Lagrange' s eequations of a suspended set of rods

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The discussion focuses on deriving Lagrange's equations for a system of three identical rods forming a square in a vertical plane, with two rods hinged to a support. The participants emphasize the importance of correctly setting up the kinetic energy (T) and potential energy (V) for the system. The potential energy for each rod is defined as PE = -mg(L/2)cosθ for the side rods and PE = -mgLcosθ for the bottom rod. The kinetic energy is expressed as T = (1/2)*m*(l*ω)^2 for the bottom rod, while the side rods are treated as rotating with T = (1/2)*I*ω^2.

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Lagrange' s equations of a suspended set of rods

Three identical rods of length l and mass m are hinged together so as to form three sides of a square in a vertical plane. The two upper free ends are hinged to a rigid support. The system is free to move in its own plane. Use Lagrange' s equations to obtain the equations of motion of the system.

I just need help in setting up T and V...I think that I can handle the rest, but any help with getting T and V would be greatly appreciated..I am really desparate here...this problem has been frustrating me a lot. I tried to use theta and phui but they led me nowhere
 
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I don' t think that you would have to use theta that' s for sure, but that' s just me.
 
Surely someone can help me. Just hlpe me to set up T and V and I can take it from there. Please I am really desparate.
 
Really just something...anything
 
student1938,
You seem desperate, so I'm going to jump in here preemptively.

First, notice that this is a 1-D problem, so your lagrangian should be 1-D (the configuration of the rods is restricted to a rombus, for which only the angle between adjacent sides causes a change in the configuration energy). The Lagrangian is a function of only 1 q, 1 q_dot, and time: L = f(q,q_dot,t).

You can use whatever you want to represent this single generalized coordinate, so long as it can completely specify the configuration and state of motion. 2 suggestions that immediately come to mind are the angle that the rods attached to the ceiling make with some reference plane or the height of the horizontal rod.
 
I'll just add my two cents to turin's response:

Assuming I have the picture correct, this assembly can swing from side to side (in its plane), but that's it. The bottom rod is always horizontal. There is only one degree of freedom. (I'll use the angle with respect to vertical as my coordinate.)

Start by writing the potential energy of each rod as a function of the angle of the side rods to vertical: PE of each side rod is -mg(L/2)cosθ; the PE of the bottom rod is -mgLcosθ. (I'm measuring PE from the top support--you can measure from any point you choose.)

That should get you started... :smile: (Now you work out the KE.)
 
Thanks for the PE ..it provided a lot of insight...I even think that I might be onto something for the KE:

ok for the KE, I get (1/2)*I*theta^2...now I will be the same for the two side rods and different for the bottom rod. For a rod I get (1/3)*m*l^2 but for the bottom what do I use?
 
Any suggestions?
 
student1938 said:
ok for the KE, I get (1/2)*I*theta^2...now I will be the same for the two side rods and different for the bottom rod. For a rod I get (1/3)*m*l^2 but for the bottom what do I use?
I assume you mean KE = 1/2*I*ω^2. The side rods can be treated as purely rotating. Here's a hint for the bottom rod: does it rotate?
 
  • #10
So then KE is just (1/2)*I*omega^2. The bottom rod does not rotate so there is no KE like you said. IS that all that is needed for the KE? how would I get omega? I think that I should just leave it the way it is
 
  • #11
No wait the rod moves so there must be some KE...(1/2)*m*(dl/dtheta)^2 if you take it from the tip?
 
  • #12
student1938 said:
The bottom rod does not rotate so there is no KE like you said.
I never said that! :eek: It's just pure translation, not rotation. It's center of mass moves along with the bottom tip of the side rods.
how would I get omega? I think that I should just leave it the way it is
Omega is just your generalized velocity d(theta)/dt. Leave it.
 
  • #13
For some reason, I am just unable to figure it out...according to me, based on the last hint, it should be (1/2)*m*(l*cos(theta))^2...I am saying l cos theta because that is the distance to the CM from the tip of the side rods...does that make sense? Any suggestions?
 
  • #14
Anything please!
 
  • #15
velocity of the bottom rod

student1938 said:
For some reason, I am just unable to figure it out...according to me, based on the last hint, it should be (1/2)*m*(l*cos(theta))^2...I am saying l cos theta because that is the distance to the CM from the tip of the side rods...does that make sense? Any suggestions?
Ask yourself: When the tip of the side rod moves, how does the bottom rod move? I say it tracks it perfectly, just displaced horizontally. So the velocity of the bottom rod equals the velocity of the tip of the side rod. Make sense? If so, then what is the speed of the tip of the side rod?
 
  • #16
the speed should just be d(lcos(theta))/dt right? I think thta lcos(theta) is the horizontal distance and we are dividing by time here to get speed right?
 
  • #17
total speed

student1938 said:
the speed should just be d(lcos(theta))/dt right? I think thta lcos(theta) is the horizontal distance and we are dividing by time here to get speed right?
Where does the cos(theta) term come in? I want the total speed of the bottom rod, not a particular component. What's the total speed of the bottom tip of a side rod?
 
  • #18
Then it should just be l/t but then wht happens to t?
 
  • #19
student1938 said:
Then it should just be l/t but then wht happens to t?
Not sure what you are talking about. The tip moves at a speed of lω.
 
  • #20
oh did you use v = omega X r where r = l and the angle is 90 degrees...now that makes sense. So then T = (1/2)*m*(l*omega)^2 and I can now use Lagrange' s Eqs right?
 
  • #21
Now, you used your coordinate as omega so when differentiating I have to differentiate with respect to just omega since omegadot is zero right? The reason I ask is cause it might me help me out with one of my problems.
 
  • #22
student1938 said:
oh did you use v = omega X r where r = l and the angle is 90 degrees...now that makes sense. So then T = (1/2)*m*(l*omega)^2 and I can now use Lagrange' s Eqs right?
Right... finally! (That's just the KE of the bottom rod... don't forget the others.)
 
  • #23
So for the side rods, then it should just be l/2*omega ^ 2 right?
 
  • #24
Or shoul it be the same?
 
  • #25
Anything,,,I just wana finish this off please.
 
  • #26
student1938 said:
So for the side rods, then it should just be l/2*omega ^ 2 right?
We covered the KE of the side rods in posts # 9 & 10 of this thread.
 
  • #27
got it..thanks for all the help
 

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