All Questions
78
questions
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Getting an opposite sign for the centrifugal potential energy in the effective potential [duplicate]
Consider a system whose Lagrangian is
$$L = \frac12 \mu\left( \dot r^2 + r^2 \dot\theta^2 \right) -U(r) $$
By the Euler-Lagrange equation,
$$\frac{\partial L}{\partial\theta}=\frac{d}{dt}\frac{\...
- 115
6
votes
3
answers
1k
views
In equation (3) from lecture 7 in Leonard Susskind’s ‘Classical Mechanics’, should the derivatives be partial?
Here are the equations. ($V$ represents a potential function and $p$ represents momentum.)
$$V(q_1,q_2) = V(aq_1 - bq_2)$$
$$\dot{p}_1 = -aV'(aq_1 - bq_2)$$
$$\dot{p}_2 = +bV'(aq_1 - bq_2)$$
Should ...
0
votes
1
answer
56
views
Question about Problem $12$ in Chapter $11$ from Kibble & Berkshire's book
I write again the problem for convinience:
A rigid rod of length $2a$ is suspended by two light, inextensible strings of length $l$ joining its ends to supports also a distance $2a$ apart and level ...
0
votes
1
answer
43
views
What is the physical significance of this generalised potential?
Consider a generalised potential of the form $U=-f\vec{v}\cdot\vec{r}$ where $f$ is a constant. This potential should not contribute any internal forces between particles as
\begin{equation}
\vec{F}=-\...
- 838
0
votes
0
answers
89
views
Classification of equilibrium configurations for particles subject to elastic force constrained on a circle
I am interested in classifying all the possible equilibrium configurations for an arrangement of $l$ equal point particles $P_1, P_2, . . . , P_l$ $(l > 2)$ on a circle of radius $R$ and centre $O$....
- 130
0
votes
1
answer
55
views
Why is gravitational potential energy negative in this Lagrangian? [closed]
The question is given as follows:
From (6.109) shouldn't the Lagrangian be K(kinetic) - U(potential), but here its K + U? Unless the potential energy is negative, if so I'm struggling to come to ...
1
vote
1
answer
86
views
Velocity-Dependent Forces and Generalized Potential
Is there any theorem about for which velocity-dependent forces a velocity-dependent generalized potential of the form
$$F_k=\frac d {dt} \left(\frac {\partial U}{\partial \dot q_k}\right)-\frac {\...
- 193
0
votes
0
answers
59
views
How can the equation for generalized force be derived assuming the system is in equilibrium?
I have been going over sections 1.4-1.5 of Goldstein's Classical Mechanics where the equation for generalized force
\begin{equation}
Q_j=-\frac{\partial U}{\partial q_j} +\frac{d}{dt}\frac{\partial U}{...
1
vote
1
answer
87
views
Is Hamilton’s principle valid for systems that are not monogenic?
I have read in Goldstein that Hamilton’s principle works only for monogenic systems. Is it true? I thought that the action principle is universal?
1
vote
1
answer
81
views
On generalised potential in Electrodynamics
I'm studying Lagrangian Mechanics from Goldstein's Classical Mechanics. My question concerns Section 1.5 which talks about velocity-dependent potentials.
I am actually unsure about how Equation 1-64' ...
- 1,640
0
votes
2
answers
101
views
How to understand "the potential energy in an EM field is determined by $\phi$ alone"?
Goldstein page 342,
Consider a single particle (non-relativisitic) of mass $m$ and charge $q$ moving in an electromagnetic field.
The Lagrangian is
$$ L = T-V = \frac{1}{2}mv^2-q\phi +q\vec{A}\cdot \...
- 1
1
vote
2
answers
158
views
Why does the Lagrangian not show particle-interaction? Why are normal/tension forces not considered?
(1) For formulating a lagrangian for a system of particles compared to one free particle, we start with the kinetic energy and formulate a potential energy term that is in terms of each of the radius ...
- 17
0
votes
1
answer
155
views
Lagrange versus Euler-Lagrange equation for inverted pendulum
On the wikipedia page of inverted pendulum, in the section "inverted pendulum on a cart" (https://en.wikipedia.org/wiki/Inverted_pendulum#Inverted_pendulum_on_a_cart), the equations of ...
1
vote
0
answers
38
views
Why does the Lagrangian equal $T-V$? [duplicate]
When defining $L=T-V$, and using Euler-Lagrange equations ($\partial_x L = \frac{d}{dt} \partial_{\dot{x}} L$), we get back $m \ddot{x} = - \frac{dV}{dx}$ ONLY WHEN ASSUMING that $V(x, \dot{x}) = V(x)$...
- 31
1
vote
2
answers
84
views
Question from Classical mechanics Goldstein H [closed]
When we are given a Lagrangian as $L \equiv L(\dot{x}, V)$ where $V=V(x)$, while differentiating why do we set $\frac{dV}{d\dot{x}} = 0$?
Is it just because $V$ is a function of only $x$? Since it’s ...
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