# Help with rotational kinetic energy, interia, etc. problem

1. Feb 3, 2010

### fhwing12

1. The problem statement, all variables and given/known data

For a physics lab we need to calculate the final linear velocity of a disc and hoop. The only thing we are given is the length and height of the ramp and the mass and radius of the disc and hoop.

2. Relevant equations

I thought before that with conservation of energy, I could just set gPE = KE + KErot.

3. The attempt at a solution

It seems like all of the radius's and mass's cancel out. I had mgh = .5mv^2 + .5kmr^2(w^2) where w is omega. I then substituted in (v/r) for w and and cancelled out the m's from all of the terms and ended up with gh = .5v^2 + .5kv^2. This doesn't prove to be at all helpful though because I'm nearly positive that not every disc, regardless of its mass and radius, is going to have the same final velocity. Is there some other way I can solve this problem using the work-energy theorem or something with forces? I was thinking that W = change in KE = Fx but does the KE include both linear and rotational KE? Sorry for all of the information, but thanks to anyone that responds!!

2. Feb 3, 2010

### JaWiB

What makes you so sure of this?

Yes, KE includes linear and rotational energy. Work is actually the dot product of force times displacement, or in other words, the force times the displacement times the cosine of the angle between them.

You could probably also analyze the problem by using kinematic equations--i.e., calculate the net force and net torque, then angular acceleration and linear acceleration and use the standard equations like v = at + v_0.

3. Feb 4, 2010

### mdoucette225

I'm just throwing suggestions out there (mind you I'm still in high school, but we just got finished with these concepts in my AP Physics class).

Are you completely sure that the I(moment of inertias) are the same for the disc and the hoop? If my memory doesn't fail me, I believe the I for a disc is MR^2 where as for the hoop it is 1/2MR^2.

Depending on whether I completely misunderstood what exactly you are doing in the lab, that could either be helpful or completely useless.

I hope it helps! Good luck!