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QuestionPhysicsClass 11

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) $$1 : 2$$

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

As the angular velocity of the both the masses would be same we get the ratio of kinetic energy as
$$\dfrac{4m(\dfrac{r}{5})^2}{m(\dfrac{4r}{5})^2}=\dfrac{1}{4}$$
4.6
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4.6
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Solution

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
$$\frac{{4m{{(\frac{r}{5})}^2}}}{{m{{(\frac{{4r}}{5})}^2}}} = \frac{1}{4}$$