1

Rotation of a Rigid Body

Readings: Chapter 13

2

How can we characterize the acceleration during rotation?

- translational acceleration and

- angular acceleration

F ma

Newton’s second law:

3

Angular acceleration

1v

2v

Center of rotation

Both points have the same angular velocity

t

1 1v r 2 2v r

1r

Linear acceleration:

11 1

va r

t t

22 2

va r

t t

Both points have the same angular acceleration t

4

Rotation of Rigid Body:

Every point undergoes circular motion with the same angular velocity and the same angular acceleration

t

v r

5

The relation between angular velocity and angular acceleration is the same as the relation between linear velocity and linear acceleration

t

va

t

6

The Center of Mass

1 1 2 2

1 2cm

m x m xx

m m

For Rigid Body sometimes it is convenient to describe the rotation about the special point– the center of mass of the body.

Definition: The coordinate of the center of mass:

Rigid body consisting of two particles:

1 2m mIf then

1 2

2cmx x

x

2 0 0.5 0.5 0.12.0 0.5cmx m

7

The Center of Mass

i ii

cm

x m xdmx

M M

Definition: The coordinate of the center of mass:

i ii

cm

y m ydmy

M M

8

The Center of Mass: Example

220 0

0

1 12 2 2

L L

L

cm

Mx dx M xdxxdm L L Lx x

M M LM L L

The center of mass of a disk is the center O of the disk

O

9

Torque: Rotational Equivalent of Force

10

Torque

The rotation of the body is determined by the torque

sinFr

11

Torque

sinFr

090 Torque is maximum if

Torque is 0 if 0 00 180or

0

1 3

1 4

12

Torque sinFr

Torque is positive if the force is trying to rotate the body counterclockwise

Torque is negative if the force is trying to rotate the body clockwise

axis

1 0

2 0

3 0

4 0

The net torque is the sum of the torques due to all applied forces:

1F

2F

3F

4F

1 2 3 4net

13

Torque: Example sinFr Find the net torque

axis

030

1 5 3sin(30) 7.5N m

1F

2F

3F

4F

1 2 3 4 4 7.5 17.3 3.5 7.1 6.2net N m

5F

1m

2m

3m

4m

1 2 3 4 5 5F F F F F N

060

045

045

2 5 4sin(60) 17.3N m 3 5 1sin(45) 3.5N m

4 5 2sin(45) 7.1N m

5 5 0 0N m

14

Torque: Relation between the torque and angular acceleration:

v

rv

a r rt t

1F

F ma mr

2Fr mr

2net i i

i

m r I 2

i ii

I m r - moment of inertia

15

Moment of Inertia

2112

I MLThin rod, about center

Thin rod, about end 213

I ML

Cylinder (or disk), about center 212

I MR

Cylindrical loop, about center2I MR

16

Moment of Inertia: Parallel-axis Theorem

If you know the moment of Inertia about the center of mass (point O)

then the Moment of Inertia about point (axis) P will be 2

P OI I Md

OPd

2 2

2 2 2

2 2 2

( ) ( )

( ) ( ) 2 ( )

( ) 2 ( )

P i P i i O O P ii i

i O i i O i O ii i

i O i i O i Oi i

I x x m x x x x m

x x d m r r d r r d m

r r m d r r m Md I Md

17

Parallel-axis Theorem: Example 2P OI I Md

2112

I ML

Thin rod, about center of mass

22 2 2 21 1 1 1

12 2 12 4 3L

I ML M ML ML ML

18

net I Equilibrium: 0

0net

2d m

1 2m kg

?x

2 10m kg

Massless rod1n

2n

Two forces (which can results in rotation) acting on the rod

1 1 1n w m g

2 2 2n w m g 1 1 1n d m gd

2 2 2n x m gx

Equilibrium:1 2 0net

1 2 0m gd m gx 1

2

0.4m

x d mm

19

Rotational Energy:

2 2 21 1 2 2 3 3

2 2 2 2 2 21 1 2 2 3 3

2 2 2 2 21 1 2 2 3 3

1 1 1 ...2 2 2

1 1 1 ...2 2 21 1( ...)2 2

rotK m v m v m v

m r m r m r

m r m r m r I

Conservation of energy (no friction):

2 21 12 2

Mgy Mv I const

20

Kinetic energy of rolling motion 212rotK I

212rot PK I 2

P cmI I MR

2 2 2 2 22

1 1 1 12 2 2 2rot cm cm cm cmK I MR I v Mv

R

cmv R

Cylinder: 212cmI MR 2 2 21 1 1 3

2 2 2 4rot cm cm cmK Mv Mv Mv

Cylindrical loop: 2cmI MR 2 2 21 1

2 2rot cm cm cmK Mv Mv Mv

Solid sphere: 225cmI MR

2 2 21 2 1 72 5 2 10rot cm cm cmK Mv Mv Mv