Question

A disc of circumference 's' is 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. Given that 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 aceleration 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 a further time $T$

Answer 1

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

The correct options are
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 a 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 $m{a}_2=F$
$\Rightarrow a_1=\frac{F-f}{m}$ and $a_2=\frac{F}{m}$
Thus $a_2>a_1$ i.e linear acceleration will increase.
Now for one rotation between A and B, $2\pi =0+\frac{1}{2}\alpha T^2$
$\Rightarrow 2\pi =\frac{1}{2}\alpha T^2$ ........(1)
Also $w=0+\alpha T\Rightarrow w=\alpha T$
For one rotation after B $2\pi =wt$
$2\pi =\left(\alpha T\right)t\Rightarrow t=\frac{2\pi }{\alpha T}$
For one rotation after B $2\pi =wt$
$2\pi =\left(\alpha T\right)t\Rightarrow t=\frac{2\pi }{\alpha T}$
$t=\frac{2\pi }{\frac{4\pi }{T}}=\frac{T}{2}$ (from 1)
Now, $s=0+\frac{1}{2}{\alpha }_1{T}^2\Rightarrow s=\frac{1}{2}{\alpha }_1{T}^2$
and $v=a_{1}T$
Also, where $s^{\prime }$ is the distance covered after the point B $s^{\prime }=vT+\frac{1}{2}{\alpha }_2{T}^2={\alpha }_1{T}^2+\frac{1}{2}{\alpha }_2{T}^2$
$s^{\prime }=2s+\frac{1}{2}{\alpha }_2{T}^2\Rightarrow s^{\prime }>s$