Question

A uniform circular cylinder of mass m and radius r is given an initial angular velocity ${\omega }_0$ and no initial transnational velocity.It is placed in contact with a plane inclined at an angle $\alpha$ to the horizontal .If there is a coefficient of friction $\mu$ for sliding between the cylinder and plane.Find the distance the cylinder moves up before sliding stops.Also, calculate the maximum distance it travels up the plane.Assume $\mu >tan\alpha$

Answer 1

On inclined plane, the friction applies in up-the-plane direction,

So linear acceleration of centre of mass is $a=\mu gcos\alpha -gsin\alpha$

So velocity of CoM is, $v=at$

The torque due to friction about CoM is, $\tau =\mu mgrcos\alpha$

So, angular acceleration about CoM is, $\beta =\frac{\tau }{I}=\frac{\mu mgrcos\alpha }{0.5m{r}^2}=\frac{2\mu gcos\alpha }{r}$

As initial angular velocity and angular acceleration are in opposite direction. So, $\omega ={\omega }_o-\beta t$

Now, for sliding to stop, $v=\omega r$

$⟹(\mu gcos\alpha -gsin\alpha )t=({\omega }_{o}r-2\mu gcos\alpha t)$

$⟹t=\frac{{\omega }_{o}r}{g(3\mu cos\alpha -sin\alpha )}$

Thus distance traveled before sliding stops is, $s=\frac{1}{2}a{t}^2$

$⟹s=\frac{{\omega }_o^2{r}^2(\mu cos\alpha -sin\alpha )}{2g(3\mu cos\alpha -sin\alpha {)}^2}$