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[Note: Post restored from archive - gneill, PF Mentor]

A pendulum consists of a uniform circular disk of radius r which is free to turn about a horizontal axis perpendicular to it's plane. Find the position of the axis for which the periodic time ( for small amplitude oscillations) is a minimum.

τ: torque

Ι (i): moment of inertia

m: mass

g:gravity

θ: angle between force mg and the radius length

r: radius

ω: angular frequency

T: period

Sinθ=θ for small amplitudes.

L: length (offset for parallel axis theorem.) (unknown?)

Equation (1): τ= -mgL θ

Equation (2): θ'' + (mgL/ I ) θ = 0

Equation (3): ω = (gL / (.5 r^2 + L^2)) ^.5

So I solved for ω and my reasoning was that in order to have the minimum periodic time ω has to be at it's maximum. So I used applied optimization and set dω/dL = 0. However, I am not sure if this was the right way to do it since I squared both sides of the equation 3 to get rid of the square root and then used implicit differentiation. But I think this might be wrong because it's like if I differentiates the inside of Eq. 1 and that differentiation does not take into account the square root. So I am not sure if I did it right. I differentiated directly but the derivative was a monster and I couldn't factor out the L to solve for it. Is this a right way to do it correctly?

## Homework Statement

A pendulum consists of a uniform circular disk of radius r which is free to turn about a horizontal axis perpendicular to it's plane. Find the position of the axis for which the periodic time ( for small amplitude oscillations) is a minimum.

τ: torque

Ι (i): moment of inertia

m: mass

g:gravity

θ: angle between force mg and the radius length

r: radius

ω: angular frequency

T: period

Sinθ=θ for small amplitudes.

L: length (offset for parallel axis theorem.) (unknown?)

## Homework Equations

Equation (1): τ= -mgL θ

Equation (2): θ'' + (mgL/ I ) θ = 0

Equation (3): ω = (gL / (.5 r^2 + L^2)) ^.5

## The Attempt at a Solution

So I solved for ω and my reasoning was that in order to have the minimum periodic time ω has to be at it's maximum. So I used applied optimization and set dω/dL = 0. However, I am not sure if this was the right way to do it since I squared both sides of the equation 3 to get rid of the square root and then used implicit differentiation. But I think this might be wrong because it's like if I differentiates the inside of Eq. 1 and that differentiation does not take into account the square root. So I am not sure if I did it right. I differentiated directly but the derivative was a monster and I couldn't factor out the L to solve for it. Is this a right way to do it correctly?

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