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-   -   Quick Lagrangian of a pendulum question (http://www.physicsforums.com/showthread.php?t=584208)

Lucy Yeats Mar5-12 09:38 PM

Quick Lagrangian of a pendulum question
 
1. The problem statement, all variables and given/known data

Use the E-L equation to calculate the period of oscillation of a simple pendulum
of length l and bob mass m in the small angle approximation.

Assume now that the pendulum support is accelerated in the vertical direction at a rate
a, find the period of oscillation. For what value of a the pendulum does not
oscillate? Comment on this result.

2. Relevant equations



3. The attempt at a solution

I've got the first bit:
L=(m/2)(l^2)(dθ/dt)^2-mgl(1-cosθ)
E.O.M.: d2θ/dt2+(g/l)sinθ=0
d2θ/dt2+(g/l)θ=0 in the small angle approximation,
which is S.H.M. with ω^2=√(g/l) (though I'm not sure about this as there's no minus sign in the E.O.M.) so T=2pi√(l/g)

For the next bit, I just need help setting up the equations:
So the generalized coordinates are θ and a.
Are the following correct?:
x=lsinθ
y=-lcosθ+at
(taking the origin as the point from which the pendulum is swinging)

jambaugh Mar5-12 09:45 PM

Re: Quick Lagrangian of a pendulum question
 
Quote:

Quote by Lucy Yeats (Post 3800525)
I've got the first bit:
L=(m/2)(l^2)(dθ/dt)^2-mgl(1-cosθ)
E.O.M.: d2θ/dt2+(g/l)sinθ=0
d2θ/dt2+(g/l)θ=0 in the small angle approximation,
which is S.H.M. with ω^2=√(g/l) (though I'm not sure about this as there's no minus sign in the E.O.M.) so T=√(l/g)

Careful here, [itex]\omega[/itex] is that angular velocity in radians per time unit. Your T is the time to travel one radian (of the oscillatory cycle) not time per cycle. You need to multiply by [itex]2 \pi[/itex].

Lucy Yeats Mar5-12 10:00 PM

Re: Quick Lagrangian of a pendulum question
 
Thanks for pointing that out, I'll correct that in the first post. :-)

Lucy Yeats Mar6-12 03:17 AM

Re: Quick Lagrangian of a pendulum question
 
Any help would be great. :D

Lucy Yeats Mar6-12 04:22 AM

Re: Quick Lagrangian of a pendulum question
 
Should y be -lcosθ+0.5at^2 instead?

I like Serena Mar6-12 12:17 PM

Re: Quick Lagrangian of a pendulum question
 
Quote:

Quote by Lucy Yeats (Post 3800977)
Should y be -lcosθ+0.5at^2 instead?

Yes.

Quote:

So the generalized coordinates are θ and a.
No.
"a" is not a coordinate.
Actually only θ is a generalized coordinate since the y coordinate of the support is constrained to be y=0.5at^2.
The angle θ is the only freedom that the system has.

Lucy Yeats Mar13-12 08:17 AM

Re: Quick Lagrangian of a pendulum question
 
Great, I've got it now.

Thanks! :-)


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