How Does a Leaking Cubical Pendulum Affect Its Period?

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SUMMARY

The discussion focuses on determining the period of a leaking cubical pendulum with a side length of 2a, filled with water. The period formula is derived as T = 2π√((li - (rt/(8a²)))/g), where li is the initial length and r represents the rate of volume change. The volume of the cube is calculated using v = s³, leading to the conclusion that the length of the pendulum varies with the new center of mass as h/2. The approach taken to differentiate the volume with respect to time is confirmed as correct.

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Homework Statement


There is a cubical pendulum with a side of 2a filled with water that is leaking at a constant rate. Determine the period of the pendulum in terms of variables.

Homework Equations


T = 2π√(l/g)
v = s³ for a cube
v = s²h for a rectangle

The Attempt at a Solution


v=s²h
v=4a²h
The length varies with the new center of mass which is h/2, so l = h/2.
v=8a²l
differentiate with respect to time and
dv/dt = 8a²dl/dt
dl/dt = 1/(8a²)dv/dt
to avoid the problem from looking too complicated, let's set dv/dt = r
li= initial length
T = 2π√((li-(rt/(8a²)))/g)

I multiplied by time so that we would be left with the change in length, and I subtracted it because dl/dt is negative, and the length should obviously be increasing. Is this right?
 
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It is correct.
 

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