A sphere of radius r and mass m has a linear velocity v0 directed to the left and no angular velocity as it is placed on a belt moving to the right with a constant velocity v1. If after sliding on the belt the sphere is to have no linear velocity relative to the ground as it starts rolling on the belt without sliding. in terms of v1 and the coefficient of kinetic friction muk between the sphere and the belt then determine,
66When the sphere is placed on the conveyor, initially there will be slipping since the point of contact has a velocity towards the left. According, the friction (which will be that of sliding) will act on the sphere towards the right. This friction retards the center-of-mass velocity and accelerates the angular speed. Finally when the rolling begins, the point of contact will have a velocity v1 towards the right so that it may be relatively at rest w.r.t. the belt. So long the rolling has not started, the friction force f = μ mg. Accordingly, the center-of-mass acceleration ac = μg towards the right, while the angular acceleration β is obtained by
f R = 25 mR2 β => μmg R = 25 mR2 β
which gives
β = 5μg2R
At any instant t, the velocity of c.m. towards the right is given by
vc = -v0 + μg t -------- (1)
while the angular speed is ω = 5μg2R t
Rolling begins when
vc + ωR = v1
which gives the required time t1 as
t1= 27 (v1 + v0)μg --- (2)
with this value of t1, using (1), the center of mass velocity vc becomes
vc,roll = 2v1 - 5v07
Since this velocity must be zero, we get
v0 = 25 v1
Using this value of v0 and (2), we get
t1 = 25 v1μg
The displacement of the sphere towards the right by this time is
Δx = -v0 t1 +12 μg t12 = &ndash 225 v12μg
That means the sphere covers a distance of 225 v12μg towards the left.
