Quick Lagrangian of a pendulum question

Click For Summary

Homework Help Overview

The problem involves calculating the period of oscillation of a simple pendulum using the Euler-Lagrange equation, considering both the standard case and a scenario where the pendulum support is accelerated vertically. The original poster seeks to understand the implications of this acceleration on the pendulum's motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the formulation of the Lagrangian and the equations of motion for the pendulum. There is an exploration of the small angle approximation and its implications for simple harmonic motion. Questions arise regarding the correct setup of generalized coordinates and the impact of vertical acceleration on the pendulum's behavior.

Discussion Status

Participants are actively engaging with the problem, providing corrections and clarifications regarding the definitions of coordinates and the nature of the motion. Some guidance has been offered on the correct interpretation of the period of oscillation and the role of acceleration in the system.

Contextual Notes

There is a discussion about the constraints on the generalized coordinates, particularly the distinction between the angle θ and the acceleration a, with an emphasis on the constrained nature of the pendulum's support. Participants are also considering the implications of the vertical acceleration on the pendulum's oscillation.

Lucy Yeats
Messages
117
Reaction score
0

Homework Statement



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.

Homework Equations


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)
 
Last edited:
Physics news on Phys.org
Lucy Yeats said:
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, \omega 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 2 \pi.
 
Thanks for pointing that out, I'll correct that in the first post. :-)
 
Any help would be great. :D
 
Should y be -lcosθ+0.5at^2 instead?
 
Lucy Yeats said:
Should y be -lcosθ+0.5at^2 instead?

Yes.

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.
 
Great, I've got it now.

Thanks! :-)
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
10K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K