Question

A disc is rotated about its axis with a certain angular velocity and lowered gently onto an inclined plane as shown in the figure. Then,

• A. It will rotate at the position where it was placed and then will move downwards.
• B. It will go downwards just after it is placed.
• C. It will move downwards first and then climb up along the inclined plane.
• D. It will climb upwards along the inclined plane and then move downwards.

Answer 1

The correct option is A It will rotate at the position where it was placed and then will move downwards.

Maximum value of friction is $f=\mu N=\mu mg\cos {30}^{\circ }$
$f=\frac{1}{\sqrt{3}}\times mg\times \frac{\sqrt{3}}{2}=\frac{mg}{2}$
Since the point of contact will have tangential velocity down the incline, hence slippping will happen and kinetic friction will act in upward direction to prevent slipping.
Component of gravity down the inclined plane $=mg\sin {30}^{\circ }=\frac{mg}{2}$
Hence, maximum friction force will exactly balance the component of gravity down inclined plane.
Also, torque due to friction will act opposite to the direction of initial angular velocity i.e disc will slow down.


$\therefore$ Disc will rotate at its initial position and will not move upwards or downwards till its angular velocity becomes zero (friction also becomes zero). After that, it will move downwards under the influence of component of gravity$\frac{mg}{2}$.
i.e. option (a) is correct.