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Ch 9. Rotational DynamicsIn pure translational motion, all points on anobject travel on parallel paths.

The most general motion is a combination oftranslation and rotation.

Forces and Torques

Net force acceleration.

What causes an object to have an angular acceleration? TORQUE

The amount of torque depends on where and in what direction the force is applied, as well as the location of the axis of rotation.

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9.1 The Action of Forces and Torques on Rigid Objects

DEFINITION OF TORQUE

Magnitude of Torque = (Magnitude of the force) x (Lever arm)

FDirection: The torque is positive when the force tends to produce a counterclockwise rotation about the axis.

SI Unit of Torque: newton x meter (N·m)

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Action of Forces and Torques

Example 2 The Achilles Tendon

The tendon exerts a force of magnitude 790 N. Determine the torque (magnitude and direction) of this force about the ankle joint.

F

m106.355cos 2

mN 15

55cosm106.3N 720 2

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Rigid Objects in Equilibrium

EQUILIBRIUM OF A RIGID BODY

A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium,the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero.

0

0yF0 xF

0 yx aa

0

Reasoning Strategy1. Select the object to which the equations

for equilibrium are to be applied.

2. Draw a free-body diagram that shows all of the external forces acting on the object.

3. Choose a convenient set of x, y axes and resolve all forces into components that lie along these axes.

4. Apply the equations that specify the balance of forces at equilibrium. (Set the net force in the x and y directions equal to zero.)

5. Select a convenient axis of rotation. Set the sum of the torques about this axis equal to zero.

6. Solve the equations for the desired unknown quantities.

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Example 3 A Diving BoardA woman whose weight is 530 N is poised at the right end of a diving board with length 3.90 m. The board has negligible weight and is supported by a fulcrum 1.40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board. 022 WWF

N 1480m 1.40

m 90.3N 5302 F

22

WWF

021 WFFFy

0N 530N 14801 F

N 9501 F

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Example 5 Bodybuilding

The arm is horizontal and weighs 31.0 N. The deltoid muscle can supply 1840 N of force. What is the weight of the heaviest dumbbell he can hold?

0 Mddaa MWW

0.13sinm 150.0M

d

Maad

MWW

m 620.0

0.13sinm 150.0N 1840m 280.0N 0.31

N 1.86

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Center of Gravity

The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated.

21

2211

WWxWxWxcg

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Example 6 Center of Gravity of an Arm

The horizontal arm is composed of three parts: the upper arm (17 N), the lower arm (11 N), and the hand (4.2 N). Find the center of gravity of the arm relative to the shoulder joint.

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2211

WWxWxWxcg

m 28.0

N 2.4N 11N 17

m 61.0N 2.4m 38.0N 11m 13.0N 17

cgx

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9.3 Center of Gravity

Conceptual Example 7 Overloading a Cargo Plane

This accident occurred because the plane was overloaded towardthe rear. How did a shift in the center of gravity of the plane cause the accident?

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Newton’s 2nd Law for Rotational Motion

TT maF raT

rFT

2mrMoment of Inertia, I

2mrNet externaltorque

Moment of inertia

onacceleratiAngular

inertia ofMoment

torqueexternalNet

I 2mrI

Angular acceleration must be expressed in radians/s2.

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9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

Example 9 The Moment of Inertial Depends on Where the Axis Is.

Two particles each have mass and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at(a) one end and (b) the center.

(a)

22222

211

2 0 LmmrmrmmrI

Lrr 21 0mmm 21

2mLI

22222

211

2 22 LmLmrmrmmrI

22 21 LrLr mmm 21

221 mLI

(b)

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Newton’s 2nd Law Rotational Motion Example 12 Hoisting a Crate

The combined moment of inertia of the dual pulley is 50.0 kg·m2. The crate weighs 4420 N. A tension of 2150 N is maintained in the cable attached to the motor. Find the angular acceleration of the dual pulley.

ITT 2211 yy mamgTF 2

ymamgT 2

2ya

ImamgT y 211

ImmgT 2211 22

211

mImgT

222

2

srad3.6m 200.0kg 451mkg 46.0

m 200.0sm80.9kg 451m 600.0N 2150

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9.5 Rotational Work and Energy

FrFsW

rs

Fr

W

DEFINITION OF ROTATIONAL WORK

Rotational work done by a constant torque in turning an object through an angle is

RW

The angle must be in radians.

Units: joule (J)

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9.5 Rotational Work and Energy

2221 mrKE

22212

21 mrmvKE T

rvT

22122

21 Imr

221 IKER

DEFINITION OF ROTATIONAL KINETIC ENERGY

The rotational kinetic energy of a rigid rotating object is

The angular speed must be in rad/s.

Units: joule (J)

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9.5 Rotational Work and Energy

Example 13 Rolling Cylinders

A thin-walled hollow cylinder (mass = mh, radius = rh) and a solid cylinder (mass = ms, radius = rs) start from rest at the top of an incline. Determine which cylinder has the greatest translational speed upon reaching the bottom.

mghImvE 2212

21

fff mghImv 2212

21

ENERGY CONSERVATION

iii mghImv 2212

21

iff mghImv 2212

21

rv ff

iff mghrvImv 22212

21

2

2rIm

mghv of

The cylinder with the smaller moment of inertia will have a greater final translational speed.

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9.6 Angular Momentum

DEFINITION OF ANGULAR MOMENTUM

The angular momentum L = body’s moment of inertia * angular velocity

IL The angular speed must be in rad/s.Units: kg·m2/s

CONSERVATION OF ANGULAR MOMENTUM

Angular momentum remains constant (is conserved) if net external torque is zero.

Conceptual Example 14 A Spinning Skater

An ice skater is spinning with botharms and a leg outstretched. Shepulls her arms and leg inward andher spinning motion changesdramatically.

Use the principle of conservationof angular momentum to explainhow and why her spinning motionchanges.

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9.6 Angular Momentum

Example 15 Satellite in Elliptical Orbit

A satellite in elliptical orbit around earth. Its closest approach is 8.37x106m from earth-center. Its greatest distance is 25.1x106m from earth-center. Speed of satellite at perigee is 8450 m/s. Find speed at apogee.

IL

angular momentum conservation

PPAA II

rvmrI 2

P

PP

A

AA r

vmrrvmr 22

PPAA vrvr

m 1025.1

sm8450m 1037.86

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A

PPA r

vrv

sm2820