kinetic energy of a rotating rigid body • Rigid body can be consider as a collection of

particles rotating about a fixed z axis • Each particle has kinetic energy.If the mass of

the i th particle is mi and its linear speed is vi ,its kinetic energy is

Rotational K.E. & Moment of Inertia

For this point mass m, the kinetic energy is:

2I mr

Define the moment of inertia of a point mass as: Then the

kinetic energy is:

1 22

K I

•Total Rotational kinetic energy is the sum of the kinetic energies of the individual particles

Now

• Every particle in the rigid object has the same angular speed , But it’s linear speeds depend on the distance ri from the axis of rotation.

(KR is called rotational KE)

Rotational Kinetic Energy & Moment of Inertia

We generalize this to any rotating object, which will have a kinetic energy:

Here I, the moment of inertia, is given by

Angular MomentumLike linear momentum =MV

Angular momentum

For linear motionForce = rate of change of momentum

For Angular motionTorque = rate of change of angular momentum

Direction of Angular Momentum

Rotational inertia or movement of inertia

• The quantity, is defined as the moment of inertia or rotational inertia. It's value depend on shape of body & position of rotation axis

• The term is defined as angular movement• In absence of external force linear movement &

KE is conserved • Similarly absence of torque angular movement

& rotational KE is conserved

Some important points about rotational kinetic energy KR or KE

Where

• It is sum of KE of all particles in the rigid object.

It is not new form of energy• Equation of the KE is a convenient form when

we deal with rotational motion• The KE of rotational motion is analogous to

(1/2 x m x v2 ) KE of linear motion• takes the place of ‘m’ & takes place of ‘v’

when we compare a linear & rotational motion

ω

Where

Newton's laws are applicable for rotational motion also,where

• Force (F) is replaced by torque (T)• Mass (m) is replaced by movement of inertia (I)• Acc. (a) is replaced by angular Acc.()

Example: A Dumbbell

Use the definition of momentof inertia to calculate that of adumbbell-shaped object withtwo point masses m separatedby a distance of 2r and rotatingabout a perpendicular axis throughtheir center of symmetry.

2 2 2 21 1 2 2 2i iI m r m r m r mr

Example:Starting a Microhematocrit

Suppose a microhematocrit centrifuge is starting up with a constant angular acceleration of = 95.0 rad/s2.

(a) What is the magnitude of the centripetal, tangential, and total acceleration of the bottom of a tube when the angular speed is = 8.00 rad/s?

(b) What angle does the total acceleration vector make with the direction of motion?2 2 2(0.0907 m)(8.00 rad/s) 5.80 m/scpa r

2 2(0.0907 m)(95.0 rad/s ) 8.62 m/sta r 2 2 2 2 2 2 2(5.80 m/s ) (8.62 m/s ) 10.4 m/scp ta a a

2 2arctan( / ) arctan (5.80 m/s ) / (8.62 m/s ) 33.9cp ta a

Rolling MotionRolling Motion If a round object rolls without slipping, there is a fixed relationship between the translational and rotational speeds:

Rolling MotionRolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion:

Rolling MotionRolling Motion We may also consider rolling motion at any given instant to be a pure rotation at rate about the point of contact of the rolling object.

Example: A Rolling Tire A car with tires of radius 32 cm drives on a highway at a speed of 55 mph.

(a) What is the angular speed of the tires?

(b) What is the linear speed vtop of the top to the tires?

(55 mph)(0.447 m/s/mph)77 rad/s

(0.320 m)

v

r

2 110 mphtopv v

(77 rad/s) / (2 rad/rev) 12.25 rev/s

n = w / cos   

Fnet,x = n sin = Fc = m v2 / r

x

Fc = mv 2 / r = n sin

= [w / cos ] sin

Fc = mv 2 / r = w [ sin / cos ]

Fc = mv 2 / r = w tan

m v 2 / r = m g tan

Tan = v 2 / g r

Banking of road provide centrifugal force.Let Angle of banking be radians

ExampleAt what angle should a curve of 200 m radius be banked so that no friction is required when a car travels at 60 kilometers per hour around the curve?

For a banked curve with no friction, the only forces acting on a car are the normal force and the weight. We have just looked at this situation. The centripetal force must be supplied by the horizontal component of the normal force. We have just analyzed this situation in arriving at

Tan = v 2 / g r

v = 16.7 m/s

tan = v 2 / g r = (16.7 m/s) 2 / (10 m/s2)(200 m)

= 7.91o

Rotational vs. Linear Kinematics

Rotational vs. Linear KinematicsAnalogies between linear and rotational

kinematics:

Sources of Centripetal ForceSources of Centripetal Force This centripetal force may be provided by the tension in a string, the normal force, or friction, among other sources.

Banked Curves

Example: Bank on It

If the road is banked at the proper angle , a car can round a curve without the assistance of friction between the tires and the road and without skidding. What bank angle is needed for a 900 kg car travelling at 20.5 m/s around a curve of radius 85.0 m?

0 cos cosyF N W N mg 2sin /x cpF N ma mv r

2 2sin /tan

cos

N mv r v

N mg gr

2 2

2

(20.5 m/s)arctan arctan 26.7

(9.81 m/s )(85.0 m)

v

gr

Notice that there is only one speed at which gravity exactly provides the needed centripetal force.

A satellite moves at constant speed in a circular orbit about the center of the Earth and near the surface of the Earth. If the magnitude of its acceleration is g = 9.81 m/s2 and the Earth’sradius is 6,370 km, find:(a) its speed v; and(b) the time T required for one completerevolution.

Example: A Satellite’s Motion

2

cp

va g

r

3 2 3(6,370 10 m)(9.81 m/s ) 7.91 10 m/s 17,700 mi/hv rg

3 32 / 2 (6,370 10 m) /(7.91 10 m/s) 5,060 s 84.3 minT r v