Question

A rod of length $l$,and mass $M$ is rotating about a vertical axis with angular velocity $\omega$ but rod makes an angle of 30 degrees with the vertical then its angular momentum is given as $\frac{M{l}^2}{3k}\omega$ then the value of $k$ is

Answer 1

MOI of rod axis from end $=\frac{1}{3}m(l_{eff}{)}^2$

Here ,

$l_{eff}=lsin\theta =\frac{l}{2}\therefore I=\frac{1}{3}m{\left(\frac{l}{2}\right)}^2=\frac{1}{12}m{l}^{2}L=I\omega =\frac{1}{12}m{l}^2\omega =\frac{1}{3}\frac{m{l}^2\omega }{k}k=4$