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 $v_0$ . The kinetic energy of the system is (slipping is absent),

• A. $6m{v}_o^2$
• B. $12m{v}_0^2$
• C. $4m{v}_0^2$
• D. $8m{v}_0^2$

Answer 1

Answer: (A)

As the ring is rolling without slipping,
$v_0=\omega r$ .
The velocity of particle at point $A$ can be given by,
$v_A=\sqrt{v_0^2+v_0^2}=\sqrt{2}{v}_0$ .
The velocity of particle at point $B$ can be given by,
$v_B=2v_0$ .
The velocity of particle at point $C$ can be given by,
$v_C=\sqrt{v_0^2+v_0^2}=\sqrt{2}{v}_0$ .
Now, the total kinetic energy of the system can be calculated by summing up kinetic energy of the particles and the total kinetic energy of the ring.
$KE=\frac{1}{2}m{\left(v_0\right)}^2+\frac{1}{2}I{\left(\omega \right)}^2+\frac{1}{2}\left(2m\right){\left(v_A\right)}^2+\frac{1}{2}\left(m\right){\left(v_B\right)}^2+\frac{1}{2}\left(m\right){\left(v_C\right)}^2$
$KE=\frac{1}{2}m{\left(v_0\right)}^2+\frac{1}{2}\left(m{r}^2\right)\left(\frac{{\left(v_0\right)}^2}{r^2}\right)+\frac{1}{2}\left(2m\right)\left(2{\left(v_0\right)}^2\right)+\frac{1}{2}\left(m\right)\left(4{\left(v_0\right)}^2\right)+\frac{1}{2}\left(m\right)\left(2{\left(v_0\right)}^2\right)$
$KE=6m{v}_0^2$ .