Angular Kinetic Energy

1. Jan 3, 2009

tachu101

1. The problem statement, all variables and given/known data
A hoop of mass 1kg and radius 2m is rotating about its center with an angular speed of 3rad/sec. A force of 10N is applied tangentially at the rim.

a. The rotational kinetic energy of the hoop is?
b. The instantaneous rate at which the kinetic energy is changing is?

2. Relevant equations

KE= 1/2Iw^2

3. The attempt at a solution
For the first part I can find I= MR^2 for a hoop, so I= (1)(2^2) so I=4. So KE= 1/2Iw^2 = (1/2)(4)(3^2)= 18. What do I do with the extra 10N applied (do I just add it in or does it not factor into the rotational kinetic energy?) I don't know what equation to use for the second part.

2. Jan 3, 2009

tiny-tim

Hi tachu101!

KE is geometric … it only depends on the motion, not on what's causing it.

So you only need the 10N for the second part …

and for that, what equation do you know that connects force and rotation?

3. Jan 3, 2009

tachu101

For the second part would I use torque=Ia which would then be torque/I=a? Would that be (10)(2)/((1)(2^2))= 5 rad/sec^2 does that help me?

4. Jan 3, 2009

tiny-tim

(have an omega: ω and an alpha: α and a delta: ∆ )
Yup!

That gives you dω/dt, from which you could find dKE/dt.

But there's a more direct way …

remember, ∆KE = ∆(work done) …

so what is the equation for work done by a torque ?

5. Jan 3, 2009

tachu101

W= torque* $$\Delta$$$$\Theta$$ which would then be (10)(2)*(3)= 60J ? And how would I get dKe/dt from dW/dt

6. Jan 3, 2009

tiny-tim

uh … W= torque*∆θ, where ∆θ is the total change in angle, is correct

but 3 is not the total change in angle, it's only the rate-of-change of angle, dθ/dt …

you need dW/dt = d(torque*θ)/dt

7. Jan 3, 2009

tachu101

hmmm... am I missing an equation because I am not sure how to find the d(torque*θ)/dt? Also, given that I found 5 rad/sec^2 how could I use that to find a change in KE?

8. Jan 3, 2009

tiny-tim

The torque here is constant, so d(torque*θ)/dt = torque * dθ/dt = torque * ω.

(you could use the dω/dt = 5 if you differentiate 1/2 Iω2

but it's more direct if you use the work done method)

9. Jan 3, 2009

tachu101

So is (10)(2)*(3)= 60J Correct?

10. Jan 3, 2009

tiny-tim

This would be easier to understand if you wrote it out in full …

but it's not joules anyway, is it?

11. Jan 3, 2009

tachu101

Does the instantaneous change in KE equal torque * ω? Is torque (10)*(2)=20?

12. Jan 3, 2009

tiny-tim

Instantaneous rate of change in KE equal torque * ω.

And yes, torque (10)*(2)=20.

13. Jan 3, 2009

tachu101

In summary...

I think that part a is 9J because I think it is (1/2)(I)(w^2)= 9
Part b is then torque*W which is then (10)(2)*(3)= 60

14. Jan 3, 2009

tiny-tim

I thought you got 18J for a?

Yup, b is 60.

15. Jan 3, 2009

tachu101

I looked back and I think I missed a 1/2 in part a. It should be (1/2)((1/2)(1)(2^2))(3^2)= 9 I think I missed the 1/2 in the I value.

16. Jan 3, 2009

tiny-tim

What 1/2? … it's a hoop, not a disc.

Going to bed now … :zzz: