Question

A uniform ring of mass M and radius R is rolling
without slipping over a smooth horizontal surface. If
centre of mass of ring is moving with speed vas
shown, then total kinetic energy of ring would be​

Answer 1

The kinetic energy of the ring would be K = 5 / 3 mv^2

Explanation:

The kinetic energy is given as:

k = 1/2 m.v^2

The total kinetic energy of the system is:

K = K(trans) + k(rotational)

K = 1/2 x 2 m x v^2  

[ Since v = ωR]

[ω = V / R]

K = 1/2 [ mR^2 + M(2R)^2 /2 ]( V / R)^2

K = 10 / 6 mv^2

K = 5 / 3 mv^2

Thus the kinetic energy of the ring would be K = 5 / 3 mv^2