Recent content by larkmen

Ohhhh, I see. I was reading the question wrong. Thank you so much for your help!

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...