A uniform thin rod $\mathrm{AB}$ of length $L$ and mass $m$ is undergoing rotation about a fixed axis passing through end $\mathrm{A}$ and perpendicular to the rod, such that end $\mathrm{A}$ remains stationary as shown. The kinetic energy of section $\mathrm{AP}$ of the rod is equal to the kinetic energy of section $\mathrm{BP}$ of the rod at an instant. Then what is the value of ${\left(\frac{\mathrm{A}\mathrm{B}}{\mathrm{A}\mathrm{p}}\right)}^3$?($\mathrm{AB}$ and $\mathrm{AP}$ are lengths of respective parts of the rod)

Solution

The KE of given section AP and PB will be equal if the moment of inertia of each section AP and section PB about A is same.



Moment of inertia of segment AP w.r.t given axis ${\mathrm{I}}_1=\left(\mathrm{\lambda }\mathrm{x}\right)\frac{{\mathrm{x}}^2}{3}\mathrm{ }$

Moment of inertia of segment PB w.r.t given axis ${\mathrm{I}}_2={\mathrm{I}}_0-{\mathrm{I}}_1=\left(\mathrm{\lambda }\mathrm{L}\right)\frac{{\mathrm{L}}^2}{3}-\left(\mathrm{\lambda }\mathrm{x}\right)\frac{{\mathrm{x}}^2}{3}$

(Moment of inertia or rod AB w.r.t. given axis is ${\mathrm{I}}_0$ )

$\therefore {\mathrm{I}}_1={\mathrm{I}}_2\Rightarrow \frac{\mathrm{\lambda }{\mathrm{x}}^3}{3}=\frac{\mathrm{\lambda }{\mathrm{L}}^3}{3}-\frac{\mathrm{\lambda }{\mathrm{x}}^3}{3}$

Or ${\mathrm{x}}^3=\frac{{\mathrm{L}}^3}{2}$

${\left(\frac{\mathrm{A}\mathrm{B}}{\mathrm{A}\mathrm{P}}\right)}^3=\frac{{\mathrm{L}}^3}{\frac{{\mathrm{L}}^3}{2}}=2$

Hence, the correct answer is 2.