Question

A uniform rod of mass m and length L performs small oscillations about a horizontal a is passing through its upper end. Find the mean kinetic energy of the rod during its oscillation period , If at t=0 it s deflected from its vertical by an angle ${\theta }_0$ and imparted an angular velocity ${\omega }_0$.

• A. $\frac{1}{4}mgL[\frac{L{\omega }^2}{3g}+{\theta }_0^2]$
• B. $mgL[\frac{2L{\omega }^2}{3g}+{\theta }_0^2]$
• C. $\frac{1}{8}mgL[\frac{2L{\omega }_0^2}{3g}+{\theta }_0^2]$
• D. $\frac{1}{4}mgL[\frac{2L{\omega }^2-0}{3g}+{\theta }_0^2]$

Answer 1

The correct option is A $\frac{1}{4}mgL[\frac{L{\omega }^2}{3g}+{\theta }_0^2]$

Center of mass shifts upward by a distance $\frac{L}{2}\times 2{sin}^2\frac{\theta }{2}=\frac{L{\theta }^2}{4}$

$[sin\theta \approx \theta$ for small $\theta$]

As mechanical energy of the rod is conserved,

$KE+PE=const$

$\frac{1}{2}I{\omega }^2+mgx=const\frac{1}{2}\left(\frac{{mI}^2}{3}\right){\left(\frac{d\theta }{dt}\right)}^2+\frac{1}{2}\left(\frac{mgI}{3}\right){\theta }^2=const$

differentiate it wrt time as its derivative must be zero,

$\frac{d^2\theta }{{dt}^2}=(-\frac{3g}{2l})\theta \alpha ={\omega }^2\theta (\alpha$-angular acceleration)

$\omega =\sqrt{\frac{3g}{2I}}$

Consider $A$ general equation,

$\theta =Acos\omega t+Bsin\omega t$

As $\theta ={\theta }_0$ at $t=0$, so $A={\theta }_o$

and $B=\frac{{\theta }_o}{\omega }$

thus,

$\theta ={\theta }_{0}cos\omega t+\frac{{\theta }_0}{\omega }sin\omega tKE=\frac{1}{2}I{\omega }^2=-{[\omega {\theta }_{0}sin\omega t+{\theta }_{0}cos\omega t]}^2$

Averaging over time period,

$K_{Av}=\frac{1}{12}{mI}^2{{\theta }_0}^2+\frac{1}{8}{mgI}^2{{\theta }_0}^2$