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i don't understand this at all.... if we tip over a rectangular block (such that a corner is its pivot), will it have rotational kinetical energy, translational kinetic energy, or both when it hits the ground???
That seems the right idea. But there would be a definite difference if there were no friction. What would the block rotate around if the corner were allowed to slip? Would the cm ever be moving horizontally at all? Would the block take the same time to fall with and without friction?so the translational kinetic energy is based on its center of mass' velocity and the rotational energy is based on the I of the whole block and w?
and, just to be clear, the weight vector will always come from the center of mass, correct?
also, how would friction play into this? would it be at the axis of rotation, thus causing not torque, but preventing it from making a translational acceleration?
okay, it's good to know that i understand this. one more question, perhaps.That seems the right idea. But there would be a definite difference if there were no friction. What would the block rotate around if the corner were allowed to slip? Would the cm ever be moving horizontally at all? Would the block take the same time to fall with and without friction?
Your statements about mvsquared/2 and Iωsquared/2 are obviously correct. I think the only way that there would be no translational energy would be if it were just a couple that was applied to the block so that the cm was not moving in any direction. This is definitely not the case here.