Question

A ring of mass $m$ and radius $R$ has three particles attached to the ring as shown in the figure. The centre of the ring has a speed $v_0$. If the kinetic energy of the system is $xm{v}_0^2$. Find $x$ (Slipping is absent)

Answer 1

Since slipping is absent $v_o$ = R${\omega }_o$

Speed of the 2m particle : $\sqrt{{v_o}^2+{R{\omega }_o}^2}$ = $\sqrt{2}{v}_o$

Speed of the m(right) particle : $\sqrt{{v_o}^2+{R{\omega }_o}^2}$ = $\sqrt{2}{v}_o$

Speed of the m(top) particle : $v_o+R{\omega }_o$ = $2v_o$

KE(total) = sum of individual KE

KE = $\frac{1}{2}m{v}_o^2+\frac{1}{2}\left(m{R}^2\right)(\frac{v_o}{R}{)}^2+\frac{1}{2}(3m\left)\right(\sqrt{2}{v}_o{)}^2+\frac{1}{2}\left(m\right)(2v_o{)}^2$ = $6m{v}_o^2$

$\therefore x=6$