Question

A uniform disc of mass $m$ and radius $R$ rolls without slipping up a rough inclined plane at an angle of ${30}^o$ with the horizontal. If the coefficient of static and kinetic friction are each equal to $\mu$ and the forces acting on the disc are gravity and contact force, then find the direction and magnitude of the friction force acting on it.

Answer 1

Since disc does not slip, friction is static and static friction can have any value between $0$ and $\mu N$: Component of $mg$ parallel to the plane is $mg\sin \theta$ which is opposite to the direction of motion of the centre of mass decreases. For pure rolling the relation $v_{c.m}=\omega R$ must be obeyed. Therefore $\omega$ must decreases. Only friction can provide a torque about the centre.
Toque sue to friction must 0be opposite to the $\overrightarrow{\omega }$. Therefore, friction force will act up the plane.
Now, for translation motion
$mg\sin \theta -f=m{a}_{cm}$ $\left(i\right)$
For rotational motion
$fR=I\alpha$, where $I=M.I.$ of the disc about centre.
$I=\frac{a_{cm}}{R}$ as $a_{cm}=\alpha R$
$\Rightarrow a_{cm}=\frac{f{R}^2}{I}$ $\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$ we get,
$f=\frac{mg\sin \theta }{1+\frac{m{R}^2}{I}}$
Putting the value of $\theta$ and $I$, we get
$f=mg/6$