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 make one rotation in time T / 2

Answer

B, C, D

Explanation

To the rightof B, net force increase so linear acceleration is also increases A to B: F-f = `ma_(1)`
`fR = 1/2 mR^(2) alpha`, `a_(1) = R alpha rArr a_(1) = (2F)/(3m) rArr alpha propto (2F)/(3mR)`
`2pi =1/2 alpha T^(2) rArr T = sqrt((6pi mR)/F)`
`omega_(B) = alpha T = (2F)/(3mR) = sqrt((6pi mR)/F) =2/3 sqrt((6pi F)/(mR))` remains same after crossing B
Time to complete one revolutions after crossing point `B=(2pi)/(omega_(B)) = 1/2sqrt((6pi mR)/(F)) = T/2`
As linear acceleration increases, displacement also increases in time T.