Question

In the figure shown, the spherical body and block, each have a mass $m$. The moment of inertia of the spherical body about centre of mass is $2m{R}^2$. The spherical body rolls without slipping on the horizontal surface. The ratio of kinetic energy of the spherical body to that of block is:

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{2}{3}$

Given that,

Mass = m

Moment of inertia $I=2m{R}^2$

Velocity of block$v=v_{1}m/s$

Velocity of body $v=v_{2}m/s$

Angular acceleration =$\omega$

For spherical body

Now, for pure rolling motion

$v_2-\omega \left(2R\right)=0$

$v_2=\omega 2R$

Now, for the point where the string is attached to the spherical body

So,

$v_2+\omega R=v_1$

$v_1=3\omega R$

Now, the total kinetic energy of the spherical body its sum of the rotational kinetic and translational kinetic energy

$K.E=\frac{I{\omega }^2}{2}+\frac{m{v}_2^2}{2}$

$K.E=\frac{2m{R}^2{\omega }^2}{2}+\frac{m4R^2{\omega }^2}{2}$

$K.E=3m{R}^2{\omega }^2$

Now, the kinetic energy of the block is

$K.E=\frac{1}{2}m{v}_1^2$

$K.E=\frac{1}{2}m9{\omega }^2{R}^2$

Now, the ratio of their kinetic energies will be

$=\frac{3m{R}^2{\omega }^2\times 2}{9m{R}^2{\omega }^2}$

$=\frac{2}{3}$

Hence, the ratio of kinetic energy of the spherical body to that of block is $\frac{2}{3}$