physics 218: lecture 15

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Physics 218 Fall 2010 Slide 1 Physics 218: Lecture 15 Chapter 8: Rotational Motion, cont Torque and Newton’s Law Moment of inertia Momentum conservation Law

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Page 1: Physics 218: Lecture 15

Physics 218 Fall 2010 Slide 1

Physics 218: Lecture 15

Chapter 8: Rotational Motion, cont

Torque and Newton’s Law Moment of inertia Momentum conservation Law

Page 2: Physics 218: Lecture 15

Physics 218 Fall 2010

distance along arc

rs

raacceleration

velocity rv

Rotational motion

tv

angle

angularvelocity

angularacceleration

t

Basically, to get the linear quantity, take the angular quantity (in radians) and multiply by r.

Linear motion

s

Page 3: Physics 218: Lecture 15

Physics 218 Fall 2010

Torque

F

Force causes a torque. Amount of torque depends on:

magnitude of force

distance from pivot point

direction of force

r

F

Page 4: Physics 218: Lecture 15

Physics 218 Fall 2010

Torque – Simplest case, F perpendicular to r

F

r

rF

Page 5: Physics 218: Lecture 15

Physics 218 Fall 2010

Torque – General case

F

rrF

rF sin

sinFF F

Page 6: Physics 218: Lecture 15

Physics 218 Fall 2010

Quiz

At what angle is the maximum torque produced?

A.

= 90 degreesB.

= 45 degrees

C.

= 0 degreesF

r

Page 7: Physics 218: Lecture 15

Physics 218 Fall 2010

Torque is the rotational analog to force

Force effects motion through accelerations

amF

Torque effects motion through

rotational accelerations

I

equilibrium

0 0

0F

0

Page 8: Physics 218: Lecture 15

Physics 218 Fall 2010

Rotational Equilibrium – The leverWhat is the force needed to support the

elephant (neglect the mass of the plank)?

forces on plank

0F

F

0F

eF

eF pF

pF torques on plank

0r

er

0 eerFFr

ee FrrF

force amplification

Page 9: Physics 218: Lecture 15

Physics 218 Fall 2010

Rotational Equilibrium – The lever - Summary

F

rer

ee FrrF

0

0 eerFFr

eF

Page 10: Physics 218: Lecture 15

Physics 218 Fall 2010

Rotational Equilibrium – The leverThis time, the lever itself has mass.

Treat the mass as centered at one point. For objects of uniform mass, that point is at the geometric center of the object. For

non-uniform objects, see Section 7.6.

Page 11: Physics 218: Lecture 15

Physics 218 Fall 2010

Rotational Equilibrium – The leverThis time, the lever itself has mass, mp

Treat the mass as centered at one point.

mp

pr er

me

Page 12: Physics 218: Lecture 15

Physics 218 Fall 2010

Rotational EquilibriumA plank has a length of 5 m and has

mass 40 kg. It protrudes from the table 1 m and can support an elephant. What is

the elephant’s weight?

Page 13: Physics 218: Lecture 15

Physics 218 Fall 2010

In summary, if the system is in rotational equilibrium, look for a balance of torques.

Constant angular velocity, including zero angular

velocity.

0

Page 14: Physics 218: Lecture 15

Physics 218 Fall 2010

Rotational Dynamics

I

0 produces non-zero

Moment of Inertia, I, impedes changes in rotational acceleration, just as the mass, m, impedes changes in linear acceleration.

Page 15: Physics 218: Lecture 15

Physics 218 Fall 2010

Moment of Inertia

I miri2

More mass, more moment of inertia.

Mass further away from axis of rotation, more moment of inertia

Page 16: Physics 218: Lecture 15

Physics 218 Fall 2010 16

Calculating Moment of Inertia

I miri2

i1

N

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

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:

mm

mm

L

Page 17: Physics 218: Lecture 15

Physics 218 Fall 2010 17

Calculating Moment of Inertia

mm

mm

Lr

L/2The squared distance from each point mass to the axis is:

r2 L2

2

L2

2

2L2

4

L2

2

Using the Pythagorean Theorem

I miri2

i1

N m

L2

2 m

L2

2 m

L2

2 m

L2

2 4m

L2

2

So:

I = 2mL2

Page 18: Physics 218: Lecture 15

Physics 218 Fall 2010 18

Calculating Moment of InertiaNow calculate I for the same object about an axis through the center, parallel to the plane (as shown):

So:

I = mL2

mm

mm

L

r

I miri2

i1

N m

L2

4 m

L2

4 m

L2

4 m

L2

4 4m

L2

4

Page 19: Physics 218: Lecture 15

Physics 218 Fall 2010 19

Calculating Moment of InertiaFor a single object, I clearly depends on the rotation axis.

I = mL2

mm

mm

L

r

mm

mm

L

mm

mm

L

L

r

L/2

I = 2mL2 I = 2mL2

I miri2

i1

N

Page 20: Physics 218: Lecture 15

Physics 218 Fall 2010

Moment of Inertia

I miri2

Neglect the mass of the hoop. Which will have a greater moment of inertia?

Page 21: Physics 218: Lecture 15

Physics 218 Fall 2010

Quiz

The two rods are identical, the axis of rotation is different. Which configuration has

a greater moment of inertia?

A. B.

Page 22: Physics 218: Lecture 15

Physics 218 Fall 2010 22

Moment of Inertia (from your book, don’t bother copying)

Page 23: Physics 218: Lecture 15

Physics 218 Fall 2010

Combined Linear and Rotational Motion

r

mw

mp

A weight is attached to a massive frictionless pulley. What is its acceleration when dropped?

forces on mass

amF w

amwT

torques on pulley

Igmw Tr

ra /2

21 rmI p

rarmTr p

2

21 amT p2

1

amgmam wwp 21

wp

w

mm

gma

21

gmw

T

Page 24: Physics 218: Lecture 15

Physics 218 Fall 2010

Combined Linear and Rotational Motion

r

mw

mp

A weight is attached to a massive frictionless pulley. What is its acceleration when dropped?

wp

w

mm

gma

21

mass of pulley slows down acceleration

pw mm What if ?

wp

w

mm

gma

21 0

ga

(passes sanity check)

Page 25: Physics 218: Lecture 15

Physics 218 Fall 2010

For Next Time

Read Chapter 9.1-9.6