Question

A uniform rod of mass M and length L hinged about top point is rotating at angular velocity $\omega$ in such a way that it carves out conical shape while moving as shown in figure. Magnitude of angular momentum of rod about hinge point is

• A. $\frac{M\omega L^2}{12}$
• B. $\frac{M\omega L^2}{6}$
• C. $\frac{3}{4}M\omega L^2$
• D. $\frac{M\omega L^2}{4}$

Answer 1

The correct option is B $\frac{M\omega L^2}{6}$


$\overrightarrow{dL}=\overrightarrow{r}\times \overrightarrow{P}\Rightarrow dL=l\left(dm\right)v=l\left(\lambda dl\right)r\omega$

$dL=l\left(\lambda dl\right)lsin\theta \omega =\lambda l^2\omega sin\theta dl$

$|\overrightarrow{L}|=\int dL=\lambda \omega sin\theta {\int }_0^L{l}^{2}dl=\lambda \omega sin\theta \frac{L^3}{3}$

$|\overrightarrow{L}|=\frac{M}{L}\omega sin{30}^{\circ }\frac{L^3}{3}=\frac{M\omega L^2}{6}$