Physics 151: Lecture 20 Today’s Agenda



Topics (Chapter 10) :

 Rotational Kinematics  Rotational Energy  Moments of Inertia Ch. 10.1-3 Ch. 10.4

Ch. 10.5

Physics 151: Lecture 20, Pg 1

Rotation

 Up until now we have gracefully avoided dealing with the rotation of objects.  We have studied objects that slide, not roll.

 We have assumed wheels are massless.

 Rotation is extremely important, however, and we need to understand it !

 Most of the equations we will develop are simply rotational versions of ones we have already learned when studying linear kinematics and dynamics.

Physics 151: Lecture 20, Pg 2

Recall Kinematic of Circular Motion:

y

x y



= R = R cos(



) sin(



) = R = R = tan -1 (y/x) cos(



sin(



t) t)

v R

  

t (x,y) s

x 

=



t s s = v t = R



= R



t

 is angular velocity

v =



R

Animation For uniform circular motion:

a



v 2 R

Physics 151: Lecture 20, Pg 3

See text: 10.1

Example:

 The angular speed of the minute hand of a clock, in rad/s , is: a. b. c. d. p/ 1800 p /60 p /30 p e. 120 p Physics 151: Lecture 20, Pg 4

See text: 10.1

Rotational Variables

 Rotation about a fixed axis:  Consider a disk rotating about an axis through its center:  First, recall what we learned about Uniform Circular Motion:  

d



dt v



dx

(Analogous to )

dt

  Physics 151: Lecture 20, Pg 5

See text: 10.1

Rotational Variables...

  Now suppose  can change as a function of time: We define the angular acceleration:  

d



dt



d 2



dt 2

 Consider the case when  is constant.

 We can integrate this to find  and  as a function of time:       

constant

   0  0  

t

  0

t

 1 2 

t

2 Physics 151: Lecture 20, Pg 6

See text: 10.1

Example:

 The graphs below show angular velocity as a function of time. In which one is the magnitude of the angular acceleration constantly decreasing ?

Physics 151: Lecture 20, Pg 7

See text: 10.2

Rotational Variables...

  

constant

 

0

 

t

  

0

 

0 t



1 2



t 2

 Recall also that for a point a distance

R

away from the axis of rotation: 

x =



R



v =



R

And taking the derivative of this we find 

a =



R v R

 

x

 Animation Physics 151: Lecture 20, Pg 8

 See text: 10.3

Summary (with comparison to 1-D kinematics) Angular Linear

a



constant

 

constant

   0  

t

 

0

 

0 t



1 2



t 2 v



v 0



at x



x 0

 

1 2 at 2

And for a point at a distance

R

from the rotation axis:

x = R



v =



R a =



R

Physics 151: Lecture 20, Pg 9

See text: 10.1

Example: Wheel And Rope

 A wheel with radius from rest at

t = 0 R = 0.4m

rotates freely about a fixed axle. There is a rope wound around the wheel. Starting , the rope is pulled such that it has a constant acceleration

a = 4m/s 2

. How many revolutions has the wheel made after

10

(One revolution = 2 p radians) seconds?

a

R

 500 rad x 1 2 p 

80 rev rev rad

Physics 151: Lecture 20, Pg 10

See text: 10.1

Example:

 The turntable of a record player has an angular velocity of 8.0 rad/s when it is turned off. The turntable comes to rest angular acceleration.

2.5 s after being turned off. Through how many radians does the turntable rotate after being turned off ? Assume constant a. 12 rad b. 8.0 rad c. 10 rad d. 16 rad e. 6.8 rad Physics 151: Lecture 20, Pg 11

Rotation & Kinetic Energy

  Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods).

The kinetic energy of this system will be the sum of the kinetic energy of each piece:

m 4

r

4



r

1 m 1 m 3

r

3

r

2 m 2

Recall text 9.6, systems of particles, CM Physics 151: Lecture 20, Pg 12

Rotation & Kinetic Energy...

 So:

K



i



1 2 m v 2

but

K

 

2 i m i

 

2



1 2



2



i v i =



r i 2

which we write as:

K



1 2 I



2

v

4 m 4

r

4



I

 

m r i 2

Define the

moment of inertia

about the rotation axis

m 3

r

3

v

3

Recall text 9.6, systems of particles, CM

r

2 m 2

v

2

v

1

r

1 m 1 I

has units of

kg m 2

.

Physics 151: Lecture 20, Pg 13

Lecture 20,

Act 1

Rotational Kinetic Energy

 I have two basketballs. BB#2 is on a 0.2m

BB#1 is attached to a 0.1m

long rope. I spin around with it at a rate of 2 revolutions per second. long rope. I then spin around with it at a rate of 2 revolutions per second. What is the ratio of the kinetic energy of BB#2 to that of BB#1?

A) 1/4 B) 1/2 C) 1 D) 2 E) 4 BB#1 BB#2 Physics 151: Lecture 20, Pg 14

Rotation & Kinetic Energy...

 The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System

K



1 2 mv 2 v

is “linear” velocity

m

is the mass.



K



1 2 I



2

is angular velocity

I

is the moment of inertia about the rotation axis.

I



i



m r 2

Physics 151: Lecture 20, Pg 15

See text: 10.4

 So

K



1 2 I



2

Moment of Inertia

where

I



i



m r 2

 Notice that the moment of inertia

I

distribution of mass in the system.

depends on the  The further the mass is from the rotation axis, the bigger the moment of inertia.

 For a given object, the moment of inertia where we choose the rotation axis (unlike the center of mass).

will depend on  We will see that in rotational dynamics, the moment of inertia

I

appears in the same way that mass

m

does when we study linear dynamics !

Physics 151: Lecture 20, Pg 16

See text: 10.5

Calculating Moment of Inertia

 We have shown that for

N

discrete point masses distributed about a fixed axis, the moment of inertia is:

I



i N



m r



1 2

where

r

is the distance from the mass to the axis of rotation.

Example: Calculate the moment of inertia of four point masses (

m

) on the corners of a square whose sides have length

L

, about a perpendicular axis through the center of the square:

m L

See example 10.4 (similar)

m m m

Physics 151: Lecture 20, Pg 17

See text: 10.5

Calculating Moment of Inertia...

 The squared distance from each point mass to the axis is:

m L/2 r m L m m I = 2mL 2

See example 10.4 (similar) Physics 151: Lecture 20, Pg 18

See text: 10.5

Calculating Moment of Inertia...

 Now calculate

I

for the same object about an axis through the center, parallel to the plane (as shown):

I



i N

 

1 m i r i 2



m L 2 4



m L 2 4



m L 2 4



m L 2 4



4 m L 2 4 r m I = mL 2 m L m m

See example 10.4 (similar) Physics 151: Lecture 20, Pg 19

See text: 10.5

Calculating Moment of Inertia...

 Finally, calculate

I

for the same object about an axis along one side (as shown):

I



i N

 

1 m i r i 2



mL 2



mL 2



m 0 2



m 0 2 r m m I = 2mL 2 L m m

See example 10.4 (similar) Physics 151: Lecture 20, Pg 20

See text: 10.5

Calculating Moment of Inertia...

 For a single object,

I

clearly depends on the rotation axis !!

I = 2mL 2 L m m m m

See example 10.4 (similar)

I = mL 2 I = 2mL 2

Physics 151: Lecture 20, Pg 21

Lecture 20,

Act 2

Moment of Inertia

 A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the

a

,

b

, and

c

axes is

I

a ,

I

b , and

I

c respectively.

 Which of the following is correct: (a) (b) (c)

I

a >

I

b >

I

c

I

a >

I

c >

I

b

I

b >

I

a >

I

c a b c Physics 151: Lecture 20, Pg 22

Lecture 20,

Act 2

Moment of Inertia

 Label masses and lengths:  Calculate moments of inerta:

I a



m

 

2



m

 

2



8 mL 2 I b



mL 2



mL 2



mL 2



3 mL 2 I c



m

 

2



4 mL 2 m

So (b) is correct:

I

a >

I

c >

I

b

m m L L

a b c Physics 151: Lecture 20, Pg 23

See text: 8-5

Calculating Moment of Inertia...

 For a discrete collection of point masses we found:

N I



i

 

1 m r 2

 For a continuous solid object we have to add up the

mr 2

contribution for every infinitesimal mass element

dm

.

dm

 We have to do an integral to find

I

:

I

 

r dm r

Physics 151: Lecture 20, Pg 24

See text: 10.5

Moments of Inertia

 Some examples of

I

for solid objects:

I



MR 2 R

Thin hoop (or cylinder) of mass

M

and radius

R

, about an axis through its center, perpendicular to the plane of the hoop.

I I    

r

2

dm R

2

dm



R

2 

dm



MR

2

R I

 

1 2 MR 2

Thin hoop of mass

M

and radius

R

, about an axis through a diameter.

see Example 10.5 in the text Physics 151: Lecture 20, Pg 25

Moments of Inertia

L  Some examples of

I

for solid objects: dr Solid disk or cylinder of mass

M

and radius

R

, about a perpendicular axis through its center.

r

R

I  

r

2

dm

I  1 2

MR

2   Physics 151: Lecture 20, Pg 26

See text: 10.5

Moments of Inertia...

 Some examples of

I

for solid objects:

R I



2 MR 2 5

Solid sphere of mass

M

and radius

R

, about an axis through its center.

R

I  2 3

MR

2 Thin spherical shell of mass

M

and radius

R

, about an axis through its center.

 See Table 10.2, Moments of Inertia Physics 151: Lecture 20, Pg 27

Recap of today’s lecture

 Chapter 9,  Center of Mass  Elastic Collisions  Impulse Physics 151: Lecture 20, Pg 28