Question

A disc of mass $m$ and radius $R$ is placed over a plank of same mass $m$. There is sufficient friction between disc and plank to prevent slipping. A force $F$ is applied at the centre of the disc. Which option(s) is/are correct.
[$a=$ acceleration of disc and $f=$ force of friction]

• A. $a=\frac{F}{2M}$
• B. $a=\frac{2F}{3M}$
• C. $f=\frac{F}{3}$
• D. $f=\frac{2F}{3}$

Answer 1

Answer: (B), (C)

$f=\frac{F}{3}$


From the Newton's 2nd law of motion,

$F-f=ma....\left(1\right)$

Rolling without slipping condition,

$a=R\alpha ....\left(2\right)$

Friction force $f$, applies torque on disc,

$\tau =fR$

Using, $\tau =I\alpha$, we get

$I\alpha =fR$

$\therefore f=\frac{I\alpha }{R}....\left(3\right)$

Now, using equation $\left(1\right)$, we get

$F-\frac{I\alpha }{R}=ma$

$\Rightarrow F-\frac{Ia}{R^2}=ma\left[\text{Using 2}\right]$

$\Rightarrow F-\frac{ma}{2}=ma[\text{for disc,}I=\frac{1}{2}m{R}^2]$

$\Rightarrow F=\frac{3ma}{2}$

$\therefore a=\frac{2F}{3m}$

Now, using equation $\left(3\right)$,

$f=\frac{I\alpha }{R}=\frac{Ia}{R^2}=\frac{1}{2}\frac{m{R}^{2}a}{R^2}$

$=\frac{1}{2}ma=\frac{1}{2}m\left(\frac{2F}{3m}\right)$

$\therefore f=\frac{F}{3}$

Hence, $\left(\text{B}\right)$ and $\left(\text{C}\right)$ are the correct answers.

Why this question? This question involves the concept of rolling motion on plank when the surface is rough and the friction between object and surface provide torque for rolling.