Equation of motion for a pendulum using Lagrange method

morpheus343
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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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What calculations? All you have given is the final result, which does not look correct to me.
 
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?
 
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?
 
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
 
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.
 
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?
 
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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