Question

A rod of length $l$ rotating with an angular velocity ${\omega }_0$ and its center moves with a linear velocity $v_0=\frac{{\omega }_{0}l}{3}$, if the end $B$ of the rod is suddenly fixed, the angular velocity of the rod will be

• A. $\frac{4}{3}{\omega }_0$
• B. $\frac{{\omega }_0}{3}$
• C. $\frac{{\omega }_0}{4}$
• D. $\frac{3{\omega }_0}{4}$

Answer 1

The correct option is D $\frac{3{\omega }_0}{4}$

Given angular velocity $=\omega$ $\text{Liniar velocity}=\frac{\text{wd}}{3}\left(v_0\right)$ Applying conservation of angular momentum $\begin{array}{c}\frac{m{L}^2}{12}{\omega }_0+m\times \frac{{\omega }_d}{3}x\frac{1}{2}=\frac{m{\lambda }^2}{3}\omega \\ \frac{{\omega }_0}{12}+\frac{{\omega }_0}{6}=\frac{{\omega }^{\prime }}{3}\end{array}$

$\begin{array}{cc}& \frac{3{\omega }_0}{18}=\frac{{\omega }^1}{3}\\ \Rightarrow {\omega }^{\prime }& =\frac{3{\omega }_0}{4}\end{array}$