# Kinetic Energy Problem - from GRE

1. Nov 28, 2003

### yxgao

This is #32 on the GRE practice exam from www.gre.org.
Three equal masses m are rigidly connected to each other by massless rods of length l forming an equilateral triangle, as shown above. The assembly is to be given an angular velocity w about an axis perpendicular to the triangle. For fixed w, the ratio of the kinetic energy of the assembly for an axis through B compared with that for an axis through A is equal to
[A is located at the center of the triangle, B is located at a corner of the triangle]
a. 3
b. 2
c. 1
d. 1/2
e. 1/3
This is an easy problem, but for some reason I don't know why I am not getting it. Can someone else work this out and tell me if you get the right answer, and how you arrive at it?

2. Nov 28, 2003

### csmines

Solution.

Moments of Inertia of Point masses are MR^2 when you're rotating through the center of the triangle the Radius is L/(3^.5) and there are three masses when you're rotating about the corner the radius is L and there are two masses. So it works out something lize this...

B=2*M*(L)^2 : A=3*M*[L/(3^.5)]^2

which reduces to B=2ML^2 and A=ML^2
this works because Omega is fixed and the only portion of the kinetic energy changing is the Rotational which is equivalent to a change in the net moment of inertia. I think that is right...

A bored CSMPhysicist.

3. Nov 28, 2003

### yxgao

Thanks,
I misread the question. I was treating the sticks as the particles instead of them being massless. I did the problem again, and got 2.
--Ying