Chapter 8Rotational Motion

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Objectives

Distinguish between translational kinetic energy and rotational kinetic energy.

Apply the Law of Conservation of Energy to solve problems that involve rotational as well as translational kinetic energy.

8-7 Rotational Kinetic Energy

The kinetic energy of a rotating object is given by

KE = Σ(½ mv2)

By substituting the rotational quantities, we find that the rotational kinetic energy can be written:

A object that has both translational and rotational motion also has both translational and rotational kinetic energy:

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(8-16)

Rotational Energy

A rotating rigid body has kinetic energy because all atoms in the object are in motion. The kinetic energy due to rotation is called rotational kinetic energy.

Rotational Kinetic Energy

• Consider a mass M on the end of a string being spun around in a circle with radius r and angular frequency w

– Mass has speed v = w r– Mass has kinetic energy

• K = ½ M v2

• K = ½ M w2 r2

• Rotational Kinetic Energy is energy due to circular motion of object.

M

24

8-7 Rotational Kinetic Energy

When using conservation of energy, both rotational and translational kinetic energy must be taken into account.

All these objects have the same potential energy at the top, but the time it takes them to get down the incline depends on how much rotational inertia they have.

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Compare Heights

A ball is released from rest on a no-slip surface, as shown. After reaching the lowest point, it begins to rise again on a frictionless surface.

When the ball reaches its maximum height on the frictionless surface, it is higher, lower, or the same height as its release point?

The ball is not spinning when released, but will be spinning when it reaches maximum height on the other side, so less of its energy will be in the form of gravitational potential energy. Therefore, it will reach a lower height.

8-7 Rotational Kinetic Energy

The torque does work as it moves the wheel through an angle θ:

© 2014 Pearson Education, Inc.

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video

Like a Rolling Disk

A 1.20 kg disk with a radius 0f 10.0 cm rolls without slipping. The linear speed of the disk is v = 1.41 m/s.

(a) Find the translational kinetic energy.

(b) Find the rotational kinetic energy.

(c) Find the total kinetic energy.

1 12 22 2

(1.20 kg)(1.41 m/s) 1.19 JtK mv 1 1 1 12 2 2 22 2 2 4

( )( / ) (1.20 kg)(1.41 m/s) 0.595 JrK I mr v r

(1.19 J) (0.595 J) 1.79 Jt rK K K

Practice Problem #1

Spinning WheelA block of mass m is attached to a string that is wrapped around the circumference of a wheel of radius R and moment of inertia I, initially rotating with angular velocity w that causes the block to rise with speed v . The wheel rotates freely about its axis and the string does not slip.

To what height h does the block rise?

i fE E1 1 1 1 12 2 2 2 2 22 2 2 2 2

( / ) (1 / )iE mv I mv I v R mv I mR fE mgh

2

21

2

v Ih

g mR

A Bowling Ball A bowling ball that has an 11 cm radius and a 7.2 kg mass is rolling without slipping at 2.0 m/s on a horizontal ball return. It continues to roll without slipping up a hill to a height h before momentarily coming to rest and then rolling back down the hill.

Model the bowling ball as a uniform sphere and calculate h.

ext mech therm mech0 0W E E E

2

1 1 2 72 2 2cm cm cm 22 2 5 10

ii i

vMgh Mv MR Mv

R

2 2cm

2

7 7(2.0 m/s)0.29 m

10 10(9.8 m/s )iv

hg

Practice Problem 2

Practice Problem 3

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Practice Problem 3 Solution

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Homework

• Chapter 8 Problems• 43, 45, 47, 49

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Closure

• Kahoot 8.7 Rotational Kinetic Energy

© 2014 Pearson Education, Inc.