A certain solid uniform ball reaches a maximum height h0 when it rolls up a hill without slipping. What maximum height (in terms of h0) will it reach if you (a) double its diameter, (b) double its mass, (c) double both its diameter and mass, (d) double its angular speed at the bottom of the hill?

Solution 15DQ Step 1: Mass of the solid ball m Maximum height reached by the ball h o Let the angular speed of the ball up the hill be By using law of conservation of energy KE = PE 2 2 We have KE = (1/2 ) × I + (1/2) × mv PE = mgh 0 Using this we can write 2 2 (1/2 ) × I + (1/2) × mv = mgh 0 (1/2 ) × ((2/5)mr ) × + (1/2) × mv = mgh 0 2 2 2 2 (1/5)mr + (1/2) m(r ) = mgh 0 7/10 r = gh 0 2 2 h 0 (7r )/10g h 0 (7v )/10g Step 2: When the diameter is doubled we have 2 2 Using h 0 (7r )/10g When r = 2r We have h 0 (7(2r) )/10g h = (7(4r )/10g 0 2 2 h 0 4 (7r )/10g Hence when the diameter doubled the height reached will be h = 4 (7r )/10g2 2 0