Question

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?

Answer 1

Answer:

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

$\omega=2\pi f\\\Rightarrow \omega=2\pi 0.66$

For a compound pendulum the angular velocity is given by

$\omega=\sqrt{\dfrac{mgd}{I}}\\\Rightarrow d=\dfrac{\omega^2I}{mg}\\\Rightarrow d=\dfrac{(2\pi 0.66)^2\times 1.28}{2\times 9.81}\\\Rightarrow d=1.12191\ m$

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