Frequency of Oscillation of a monkey hanging from a tree

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SUMMARY

The frequency of oscillation of a gibbon hanging from a tree branch is calculated using the formula f = 1/T, where T is the period derived from T = 2π * SQRT(I/mgd). Given the values I/m = 0.289 m², d = 0.205 m, and g = 9.8 m/s², the correct calculation yields a frequency of approximately 0.209 Hz. The initial confusion stemmed from the interpretation of the rotational inertia value, but the problem was ultimately resolved by correctly applying the equations.

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Homework Statement



A gibbon, hanging onto a horizontal tree branch with one arm, swings with a small amplitude. The gibbon's center of mass is 0.410 m from the branch and its rotational inertia divided by its mass is I/m = 0.289 m2. Estimate the frequency of oscillation.


Homework Equations



d = 1/2L
T = 2π * SQRT(I/mgd)
f = 1/T

The Attempt at a Solution



Okay, they have given me I/m, and I need to find f using the last equation. In order to find T, I need to use the first and second equation.

d = 1/2(.410)
d = .205

I/m = .289 m2

T = 2π * SQRT([.289][9.8][.205])
T = 2π * .762
T = 4.78

f = 1/4.78
f = .209 Hz

Which is, of course, wrong. I'm thinking I'm getting confused by the I/m = 0.289 m2 value which is throwing off how I plug and chug. I may even have the wrong equation. Thanks for all your help.
 
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Nevermind, I solved it!
 

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