# Easy Rotational Motion Question

1. Oct 8, 2011

### GreenPrint

1. The problem statement, all variables and given/known data

A grindstone has a moment of inertia of 1.6*10^-3 kg m^2. When a constant torque is applied, the flywheel reaches an angular velocity of 1200 rev/min in 15 s. Assuming it started from rest, find (a.) the angular acceleration; (b.) the torque applied; (c.) the angle turned through the 15 s; (d.) the work W done on the flywheel by the torque.

2. Relevant equations

3. The attempt at a solution

I'm having some issues with part (a).
I used ω = ω_0 + alpha*t
and solved for alpha
alpha = (ω-ω_0)/t
alpha = (1200 rev/min * (2 pi)/rev * 60 min/s)/15 s
I get alpha is about 30,159 1/s^2

I don't see what I'm doing wrong. Thanks for any help which you can provide me.

2. Oct 8, 2011

### WJSwanson

For part (a), you're on the right track. Because the applied torque is constant, this means that angular velocity increases linearly. This should allow you to work backwards using $\omega_{0} , \omega_{f} ,$ and $\Delta t$.

Your problem is in the calculations. You've got 60min/s, which is incorrect. It should be 1min/60s. See if that puts you closer to your expected value and then substitute it back in to check the validity of your answer by solving for $\omega_{f}$ from $\omega_{0} , \delta t ,$ and $\alpha$.

3. Oct 8, 2011

### Staff: Mentor

You messed up the conversion from rev/min to rad/sec.

4. Oct 8, 2011

### GreenPrint

ah i feel dumb thanks

5. Oct 8, 2011

### GreenPrint

for part (c) I used
θ = θ_0 + ω_0*t + (alpha*t^2)/2
θ = 1200 rev/min * (2pi)/rev * min/(60 s) * 15 s + (8.38 1/s^2 * (15 s)^2)/2 ≈ 2,828
which I guess is wrong because the answer is 942
I don't see what I'm doing wrong here, I checked that the units canceled out

6. Oct 8, 2011

### WJSwanson

Well the answer you derived is just about 3x the answer the book gives you. See if you can find where you might have accidentally multiplied something by 3 (or got a constant 3x its correct value or forgotten to divide by 3. Your approach looks right as best I can tell so just double check your inputs and whatnot.

7. Oct 8, 2011

### GreenPrint

I have checked it and can't seem to find anything wrong with it, yet the answer is wrong.

8. Oct 8, 2011

### WJSwanson

Hmm. I can't think of what might be the issue. Maybe try integrating over the specified time interval and using your known initial values and whatnot? I don't know why that would change since your alpha is constant, but it couldn't hurt. Beyond that, I can't think of anything other than just making sure you're using the right values everywhere.

9. Oct 8, 2011

### Staff: Mentor

For one thing, it starts from rest so ω_0 = 0.

10. Oct 8, 2011

### GreenPrint

Ya I don't think integrating is needed because the angular acceleration is constant. And ya I checked to make sure all my values were correct and they appear to be. Do you think the supplied answer is wrong?

11. Oct 8, 2011

### GreenPrint

thanks

12. Oct 8, 2011

### WJSwanson

Derp. I can't believe I missed that too!