Finding Frequency of an Oscillating Disc with a Hole

AI Thread Summary
The discussion centers on calculating the oscillation frequency of a circular disk with a rectangular hole, suspended at its perimeter. The moment of inertia and center of mass are provided, and the initial attempt to find the period of oscillation yielded a result of 1.93 seconds. However, confusion arises regarding the scenario where the pivot point is at the center of mass, as this would result in no torque and thus no oscillation. The conclusion reached is that the frequency in this case is zero, indicating that the problem may be designed to illustrate the lack of movement when the pivot is at the center of mass. The discussion highlights the importance of understanding torque and pivot points in oscillatory motion.
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Homework Statement

A circular disk with a rectangular hole has a radius of 0.600 m and mass of 0.390 kg. It is suspended by a point on its perimeter as shown in the figure. The moment of inertia about this point is I_p = 2.60E-1 kgm2. Its center of mass is located at a distance of s=0.120 m from the center of the circle as shown. [Photo attached]

What is the oscillation frequency if the pivot point is at the center of mass?

The attempt at a solution

In truth, there were two parts to this question. The first, I had to find the periodof oscillations if it was allowed to oscillate side to side as a pendulum.

I used the equation T = 2(pi)√(I/mgd) and found that the answer was 1.93 s.

However, I'm at a loss as to how to try to solve for the frequency with the pivot point at the center of mass. Obviously, taking the inverse of the above answer resulted in an incorrect answer.

Any and all assistance would be hugely appreciated.
 

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The problem as stated doesn't make sense to me. If the pivot point is through the center of mass, it won't oscillate because there will be no torque on the disc regardless of its orientation.
 
vela said:
The problem as stated doesn't make sense to me. If the pivot point is through the center of mass, it won't oscillate because there will be no torque on the disc regardless of its orientation.

I realized that not long after I asked. I'm now assuming it's a trick question. 0 is the correct answer. I think it was trying to show that because distance from the center of mass was zero, all the rest was zero as well.
 
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