An irregularly shaped flat object of mass 2.00 kg is suspended from a point at a distance d from its center of mass and allowed to undergo simple harmonic motion in the vertical plane. the object has moment of inertia i = 1.28 kg · m2 about an axis passing through the point of suspension and perpendicular to the plane of the object. the frequency of this oscillatory motion is 0.660 hz. what is the distance d of the pivot point from the center of mass of the object?

1.12191 m

Explanation:

m = Mass of point = 2 kg

f = Frequency = 0.66 Hz

I = Moment of inertia = 1.28 kgm²

g = Acceleration due to gravity = 9.81 m/s²

For a compound pendulum the angular velocity is given by

The distance of the pivot point from the center of mass of the object is 1.12191 m

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

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