Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $\omega_1$ and $\omega_2$ They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is
$I{\left( {{\omega _1} - {\omega _2}} \right)^2}$
$\frac{I}{8}{\left( {{\omega _1} - {\omega _2}} \right)^2}$
$\;\frac{I}{2}{\left( {{\omega _1} + {\omega _2}} \right)^2}$
$\;\frac{I}{4}{\left( {{\omega _1} - {\omega _2}} \right)^2}$
A solid sphere of mass $1\,kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}\,J$. The speed of the centre of mass of the sphere is $.........cm s ^{-1}$.
Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are $0.1 \;kg - m ^{2}$ and $10\; rad \,s^{-1}$ respectively while those for the second one are $0.2 \;kg - m ^{2}$ and $5\; rad \,s ^{-1}$ respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The Kinetic energy of the combined system is ...........$J$
Write the formula for power and angular momentum in rotational motion.
A body rolls down an inclined plane without slipping. The kinetic energy of rotation is $50 \,\%$ of its translational kinetic energy. The body is :
A solid sphere is rolling down an inclined plane. Then the ratio of its translational kinetic energy to its rotational kinetic energy is