In the figure shown,the plank is being pulled to the right with a constant speed $v$. If the cylinder does not slip,then:

$A$ uniform circular disc of radius $R$, lying on a frictionless horizontal plane, is rotating with an angular velocity $\omega$ about its own axis. Another identical circular disc is gently placed on top of the first disc coaxially. The loss in rotational kinetic energy due to friction between the two discs, as they acquire a common angular velocity, is ($I$ is the moment of inertia of the disc).

$A$ particle performs uniform circular motion with an angular momentum $L$. If the angular frequency of the particle is doubled and its kinetic energy is halved,what will be its new angular momentum?

$A$ uniform disc is acted upon by two equal forces of magnitude $F$. One of them acts tangentially to the disc,while the other acts at the central point of the disc. The friction between the disc surface and the ground surface is $nF$. If $r$ is the radius of the disc,then the value of $n$ would be:

$A$ rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. $A$ particle of mass $m$ travelling along the surface hits at one end of the rod with velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After the collision, the particle comes to rest. The ratio of masses $\left(\frac{m}{M}\right)$ is $\frac{1}{x}$. The value of $x$ will be ..... .

According to the parallel axis theorem,$I = I_{cm} + Mx^2$. What will be the graph between $I$ and $x$?