How Does Reducing Balance Wheel Dimensions Affect Angular SHM Frequency?

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Reducing the dimensions of the balance wheel to one-third results in a significant change in both mass and moment of inertia, affecting the frequency of angular simple harmonic motion (SHM). The moment of inertia decreases to one-ninth of the original due to the radius reduction, while the mass decreases to one-twenty-seventh. The frequency formula indicates that frequency is inversely proportional to the square root of the moment of inertia. Therefore, the frequency increases, but the exact factor requires careful calculation considering both the new mass and moment of inertia. Ultimately, the frequency of angular SHM will increase, but the precise factor needs to be determined through correct application of the equations.
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Homework Statement



Your boss at the Cut-Rate Cuckoo Clock Company asks you what would happen to the frequency of the angular SHM of the balance wheel if it had the same density and the same coil spring (thus the same torsion constant), but all the balance wheel dimensions were made one-third as great to save material. By what factor would the frequency change?


Homework Equations



frequency = 1/2(pi) * square root of (torsion constant/I)

The Attempt at a Solution



I tried finding the factor by setting up a second equation where frequency = 1/2(pi) * square root of (torsion constant/ m(r/3)^2 since I=mr^2 and got square root 3 but it didn't work. Is it because I'm using the wrong moment of inertia equation or is it something else? Please help and many thanks in advance.
 
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I thought for a thin disk the I=1/2Mr^2, if you reduce the radius by a factor of three, the I shoud will be 1/9th the original, hence...
 
You also have to take into account the mass that change. Since the balance wheel dimensions were made 1/3 the original, you should have 1/27 the original mass.
 
thanks for the help denverdoc and enter260!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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