Question

Uniform thin rod $AB$ of length $L$ and mass $m$ is undergoing fixed axis rotation whose axis of rotation passes through $A$ , such that end $A$ remains stationary as shown. The kinetic energy of section $AP$ of rod is equal to kinetic energy of section $BP$ of the rod at a given instant. Then the ratio of length $AP$ and length $AB$ , that is, $\frac{AP}{AB}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{2^{1 / 3}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2 \sqrt{2}}$

Answer 1

Answer: (B)

The kinetic energy $\left(K E = \frac{1}{2} I \left(\omega \right)^{2}\right)$ of given sections $AP$ and $PB$ will be equal if Moment of Inertia of each section $AP$ and section $PB$ about $A$ is same.

$I_{AP}=\left(\right.\lambda x\left.\right)\frac{x^{2}}{3}$
$I_{PB}=I_{AB}-I_{AP}=\left(\right.\lambda L\left.\right)\frac{L^{2}}{3}-\left(\right.\lambda x\left.\right)\frac{x^{2}}{3}$
$\therefore I_{AP}=I_{PB}\Rightarrow \frac{\lambda x^{3}}{3}=\frac{\lambda L^{3}}{3}-\frac{\lambda x^{3}}{3}$
or $x^{3}=\frac{L^{3}}{2}orx=\frac{L}{2^{1 / 3}}$
Now, $\frac{A P}{A B}=\frac{x}{L}=\frac{1}{2^{1/3}}$