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## Homework Statement

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?

## Homework 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

## 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.