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

1.12191 m

Step-by-step 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{(mgd)/(I)}\\\Rightarrow d=(\omega^2I)/(mg)\\\Rightarrow d=((2\pi 0.66)^2* 1.28)/(2* 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