Torque problem - game show wheel

1. Apr 21, 2006

FlipStyle1308

A solid wheel on a game show is given an initial angular speed of 1.25 rad/s in the counterclockwise direction. It comes to rest after rotating through 3/4 of a turn.

Find the torque exerted on the wheel given that it is a disk of radius 0.73 m and mass 6.4 kg.

I got 0.5654 as the answer, but it's not correct. Anyone able to help?

2. Apr 21, 2006

dav2008

Can you show how you obtained that answer so we can see where you made your mistake?

3. Apr 21, 2006

FlipStyle1308

I used mr^2 = (6.4)(0.73)^2 = 3.41056. Then I plugged that into Wf^2 = Wi^2 + 2 (angular acceleration) (delta theta) => 0 = (1.25)^2 + (angular acceleration) (1.5 pi) => 0.1658. Then I multiplied that by 3.41056 to get 0.5654 as my torque.

4. Apr 21, 2006

dav2008

mr2 is the moment of inertia for a point object about a point a distance r away.

You are not dealing with a point, but with an entire disc. The moment of ineria of a disc about the central axis is a slightly different expression.

Here is a helpful website that shows the moments of inertia of various objects: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi

You can either memorize the formulas or derive them by integrating, depending what your teacher/professor requires you to do.

5. Apr 21, 2006

FlipStyle1308

Okay, so I use T = 1/2 mr^2. Then once I get the inertia, do I still multiply it by 0.1658, the angular acceleration I got, in order to get torque?

6. Apr 21, 2006

dav2008

Yes, that looks right.

7. Apr 21, 2006

FlipStyle1308

Wait...how do I incorporate the 3/4 of a turn into this problem? Do I translate that to 1.5 pi?

8. Apr 21, 2006

dav2008

All of that work is right, except that you used mr2 instead of 1/2mr2

Edit: Your answer for acceleration of .1658 looks right but you probably made a typo when posting the equation that's underlined, since your previous equation of "Wf^2 = Wi^2 + 2 (angular acceleration) (delta theta)" is right.

Last edited: Apr 21, 2006
9. Apr 21, 2006

FlipStyle1308

Will my answer end up negative or positive? Cuz my result for the angular acceleration is negative.

10. Apr 21, 2006

dav2008

If you define clockwise as negative and counterclockwise as positive it makes sense that the torque is negative because it is acting in the clockwise direction, causing the wheel to stop. You just have to be consistant as to the signs you use.

You are finding the torque excerted that causes the wheel to stop, right? That's what you solved for. If you are trying to find the torque that the person who spun the wheel applied then I don't think you have enough information.

Last edited: Apr 21, 2006
11. Apr 21, 2006

FlipStyle1308

Never mind, I got it. Thank you for your help!

Last edited: Apr 21, 2006