Question

A disc of circumference s is at rest at a point A on a horizontal surface when a constant horizontal force towards right begins to act on its centre. Between A and B on the horizontal surface there is sufficient friction to prevent 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:

• 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 $\frac{T}{2}$
• D. The disc will cover a distance greater than s in further time T

Answer 1

Answer: (B), (C), (D)

The disc will cover a distance greater than s in further time T

Let P = external force, f = force of friction between A and B,
$a_1$ = acceleration between A and B
$a_2$ = acceleration beyond B
$P-f=m{a}_1$
$P=m{a}_2$
$a_2>a_1$
So, the disc will cover a distance greater than s in further time T
Let $\alpha$ = angular acceleration between A and B
For one rotation, $\theta =2\pi =\frac{1}{2}\alpha T^2$
$T=\sqrt{\frac{4\pi }{\alpha }}$ = time of travel from A to B
Angular velocity at $B={\omega }_B=\alpha T$
For one rotation to the right of B,
$\theta =2\pi ={\omega }_{B}t$
$t=\frac{2\pi }{\alpha T}=\frac{\frac{1}{2}\alpha T^2}{\alpha T}=\frac{T}{2}$