Question

A bowler throws a bowling ball of radius R = 11 cm down the lane with initial speed v0 = 6.5 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 0.20. (a) For what length of time does the ball skid? (Hint: As the ball skids, its speed v decreases and its angular speed ω increases; skidding ceases when v = Rω.) s (b) How far down the lane does it skid? 4.702 Incorrect: Your answer is incorrect. m (c) How fast is it moving when it starts to roll? m/s

please show work and answer thank you!

Answer 1

a)The forces on the ball will be:

X direction: F = M a(com)

Along Y : N - mg = 0 =. N = mg

Ff = mu N

Ff x R = I x alpha

mu mg R = 2/5 mR^2 alpha

alpha = - 5 mu g / 2R

a(com) = - mu g

V(com) = v - at = v - mu g t (1)

omega = -5 mu g/2R t (2)

Rolling without slipping startes when:

V(com) = -R omega

v - mu g t = (-5 mu g/2 ) t

t = 2v/7 mu g

t = 2 x 6.5 / 7 x 0.2 x 9.8 = 0.95 s

Hence, t = 0.95 s

b)From eqn of motion:

S = ut + 1/2 at^2

X = vt -1/2 x mu x g t^2

X = 6.5 x 0.95 - 0.5 x 0.2 x 9.8 x 0.95 x 0.95

X = 5.29 m

c)we have derived

v = 5mu g t/2

v = 5 x 0.2 x 9.8 x 0.95 / 2 = 4.66 m/s

Hence, v = 4.66 m/s