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.
Answer: (a.) 8.38 rad/s^2 (b.) 0.0134 N m (c.) 942 rads (d.) 12.6 J
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.