Recent content by bpru

You're right about MI being (1/3)mr^2 for a rod; sorry about that. But shouldn't there be an L on the left side of your equation, since torque is Fxr? (I accidentally left it out when I typed up my equation too.) If there is, then the Ls cancel out completely.

Thanks. The way I figure it, to find the linear acceleration for any of the rods, I would use the formula for torque: T=I(a/r) Fr=I(a/r) mgcos(theta)r=(2/3)(mr^2)(a/r) After canceling terms, I get: a=(3/2)gcos(theta). I took that to mean that the linear acceleration does not depend on the...

Homework Statement The three thin rods shown below are initially at the same angle, with one rod twice as long as each of the others two. Rod X is made of brass (density = 8.6 g/cm3), the others are made of aluminum (density = 2.7 g/cm3). All rods are released from rest at the same time...