An irregularly shaped flat object of mass 2.40 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 = 1.24 kg Â· 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 0.640 Hz. What is the distance d of the pivot point from the center of mass of the object?
_____m

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faizahmihani faizahmihani · Answer 1 · 2022-12-16T07:49:43+01:00

The distance of the pivot point from the center of mass of the object = 0.85 m

Oscillatory motion is any motion that is repeated at the same time interval and through the same trajectory in its motion.

The simplest oscillatory motion is harmonic motion or simple harmonic motion (SHM).

The simple harmonic motion is a periodic motion that occurs at the same time interval.

We have,

Mass of point = 2.40 kg ⇒ m

Moment of inertia = 1.24 kgm² ⇒ I

Frequency = 0.640 Hz ⇒ f

Acceleration due to gravity = 9.8 m/s² ⇒ g

Now, find the angular velocity first,

ω = 2[tex]\pi[/tex]f

= 2[tex]\pi[/tex] (0.640)

= 4.0192 rad/s

So, the distance of the pivot point from the center of mass of the object is:

ω = [tex]\sqrt{\frac{mgd}{I} }[/tex]

4.0192 = [tex]\sqrt{\frac{(2.40)(9.8)(d)}{1.24} }[/tex]

d = (4.0192)² (1.24) / (2.40)(9.8)

= 20.0309211/23.52

= 0.85 m

Learn more about oscillatory here: https://brainly.com/question/29765470

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