Question

A solid sphere of radius r starts rotating on a rough horizontal surface with translation velocity v and

Answer 1

# 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%