Consider a point $P$ at the contact point of a wheel on the ground,which rolls on the ground without slipping. Find the displacement of point $P$ when the wheel completes half of a rotation (given the radius of the wheel is $1 \ m$).

$A$ disc of radius $1\,m$ and mass $4\,kg$ rolls on a horizontal plane without slipping in such a way that its centre of mass moves with a speed of $10\,cm/s$. Its rotational kinetic energy is (in $,J$)

$A$ disc of radius $2\; m$ and mass $100\; kg$ rolls on a horizontal floor. Its centre of mass has a speed of $20\; cm/s$. How much work is needed to stop it?

$A$ disc of radius $R$ is rotating with an angular velocity $\omega_0$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $\mu_k$.
$(a)$ What was the velocity of its centre of mass before being brought in contact with the table?
$(b)$ What happens to the linear velocity of a point on its rim when placed in contact with the table?
$(c)$ What happens to the linear speed of the centre of mass when the disc is placed in contact with the table?
$(d)$ Which force is responsible for the effects in $(b)$ and $(c)$?
$(e)$ What condition should be satisfied for rolling to begin?
$(f)$ Calculate the time taken for the rolling to begin.

$A$ wheel is rolling straight on ground without slipping. If the axis of the wheel has speed $v$,the instantaneous velocity of a point $P$ on the rim,defined by angle $\theta$ relative to the vertical,relative to the ground will be:

$A$ solid sphere of mass $500 \ g$ and radius $10 \ cm$ rolls without slipping with a velocity of $20 \ cm/s$. The total kinetic energy of the sphere will be ........ $J$.