Question

Three particles are attached to a ring of mass $m$ and radius $R$ as shown in the figure. The center of mass the ring has a speed $V_0$ and rolls without slipping on a horizontal surface. The kinetic energy of the system in the position shown in the figure is:

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

Answer 1

The correct option is A $6m{V}_0^2$

Since body is in pure rolling,

$V_o=R\omega$

Kinetic energy of the ring is:

$E_{ring}=\frac{1}{2}m{V}_o^2+\frac{1}{2}I{\omega }^2$

$=\frac{1}{2}m{V}_o^2+\frac{1}{2}m{R}^2{\omega }^2$

$=\frac{1}{2}m{V}_o^2+\frac{1}{2}m{V}_o^2$

$=m{V}_o^2$

Speed of the particle on the left is:

$v_1=\sqrt{V_o^2+\left(R{\omega }^2\right)}$

$=\sqrt{V_o^2+V_o^2}$

$=V_o\sqrt{2}$

Its kinetic energy is:

$E_1=\frac{1}{2}\times 2m\times v_1^2$

$=2m{V}_o^2$

Speed of the particle on the right is:

$v_2=\sqrt{V_o^2+\left(R{\omega }^2\right)}$

$=\sqrt{V_o^2+V_o^2}$

$=V_o\sqrt{2}$

Its kinetic energy is:

$E_2=\frac{1}{2}\times m\times v_2^2$

$=m{V}_o^2$

Speed of the particle on the top is:

$v_3=V_o+R\omega$

$=2V_o$

Its kinetic energy is:

$E_3=\frac{1}{2}\times m\times v_3^2$

$=2m{V}_o^2$

Hence, total kinetic energy of the system is:

$E=E_{ring}+E_1+E_2+E_3$

$E=6m{V}_o^2$