CP? A solid uniform ball rolls without slipping up a hill (?Fig. P10.70?). At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff. (a) How far from the foot of the cliff does the ball land, and how fast is it moving just before it lands? (b) Notice that when the balls lands, it has a greater translational speed than when it was at the bottom of the hill. Does this mean that the ball somehow gained energy? Explain!

Solution 82P Step 1: Total energy of the ball at the bottom of the cliff = KE + PE KE of the ball at bottom of the cliff = KE translationKE rotation PE of the ball at bottom of the cliff = PE bottom= 0 KE = ½ mv 2 translation Provided, v = 25 m/s 2 2 2 2 2 Therefore, KE translation m × (25 m/s) = ½ m × 625 m /s = 312.5 m (m /s ) KE rotation ½ I 2 Where, I - Moment of inertia of the ball - angular velocity of the ball 2 Moment of inertia of the ball, I = mR Where, m - mass of the ball and R - radius Angular velocity, = v/ R KE = ½ I = ½ × mR × (v/R) = mv = m × (25 m/s) = m × 625 m /s 2 2 2 rotation KE rotation 125 m (m /s ) 2 Total kinetic energy of the ball at the bottom of the cliff, KE = 312.5 m (m /s ) + 125 bottom 2 2 m (m /s ) KE = 437.5 m (m /s ) 2 2 bottom Therefore, total energy of the ball at the bottom of the cliff, TE = 437.5 m (m /s ) 2 2 bottom Step 2: At the top, the potential energy of the ball, PE top= mgh = m × 9.8 m/s × 28 m 2 2 2 PE top= 274.4 m (m /s ) At the top, the consider, the velocity as v’ . Therefore, the KE topwill be, 2 KE’ translation½ mv’ KE’ = mv’ 2 rotation Therefore, KE = ½ mv’ + mv’ = 7/10 mv’ 2 top TE top= PE top+ KE top= 274.4 m (m /s ) + 7/10 mv’ 2 Step 3: According to the law of conservation of energy we can write, TE bottom = TE top 2 2 2 2 2 That is, 437.5 m (m /s ) = 274.4 m (m /s ) + 7/10 mv’ Cancel the term “m” in RHS and LHS 437.5 (m /s ) = 274.4 (m /s ) + 7/10 v’ 2 2 2 2 2 2 Rearranging, we get, 7/10 v’ = 437.5 (m /s ) - 274.4 (m /s ) 2 2 2 Or, 7/10 v’ = 163.1 m /s v’ = (10 × 163.1 m /s ) / 7 = 233 m /s 2 2 Taking square root on both sides, v’ = 15.26 m/s