A uniform rod of mass `m` and length `l` performs small oscillations about the horizontal axis passing throughits upper end. Final the mean kinetic energy of the rod averaged over one oscillation period it at the initial mment it was deflected from the vertical by an angle `theta_(0)` and then imparted an angular velocity `theta_(0)`.

Explanation

Moment of inertia of the rod equals `(ml^(2))/(3)` about its one end and perpendicular to its length. Thus rotational kinetic of the rod `=(1)/(2) ((ml^(2))/(3))dot(theta^(2))=(ml^(2))/(6)dot (theta^(2))`
when the rod is displaced by an angle `theta` its C.G. goes up by a distance `(l)/(2)(1-cos theta)~=(l theta^(2))/(4)` for small `theta`
Thus the P.E. becomes `: mg((l theta^(2))/(4))`
As the mechanial energy of oscillation fo the rod is conserved .
`(1)/(2)((ml^(2))/(3))dot(theta^(2))+(1)/(2)((mgl)/(2))theta^(2)=` Constant
on differenticating w.r.t. time and for the simplifies we get `: ddot (theta)=-(3g)/(2l) theta` for small `theta`.
we see tha the angular frequency `omega` is
`=sqrt(3g//2l)`
we write the general solution of the angular oscillation is `:`
`theta=A cos omega t + B sin omega t `
But ` theta=theta_(0)` at `t=0`, so `A=theta_(0)`
and ` theta=theta_(0)` at `t=0, ` so
`B=theta_(0)//omega`
Thus `theta=theta_(0)cos omegat+(theta_(0))/(omega)sin omegat`
Thus the K.E. of the rod
`T=(ml^(2))/(6) dot(theta^(2))=[-omega theta_(0)sin omegat +dot(0_(0))cos omegat ]^(2)`
`=(ml^(2))/(6)[dot(theta_(0)^(2))cos ^(2)omegat +omega^(2)theta_(0)^(2)sin ^(2) omega t - 2 omega 0_(0)dot (0_(0)) sin omegat cos omegat ]`
On averaging over one time period the last term vanishes and `lt sin^(2)omegat gt =lt cos ^(2)omegat lt = 1//2.` Thus
`lt T gt =(1)/(12 ml^(2)) dot(theta_(0)^(2))+(1)/(8)mgl^(2)theta_(0)^(2)` ( where `omega^(2)=3g//2l)`