Chapter 10

Rotation(Part 3)

Outline for today

• Torque

• Newton’s Second Low for Rotation

• Work and Rotational Kinetic Energy

• Example Problems

RECAP: Rotational Inertia• Translation

• M does NOT dependon the motion!

• Rotation

• I depends on:• Shape of the body• Mass distribution• The axis of rotation

• Parallel-Axis Theorem

!

K =1

2Mv

2

!

K =1

2I" 2

!

M = mi"

!

I = Ii" = m

iri

2"

!

M = dm"

!

I = r2" dm

Parallel-Axis Theorem• If the rotational inertia (Icom) of the body about

a parallel axis that extends through the body'scenter of mass is known we can calculate therotational inertia Inew around ANY parallel axis

Inew = Icom + Mh2

where h is the perpendicular distance between thegiven axis and the axis through the center of mass

= +IcomInew

Mh2

com

Example problem• Find the rotational inertia of a thin uniform rod

with length L and mass M rotating around an axispassing through one of its end.

= +IcomInew

Mh2

com

!

Inew

= Icom

+ Mh2

!

Inew

=1

12ML

2+ M

L

2

"

# $ %

& '

2

=1

12+1

4

"

# $

%

& ' ML

2=4

12ML

2=1

3ML

2

Quiz Question• Consider a thin rotating rod with length L and

mass M. It rotates with angular velocity ω. Ifwe double the length L the rotational kineticenergy will:

A) Remain the same.B) Double.C) be half the original value.D) be four times the original value

How hard is it to open a door?

• It depends!• How hard we push (magnitude of the force)• Where we push (where the force is applied)• In what direction we push (the direction of the force)

Torque (“twist”)

• It shows the ability of a force torotate an object.

• Units: N.m

τ

!

" = rF sin#

!

" = r#F

!

" = rFt

!

r " =

r r #

r F

Torque

• Positive: counterclockwise rotation• Negative: clockwise rotation• Zero: the extension of the force passes

through the axis of rotation.

τ > 0 τ < 0 τ = 0

F

F

F

Quiz Question• The length of a bicycle pedal arm is 0.1 m and a

downward force of 200 N is applied to the pedalby a rider. What is the magnitude of the torqueabout the pedal arm’s pivot when the arm is 30o

with respect to the vertical ? A) 10 B) 50 C) 100 D) 150 E) 200

!

" = Frsin#

!

" = 200 # 0.1# sin30 =10N .m

F r

Newton’s Second Law for Rotation• Newton’s II law: The net force applied

to a body results in a acceleration so that

• What about rotation? What is the resultwhen we apply a torque?

!

r F

net= m

r a

!

" = Frsin# = (ma)rsin# = matr

" = m($r)r = mr2$

" = I$

x

y F

FtFr

φ

!

r " net

=r " i# = I

r $

Example problem• Two particles of equal mass m are attached to the

ends of a rigid, massless rod of length L1+L2. Therod is held horizontally on a fulcrum and thenreleased. What are the magnitudes of the angularacceleration of particle 1 and 2 ?

L1 L2

Fg Fg

m m

!

"net

= I#

!

" net = "1

+ "2

= Fg1L1 # Fg2L2

" net = mg(L1# L

2)

!

I = m1L1

2+ m

2L2

2= m(L

1

2+ L

2

2)!

" = #net/I

!

" = gL1# L

2

L1

2+ L

2

2

Work and Rotational Kinetic Energy

!

W = "K = K f #Ki

W =1

2mv f

2#1

2mvi

2

W =1

2mr

2$ f

2#1

2mr

2$ i

2

W =1

2I$ f

2#1

2I$ i

2

x

y

!

r v

mr

• Work

• Power

!

P =dK

dt=1

2Id

dt(" 2

) = I"d"

dt

P = I"# = $"

• Variable torque.

• Constant torque.

• Power (2)

Work and Torque

x

y FFt

Fr

φ

!

dW =r F " d

r s

dW = Ft ds = Ftr d# = $ d#

W = $# i

# f

% d#Δθ

!

W = "#$ , #$ = $ f %$i

!

P =dW

dt= "

d#

dt= "$

Example problem• A yo-yo-shaped device mounted on a horizontal frictionless axis is

used to lift a 30 kg box. The outer radius R of the device is 0.5 m andthe radius r of the hub is 0.2 m. When horizontal force Fapp=140N isapplied the box has an upward acceleration of 0.8 m/s2. What is therotational inertia of the devise about its axis of rotation?

m=30 kgR=0.5 mr=0.2 mFapp=140 Na=0.8 m/s2

_________I=?

!

" net = I# $ I =" net

#

" net = FappR %Tr

for the hanging body : T %mg = ma

T = m(g + a) and # = a /r

I =r

a(FappR % (g + a)mr)

I =0.2

0.8(140 & 0.5 % (9.8 + 0.8) & 30 & 0.2)

I =1.6 kg.m2

mg

T

T