Question

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 3.76 s. Find d, in cm.

Answer 1

Let:
I be the moment of inertia of the rod about the pivot,
L be the length of the rod,
g be the acceleration due to gravity,
a be the small angular displacement of the rod,
d be the distance of the pivot above centre,
m be the mass of the rod,
w be the angular frequency of the motion.

Taking moments about the pivot:
mgd sin(a) = - Ia'' ...(1)

Using the parallel axis theorem:
I = m(L^2 / 12 + d^2)
I = m(L^2 + 12d^2) / 12

Substituting this in (1):
gd sin(a) = - (L^2 + 12d^2)a'' / 12

Using the small angle approximation:
gda = - (L^2 + 12d^2)a'' / 12

a'' = - 12gda / (L^2 + 12d^2)
w^2 = 12gd / (L^2 + 12d^2)

(L^2 + 12d^2)w^2 = 12gd
(L^2 + 12d^2)pi^2 = 3gd T^2
12pi^2 d^2 - 3gT^2 d + pi^2 L^2 = 0

d = [ 3gT^2 +/- sqrt(9g^2 T^4 - 48pi^4 L^2) ] / (24pi^2)
= [ 3 * 9.81 * 3.76^2 +/- sqrt(9 * 9.81^2 * 3.76^4 - 48pi^4) ] / (24pi^2)

The larger root exceeds 1m. The smaller root is:
d = 0.02388 m = 2.388 cm