Chapter 9

Chapter 9

Rotational Dynamics

9.1 The Action of Forces and Torques on Rigid Objects

In pure translational motion, all points on anobject travel on parallel paths.

The most general motion is a combination oftranslation and rotation.


According to Newton’s second law, a net force causes anobject to have an 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.


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)


Example 2 The Achilles Tendon

The tendon exerts a force of magnitude790 N. Determine the torque (magnitudeand direction) of this force about the ankle joint.


790 N

F

m106.355cos

2

mN 15

55cosm106.3N 720 2

9.2 Rigid Objects in Equilibrium

If a rigid body is in equilibrium, neither its linear motion nor its rotational motion changes.

0 xF 0yF 0

0 yx aa 0


EQUILIBRIUM OF A RIGID BODY

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

0 0yF0 xF


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 theobject.

3. Choose a convenient set of x, y axes and resolve all forces into componentsthat 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 thisaxis equal to zero.

6. Solve the equations for the desired unknown quantities.


Example 3 A Diving Board

A woman whose weight is 530 N is poised at the right end of a diving boardwith length 3.90 m. The board hasnegligible weight and is supported bya fulcrum 1.40 m away from the leftend.

Find the forces that the bolt and the fulcrum exert on the board.


022 WWF

N 1480

m 1.40

m 90.3N 5302 F

22

WWF


021 WFFFy

0N 530N 14801 F

N 9501 F


Example 5 Bodybuilding

The arm is horizontal and weighs 31.0 N. The deltoid muscle can supply1840 N of force. What is the weight of the heaviest dumbell he can hold?


0 Mddaa MWW

0.13sinm 150.0M


N 1.86

m 620.0

0.13sinm 150.0N 1840m 280.0N 0.31

d

Maad

MWW

9.3 Center of Gravity

DEFINITION OF CENTER OF GRAVITY

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


When an object has a symmetrical shape and its weight is distributed uniformly, the center of gravity lies at its geometrical center.


21

2211

WW

xWxWxcg


Example 6 The Center of Gravity of an Arm

The horizontal arm is composedof three parts: the upper arm (17 N),the lower arm (11 N), and the hand (4.2 N).

Find the center of gravity of thearm relative to the shoulder joint.


21

2211

WW

xWxWxcg

m 28.0

N 2.4N 11N 17

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

cgx


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?


Finding the center of gravity of an irregular shape.

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

TT maF

raT rFT

2mr

Moment of Inertia, I


2

2222

2111

NNN rm

rm

rm

2mr

Net externaltorque Moment of

inertia


ROTATIONAL ANALOG OF NEWTON’S SECOND LAW FORA RIGID BODY ROTATING ABOUT A FIXED AXIS

onaccelerati

Angular

inertia

ofMoment torqueexternalNet

I

2mrIRequirement: Angular acceleration must be expressed in radians/s2.


Example 9 The Moment of Inertial Depends on Wherethe Axis Is.

Two particles each have mass and are fixed at theends 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


(b) 22222

211

2 22 LmLmrmrmmrI

22 21 LrLr mmm 21

221 mLI



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 cableattached to the motor. Find the angular acceleration of the dualpulley.


ITT 2211 yy mamgTF 2

equal

ymamgT 22ya


2ya

ImamgT y 211

ImmgT 2211

222

2

22

211

srad3.6m 200.0kg 451mkg 46.0

m 200.0sm80.9kg 451m 600.0N 2150

mI

mgT

9.5 Rotational Work and Energy

FrFsW

rs

Fr

W


DEFINITION OF ROTATIONAL WORK

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

RW

Requirement: The angle mustbe expressed in radians.

SI Unit of Rotational Work: joule (J)


22122

2122

21 ImrmrKE

22212

21 mrmvKE T

rvT


221 IKER

DEFINITION OF ROTATIONAL KINETIC ENERGY

The rotational kinetic energy of a rigid rotating object is

Requirement: The angular speed mustbe expressed in rad/s.

SI Unit of Rotational Kinetic Energy: joule (J)


Example 13 Rolling Cylinders

A thin-walled hollow cylinder (mass = mh, radius = rh) anda solid cylinder (mass = ms, radius = rs) start from rest atthe top of an incline.

Determine which cylinder has the greatest translationalspeed upon reaching the bottom.


mghImvE 2212

21

iiifff mghImvmghImv 2212

212

212

21

iff mghImv 2212

21

ENERGY CONSERVATION

rv ff


iff mghrvImv 22212

21

2

2

rIm

mghv of

The cylinder with the smaller momentof inertia will have a greater final translationalspeed.

9.6 Angular Momentum

DEFINITION OF ANGULAR MOMENTUM

The angular momentum L of a body rotating about a fixed axis is the product of the body’s moment of inertia and its angular velocity with respect to thataxis:

IL

Requirement: The angular speed mustbe expressed in rad/s.

SI Unit of Angular Momentum: kg·m2/s


PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM

The angular momentum of a system remains constant (is conserved) if the net external torque acting on the system 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.


Example 15 A Satellite in an Elliptical Orbit

An artificial satellite is placed in an elliptical orbit about the earth. Its pointof closest approach is 8.37x106mfrom the center of the earth, andits point of greatest distance is 25.1x106m from the center ofthe earth.

The speed of the satellite at the perigee is 8450 m/s. Find the speedat the apogee.


IL

angular momentum conservation

PPAA II

rvmrI 2

P

PP

A

AA r

vmr

r

vmr 22


PPAA vrvr

sm2820

m 1025.1

sm8450m 1037.86

6

A

PPA r

vrv

P

PP

A

AA r

vmr

r

vmr 22

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Chapter 9