Inclined Plane Rotational Motion

1. The problem statement, all variables and given/known data
A cyclinder of mass m and radius r is rotated about its axis by an angular velocity ω and then lowered gently on an inclined plane as shown in figure. Then:

(a) It will start going upward.
(b) It will first go up and then downward.
(c) It will go downward just after it is lowered.
(d) It can never go upward.

2. The attempt at a solution

Taking torque that is due to sine component of weight about the point of contact,
$$\tau\ =\ I\alpha\\ \implies \tau\ =\ \frac{3}{2}mr^2(\alpha)\\ \implies r.mgsin(30°)\ =\ \frac{3}{2}mr^2(\alpha)\\ \implies \alpha\ =\ \frac{g}{3r}~~…(i)$$
The sense of rotation of angular acceleration and of angular velocity is opposite, therefore soon the cyclinder will stop rotating.

More Equations, from newtons laws of motion
$$mgsin(30°)\ -\ f\ =\ ma~~…(ii),~~~~f\ =\ frictional force,~~a\ =\ net acceleration.$$
Taking torque about center of mass
$$\tau\ =\ I\alpha\\ \implies f\ =\ \frac{1}{2}mr^2.\frac{g}{3r}~~~~from (i)\\ \implies f\ =\ \frac{mg}{6}~~…(iii)$$

Substituting (iii) in (ii), We get
$$a\ =\ \frac{g}{3}$$

Therefore, we can say that the body when lowered will begin to move down with its rotational motion getting reduced. Therefore, the answer should be (c)

But,
Solution Given in Book

(d) It can never go upward.

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