Work in a Rotational Motion

1. Oct 16, 2006

bearhug

A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m is rotating at 280 rev/min. It must be brought to a stop in 15.0s (a) How much work must be done to stop it?
t=15.0s
m=32.0kg
r=1.20m
I need to find W

I think I need to find torque in order to find work. Would torque τ=Fr
F=ma so F=32.0kg(9.8m/s^2)= 313.6 N τ=313.6(1.20m)=376.3
Is this the right approach, if it is what equation should I use to find W. The equation in my book is an integral does that sound right? Thanks

2. Oct 16, 2006

BishopUser

I'm not sure on this one, but I know that $$KE = \frac {1} {2} I\omega^2$$ where I is the moment of inertia (in the case of a thin hoop it is mr^2).

I also imagine that Wnet = change in KE still applies in this situation, but I'm not sure.

Last edited: Oct 16, 2006
3. Oct 17, 2006

bearhug

Thanks for writing that out I was able to figure it out using what you wrote.

4. Oct 17, 2006

bearhug

What is the required average power to do this? The answer to the previous question is 1.98e4 J.
Ρ= τω for power
so I have to find torque which is τ= Fd