An engineer wishes to determine the moment of inertia oa a machine part mass 1.2 kilogram about a particular axis (the X-X' axis) through the CM. He locates the CM by suspending the object motionless from several different points around its periphery. He the suspends the objects from a pivot distance o.25 meter from the CM and observes that it undergoes small oscillations with a period of 1.50 seconds about an axis parallel to th X-X' axis. What is the moment of inertia of the object about the X-X' axis?

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To calculate the moment of inertia of the object about the X-X' axis, we can use the formula for the period of oscillation of a physical pendulum.

The formula for the period of oscillation (T) of a physical pendulum is given by:

T = 2π√(I/mgd)

Where:
I is the moment of inertia of the object about the axis of rotation
m is the mass of the object
g is the acceleration due to gravity
d is the distance between the axis of rotation and the center of mass (CM)

In this case, the object is suspended from a pivot distance (r) of 0.25 meters from the CM and undergoes small oscillations with a period of 1.50 seconds. Therefore:

T = 1.50 seconds
r = 0.25 meters

Using the given values and rearranging the formula, we can solve for the moment of inertia (I):

I = m * g * (T / 2π)^2 * r

Given:
m = 1.2 kilograms (mass of the object)
g = 9.8 meters per second squared (acceleration due to gravity)

Plugging in the values, we get:

I = 1.2 kg * 9.8 m/s^2 * (1.5 s / 2π)^2 * 0.25 m

Simplifying the equation, we find:

I ≈ 0.57 kg⋅m²

Therefore, the moment of inertia of the object about the X-X' axis is approximately 0.57 kg⋅m².