Question

A disc of circumference of $s$ 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 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 $T/2$
• D. the disc will cover a distance greater than $s$ in further time $T$

Answer 1

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

The correct options are
B the disc will make one rotation in time $T/2$
C linear acceleration of the disc will increase
D the disc will cover a distance greater than $s$ in further time $T$

Between A and B, friction force will act backwards because the contact point tends to slide in forward direction due to force $F$ and also after B there is no friction and disc will do only translational motion.

$m{a}_1=F-f$ and $m{a}_2=F$

$⟹a_1=\frac{F-f}{m}$ and $a_2=\frac{F}{m}$

Thus $a_2>a_1$ i.e linear acceleration will iincrease.

Now for one rotation between A and B, $2\pi =0+\frac{1}{2}\alpha T^2$

$⟹2\pi =\frac{1}{2}\alpha T^2$ ..............(1)

Also $w=0+\alpha T⟹w=\alpha T$

For one rotation after B $2\pi =wt$

$2\pi =\left(\alpha T\right)t⟹t=\frac{2\pi }{\alpha T}$

$t=\frac{2\pi }{\frac{4\pi }{T}}=\frac{T}{2}$ (from 1)

Now, $s=0+\frac{1}{2}{a}_1{T}^2⟹s=\frac{1}{2}{a}_1{T}^2$

and $v=a_{1}T$

Also, $s^{\prime }=vT+\frac{1}{2}{a}_2{T}^2=a_1{T}^2+\frac{1}{2}{a}_2{T}^2$

$s^{\prime }=2s+\frac{1}{2}{a}_2{T}^2⟹s^{\prime }>s$