Applied optimization to SHM problem? Hard?

  • #1
[Note: Post restored from archive - gneill, PF Mentor]

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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Answers and Replies

  • #2
BvU
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You removed the post ? I thought you did just fine!
 
  • #4
Null
 
Last edited:
  • #5
You removed the post ? I thought you did just fine!

It was correct?
 
  • #6
BvU
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Can't find no mistakes :smile: (PF isn't really meant for stamp approving, but I think you're doing fine).

Don't know of a different path to the same answer. Your squaring is justified: ##\omega## is known to be > 0 , so no problem.
 
  • #7
Can't find no mistakes :smile: (PF isn't really meant for stamp approving, but I think you're doing fine).

Don't know of a different path to the same answer. Your squaring is justified: ##\omega## is known to be > 0 , so no problem.
Oh alright. Got it. Thanks for your guidance then ;p
 

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