- #1

phoenixXL

- 49

- 3

## Homework Statement

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,

[tex]

\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)

[/tex]

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

[tex]

mgsin(30°)\ -\ f\ =\ ma~~...(ii),~~~~f\ =\ frictional force,~~a\ =\ net acceleration.[/tex]

Taking torque about center of mass

[tex]

\tau\ =\ I\alpha\\

\implies f\ =\ \frac{1}{2}mr^2.\frac{g}{3r}~~~~from (i)\\

\implies f\ =\ \frac{mg}{6}~~...(iii)

[/tex]

Substituting (iii) in (ii), We get

[tex]

a\ =\ \frac{g}{3}

[/tex]

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.Please help me out, I am not able to clear my concepts of rotational motion.

Thanks for your time