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) \(1 : 2\)
(A) \(1 : 4\)
(B) \(1 : 5\)
(C) \(1 : 1\)
(D) \(1 : 2\)
Answer
Answer: A
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
\(\dfrac{\dfrac{1}{2}[I\omega^2]_{4m}}{\dfrac{1}{2}[I\omega^2]_m}\)
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
\(\dfrac{4m(\dfrac{r}{5})^2}{m(\dfrac{4r}{5})^2}=\dfrac{1}{4}\)
\(\dfrac{4m(\dfrac{r}{5})^2}{m(\dfrac{4r}{5})^2}=\dfrac{1}{4}\)
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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
\(\frac{{4m{{(\frac{r}{5})}^2}}}{{m{{(\frac{{4r}}{5})}^2}}} = \frac{1}{4}\)
\(\frac{{4m \times 0 + m \times r}}{{4m + m}} = \frac{r}{5}\)
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
\(\frac{{4m{{(\frac{r}{5})}^2}}}{{m{{(\frac{{4r}}{5})}^2}}} = \frac{1}{4}\)
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