 #1
 95
 1
[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?
Attachments

23.7 KB Views: 327

28.7 KB Views: 304
Last edited by a moderator: