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Q.
If the angular momentum of a rotating body about a fixed axis is increased by $10 \% .$ Its kinetic energy will be increased by
System of Particles and Rotational Motion
Solution:
Kinetic energy of rotation, $K_{R}=\frac{L^{2}}{2 I}$
For a given $I$ (moment of inertia of a body), $K_{R} \propto L^{2}$
When the angular momentum of a body is increased by $10 \%$
i.e., $\frac{L}{10}$, its angular momentum will become
$L^{\prime}=L+\frac{L}{10}=\frac{11 L}{10}$
$\therefore \frac{K_{R}^{\prime}}{K_{R}}=\frac{L^{\prime^{2}}}{L^{2}}=\frac{\left(\frac{11 L}{10}\right)^{2}}{L^{2}}=\frac{121}{100} ; \text { or } K_{R}^{\prime}=\frac{121}{100} K_{R}$
Percentage increase in the kinetic energy
$\frac{K_{R}^{\prime}-K_{R}}{K_{B}} \times 100 \%=\frac{\frac{121}{100} K_{R}-K_{R}}{K_{p}}\times 100 \%=21 \%$