A disc of radius $20 \mathrm{~cm}$ and mass $1 / 2 \mathrm{~kg}$ is rolling on an inclined plane. Find out the frictional force such that disc performs pure rolling.

• A

  $\frac{5 \sqrt{2}}{3} N$
• Correct answer
  B

  $\frac{5}{3 \sqrt{2}} \mathrm{~N}$
• C

  $\frac{5}{\sqrt{2}} \mathrm{~N}$
• D

  $\frac{5}{2 \sqrt{3}} \mathrm{~N}$

Solution

Let $m$ be the mass and $R$ be the radius of the disc. $m g \sin \theta-f=m a_{c m}$ $\begin{aligned} & \tau_{c m}=I_{c m^\alpha} \\ & f R=\frac{m R^2}{2}\left(\frac{a_{c m}}{R}\right)\end{aligned}$ $f=\frac{m a_{c m}}{2}$ or $a_{c m}=\frac{2 f}{m}$ So, $m g \sin \theta-f=m\left(\frac{2 f}{m}\right)$ $f=\frac{m g \sin \theta}{3}=\frac{1}{2} \times \frac{10}{3} \times \frac{1}{\sqrt{2}}=\frac{5}{3 \sqrt{2}} \mathrm{~N}$

Hence, the correct answer is option: 'B'.