Question

In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 4.07 kg. They then hang the object on a pivot located 0.155 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 247 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis

Answer 1

I = 0.65 kgm²

Step-by-step explanation:

Since the mass is an inertial pendulum, we use the formula for the period, T of an inertial pendulum.

T = 2π√(I/mgh) where I = moment of inertia of object about pivot point, m = mass of object5 = 4.07 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.155 m.

Given that the team measures 113 cycles of oscillation in 247 s, the period, T = time of oscillations/total number of oscillations = 247 s/113 oscillations = 2.186 s/oscillation

So, T = 2.186 s

We now find I by making it subject of the formula in the equation for T.

So,

T = 2π√(I/mgh)

dividing both sides by 2π, we have

T/2π = √(I/mgh)

squaring both sides, we have

(T/2π)² = [√(I/mgh)]²

T²/4π² = I/mgh

multiplying both sides by mgh, we have

T²mgh/4π² = I

I = T²mgh/4π²

substituting the values of the variables into the equation, we have

I = T²mgh/4π²

I = (2.186 s)² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²

I = 4.778 s² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²

I = 29.539 kgm²/4π²

I = 0.748 kgm²

Now I = I' + mh² (parallel axis theorem) where I' = moment of inertia of object about its center of mass, m = mass of object = 4.07 kg and h = distance of center of mass object from pivot point.

So, I' = I - mh²

Substituting the values of the variables into the equation, we have

I' = I - mh²

I' = 0.748 kgm² - 4.07 kg × (0.155 m)²

I' = 0.748 kgm² - 4.07 kg × 0.02403 m²

I' = 0.748 kgm² - 0.098 kgm²

I = 0.65 kgm²