A disc rotates freely with rotational kinetic energy E about a normal axis through centre. A ring having the same radius but double the mass of disc is now, gently placed on the disc. The new rotational kinetic energy of the system would be

# A disc rotates freely with rotational kinetic energy E about a normal axis through centre. A ring having the same radius but double the mass of disc is now, gently placed on the disc. The new rotational kinetic energy of the system would be

1. A

E/2

2. B

2E/5

3. C

E/5

4. D

E/3

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### Solution:

From Law of conservation of angular momentum :
$\frac{{\mathrm{MR}}^{2}}{2}\mathrm{\omega }=\left[\frac{{\mathrm{MR}}^{2}}{2}+\left(2\mathrm{M}\right){\mathrm{R}}^{2}\right]{\mathrm{\omega }}_{2}$

${\mathrm{\omega }}_{2}=\frac{\frac{{\mathrm{MR}}^{2}}{2}\mathrm{\omega }}{\frac{5{\mathrm{MR}}^{2}}{2}}=\frac{\mathrm{\omega }}{5}$

The ratio of energies in both the cases is $\frac{{\mathrm{E}}_{1}}{{\mathrm{E}}_{2}}=\frac{\frac{1}{2}\mathrm{I\omega }2}{\frac{1}{2}\left(5\mathrm{I}\right){\left(\frac{\mathrm{\omega }}{5}\right)}^{2}}=5$

$\therefore {\mathrm{E}}_{2}=\frac{\mathrm{E}}{5}$

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