A uniform rod of length `l` performs small oscillations about the horizontal axis `OO^(')` perpendicular to the rod and passing through one of its points. Find the distance the centre of inertia of the rod and the axis `OO^(')` at which the oscillation period is the shortest. What is it equal to ?

Explanation

Let us locate the rod when it is at small angular position `theta` relative to its equlibrium position. If `a` be the sought distance, then from the conservation of mechanical energy of oscialltion
`mga(1-cos theta)+(1)/(2)I_(o o^('))(dot(theta))^(2)=` constant
Differentiating w.r.t. time we get `:`
`mg a sin theta dot(theta)+(1)/(2)I_(O O ^('))2 dot(theta)ddot (theta)=0`
But `I_(O O^('))=(ml^(2))/(12)+ma^(2)` and for small `theta, sin theta=theta`, we get
`ddot (theta)=-((ga)/((l^(2))/(12)+a^(2)))theta`
Hence the time periof of one full oscillation becomes
`T=2pi sqrt(((l^(2))/(12)+a^(2))/(ag))` or `T^(2)=(4pi^(2))/(g)((l^(2))/(12 a )+a)`
For `T_(min ), ` obviously `(d)/(da)((l^(2))/(12 a )+a)=0`
So, `-(l^(2))/(12a^(2))+1=0` or `a=(l)/(2 sqrt(3))`
Hence `T_(min)=2pisqrt((l)/(gsqrt(3)))`