CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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).
741643_66fbd7e8aec84f669fdf356d0785f6d1.png


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
$$\implies 5mv_0^2 +mv_0^2 = 6mv_0^2$$

957027_741643_ans_1d8aceeb4ac24910a62858e7fec8bde0.png

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image