Ah, so yea we do probably have to include gravity. If we do include gravity then the equation for
\Gamma = + mgRsin\theta - 2mg\frac{1}{2}Rsin\theta - 3mgRsin\theta = -3mgRsin\theta. (measuring positive torque as counter-clock wise)
Shouldn't this depend on v however? Since if the 3m mass is...
But if we ignore the force of the collision wouldn't the free-body diagram have no forces on it, since there is no gravity? Or is there gravity and I'm reading the problem wrong?
Thanks for the reply TSny!
I see what you mean by it being an inelastic collision and so the KE isn't conserved. I hadn't thought about that before, that definitely helps clear a lot up.
As for part b, after drawing the free body diagram, I get this http://i.imgur.com/sn06rfF.png...
Homework Statement
Figure: http://i.imgur.com/E2D1hkW.png
A massless rod of length 2R is attached at the middle to a pivot point that allows it to rotate in the vertical plane. Masses m and 2m are attached to the rod at the locations depicted in the figure. Initially the rod makes an angle of...