A physical pendulum consists of a uniform solid disk (of radius R 2.35 cm) supported in a vertical plane by a pivot located a distance d 1.75 cm from the center of the disk. The disk is dis - placed by a small angle and released. What is the period of the resulting simple harmonic motion?

T = 0.3658

Explanation:

The expression to use to calculate the period is the following:

T = 2π √I/Mgd (1)

Where:

I: moment of Innertia of pendulum

g: gravity acceleration (9.81 m/s²)

d: distance of the pivot

M: mass of the disk.

Before we do anything, we will find first the moment of Innertia of the pendulum.

This can be calculated with the following expression:

I = 1/2 MR² + Md² (2)

At the moment we don't have the mass of the disk, but we don't need it, we will express I in function of M, and then, it will be canceled with the M of expression (1). Calculating M we have (Remember that the units of radius and distance should be in meter):

I = 1/2 M (0.0235) ² + M (0.0175) ²

I = (2.76x10^-4) M + (3.06x10^-4) M

I = (5.82x10^-4) M (3)

Now, we will replace this value in equation (1):

T = 2π √ (5.82x10^-4) M / (9.81) * (0.0175) M - --> Here M cancels out

T = 2π √ (5.82x10^-4) / (9.81) * (0.0175)

T = 2π * 0.0582

T = 0.3658 s

This is the period of the pendulum