posted by Anonymous on .
A solid sphere of radius R is placed at a height of 32cm on a 15degree slope. It is released and rolls, without slipping, to the bottom. From what height should a circular hoop of radius R be released on the same slope in order to equal the sphere's speed at the bottom?
Figure the rolling and translational Ke for the sphere at the bottom, from knowing that sum of those is equal to the PE at the beginning. You should then be able to determine speed of the sphere. Then, do the same speed (and angular momentum) for the hoop (recalling the only thing different is momentofInertia), set that equal to mgh, and you have the new height.
Nice problem. Watch algebra.