A disc of circumference `s` is at rest at a point `A` on a horizontal surface when a constant horizontal force begins to act on its centre. Between `A` and `B` there is sufficient friction toprevent slipping, and the surface is smooth to the right of `B.AB = s`. The disc moves from `A` to `B` in time `T`. To the right of `B`,
.

Options

A. the angular acceleration of the disc will disappear, linear acceleration will remain unchanged.
B. linear acceleration of the disc will increase
C. the disc will make one rotation in time `T//2`
D. the disc will cover a distance greater than `s` in further time `T`.

Answer

B, C, D

Explanation

(B) To the right of `B`, friction vanishes, so linear acceleration of disc will increase.
( c) From `(A)` to `B` it complete one rotation in time `T`
`theta = (1)/(2) alpha t^(2), 2 pi = (1)/(2) alpha T^(2), alpha = (4 pi)/(T^(2))` ....(1)
& `omega = alpha T`...(2)
from (1) & (2)
`omega = (4 pi)/(T)`
after `B, omega` becomes constant there fore time taken to complete one rotation after `B`
`theta = omega t, 2 pi = (4 pi)/(T) t, t = (T)/(2)`
(D) After `B` disc is linearly accelerated so it will cover greater distance than `s` in further time `T`.