rotational mechanics

When 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 a_c = 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
v_c = -v₀ + μg t -------- (1)
while the angular speed is ω = 5μg2R t
Rolling begins when
v_c + 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 v_c becomes
v_c,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 v₁²Î¼g
That means the sphere covers a distance of 225 v₁²Î¼g towards the left.

@ kaymant
sir its a sphere
rst all i hav done same

ball shld be rel at rest wrt grnd accrdng 2 que.......???

lol @ akshay
mast photo hai end mein