Question

A thin rod of length $L$ and mass $M$ is held vertically with one end on the floor and s allowed to fall. Find the velocity of the other end when it hits the floor, assuming that the end on the floor not slip

• A. $\sqrt{\frac{3g}{L}}$
• B. $\sqrt{3gL}$
• C. $\sqrt{\frac{L}{3g}}$
• D. $\sqrt{\frac{8}{3L}}$

Answer 1

Answer: (B)

$\sqrt{3gL}$

$\begin{array}{c}\text{On applying the conservation of mechanical energy}\\ \text{displacement of centre of mass}=\frac{L}{2}\text{,}\\ \text{so,}{W}_{\text{done}}=\Delta k\cdot E\\ \Rightarrow \mathrm{mg}\frac{L}{2}=\frac{1}{2}I{\omega }^2\begin{array}{c}\text{(I}\to \text{moment of}\\ \text{inertia about end}0\end{array}\end{array}$

$\begin{array}{cc}\text{Here,}& I=\frac{M{L}^2}{3}\\ \text{S0,}& MgL/2=\frac{1}{2}\times \frac{m{L}^2}{3}\cdot {\omega }^2\\ & \Rightarrow \omega L=\sqrt{3gL}\end{array}$

$\begin{array}{c}\text{Note that, velocity (linear) of end}A=\omega L=\sqrt{3gL}\text{.}\\ \text{So, the correct option is (B).}\end{array}$