For a rigid body in rotational motion,if a particle at a distance of $5 \, cm$ from the axis of rotation has an angular velocity of $10 \, rad/s$,what is the linear velocity of a particle at a distance of $10 \, cm$ from the axis of rotation (in $, cm/s$)?

The pulley shown in the figure is made using a thin rim and two rods of length equal to the diameter of the rim. The rim and each rod have a mass of $M$. Two blocks of mass $M$ and $m$ are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in a vertical plane about its centre. The magnitude of the acceleration experienced by the blocks is . . . . . . (assume no slipping of the string on the pulley.)

In pure translational motion,the velocity of every particle of a body at any instant is what? Equal or unequal?

The linear velocity of a rotating body is given by $\overrightarrow v = \overrightarrow \omega \times \overrightarrow r,$ where $\overrightarrow \omega$ is the angular velocity and $\overrightarrow r$ is the radius vector. If the angular velocity of a body is $\overrightarrow \omega = \hat i - 2\hat j + 2\hat k$ and the radius vector is $\overrightarrow r = 4\hat j - 3\hat k,$ then find the magnitude of linear velocity $|\overrightarrow v |$.

$A$ mass $m$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $m$ and radius $R$. Assuming the pulley to be a perfect uniform circular disc,the acceleration of the mass $m$,if the string does not slip on the pulley,is

$A$ $5\, m$ long pole of $3\, kg$ mass is placed against a smooth vertical wall as shown in the figure. Under equilibrium condition,if the pole makes an angle of $37^{\circ}$ with the horizontal,the frictional force between the pole and the horizontal surface is