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?
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
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.