Final answer:
The moment of inertia of a physical pendulum undergoing simple harmonic motion with a frequency of 0.250 Hz, mass of 2.00 kg, and a pivot 0.240 m from the center of mass is approximately 0.596 kg·m².
Step-by-step explanation:
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.250 Hz. The pendulum has a mass of 2.00 kg, and the pivot is located 0.240 m from the center of mass. To determine the moment of inertia (I) of the pendulum about the pivot point, we use the relationship between the period of oscillation (T), the mass (m), the distance from the pivot to the center of mass (r), and the gravity acceleration (g).
The formula for the period of a physical pendulum is
T = 2π√(I/mgr)
First, solve for
T
using the given frequency (f).
Let's start by calculating the period using the formula: T = 1/f,
which gives us T = 1/0.250 Hz = 4 s.
Now, plugging the values into the period formula for a physical pendulum T = 2π√(I/mgr) and isolating I, we get:
I = (T^2 * m * g)/(4π^2 * r)
Where:
T = 4 s
m = 2.00 kg
g = 9.81 m/s²
r = 0.240 m
Substitute the values into the equation for I:
I = [(4 s)^2 * 2.00 kg * 9.81 m/s²] / [4π^2 * 0.240 m]
After performing the calculations, we find that the moment of inertia of the pendulum about the pivot point is approximately 0.596 kg·m².