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