Set 8 Problem number 16

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s); Extension

• Figure

Problem

What is the moment of inertia of a disk that accelerates from 2.4 radians/second to 7.9 radians/second in 7 seconds when subject to a torque of 6.4 meter Newtons?

Solution

The information given implies an angular acceleration of .7857143 radians/second ^ 2.

• Newton's Second Law in rotational form says that `tau = I `omega , where I is the moment of inertia, `tau is the torque and `omega is the angular acceleration.
• Since we know the angular acceleration and the torque, it is easy to find the moment of inertia.

We obtain

I = `tau / `omega

= ( 6.4 meter Newtons)/( .7857143 radians/second ^ 2)

= 8.145454 kg m ^ 2/s ^ 2/(rad/s ^ 2)

= 8.145454 kg m ^ 2.