A hoop of radius $r$ and mass $m$ 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. What will be the velocity of the centre of the hoop when it ceases to slip?
$\frac{{r{\omega _0}}}{4}$
$\frac{{r{\omega _0}}}{3}$
$\frac{{r{\omega _0}}}{2}$
${r{\omega _0}}$
The given figure shows a disc of mass $M$ and radius $R$ lying in the $x-y$ plane with its centre on $x$ axis at a distance a from the origin. then the moment of inertia of the disc about the $x-$ axis is
Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed $\omega _1$ and $\omega _2$ are brought into contact face to face with their axes of rotation coincident. What is the loss in kinetic energy of the system in the process?
If the equation for the displacement of a particle moving on a circular path is given by:
$\theta = 2t^3 + 0.5$
Where $\theta $ is in radian and $t$ in second, then the angular velocity of the particle at $t = 2\,sec$ is $t=$ ....... $rad/sec$
A disc of mass $M$ and radius $R$ is rolling with angular speed $\omega $ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origion $O$ is
Five masses each of $2\, kg$ are placed on a horizontal circular disc, which can be rotated about a vertical axis passing through its centre and all the masses be equidistant from the axis and at a distance of $10\, cm$ from it. The moment of inertia of the whole system (in $gm-cm^2$) is (Assume disc is of negligible mass)