A uniform rod of length l is pivoted distance x from the top of the rod . Neglecting friction find the (a) value of x for minimum period of oscillation, (b) minimum period of oscillation of the rod.

Explanation

a. For minimum time period, we have
`r = k and k = sqrt(l_(C)//m)`
Substituting `l_(C) = ml^(2)//12` for the rod, we have
`r = (1)/(2 sqrt3)`
Then, `x = 1//2 - r`.
where `r = l//2sqrt3`
This gives `x = (1)/(2) (1 - (1))/(sqrt3)`
b. `T_(min) = 2 pi sqrt((2k)/(g))`
where `k = (1)/(2sqrt3)`
This gives
`T_(min) = 2 pi sqrt((l)/(sqrt3 g))`