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## Homework Statement

## Homework Equations

L = T-V

## The Attempt at a Solution

I got a forumla for the lagrangian as

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- Thread starter dynamicskillingme
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If θ increases, both the spring potential energy and the gravitational potential energy increase.(3) The distance from the pivot to the center of mass of a ball.f

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L = T-V

I got a forumla for the lagrangian as

- #2

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check the L written above and set up the Lagrange's equations of motion...I got a forumla for the lagrangian as

- #3

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Its more I don't know if the L is correct and wanted to check what the derivation is with someone else's because I'm pretty sure my derivation is wrong.check the L written above and set up the Lagrange's equations of motion...

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Its more I don't know if the L is correct and wanted to check what the derivation is with someone else's because I'm pretty sure my derivation is wrong.

the lagrangian is T- V , where T is kinetic energy and V is potential energy of the system.

The choice of generalized coordinate is important- you may consult introductory book /material on the Langrangian Method and try to set up equations for simple system like a conical pendulum and then move on to the present problem .

you can get a short introduction in the following;

http://ocw.mit.edu/courses/aeronaut...namics-spring-2003/lecture-notes/lecture7.pdf

- #5

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Its more I don't know if the L is correct and wanted to check what the derivation is with someone else's because I'm pretty sure my derivation is wrong.

your regulator has two degrees of freedom one described by angle si and the other by angle theta.

so the kinetic energy can be written using the rotational motion about si and about theta.if you take vertical axis as z then the other motion is in x-y plane and can be described by time rate of change of theta .

the potential energy can be written using the position of two messes at any theta.

- #6

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You have neglected the rotational inertia of the spheres about their centers. This will add additional terms to T. Also, check the sign of the gravitational potential energy part of L.Its more I don't know if the L is correct and wanted to check what the derivation is with someone else's because I'm pretty sure my derivation is wrong.

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- #8

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Each ball does rotate as either ##\theta## or ##\psi## changes. Consider a ball attached to the end of a stick as shown in the figure below. As the stick rotates through an angle θ, the ball also rotates through θ.I don't believe the balls are allowed to rotate about their centers.

The potential energy is subtracted in the Lagrangian: L = T - V. You got the sign correct for the spring potential energy.Why wouldn't the gravitational potential be positive?

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Also wouldn't the gravity term be the opposite direction to the spring?

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It's just a kinetic energy term due to rotation about the center of mass.Yeah I agree with that, but what other term would that introduce?

If θ increases, both the spring potential energy and the gravitational potential energy increase.Also wouldn't the gravity term be the opposite direction to the spring?

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Okay so what would the term be?

- #12

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Check your text, or do a web search on "rotational kinetic energy". It involves the moment of inertia and angular speed.

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In your Lagrangian you use the symbol ##l##. What does that stand for?

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- #15

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This is not quite the correct expression for ##l##. The ##R/2## part is incorrect. ##l## should be the distance from ##O## to the center of mass of one of the balls.

I won't write out what I get for the Lagrangian since this is your assignment. I have pointed out a few places where your expression for the Lagrangian is either incomplete or incorrect:

(1) Your first two terms represent the kinetic energy due to motion of the center of mass of the balls. In addition, you need to include terms representing the kinetic energy due to rotation of the balls about their center of mass as ##\theta## and ##\psi## change.

(2) The overall sign of the gravitational energy term needs correction.

(3) In the gravity term, you have not used the correct distance from ##O## to the center of mass of a ball.

To help with (1), consider how you would write the kinetic energy of a sphere of moment of inertia ##I## rotating with angular speed ##\omega## about an axis through its center.

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That goes to

and

(please be right, also I just saw the gravity term isn't right but its late. will change)

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That's getting very close to the correct ##L##. However, you need to include a rotational KE term associated with ##\dot{\theta}## as well as with ##\dot{\psi}##. And, as you say, the coefficient for the gravity terms needs to be corrected.

When evaluating ##\frac{\partial{L}}{\partial \theta}##, how did you get factors of ##\dot{\theta}## to appear?

I believe your expression for ##\frac{\partial{L}}{\partial \dot\psi}## is correct. In ##\frac{d}{dt}\frac{\partial{L}}{\partial \dot\psi}##, you dropped a factor of ##\dot\psi## the last term.

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Maybe

- #19

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This expression for ##L## looks correct to me as long as the distance ##l## has the correct interpretation.

Everything else looks good except for your result for ##\frac{\partial L}{\partial \theta}## :

The gravity term is not correct and I think you might have dropped a factor of 2 in the last term.

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