Period of Pendulum When Moving Axis Closer to COM

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SUMMARY

The discussion centers on the relationship between the axis of rotation of a physical pendulum and its period of rotation. When the axis is moved closer to the center of mass (COM), the period of the pendulum decreases, contrary to the initial assumption that it would increase. The correct formula for the period is T = 2π√(g/L), where L is the distance from the center of rotation to the COM. The conclusion is that as the arm length (L) decreases, the period (T) becomes shorter.

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Homework Statement



If the axis of rotation of a physical pendulum is placed closer to the COM, is the period of rotation longer? If it originally starts .5m from the COM the moves .3M from the COM the peroid shoud be longer right? I keep getting a longer period when the axis of roatation is placed closer to the COM, but it just deosnt seem right


Homework Equations



t=2pi*(sqrt(I/mgh)
I=(1/12)Ml^2

The Attempt at a Solution


mass is unknown
L= 1m
 
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chewytess said:

Homework Statement



If the axis of rotation of a physical pendulum is placed closer to the COM, is the period of rotation longer? If it originally starts .5m from the COM the moves .3M from the COM the peroid shoud be longer right? I keep getting a longer period when the axis of roatation is placed closer to the COM, but it just deosnt seem right


Homework Equations



t=2pi*(sqrt(I/mgh)
I=(1/12)Ml^2

The Attempt at a Solution


mass is unknown
L= 1m

I'm not familiar with that form of the equation. It seems to me the more common form is
T =2[tex]\pi[/tex][tex]\sqrt{g/L}[/tex]
for T = period
L = arm length (i.e. distance from center of rotation to center of mass)
g= acceleration due to gravity

As with the equation you cite, period is inversely proportional to arm length, so your expectation is wrong and your calculation is correct.

The period-arm length relationship is easy to test with any crude pendulum (say you cell phone hanging on its charging wire).
 

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