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).

Solution

## Velocity of C.M is $$v_{com}=v_0$$As this motion is pure rotational the velocity is of rotation.$$\omega = \cfrac{v_0}{R}$$Figure shows the velocities of acting on the particles.Resultant velocity particle of mass 2m equals to$$\implies \sqrt{(v_0)^2+(\omega R)^2} = \sqrt{(v_0)^2 + (v_0)^2}$$$$\implies \sqrt2 v_0$$Resultant velocity particle on top of mass m is$$\implies \omega R+v_0 =2v_0$$Resultant velocity particle of mass'm' on right$$\implies \sqrt{(\omega R)^2 + (v_0)^2} = \sqrt2 v_0$$So Kinetic energy of particles$$\implies \cfrac{1}{2}m(2v_0)^2 + \cfrac{1}{2} m(\sqrt2 v_0)^2 +\cfrac{1}{2} 2m(\sqrt2 v_0)^2$$$$\implies 2mv_0 +m(v_0)^2 + 2(v_0)^2m$$$$\implies 5m(v_0)^2$$Kinetic energy of ring$$\implies \cfrac{1}{2}m(v_{cm})^2 + \cfrac{1}{2} I\omega^2$$Here, I = Moment of inertia of ring about its centre$$=mR^2$$So, Kinetic energy of ring$$=\cfrac{1}{2}m(v_0)^2 + \cfrac{1}{2}mR^2(\cfrac{v_0}{R})^2$$$$=mv_0^2$$So Total kinetic energy is equal to$$\implies 5mv_0^2 +mv_0^2 = 6mv_0^2$$Physics

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