Question

A solid sphere is given an angular velocity $\omega$ about its centre and kept on a fixed rough inclined plane. Then choose the correct statement(s).

• A. If $\mu =\tan \theta$, then the sphere will be in translational equilibrium for some time and after that, pure rolling down the plane will start.
• B. If $\mu =\tan \theta$, then sphere will move up the plane and frictional force acting all the time will be $2mg\sin \theta$.
• C. If $\mu =\frac{\tan \theta }{2}$, there will never be pure rolling (considering inclined plane to be long enough).
• D. If inclined plane is not fixed and it is on a smooth horizontal surface, then linear momentum of the system (wedge and sphere) will be conserved in horizontal direction.

Answer 1

Answer: (A), (D)

The correct options are
A If $\mu =\tan \theta$, then the sphere will be in translational equilibrium for some time and after that, pure rolling down the plane will start.
D If inclined plane is not fixed and it is on a smooth horizontal surface, then linear momentum of the system (wedge and sphere) will be conserved in horizontal direction.

$\mu =\tan \theta$ is the condition for no slipping for a body placed on a rough inclined plane. Hence in this case, solid sphere will remain in translational equilbrium with friction $f$ balancing $mg\sin \theta$.

${\tau }_{mg}=0$ as it passes through the centre. However due to torque provided by friction about the centre i.e ${\tau }_f=fR$, sphere will have angular acceleration $\left(\alpha \right)$ in a direction opposite to $\omega$ which will decrease the angular velocity $\left(\omega \right)$ till it becomes zero. After that, it will move downwards along the inclined plane and pure rolling will start after some time due to friction coming in picture again.
Thus, option $\left(A\right)$ is correct and $\left(B\right)$ is wrong.

If $\mu =\frac{\tan \theta }{2}$, then the sphere will slip down the inclined plane immediately, so the kinetic friction force at point of contact will act upward. Torque due to this kinetic friction will be opposite to angular velocity, so it is possible that sphere will attain angular velocity ${\omega }^{\prime }$ and velocity $v^{\prime }$ such that $v^{\prime }={\omega }^{\prime }R$ i.e pure rolling. So, option C is wrong.



In the case of movable inclined plane, since there is no external force on the system of (wedge+ cylinder) in $x$ direction, therefore linear momentum of system will remain conserved along horizontal or $x-$ direction.
Option $\left(D\right)$ is correct.