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 $v_0$. Find the kinetic energy of the system.(Slipping is absent).

Answer 1

Velocity of C.M is $v_{com}=v_0$

As this motion is pure rotational the velocity is of rotation.

$\omega =\frac{v_0}{R}$

Figure shows the velocities of acting on the particles.

Resultant velocity particle of mass 2m equals to

$⟹\sqrt{(v_0{)}^2+(\omega R{)}^2}=\sqrt{(v_0{)}^2+(v_0{)}^2}$

$⟹\sqrt{2}{v}_0$

Resultant velocity particle on top of mass m is

$⟹\omega R+v_0=2v_0$

Resultant velocity particle of mass'm' on right

$⟹\sqrt{(\omega R{)}^2+(v_0{)}^2}=\sqrt{2}{v}_0$

So Kinetic energy of particles

$⟹\frac{1}{2}m(2v_0{)}^2+\frac{1}{2}m(\sqrt{2}{v}_0{)}^2+\frac{1}{2}2m(\sqrt{2}{v}_0{)}^2$

$⟹2m{v}_0+m(v_0{)}^2+2(v_0{)}^{2}m$

$⟹5m(v_0{)}^2$

Kinetic energy of ring

$⟹\frac{1}{2}m(v_{cm}{)}^2+\frac{1}{2}I{\omega }^2$

Here, I = Moment of inertia of ring about its centre

$=m{R}^2$

So, Kinetic energy of ring

$=\frac{1}{2}m(v_0{)}^2+\frac{1}{2}m{R}^2(\frac{v_0}{R}{)}^2$

$=m{v}_0^2$

So Total kinetic energy is equal to

$⟹5m{v}_0^2+m{v}_0^2=6m{v}_0^2$