Question

A ring of mass $3kg$ is rolling without slipping with linear velocity $1$ ${ms}^{-1}$ on a smooth horizontal surface. A rod of same mass is fitted along its one diameter. Find total kinetic energy of the system (in $J$).

Answer 1

Moment of inertia of the rod about O, $I_o^{\prime }=\frac{1}{2}m(2R{)}^2=\frac{1}{3}m{R}^2$

Moment of inertia of the ring about O, $I_o^{\prime \prime }=m{R}^2$

Total moment of inertia of the system about O, $I_O=I_o^{\prime }+I_o^{\prime \prime }=\frac{4}{3}m{R}^2$

Now total kinetic energy of the system, $K.E=K.E_{translational}+K.E_{rotational}$

$K.E=\frac{1}{2}\left(2m\right)v^2+\frac{1}{2}{I}_O{w}^2$

$K.E=m{v}^2+\frac{1}{2}\times \frac{4}{3}m{R}^2\times w^2$

Also $v=Rw$ (rolling without slipping)

$K.E=m{v}^2+\frac{2}{3}m{v}^2=\frac{5}{3}m{v}^2$

Given: $v=1m/sm=3kg$

$K.E=\frac{5}{3}\times 3\times 1^2⟹K.E=5J$