Work & Energy: Example

• Expressing the work of conservative forces as

a change in potential energy, the principle of

work and energy becomes

2211 VTVT

1 1 2 2

22

2

1 10 sin

2 3 2

3sin

T V T V

mlmgl

g

l

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

7

Introduction to Dynamics (N. Zabaras)

Sample Problem

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.

.sftlb5.10 2I

ft.lb60

Determine the velocity of the block

after it has moved 4 ft 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

rv

8

Introduction to Dynamics (N. Zabaras)

Sample Problem

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

1.25srad80.4

ft1.25

sft6 222

11

v

r

v

r

vrv

• Apply the 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

22

222

2

222

1222

12

09.725.1

5.102

1

2.32

240

2

1v

vv

IvmT

9

Introduction to Dynamics (N. Zabaras)

Sample Problem

lbft255212

1212

11 ImvT

22

222

1222

12 09.7 vIvmT

• Note that the 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,

10

Note the friction

does negative work

Introduction to Dynamics (N. Zabaras)

0

( ) ( )

G

G

A G

L mv

H

H d mv

G

G G

O G G G

L mv

H I ω

H I ω r mv

( )

G

G G

A G G

L mv

H I ω

H I ω d mv

Translation Rotation About a Fixed Axis General Plane Motion

2

2( )

G G

G G

O

I ω mr ω

I m r ω

I ω

Angular Momentum

11

Introduction to Dynamics (N. Zabaras)

( )

G G

A G G

H I ω

H I ω d mv

1 1 1 2 2 2G G G GG

I mv r M dt F r dt I mv r

Angular Impact and Momentum Principles

12

Introduction to Dynamics (N. Zabaras)

1 (1 2) 2

syst.angular syst.angular syst.angular

momentum impulse momentumO O O

Impact and Momentum Principles

13

Introduction to Dynamics (N. Zabaras)

1 1 2 2

linear linear linear

momentum impulse momentum

1 2

angular angular

momentum momentumO O

1 2( ) ( )m m

•Conservation of Angular Momentum

•Conservation of Linear Momentum

1 2

1 2

( ) ( )G G G G

H H

I ω m r I ω m r

Conservation of Momentum

14

Introduction to Dynamics (N. Zabaras)

Sample Problem

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.

mN6 M

mm80kg3

mm200kg10

BB

AA

km

km

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.

15

Introduction to Dynamics (N. Zabaras)

Sample Problem

SOLUTION:

• Consider each gear separately and apply impulse and momentum (use results

for moments of inertia and final angular velocities from Problem 8 of HW 7)

sN2.40

srad1.25mkg400.0m250.0

0 2

Ft

Ft

IFtr AAA

moments about A:

moments about B:

srad8.62mkg0192.0

m100.0mN6

0

2

2

Ftt

IFtrMt BBB

• Solve the angular momentum equations for the two gears

simultaneously for the unknown time and tangential force.

N 46.2s 871.0 Ft

16

Introduction to Dynamics (N. Zabaras)

1

2

m 60

radius of gyration 150

(5 ) ,

2

? t 3 sec

G

kg

k mm

M t N m

ω rad / s

M

32 2

20

60(0.15) (2) 5 60(0.15)t dt ω

1 2( ) ( )G G GH M dt H

2 18.7( / sec)rad

2

1 21

t

G G Gt

I ω M dt I ω

Example

17

Introduction to Dynamics (N. Zabaras)

1 2G G G G GI mv r M dt I mv r

Problem

2

2 2

1 50 500 (2)(1.2)(4) (0.6) (0.6 )(0.6)

2 32.2 32.2ω ω

Initially at rest, then apply P = 2 Ib

W = 50 Ib

Roll without slipping at A

What is = ? in 4 sec.

1 2A AH M dt H

2 11.4 /ω rad s

Example

18

Note A is the instant.

center of rotation

Moments around A

Izz from lecture 7,

slide 7

Introduction to Dynamics (N. Zabaras)

1 (1 2) 2

syst.angular syst.angular syst.angular

momentum impulse momentumA A A

1 1( ) Bm r I

22

( )6(2) (0.2) 0.4(10) 58.86(0.2) (3 ) 6 (0.2)( ) 0.4[ ]

0.2

BBs

2

2

41.716 3( )

( ) 13 /

B

B m s

mg r t 2 2 ( )Bm r I

Impulse and Momentum

19

Introduction to Dynamics (N. Zabaras)

Sample Problem

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 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 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.

20

𝐼 =2

5𝑚𝑅2

For a

sphere

Introduction to Dynamics (N. Zabaras)

Sample Problem

SOLUTION:

• Apply principle of impulse and

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 the linear and angular

velocities and solve for the time at

which sliding stops.

g

vt

k1

7

2

21

Introduction to Dynamics (N. Zabaras)

Sample Problem

x components: gtvv k 12

y components: mgWN

moments about G: tr

gk2

52

Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2

tr

grgtv

rv

kk

2

51

22

g

vt

k1

7

2

• Evaluate the 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

22