A basketball (which can be closely modeled as a hollow spherical shell) rolls down a mountainside into a valley and then up the opposite side, starting from rest at a height H 0 above the bottom. In ?Fig. P10.69?, the rough part of the terrain prevents slipping while the smooth part has no friction. (a) How high, in terms of H0, will the ball go up the other side? (b) Why doesn’t the ball return to height H0? Has it lost any of its original potential energy?

Solution 79P Step 1 At rough hill, there is friction thus it prevents friction which will help the ball to roll down the hill. Thus According to the conservation of energy, U=K +r t The potential energy is equal to the sum of rolling kinetic energy and the translational kinetic energy. 2 2 mgh 0½)Iw +(½)mv …………………………………………………………….(1) At the smooth side of the hill, there is no friction , thus the ball will slip and there will be no rolling kinetic energy, Therefore according to conservation of energy, U=K’ mgh=(½)m(v) …………………………………………………………………….(2)