Question

A cylinder of mass m is rotated about its axis by an angular velocity $\omega$ and lowered gently on an inclined plane as shown in figure. Then:

• A. it will start going upward
• B. it will first going upward and then downward
• C. it will go downward just after it is lowerd
• D. it can never go upward

Answer 1

Answer: (D)

it can never go upward

As soon as sphere touch inclined plane,friction force starts acting on it.

$fr=\mu N$

$N$ is normal reaction , here $N=mgcos\theta =mg\cos {30}^o$

From figure it is clear that point of contact will be moving downwards wrt surface.

so friction will act upward in the direction of inclined plane.

$fr=\frac{1}{\sqrt{3}}mg\frac{\sqrt{3}}{2}$

$fr=\frac{mg}{2}$

gravitation force towards downward in direction with inclined plane.

$F_g=mgsin\theta =mg\sin {30}^o=\frac{mg}{2}$

It is clear that $F_g=fr$

Since net force along the incline is zero, so cylinder will remain in position till it stops rotating. After that it will start moving downwards.

Hence option D is correct, it can never go upwards.