Question

A bowler throws a bowling ball of radius R = 11 cm down the lane with initial speed v₀ = 7.5 m/s. The ball is thrown in such a way that it skids for a certain distance before it starts to roll. It is not rotating at all when it first hits the lane, its motion being pure translation. The coefficient of kinetic friction between the ball and the lane is 0.29.
(a) For what length of time does the ball skid? S
(b) How far down the lane does it skid? m
(c) How fast is it moving when it starts to roll? m/s

Answer 1

Answer:

a) t = 0.74s

b) D = 4.76m

c) Vf = 5.35m/s

Explanation:

The ball starts rolling when Vf = ωf*R.

We know that:

Vf = Vo - a*t

ωf = ωo + α*t

With a sum of forces on the ball:

$Ff = m*a$

$\mu*N = m*a$

$\mu*m*g = m*a$

$a=\mu*g=2.9m/s^2$

With a sum of torque on the ball:

$Ff*R = I*\alpha$

$\mu*m*g*R = 2/5*m*R^2*\alpha$

$\alpha=5/2*\mu*g/R=65.9rad/s^2$

Replacing both accelerations:

$Vo - a*t=\alpha*t*R$

$7.5 - 2.9*t=65.9*t*0.11$

t=0.74s

The distance will be:

$D = Vo*t-1/2*a*t^2$

$D = 4.76m$

Final velocity:

$Vf=Vo-a*t$

Vf=5.35m/s