A bowling ball of mass Af and radius R is released onto the surface of a bowling

Question

A bowling ball of mass Af and radius R is released onto the surface of a bowling lane with a forward velocity center of mass translational speed upsilon_i and a backspin rotation, angular speed omega_i (A backspin means that the rotation is against the direction of motion.) There is a coefficient of kinetic friction of muk between the ball and the surface. (a) What is the angular speed of the ball when its motion becomes a pure rolling (b) How far down the bowling lane does the ball travel before this happens?

Explanation / Answer

F=ma
Ff = kMg
F= Ff
a= kg

The torque M=I
M= FfR= kMg
(2MR²/5) = kMgR
= 5kMgR /2MR²=5kg/2R

When speed v has decreased enough and angular speed has increased enough, the ball stops sliding and then rolls smoothly
vf =vi-at
f = i+ t
vf = fR
vi-at = (i+ t)R

vf = (i+ t)R + at
vf = i+ tR + at

vf = i+ (R + a)t
t= vf /[ i+ (R + a)]=

a) = i+ t = i+ [vf /(i+ (R + a))] = i+ [vf /(i+ (R + kg))]

b)s= at²/2 = a/2*[vf /(i+ (R + a))]^2 = (kg)/2*[vf /(i+ (R + (kg)))]^2

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