All Questions
156
questions
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2
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Generalized momentum
I am studying Hamiltonian Mechanics and I was questioning about some laws of conservation:
in an isolate system, the Lagrangian $\mathcal{L}=\mathcal{L}(q,\dot q)$ is a function of the generalized ...
- 59
1
vote
1
answer
53
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Confusing Goldstein Statement about Magnitude of the Lagrangian
On page 345 of Goldstein's Classical Mechanics 3rd Ed., he writes:
...the Hamiltonian is dependent both in magnitude and in functional form upon the initial choice of generalized coordinates. For the ...
- 7,388
4
votes
3
answers
147
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Analyzing uniform circular motion with Lagrangian mechanics
Consider swinging a ball around a center via uniform circular motion. The centripetal acceleration is provided by the tension of a rope. Now, is this force a constraint force? If it is, since it is ...
- 143
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1
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75
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Derivation of lagrange equation in classical mechanics
I'm currently working on classical mechanics and I am stuck in a part of the derivation of the lagrange equation with generalized coordinates. I just cant figure it out and don't know if it's just ...
- 41
6
votes
2
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327
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Generalized vs curvilinear coordinates
I am taking the course "Analytical Mechanics" (from on will be called "AM") this semester. In our first lecture, my professor introduced the notion of generalized coordinates. As ...
- 150
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2
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90
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How to change generalised coordinates in a Lagrangian without inverting the coordinate transformation?
Given a Lagrangian using the standard cartesian coordinates.
$$ \mathcal{L} = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - \frac{1}{2}k(x^2 + y^2) $$
How to move to the hyperbolic coordinates given as
$$2 x ...
- 173
2
votes
1
answer
121
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Independence of generalized coordinates in the derivation of Lagrange equations from d'Alembert's Principle
I am confused by this remark in the derivation of Lagrange equations from d'Alembert's principle in Goldstein:
I am not comfortable that I understand why, at this late stage of the derivation, they ...
- 31
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2
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289
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Lagrangian mechanics and generalized coordinates
In Lagrangian mechanics, we use what is called the generalized coordinates (gc's) as the variable of the machanics problem in hand. These gc's represent the degrees of freedom that the studied system ...
- 113
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1
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94
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Virtual work of constraints in Hamilton‘s principle
Goldstein 2ed pg 36
So in the case of holonomic constraints we can move back and forth between Hamilton's principle and Lagrange equations given as $$\frac{d}{d t}\left(\frac{\partial L}{\partial \...
- 1,260
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1
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48
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How to determine which coordinates to use for calculating the Hamiltonian? [closed]
In my classical mechanics course, I was tasked with finding the Hamiltonian of a pendulum of variable length $l$, where $\frac{dl}{dt} = -\alpha$ ($\alpha$ is a constant, so $l = c - \alpha t$.).
I ...
- 413
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0
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76
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Constraint force using Lagrangian Multipliers
Consider the following setup
where the bead can glide along the rod without friction, and the rod rotates with a constant angular velocity $\omega$, and we want to find the constraint force using ...
- 193
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45
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Can anyone in here explain to me what exactly is 'Quasi-Generalised Co-ordinates'?
This comes straight up from a certain text that I was going through, which of course is in the form of a question which asks 'A solid cylinder is rolling without slipping and how many generalized co-...
0
votes
1
answer
71
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Why are constraint forces and gradient of constraint functions perpendicular?
My question is about the general relationship between the constraint functions and the constraint forces, but I found it easier to explain my problem over the example of a double pendulum:
Consider a ...
- 193
2
votes
6
answers
236
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Lagrangian - How can we differentiate with respect to time if $v$ not a function of time?
In the Lagrangian itself, we know that $v$ and $q$ don't depend on $t$ (i.e - they are not functions of $t$ - i.e., $L(q,v,t)$ is a state function.)
Imagine $L = \frac{1}{2}mv^2 - mgq$
Euler-Lagrange ...
- 525
1
vote
1
answer
68
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Requirement of Holonomic Constraints for Deriving Lagrange Equations
While deriving the Lagrange equations from d'Alembert's principle, we get from $$\displaystyle\sum_i(m\ddot x_i-F_i)\delta x_i=0\tag{1}$$ to $$\displaystyle\sum_k (\frac {\partial\mathcal L}{\partial\ ...
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