# Rotational Dynamics/moment of inertia/frictional torque

#### xdevinx

1. Homework Statement

The combination of an applied force and a constant frictional force produces a constant total torque of 36.8 N·m on a wheel rotating about a fixed axis. The applied force acts for 5.92 s. During this time the angular speed of the wheel increases from 0 to 10.2 rad/s. The applied force is then removed, and the wheel comes to rest in 59.5 s.

a)find the moment of inertia

b)find the magnitude of the frictional torque

2. Homework Equations

torque=(inertia) (angular acceleration)
angular acceleration=change in angular speed/change in time

3. The Attempt at a Solution

for (a) I know what I have to do but for some reason I can't figure it out. I set the torque (36.8 N m) equal to the product of the moment of inertia the angular acceleration. however, I'm just stuck on finding the angular acceleration. Do I use an equation for constant angular acceleration? (e.g. omega(final)=omega(initial) + (angular acceleration)(time)) or do I have to integrate it? If so, how?

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#### nickjer

They said constant total torque, so that means constant angular acceleration since the inertia doesn't change.

#### Dr.D

Use the relation between angular impulse and angular momentum.

#### xdevinx

Awesome. I figured out the moment of inertia for the wheel by using the constant angular acceleration formulas, as well as the number of revolutions for the wheel.
First, I used Omega(final)=Omega(initial)+Angular Acceleration*time
Then I was able to find the moment of inertia by plugging it into
Torque=I*Angular Acceleration.

To find how many times the wheel revolved I used Omega(final)^2 - Omega(initial)^2= 2*Angular Acceleration*Total amount angle rotated

But now I just don't know how to find the magnitude of the frictional torque. Help!!!

#### tiny-tim

Homework Helper
But now I just don't know how to find the magnitude of the frictional torque.
Yes you do
… 10.2 rad/s. The applied force is then removed, and the wheel comes to rest in 59.5 s.

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