A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then?

Solution 29E Introduction We have to first calculate the initial kinetic energy of the ball using the rotational motion. So we have to calculate the moment of inertia of the ball. No at maximum height, the potential energy of the ball will be equal to the initial kinetic energy of the ball. So equating the initial kinetic energy and maximum potential energy we can calculate the rotational speed and then rotational kinetic energy. Step 1 The mass of the ball is m = 426 g = 0.426 kg And the diameter of the ball is d = 22.6 cm = 0.226 m Hence the radius of the ball is r = d/2 = 0.113 m The moment of inertia of hollow sphere is given by I = mr 2 2 3 2 3 Hence the moment of inertia of the ball is I = (0.326 kg)(0.113 m) = 3.626 × 10 kg.m 2 Step 2 Suppose the angular velocity of the ball is Since the ball is rotating without sleeping, the translational velocity of the ball is v = r Hence the kinetic energy of the ball is 1 2 1 2 1 2 1 2 2 K = I2+ mv = 2 + m 2 2 = (3.626 × 10 3 kg.m ) + (0.426 kg)(0.113 m) 2 2 2 2 = (3.626 × 10 3 kg.m ) + (5.439 × 10 3 kg.m ) 2 2 2 = (4.532 × 10 3 kg.m ) 2 Step 3 The maximum height of the ball is h = 5.00 m Hence the potential energy of the ball at maximum height is U = mgh = (0.426 kg)(9.8 m/s )(5.00 m) = 20.87 J