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 6.3 s. Find d. in m.

Answer 1

equivalent length
Leq = I / md
where I is the moment of inertia about the pivot
and m is the mass
and d is the distance from the pivot to the center of mass

By the parallel axis theorem,
I = mL²/12 + md², so
Leq = (mL²/12 + md²) / md = L²/12d + d

period T = 2??(Leq / g) ? square everything
T² = 4?²(L²/12d + d) / g
gT² / 4?² = L²/12d + d ? plug in g and T and drop units for ease (d is in meters):
9.8*(6.3)² / 4?² = 1/12d + d
9.852 = 1/12d + d ? multiply by d
9.852d = 1/12 + d²
0 = d² - 9.852d + 1/12

d² - 9.852d + 0.083333= 0

This quadratic has roots at 9.843 0.00846
d = 0.000846 m ? solution
and d = 9.843 m ? greater than 1 m; disregard