Question

The moment of inertia of a solid cylinder about its axis is given by $1/2M{R}^2$. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is:

• A. $1:1$
• B. $2:2$
• C. $1:2$
• D. $1:3$

Answer 1

The correct option is C $1:2$

The rotational kinetic energy is given by

$R_E=\frac{1}{2}I{\omega }^2$

The transnational kinetic energy is given by

$T_E=\frac{1}{2}M{v}^2$

Now,

$\frac{R_E}{T_E}=\frac{\frac{1}{2}I{\omega }^2}{\frac{1}{2}M{v}^2}$

$\frac{R_E}{T_E}=\frac{I{\omega }^2}{v^2}$

Since, $I=\frac{1}{2}M{R}^2$ ........given

Hence,

$\frac{R_E}{T_E}=\frac{\left(\frac{1}{2}M{R}^2\right){\omega }^2}{M{v}^2}$

using, $\omega =\frac{v}{R}$ in above equation, we get

$\frac{R_E}{T_E}=\frac{\left(\frac{1}{2}{R}^2\right)(\frac{v}{R}{)}^2}{v^2}$

$\frac{R_E}{T_E}=\frac{1}{2}$