Circular Motion:
A stone rests on the edge of a turntable whos radius is 0.35m. When the speed of the turntable is____rev/min then the centripetal acceleration is 8.2m/s^2

I originally used the equation ac= v^2 over r to find v^2.
I got the answer 2.87=v^2 but my paper says that the answer is 46 revs/min
Can someone please help me get this answer?

Question

Circular Motion:
A stone rests on the edge of a turntable whos radius is 0.35m. When the speed of the turntable is____rev/min then the centripetal acceleration is 8.2m/s^2

I originally used the equation ac= v^2 over r to find v^2.
 I got the answer 2.87=v^2 but my paper says that the answer is 46 revs/min 
Can someone please help me get this answer?

anonymous · Answer

Because the unit is rev/min, the speed wanted here is angular speed (the problem's a bit misleading). You can use $$a_c=\omega^2R$$ and find angular speed $\omega$.

anonymous · Answer

Oh I see, thank you so much. (:

anonymous · Answer

Well, bonus: For the unit, you'll get $omega$ in rad/s. You have to convert it to rev/min.

anonymous · Answer

Divide omega by 2 pi to get in revolutions per second. Then multiply by 60. $${rev \over \min} = {rad \over \sec} {rev \over 2 \pi rad} {60 \sec \over 1 \min}$$