A cue ball (solid sphere) with a mass of 5.3 kg and a radius of 0.16 m is on the

Question

A cue ball (solid sphere) with a mass of 5.3 kg and a radius of 0.16 m is on the top of a hill. The hill has a height of 23 m . The cue ball (solid sphere) is initially at rest at the top of the hill and then rolls without slipping down to the bottom of the hill.

1. What is the moment of intertia for the cue ball (solid sphere)?

2. What is the velocity of the cue ball (solid sphere) at the bottom of the hill?

3. If the cue ball (solid sphere) does not roll, but instead slips so that it slides down the hill, what is the velocity at the bottom of the hill?

Explanation / Answer

Here ,

mass ,m = 5.3 Kg

r = 0.16 m

height of hill , h = 23 m

1)

moment of inertia for the cue ball = 0.4 * m * r^2

moment of inertia for the cue ball = 0.4 * 5.3 * 0.16^2

moment of inertia for the cue ball = 0.054 Kg.m^2

2)

let the velocity of he cue ball is v

Using conservation of energy

m *g * h = 0.5 * I * w^2 + 0.5 * m * v^2

5.3 * 9.8 * 23 = 0.5 * (0.054) * (v/.16)^2 + 0.5 * 5.3 * v^2

solving for v

v = 18 m/s

the speed of ball at the bottom is 18 m/s

3)
for the ball at the bottom

v= sqrt(2 * g * h)

v = sqrt(2 * 9.8 * 23)

v = 21.23 m/s

the speed of ball at the bottom is 21.23 m/s

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