chapter 12: rotation of rigid bodies - srjcsrjcstaff.santarosa.edu/~lwillia2/40/40ch12p1_s14.pdf ·...

Chapter 12: Rotation of Rigid Bodies

Center of Mass

Moment of Inertia

Torque

Angular Momentum

Rolling

Statics

Translational vs Rotational

2

/

/

1/ 2

m

x

v dx dt

a dv dt

F ma

p mv

KE mv

Work Fd

P Fv

2

/

/

1/ 2

I

d dt

d dt

I

L I

KE I

Work

P

2

c

t

s r

v r

a r

a r

Fr

L pr

Connection

Center of Mass The geometric ‘center’ or average location of the mass.

Rotational & Translational Motion Objects rotate about their Center of Mass.

The Center of Mass Translates as if it were a point particle.

CMCM

rv

d

dt

The Center of Mass Translates as if it were a point particle

and, if no external forces act on the system, momentum is

then conserved. This means:

EVEN if the bat EXPLODED into a thousand pieces, all the

pieces would move so that the momentum of the CM is

conserved – that is, the CM continues in the parabolic

trajectory!!!! THIS IS VERY VERY IMPORTANT!

Center of Mass

System of Particles Center of Mass

• A projectile is fired into the air and suddenly explodes

• With no explosion, the projectile would follow the dotted line

• After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion!

If no external forces act on the system, then the velocity of the CM doesn’t change!!

Center of Mass: Stability

If the Center of Mass is above the base

of support the object will be stable. If

not, it topples over.

This dancer balances en pointe by having her center of mass directly over her toes, her base of support.

Balance and Stability

Slide 12-88

Center of Mass

CM i i

i

m x

xM

The geometric ‘center’ or average location of the mass.

System of Particles: Extended Body:

CM

1dm

M r r

CM Lecture Problem

1. Identical particles are placed at the 50-cm and 80-cm

marks on a meter stick of negligible mass. This rigid

body is then mounted so as to rotate freely about a

pivot at the 0-cm mark on the meter stick. A) What is

the CM of the system? B) What is the torque acting on

the system? C) What is the rotational inertia of the

system about the end? D) If this body is released from

rest in a horizontal position, what is the angular

acceleration at the release? E) What is the angular

speed of the meter stick as it swings through its lowest

position?

Prelab

A meter stick has a mass of 75.0 grams and has two masses attached to it – 50.0 grams at the 20.0cm mark and 100.0 grams at the 75.0 cm mark.

(a) Find the center of mass of the system - that is, at what mark on the meter stick should the fulcrum be placed so that the system balances?

(b) A fulcrum is then placed at the CM. Sketch and label the system.

(c) Show that the net torque about the cm is zero.

(d) Calculate the rotational inertia of the system about the CM axis. Box answers, make it neat.

Center of Mass The geometric ‘center’ or average location of the mass.

Extended Body:

CM

1dm

M r r

Center of Mass of a Solid Object

Divide a solid object into

many small cells of mass m.

As m 0 and is replaced

by dm, the sums become

Before these can be integrated:

dm must be replaced by

expressions using dx and dy.

Integration limits must be established.

Slide 12-41

Example 12.2 The Center of Mass of a Rod

Slide 12-42

Example 12.2 The Center of Mass of a Rod

Slide 12-43

Extended Body:

Example

CM

1dm

M r r

You must generate an expression for the density and the mass differential, dm, from geometry and by analyzing a strip of the sign. We assume the sign has uniform density. If M is the total mass then the total volume and density is given by:

2( )

1

2

M Mydm dV ytdx ytdx dx

ababt

1,

12

2

MV abt

abt

Where a, b and t are the width, height and thickness of the sign, respectively. Then the mass element for the strip shown is:

CM

1 x x dm

M

3

2

0 0

1 2 2 2 2( )

3 3

aaMy b x

x dx x x dx aM ab ab a a

Newton’s 2nd Law for Rotation

I

The net external torques acting on an object

around an axis is equal to the rotational

inertia times the angular acceleration.

The rotational equation is limited to rotation about a single

principal axis, which in simple cases is an axis of symmetry.

Inertia thing

Acceleration thing

Force thing

Torque: Causes Rotations

The moment arm, d, is

the perpendicular

distance from the axis

of rotation to a line

drawn along the

direction of the force

sinFr Fd

lever arm: sind r

The horizontal component of F (F cos ) has no tendency to produce a rotation

Torque: Causes Rotations

sinFr Fd

The direction convention is: Counterclockwise rotations are positive. Clockwise rotations are negative.

If the sum of the torques is zero, the system

is in rotational equilibrium.

Newton’s 1st Law for Rotation

250 3 750girl N m Nm

500 1.5 750boy N m Nm

0

Torque

Is there a difference in torque? (Ignore the mass of the rope)

NO!

In either case, the lever arm is the same!

What is it? 3m

CM Lecture Problem

1. Identical particles are placed at the 50-cm and 80-cm

marks on a meter stick of negligible mass. This rigid

body is then mounted so as to rotate freely about a

pivot at the 0-cm mark on the meter stick. A) What is

the CM of the system? B) What is the torque acting

on the system? C) What is the rotational inertia of the

system about the end? D) If this body is released from

rest in a horizontal position, what is the angular

acceleration at the release? E) What is the angular

speed of the meter stick as it swings through its lowest

position?

Newton’s 2nd Law for Rotation

I

The net external torques acting on an object

around an axis is equal to the rotational

inertia times the angular acceleration.

The rotational equation is limited to rotation about a single

principal axis, which in simple cases is an axis of symmetry.

Inertia thing

Acceleration thing

Force thing

Torque is a Vector!

= r x F

The direction is given by the right hand rule where the fingers

extend along r and fold into F. The Thumb gives the direction of .

The Vector Product

• The magnitude of C is

AB sin and is equal to the

area of the parallelogram

formed by A and B

• The direction of C is

perpendicular to the plane

formed by A and B

• The best way to determine

this direction is to use the

right-hand rule

ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k

COMPARE!

CROSS PRODUCT

F and d must be

mutually perpendicular! sin

r F

Fr Fd

sin

L r p

L mvr

cosW F d Fd

DOT PRODUCT

F and d must be

mutually PARALLEL!

CROSS PRODUCT

L and p must be

mutually perpendicular!

Two vectors lying in the xy plane are

given by the equations A = 5i + 2j and

B = 2i – 3j. The value of AxB is

a. 19k

b. –11k

c. –19k

d. 11k

e. 10i – j

ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k

CM Lecture Problem

1. Identical particles are placed at the 50-cm and 80-cm

marks on a meter stick of negligible mass. This rigid

body is then mounted so as to rotate freely about a

pivot at the 0-cm mark on the meter stick. A) What is

the CM of the system? B) What is the torque acting

on the system? C) What is the rotational inertia of the

system about the end? D) If this body is released from

rest in a horizontal position, what is the angular

acceleration at the release? E) What is the angular

speed of the meter stick as it swings through its lowest

position?

Net external torques

Find the Net Torque

( 20 )(.5 ) (35 cos30)(1.10 ) N m N m

1 1 2 2 Fd F d

( 20 )(.5 ) (35 )(1.10 sin60) N m N m

OR

23.3 CCW Nm

F and d must be

mutually

perpendicular!

sinFr Fd