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Chapter 8: Torque and Angular Momentum

•Rotational Kinetic Energy

•Rotational Inertia

•Torque

•Work Done by a Torque

•Equilibrium (revisited)

•Rotational Form of Newton’s 2nd Law

•Rolling Objects

•Angular Momentum

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§8.1 Rotational KE and Inertia

For a rotating solid body:

n

iiinn vmvmvmvmK

1

22222

211rot 2

1

2

1

2

1

2

1

For a rotating body vi=ri where ri is the distance from the rotation axis to the mass mi.

22

1

2

1

2rot 2

1

2

1

2

1 IrmrmKn

iii

n

iii

3

The quantity

n

iiirmI

1

2 is called rotational inertia or moment of inertia.

Use the above expression when the number of masses that make up a body is small. Use the moments of inertia in table 8.1 for extended bodies.

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Example: (a) Find the moment of inertia of the system below. The masses are m1 and m2 and they are separated by a distance r. Assume the rod connecting the masses is massless.

r1 r2m1 m2

r1 and r2 are the distances between mass 1 and the rotation axis and mass 2 and the rotation axis (the dashed, vertical line) respectively.

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2222

211

2

1

2 m kg 67.0

rmrmrmIi

ii

(b) What is the moment of inertia if the axis is moved so that is passes through m1?

Example continued:

2222

2

1

2 m kg 00.1

rmrmIi

ii

Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m , and r2 = 0.67 m.

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Example (text problem 8.4): What is the rotational inertia of a solid iron disk of mass 49.0 kg with a thickness of 5.00 cm and a radius of 20.0 cm, about an axis through its center and perpendicular to it?

From table 8.1:

222 m kg 98.0m 2.0kg 0.492

1

2

1 MRI

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§8.2 Torque

A rotating (spinning) body will continue to rotate unless it is acted upon by a torque.

hinge

Push

Q: Where on a door do you push to open it?

A: Far from the hinge.

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Torque method 1:

rF

Hinge end

FF

||F

r = the distance from the rotation axis (hinge) to the point where the force F is applied.

F is the component of the force F that is perpendicular to the door (here it is Fsin).

Top view of door

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The units of torque are Nm (not joules!).

By convention:

•When the applied force causes the object to rotate counterclockwise (CCW) then is positive.

•When the applied force causes the object to rotate clockwise (CW) then is negative.

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Torque method 2: Fr

r is called the lever arm and F is the magnitude of the applied force.

Lever arm is the perpendicular distance to the line of action of the force.

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Line of action of the force

Hinge end

F

r

Top view of door

Lever arm

The torque is:

sin

sin

rrr

r

sinrF

Fr

Same as before

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Example (text problem 8.12): The pull cord of a lawnmower engine is wound around a drum of radius 6.00 cm, while the cord is pulled with a force of 75.0 N to start the engine. What magnitude torque does the cord apply to the drum?

F=75 N

R=6.00 cm

Nm 5.4N 0.75m 06.0

rF

Fr

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Example: Calculate the torque due to the three forces shown about the left end of the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the bar.

4510

30

F1=25 N

F3=20 N

F2=30 N

X

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The lever arms are: m 695.010sinm4

m 73.160sinm2

0

3

2

1

r

r

r

Lever arm for F3

Lever arm for F2

4510

30

F1=25 N

F3=20 N

F2=30 N

X

Example continued:

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The torques are: Nm 9.13N 20m 0.695

Nm 9.51N 30m 73.1

0

3

2

1

The net torque is + 65.8 Nm and is the sum of the above results.

Example continued:

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§8.3 Work done by a Torque

.WThe work done by a torque is

where is the angle (in radians) the object turns through.

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Example (text problem 8.25): A flywheel of mass 182 kg has a radius of 0.62 m (assume the flywheel is a hoop).

(a) What is the torque required to bring the flywheel from rest to a speed of 120 rpm in an interval of 30 sec?

rad/sec 6.12sec 60

min 1

rev 1

rad 2

min

rev 120

f

Nm 4.2922

2

tmr

tmr

tmrrrmmarrF

fif

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(b) How much work is done in this 30 sec period?

J 560022

tt

tW

ffi

av

Example continued:

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§8.4 Equilibrium

The conditions for equilibrium are: 0

0

τ

F

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Example (text problem 8.36): A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.0 N. A cable, inclined at a 35 angle with the boom, is attached at a distance of 2.38 m from the hinge at the wall. The weight of the sign is 120.0 N. What is the tension in the cable and what are the horizontal and vertical forces exerted on the boom by the hinge?

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FBD for the bar:

Apply the conditions for equilibrium to the bar:

0sin2

)3(

0sin )2(

0cos )1(

bar

bar

xTLFL

w

TFwFF

TFF

sb

sbyy

xx

Example continued:

T

wbar

Fy

Fx

X

Fsb

x

y

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Equation (3) can be solved for T:

N 352sin

2

x

LFL

wT

sbbar

N 288cos TFxEquation (1) can be solved for Fx:

Equation (2) can be solved for Fy:N 00.2

sin

TFwF sbbary

Example continued:

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§8.5 Equilibrium in the Human Body

Example (text problem 8.43): Find the force exerted by the biceps muscle in holding a one liter milk carton with the forearm parallel to the floor. Assume that the hand is 35.0 cm from the elbow and that the upper arm is 30.0 cm long. The elbow is bent at a right angle and one tendon of the biceps is attached at a position 5.00 cm from the elbow and the other is attached 30.0 cm from the elbow. The weight of the forearm and empty hand is 18.0 N and the center of gravity is at a distance of 16.5 cm from the elbow.

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Example continued:

Fb

wFca

“hinge” (elbow joint)

N 130

0

1

32

321

x

xFwxF

xFwxxF

cab

cab

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§8.6 Rotational Form of Newton’s 2nd Law

I

Compare to aF m

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Example (text problem 8.57): A bicycle wheel (a hoop) of radius 0.3 m and mass 2 kg is rotating at 4.00 rev/sec. After 50 sec the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?

2MRI 0

rad/sec 1.25rev 1

rad 2

sec

rev00.4

f

i

2rad/s 50.0

tt

if

Nm 09.02 MRav

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§8.7 Rolling Objects

An object that is rolling combines translational motion (its center of mass moves) and rotational motion (points in the body rotate around the center of mass).

For a rolling object:

22cm

rotTtot

2

1

2

1 Imv

KKK

If the object rolls without slipping then vcm = R.

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Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp. Both objects start from rest and from the same height. Which object reaches the bottom of the ramp first?

h

The object with the largest linear velocity (v) at the bottom of the ramp will win the race.

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2

22

2222

2

2

1

2

1

2

1

2

1

2

100

R

Im

mghv

vR

Immgh

R

vImvImvmgh

KUKU

EE

ffii

fi

Apply conservation of mechanical energy:

Example continued:

Solving for v:

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2sphere

2disk

5

22

1

mRI

mRI

For the disk:

For the sphere:

ghv3

4disk

ghv7

10sphere

Since Vsphere> Vdisk the sphere wins the race.

ghv 2box Compare these to a box sliding down the ramp.

Example continued:

The moments of inertia are:

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How do objects in the previous example roll?

FBD:

Both the normal force and the weight act through the center of mass so =0. This means that the object cannot rotate when only these two forces are applied.

N

w

y

x

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FBD:

Add friction:

0cos

sin cm

wNF

maFwF

IrF

y

sx

sy

x

N

w

Fs

Also need acm = R and

The above system of equations can be solved for v at the bottom of the ramp. The result is the same as when using energy methods. (See text example 8.13.)

xavv 220

2

It is static friction that makes an object roll.

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§8.8 Angular Momentum

vp

pF

mtt

0

net lim

ωL

L

Itt

0

net lim

Units of p are kg m/s Units of L are kg m2/s

When no net external forces act, the momentum of a system remains constant (pi = pf)

When no net external torques act, the angular momentum of a system remains constant (Li = Lf).

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Example (text problem 8.70): A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a frequency of 0.500 rev/sec. What is the angular momentum? Assume a uniform disk.

rad/sec 14.3rev 1

rad 2

sec

rev500.0

/sm kg 079.02

1 22

MRIL

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Example (text problem 8.79): A skater is initially spinning at a rate of 10.0 rad/sec with I=2.50 kg m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces I to 1.60 kg m2?

He/she is on ice, so we can ignore external torques.

rad/sec 6.15rad/sec 0.10m kg 60.1

m kg 50.22

2

if

if

ffii

fi

I

I

II

LL

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§8.9 The Vector Nature of Angular Momentum

Angular momentum is a vector. Its direction is defined with a right-hand rule.

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Curl the fingers of your right hand so that they curl in the direction a point on the object moves, and your thumb will point in the direction of the angular momentum.

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Consider a person holding a spinning wheel. When viewed from the front, the wheel spins CCW.

Holding the wheel horizontal, they step on to a platform that is free to rotate about a vertical axis.

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Initially, nothing happens. They then move the wheel so that it is over their head. As a result, the platform turns CW (when viewed from above).

This is a result of conserving angular momentum.

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Initially there is no angular momentum about the vertical axis. When the wheel is moved so that it has angular momentum about this axis, the platform must spin in the opposite direction so that the net angular momentum stays zero.

Is angular momentum conserved about the direction of the wheel’s initial, horizontal axis?

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It is not. The floor exerts a torque on the system (platform + person), thus angular momentum is not conserved here.

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Summary

•Rotational Kinetic Energy

•Moment of Inertia

•Torque (two methods)

•Conditions for Equilibrium

•Newton’s 2nd Law in Rotational Form

•Angular Momentum

•Conservation of Angular Momentum