Question

If the angular momentum of a rotating body about a fixed axis is increased by $10\%.$ Its kinetic energy will be increased by

• A. $10\%$
• B. $20\%$
• C. $21\%$
• D. $5\%$

Answer 1

Answer: (C)

Kinetic energy of rotation, $K_R=\frac{L^2}{2I}$
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
$L^{\prime }=L+\frac{L}{10}=\frac{11L}{10}$
$\therefore \frac{K_R^{\prime }}{K_R}=\frac{L^{{\prime }^2}}{L^2}=\frac{{\left(\frac{11L}{10}\right)}^2}{L^2}=\frac{121}{100};\text{ or }{K}_R^{\prime }=\frac{121}{100}{K}_R$
Percentage increase in the kinetic energy
$\frac{K_R^{\prime }-K_R}{K_B}\times 100\%=\frac{\frac{121}{100}{K}_R-K_R}{K_p}\times 100\%=21\%$