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Chapter 1: Q15E (page 330)

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 N is applied to the rim of the wheel. The wheel has radius 0.120 m. Starting from rest, the wheel has an angular speed of 12 rev/s after 2.00 s. What is the moment of inertia of the wheel?

Short Answer

Expert verified

The moment of inertia of the wheel is, I=0.255鈥�鈥�kg路m2.

Step by step solution

01

To mention the given data

We have the given data:

The radius of the wheel r=0.120 m.

Force applied to the rim of the wheel F=80.0 N.

Angular speed of the wheel 蝇z=12.0 rev/s =75.40 rad/s.

The system is initially at rest.

Time interval during which the revolutions take place is,

螖t=2.00s.

02

Concept

If a rigid object free to rotate about a fixed axis has a net external torque acting on it, the object undergoes an angular acceleration 伪, where,

鈭�蟿ext=I伪鈥�鈥�鈥�鈥�鈰�鈰�1

03

To find the moment of inertia of the wheel

First, we will calculate angular acceleration which is given by,

蝇z=蝇0z+伪zt.

Since the system starts from the rest, we have, 蝇0z=0.

Therefore,

蝇z=伪zt鈬�伪z=蝇zt鈬�伪z=75.402.00鈬�伪z=37.70鈥�rad/s2

From 1, the net torque is given by,

鈭�蟿z=I伪z鈥�鈥�鈥�鈥�鈬�I=鈭�蟿z伪z鈬�I=Fr伪z鈬�I=80.00.12037.70鈬�I=0.255鈥�鈥�kg路m2

Hence, the moment of inertia of the wheel is, I=0.255鈥�鈥�kg路m2.

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