Period of a Physical Pendulum: Finding T for a Uniform Disk

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SUMMARY

The discussion focuses on calculating the period (T) of a uniform metal disk acting as a physical pendulum, with a mass (M) of 9.81 kg and a radius (R) of 8.99 m. The correct formula for the moment of inertia (I) of the disk about the edge is derived using the Parallel Axis Theorem, resulting in I = (3/2)MR². The period is calculated using T = 2π√(I/mgd), leading to an incorrect initial result of T = 3.2 seconds, prompting a request for clarification on the calculation process.

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Homework Statement


A uniform metal disk (M = 9.81 kg, R = 8.99 m) is free to oscillate as a physical pendulum about an axis through the edge. Find T, the period for small oscillations.


Homework Equations



I (uniform disk, with axis through center of mass) = (1/2)MR^2
T = 2π√(I/mgd), where d = distance from center of mass to point of rotation (axis)
I (uniform disk, with axis through the edge = (3/2)MR^2, after using Parallel Axis Theorem:
I (center of mass) + Md^2, where d = RADIUS of the disk:

The Attempt at a Solution


Plugging in for T, we have 2π√(I/(9.81)(9.81)(8.99).
After plugging in I = (3/2)*(9.88)*(8.99)^2, we can plug and chug away:

T = 3.2 seconds, however, this was incorrect.

I was wondering where I went wrong here, thanks again !
 
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You might find that fewer "finger" errors can creep in if you carry out more of the process symbolically before plugging in numbers.

Put your expression for the moment of inertia into the expression for the period and simplify before going numerical :wink:
 

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