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A disk with a rotational inertia of 5.0 kg·m2 and a radius of 0.25 m rotates on a fixed axis perpendicular to the disk and through its center. A force of 2.0 N is applied tangentially to the rim. As the disk turns through half a revolution, the work done by the force is:

This was another question on my test that I missed. I didn't know where to get started. I know the work done is the integral of the torque. However, I don't know how to get a function that I can integrate from this example. Any ideas?

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2 Answers 2

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Work is force over distance. What is force and over what distance does it apply?

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Like I said in my comment I figured it out. d can be found by using the radius*pi, and work is just W=Fd. So I figured it out. ;) –  Fat Boy Nov 19 '11 at 23:36

Think about what information is given in the problem that you can use to calculate the torque. (To say any more would be giving too much away)

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Well what I just thought of is that the distance = R*pi. From there I could use W=Fd and the answer would be 1.6. However, tha doesn't sound like what you're trying to convey. –  Fat Boy Nov 19 '11 at 22:00
That is one way to do it. –  David Z Nov 19 '11 at 22:15
Well since that looks like it's the correct answer, how would you go about it your way? I feel so dumb that I can't see how to calculate the torque from the information given. –  Fat Boy Nov 19 '11 at 22:18
What ways do you know of to calculate torque? –  David Z Nov 19 '11 at 22:19
Tnet = I a, Tnet = dl/dt, T= r x F, T=rFsin(theta). Duh. The last one? ;) So once I have the torque, is that equivalent to the force on the object? –  Fat Boy Nov 19 '11 at 22:23

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