If we want to make a pendulum of minimal period of thin uniform rod, at what point should we suspend the rod?

Answer 1

Let the rod have length L = 2*(a+b)
And the rod is suspended at a point b units from the center.

I = mass moment of inertia about the point of rotation
= 1/12 * ml^2 + mb^2 by the parallel axis theorum
= 1/3 * m*(a+b)^2 + mb^2

For a displaced angle θ, taking moments about the center of rotation
I*θ" + mgbsinθ = 0
I*θ" + mgbθ =~ 0 for small θ

ω^2 = mgb / I
(T/2π)^2 = (1/ω)^2 = I/(mgb)
= [1/3 *m*(a+b)^2 + mb^2] / [mgb]
= (a+b)/g * [1/3 + x^2] / [x]
where x = b / (a+b)

So we just need to find the minimum value of
[1 + 3x^2] / x = 1/x + 3x
which occurs at x = 1/sqrt(3)
It's also not too difficult to verify that it is a minimum rather than maximum.

b/(a+b) = sqrt(3) / 3
b / L = sqrt(3) / 6

So the rod should be suspended at a point L*sqrt(3)/6 from its center to minimize the period.

Answer 2

what do you mean?????
for minimu time period u have to take the rod length 0!!!