Question

A body rolls without slipping. The radius of gyration of the body above an axis passing through its centre of mass is $K$. The radius of the body is $R$ The ratio of rotational kinetic energy to translational kinetic energy is

• A. $\frac{V^2}{R^2}$
• B. $\frac{R^2}{v^2+R^2}$
• C. $\frac{V^2}{V^2+R^2}$
• D. $v^2+R^2$

Answer 1

Answer: (A)

First of all the question has an grammatical error like the radius of gyration of the body'about an axis.......
the ratio of rotational' kinetic energy.........
moment of inertia of the body about that axis $=m\times k^2=M.I$.
so, rotational kinetic energy of the body is $=1/2\times$ M.I. $\times {\omega }^2$ (where $\omega$ is angular velocity of the body w.r.t. the axis)
translational kinetic energy of the body $=1/2\times M\times {\left(V_{CM}\right)}^2$
Now you cannot proceed further until $u$ get the idea about the relation between $v$ and $\omega$