# Work and rotational Kinetic Energy

1. May 1, 2008

### frig0018

1. The problem statement, all variables and given/known data
A 32.0 kg wheel, essentially a thin hoop with r=1.20 m is rotating at 280 rev/min. It must be brought to stop in 15.0 s. How much work must be done to stop it?

2. Relevant equations/3. The attempt at a solution
I=mr^2=46.08 kg*m^2
W=(1/2)I$$\omega$$f^2 - (1/2)I$$\omega$$i^2
W=$$\tau$$($$\Delta$$$$\theta$$)
$$\tau$$=I$$\alpha$$
Rev in 15 sec = 70 rev => 140$$\Pi$$ radians in 15 sec.

I'm not sure which equation(s) to use. I've tried plugging numbers into all of them and getting stuck or wrong answers. Thanks!!

2. May 1, 2008

### Nabeshin

Remember, work is a change in kinetic energy. Do you know the formula for kinetic energy of a rotating object?

3. May 2, 2008

### frig0018

i thought it was
W = delta K = 1/2(I)(omega final)^2 - 1/2(I)(omega initial)^2...?
or do i use the formula
W= (integral from theta initial to theta final) torque d-theta?

4. May 3, 2008

### tiny-tim

… one step at a time …

Hi frig0018!

First step: what angular acceleration is needed to stop it in 15 s?