Question

a physical pendulum consists of a uniform solid disk (of radius R = 42.0 cm) supported in a vertical plane by a pivot located a distance d = 13.0 cm from the center of the disk. The disk is displaced by a small angle and released. What is the period of the resulting simple harmonic motion?

Answer 1

T = 2π√(I/mgr)

where I is the moment of inertia of the disk, m is the mass of the disk, g is the acceleration due to gravity, and r is the distance from the pivot to the center of mass of the disk.

I = (1/2)mR^2

m = (ρπR^2)

where ρ is the density of the disk.

r = d + R

T = 2π√((1/2)mR^2/(ρπR^2)gr)

T = 2π√((1/2)(d + R)/(ρπg))

T = 2π√((d + R)/(2ρπg))

T = 2π√((13.0 cm + 42.0 cm)/(2(1000 kg/m^3)π(9.8 m/s^2)))

T = 2.45 s