1. The problem statement, all variables and given/known data A 1 kg ball with a radius of 20 cm rolls down a 5 m high inclined plane. Its speed at the bottom is 8 m/s. How many revolutions per second is the ball making when at the bottom of the plane? 2. Relevant equations circumference = 2πr velocity = distance / time = circumference / time period (T) = time in seconds radius = 20 cm ⇒ 0.2 m mass = 1 kg velocity at bottom of plane = 8 m/s 3. The attempt at a solution The velocity and circumference of a circle is related through the equation ⇒ velocity = circumference / period ⇒velocity = 2πr / T ⇒velocity = ( 2 * 3.14 * 0.2 m ) / T ⇒8 m/s = 1.3 m / T ⇒T = 0.16 s 8 m /1 s so 8 m / 0.16 s = 50 m/s ?? I have trouble getting the revolutions/second for the ball when it is at the bottom of the plane. Is revolutions/second the equivalency of meters/second for velocity? I know my approach to this problem is incorrect, and so any help would be great. Thanks.