$A$ particle on the outer surface of a wheel,which is initially at rest,is in contact with the ground at point $P$. Find the displacement of this particle when the wheel completes half a rotation in the forward direction. (Radius of the wheel $= 5 \ m$)

The ratio of the total kinetic energy to the rotational kinetic energy for a rolling disc is:

$A$ small roller of diameter $20 \ cm$ has an axle of diameter $10 \ cm$. It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle. The scale is now pushed slowly on the axle so that it moves without slipping on the axle,and the roller starts rolling without slipping. After the roller has moved $50 \ cm$,the position of the scale will look like (figures are schematic and not drawn to scale)-

Read each statement below carefully,and state,with reasons,if it is true or false;
$(a)$ During rolling,the force of friction acts in the same direction as the direction of motion of the $CM$ of the body.
$(b)$ The instantaneous speed of the point of contact during rolling is zero.
$(c)$ The instantaneous acceleration of the point of contact during rolling is zero.
$(d)$ For perfect rolling motion,work done against friction is zero.
$(e)$ $A$ wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

$A$ uniform solid sphere of mass $m$ and radius $R$ is being pulled on a horizontal surface with a force $F$ parallel to the surface,applied at its topmost point. If the acceleration of the center of mass of the sphere is $a$ and it is rolling without slipping on the surface,the value of $F$ is:

$A$ circular hoop of radius $50 \ cm$ and mass $1 \ kg$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. Let $v$ be the velocity of the centre of the hoop when it ceases to slip. The ratio $v / \omega_0$ will be (in $cm$)