How Much Work is Required to Stop a Rotating Wheel in 15.0s?

[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
ω= 29.3 rad/s after conversion
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

 

[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.

 

[bearhug]

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

 

[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
F=ma so τ= mad
so Ρ= (mad)(29.3rad/s)
First of all is this right? What I'm getting stuck at is that I don't think a=g (9.8) in this situation so I have to calculate a using ωf = ωi + αt
0= 29.3 + α(15s)
α= -1.95rad/s^2
and plug that in for acceleration with distance= 1.2m

Can anyone let me know if there's something I'm missing.