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.

9.1 The Action of Forces and Torques on Rigid Objects

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

What causes an object to have an angular acceleration?

9.1 The Action of Forces and Torques on Rigid Objects

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

9.1 The Action of Forces and Torques on Rigid Objects

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.

9.1 The Action of Forces and Torques on Rigid Objects

DEFINITION OF TORQUEMagnitude of Torque = (Magnitude of the force) x (Lever arm)

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

Lever arm, ℓ is minimum distance between line of action and axis of rotation.

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

9.1 The Action of Forces and Torques on Rigid Objects

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.

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 ∑ = 0τ

0== yx aa 0=α

9.2 Rigid Objects in Equilibrium

1. 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.

Reasoning Strategy

9.2 Rigid Objects in Equilibrium

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.

𝐹𝐹2ℓ2 − 𝑤𝑤ℓ𝑤𝑤 = 0

∑ = 0τ

1476.4 N

9.2 Rigid Objects in Equilibrium

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 dumbbell he can hold?

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.

9.3 Center of Gravity

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

9.3 Center of Gravity

++++

=21

2211

WWxWxWxcg

9.3 Center of Gravity

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.

0.278 m

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?

9.3 Center of Gravity

Finding the center of gravity of an irregular shape.

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

TT maF =

αraT =rFT=τ

( )ατ 2mr=

Moment of Inertia, I

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

( )( )

( )ατ

ατ

ατ

2

2222

2111

NNN rm

rm

rm

=

=

=

( )ατ ∑∑ = 2mrNet externaltorque Moment of

inertia

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

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

×

=

onacceleratiAngular

inertia ofMoment

torqueexternalNet

ατ I=∑

( )∑= 2mrIRequirement: Angular acceleration must be expressed in radians/s2.

Units of I ? 𝑘𝑘𝑘𝑘 � 𝑚𝑚2

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

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

Two particles each have mass m 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.

𝐼𝐼 = 𝑚𝑚2𝑟𝑟22

𝐼𝐼 = 𝑚𝑚1𝑟𝑟12 + 𝑚𝑚2𝑟𝑟22

(a)

(b)

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

Table 9.1

Moment of Inertia for various Rigid objects

9.5 Rotational Work and Energy

θFrFsW ==

θrs =

Fr=τ

τθ=W

9.5 Rotational Work and Energy

DEFINITION OF ROTATIONAL WORK

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

τθ=RW

Requirement: The angle mustbe expressed in radians.

SI Unit of Rotational Work?

N.m = joule (J)

9.5 Rotational Work and Energy

( ) ( ) 22122

2122

21 ωωω ImrmrKE === ∑∑

22212

21 ωmrmvKE T ==

ωrvT =

9.5 Rotational Work and Energy

221 ωIKER =

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)

9.5 Rotational Work and Energy

mghImvE ++= 2212

21 ω

Total mechanical Energy is equal to sum of translational kinetic energy, rotational kinetic energy, and gravitational energy

E = KET + KER +PE

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:

ωIL =

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

SI Unit of Angular Momentum? kg·m2/s

𝒑𝒑 = 𝑚𝑚𝒗𝒗

9.6 Angular Momentum

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.

9.6 Angular Momentum

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.

Remains same Decreases Increases

For Practice

Ch 9FOC Questions:

1, 3, 6, 10, 16 and 17.Problems:

5, 10, 12, 31, 34, and 48