Rotational Dynamics/moment of inertia/frictional torque

[xdevinx]

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

Homework Equations

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

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?

 

[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]

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.