Question

Two points masses of mass $4m$ and $m$ respectively separated by distance $r$, are revolved under mutual force of attraction. Ratio of their kinetic energies will be

• A. $1:4$
• B. $1:5$
• C. $1:1$
• D. $1:2$

Answer 1

The correct option is A $1:4$

Considering the origin of the coordinate system at $4m$, we evaluate the position of the center of mass as
$\frac{4m\times 0+m\times r}{4m+m}=\frac{r}{5}$
Thus the center of mass is $\frac{r}{5}$ from $4m$ and $\frac{4r}{5}$ from m.
The ratio of their kinetic energy is given as
$\frac{\frac{1}{2}[I{\omega }^2{]}_{4m}}{\frac{1}{2}[I{\omega }^2{]}_m}$

As the angular velocity of the both the masses would be same we get the ratio of kinetic energy as
$\frac{4m(\frac{r}{5}{)}^2}{m(\frac{4r}{5}{)}^2}=\frac{1}{4}$