Question

A sphere of mass m and radius r slips on a rough horizontal plane. At some instant it has translational velocity v and rotational velocity about the centre v/2r. The translational velocity after the sphere starts pure rolling

Answer 1

Hello friend,

# Answer- 14.28%

# Explanation-
Initially, Consider v and ω be the linear and angular velocity of centre of mass of sphere.
During rolling, consider v' and ω' be the linear and angular velocity of centre of mass of sphere.

Now,
v' = v - at
v' = v - Ft/m ...(1)

Also,
τ = Iα
F.r = 2/5 mr^2.α
α = 5/2 F/mr

Now
ω' = ω + αt
v'/r = v/2r + 5/2 Ft/mr ...(2)

Solving (1) and (2)
v' = 6/7 v

Now (v-v')/v = (v-6v/7)/v
(v-v')/v = 1/7 = 14.28%

Hence percentage change in translational velocity is 14.28%.

Thanks for asking..