Chapter 8

RotationalKinematics

8.3 The Equations of Rotational Kinematics

8.4 Angular Variables and Tangential Variables

velocityl tangentia=Tvr

speed l tangentia=Tv

The relationship between the (tangential) arc length, s,at some radius, r, and the angular displacement, θ, has been shown to be

s = rθ (θ in radians)

Let’s find other relationships between the angularand tangential variables.

Consider skaters “cracking-the-whip” :

8.4 Angular Variables and Tangential Variables

!"

#$%

&===t

rtr

tsvT

'' t!

" =

rad/s)in ( !!rvT =

8.4 Angular Variables and Tangential Variables

( ) ( )t

rtrr

tvv

a ooToTT

!!!! "=

"=

"=

to!!

"#

=

)rad/sin ( 2!!raT =

The tangential acceleration, aT, is the changeof the tangential velocity per time:

8.4 Angular Variables and Tangential Variables

Example 6 A Helicopter Blade

A helicopter blade has an angular speed of 6.50 rev/s and anangular acceleration of 1.30 rev/s2.For point 1 on the blade, findthe magnitude of (a) thetangential speed and (b) thetangential acceleration.

8.4 Angular Variables and Tangential Variables

srad 8.40rev 1rad 2

srev 50.6 =!

"

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&!"

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&='

(

( )( ) sm122srad8.40m 3.00 === !rvT rω

8.4 Angular Variables and Tangential Variables

( )( ) 22 sm5.24srad17.8m 3.00 === !raT

22 srad 17.8

rev 1rad 2

srev 30.1 =!

"

#$%

&!"

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&='

(

8.5 Centripetal Acceleration and Tangential Acceleration

( ) rad/s)in ( 222

!!! rrr

rva T

c ===

In uniform circular motion, the only accelerationpresent is the centripetal acceleration.

In nonuniform circular motion, there are both acentripetal and a tangential acceleration.

)rad/sin ( 2!!raT =

8.5 Centripetal Acceleration and Tangential Acceleration

Example 7 A Discus Thrower

Starting from rest, the throweraccelerates the discus to a finalangular speed of +15.0 rad/s ina time of 0.270 s before releasing it.During the acceleration, the discusmoves in a circular arc of radius0.810 m.

Find the magnitude of the totalacceleration of the discus just before the discus is released.

8.5 Centripetal Acceleration and Tangential Acceleration

( )( )2

22

sm182

srad0.15m 810.0

=

== !rac

( )

2sm0.45s 0.270

srad0.15m 810.0

=

!"

#$%

&===t

ù-ùrra o

T '

( ) ( ) 22222 sm187sm0.45sm182 =+=+= cT aaa

ω - ωo

ac2 + aT

2

φ = tan-1 aT/ac = tan-1 (45.0)/(182) = 13.9o

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?

TORQUE

Torque tells you how effective a given force is at rotating something about some axis.

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 TORQUE

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

lF=!Direction: 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)

9.1 The Action of Forces and Torques on Rigid Objects

The lever arm, l, for some force acting about some rotation axis is defined as the distance between the rotation axis and a perpendicular to the force.

9.1 The Action of Forces and Torques on Rigid Objects

Example 2 The Achilles Tendon

The tendon exerts a force of magnitude720 N with a geometry as shown inthe figure.

Determine the torque (magnitudeand direction) of this force about the ankle joint.

9.1 The Action of Forces and Torques on Rigid Objects

720 N

lF=!

m106.355cos 2!"

=lo

( )( )mN 15

55cosm106.3N 720 2

!=

"= # o$

First, calculate the magnitude of τ

Since the force rotates the foot about the ankle joint in a clockwise direction --> τ negative

τ = - 15 N m , if the direction is included

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=!

9.2 Rigid Objects in Equilibrium

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

9.2 Rigid Objects in Equilibrium

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.

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.

9.2 Rigid Objects in Equilibrium

022 =!=" WWF ll#

( )( ) N 1480m 1.40

m 90.3N 5302 ==F

22

l

lWWF =

9.2 Rigid Objects in Equilibrium

021 =!+!=" WFFFy

0N 530N 14801 =!+! F

N 9501 =F

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 dumbell he can hold?

9.2 Rigid Objects in Equilibrium

0=+!!=" Mddaa MWW lll#

( ) ol 0.13sinm 150.0=M

9.2 Rigid Objects in Equilibrium

( )( ) ( )( ) N 1.86m 620.0

0.13sinm 150.0N 1840m 280.0N 0.31=

+!=

+!=

o

l

ll

d

Maad

MWW