Question

Find the total energy of the inclined rod of mass $m$, rotating with an angular velocity $\omega$ about a vertical axis $Y{Y}^{\prime }$ as shown in figure.

• A. $\frac{m{\omega }^2{l}^2}{12}$
• B. $\frac{m{\omega }^2{l}^2}{24}$
• C. $\frac{m{\omega }^2{l}^2}{36}$
• D. $\frac{m{\omega }^2{l}^2}{48}$

Answer 1

The correct option is B $\frac{m{\omega }^2{l}^2}{24}$

The mass of element $dx$ shown is,
$dm=\frac{mdx}{2l}$

Kinetic energy of $dm$ is,
$dE=\frac{1}{2}\left(\frac{m}{2l}dx\right)[x\sin \theta \omega {]}^2$
total energy of rod is,
$E=\int dE=\frac{m}{4l}\left[{\sin }^2\theta \right]{\omega }^2{\int }_{-l}^{+l}{x}^{2}dx\Rightarrow E=\frac{m{\omega }^2}{4l}{\sin }^2\theta [\frac{x^3}{3}{]}_{-l}^l=\frac{m{\omega }^2}{12l}{\sin }^2({30}^{\circ })[l^3-(-l{)}^3]\Rightarrow E=\frac{m{\omega }^2}{48l}\left[2l^3\right]=\frac{m{\omega }^2{l}^3}{24l}=\frac{m{\omega }^2{l}^2}{24}$