Equation of motion for a pendulum using Lagrange method

  • #1
morpheus343
29
4
Homework Statement
Find the equation of motion of the pendulum
Relevant Equations
Lagrangian L=T-V
Kinetic-Potential energy
1706764977595.png


I move the system by a small angle θ . I am not sure if my calculations of kinetic and potential energy are correct.
1706765036261.png
 
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  • #2
What calculations? All you have given is the final result, which does not look correct to me.
 
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  • #3
1706865003406.png

I do not understand if this just a problem treated like a pendulum with a simple sphere but just adding the other 2 sphere's masses. Is the potential energy supposed to have a different y for each sphere?
 
  • #4
You are asserting that the speed of the outer spheres are the same as that of B? Is this reasonable?

morpheus343 said:
Is the potential energy supposed to have a different y for each sphere?
What would you say based on your second diagram in the OP? Are all spheres at the same height? Does the answer matter?
 
  • #5
1706869811340.png

Then it would be like this. This was my first assumption but i got lost in the algebra and thought i was maybe wrong and it was supposed to be simpler
 
  • #6
Well, for the potential energy, perhaps there is an argument to make it simpler? What can you say about the center of mass?

For the kinetic energy, I did not check the math, but yes, you indeed need to take the distance to C and D using the Pythagorean theorem. Note that, since the distance appears squared in T, you should probably keep the exact rational 1+4/9 = 13/9 in the expression.
 
  • #7
Orodruin said:
Well, for the potential energy, perhaps there is an argument to make it simpler? What can you say about the center of mass?

For the kinetic energy, I did not check the math, but yes, you indeed need to take the distance to C and D using the Pythagorean theorem. Note that, since the distance appears squared in T, you should probably keep the exact rational 1+4/9 = 13/9 in the expression.
It's at sphere B which is the midpoint of CD. So i only need the vertical distance from the centre of mass when calculating the potential energy?
 
  • #8
morpheus343 said:
It's at sphere B which is the midpoint of CD. So i only need the vertical distance from the centre of mass when calculating the potential energy?
Yes. By definition, the potential energy for any mass distribution is given by
$$
V = \int U(z) \rho(\vec x) dV
$$
where ##U(z)## is the gravitational potential and ##\rho## the density. Assuming ##U(z) = gz## (homogeneous field), you’ll find
$$
V = g \int z \rho(\vec x) dV \equiv m g \bar z
$$
where ##\bar z## is the z-coordinate of the center of mass by definition:
$$
\bar z = \frac 1m \int z \rho(\vec x) dV
$$
 
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