Two masses connected by Pulley - Lagrangian problem

AI Thread Summary
The discussion focuses on the application of the Lagrangian method to a system of two masses connected by a pulley. The user defines a coordinate system based on the center of mass of one mass and derives the Lagrangian, omitting the kinetic energy of that mass. Several participants question the validity of the Lagrangian, particularly regarding the absence of mass m1 and the dimensional consistency of the potential energy term. Concerns are raised about mixing kinetic and potential energy terms, specifically the inclusion of generalized velocities in potential energy. The conversation highlights the importance of maintaining proper dimensional analysis in Lagrangian mechanics.
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1718493064484.png

My solution to (a) is,
We have constraint ##x + y = L##. There are many places we could define our (x,y) Cartesian coordinate system. However, the most easiest I think for the problem would be to attach a ##x^*## and ##y^*## coordinate system at the COM of ##m_1##. We define ##\hat x^* > 0## parallel to rightward horizontal and ##\hat y^* > 0## parallel to upwards vertical.
1718493234131.png

Since the COM of ##m_1## is not moving with respect to our defined coordinate system (orange dot denotes m_1 COM in diagram above), then we omit ##m_1## KE in the Lagrangian form of the system. The ##(x^*, y^*)## coordinates of ##m_2## is ##(x\cos \theta + y, - x\sin \theta) = (x\cos\theta + L - x, -x\sin\theta)##

The velocity coordinates are by definition the time with respect to the linear time ##t## (I think we could generalize this from being with respect to linear time to being with respect to any non-linear, higher dimensional time ##(t', t'', t''', ...)##). However, we assume classical case. Thus ##(\dot x^*, \dot y^*) = (-x\sin \theta \dot \theta + \cos \theta \dot x - \dot x, -x\cos \theta \dot \theta - \dot x \sin \theta) ##

Thus by definition of T and V, Lagrangian is ##\mathcal{L} = \frac{m_2}{2}(x^2\dot \theta^2 + 2\dot x^2 + 2x \dot x \dot \theta \sin \theta - 2 \dot x^2 \cos \theta) + m_2g(x\cos \theta \dot \theta + \dot x \sin \theta)##

However, does anybody please know whether I am correct so far?

Thanks!
 
Physics news on Phys.org
At a glance your Lagrangian does not have m1. Dimension of potential energy term has excess T^-1. Are they OK?
 
Last edited:
  • Love
Likes member 731016
anuttarasammyak said:
At a glance your Lagrangian does not have m1. Dimension of potential energy term has excess T^-1. Are they OK?
Thank you for your reply @anuttarasammyak !

The Lagrangian does not have m1 since I chose m1 COM as coordinate system. Not sure what you mean about dimension of PE term sorry.

Thanks!
 
ChiralSuperfields said:
The Lagrangian does not have m1 since I chose m1 COM as coordinate system.
I don't understand this, please explain. Are you saying that the equations of motion will be the same regardless of the size of ##m_1##? If ##m_1## is that of a battleship, the EOM of ##m_2## will be pretty close to that of a pendulum. If ##m_1## is that of a gnat, the EOM of ##m_2## will be be pretty close to that of a free-falling object..
 
  • Like
  • Love
Likes member 731016, MatinSAR and erobz
ChiralSuperfields said:
Not sure what you mean about dimension of PE term sorry.
\mathcal{L} = \frac{m_2}{2}(x^2\dot \theta^2 + 2\dot x^2 + 2x \dot x \dot \theta \sin \theta - 2 \dot x^2 \cos \theta) + m_2g(x\cos \theta \dot \theta + \dot x \sin \theta)
the second term, containing g, has physical dimension of MLT^-2LT^-1=(ML^2T^-2)T^-1 but it should have dimension of ML^2T^-2, energy.
 
Last edited:
  • Love
Likes member 731016
anuttarasammyak said:
\mathcal{L} = \frac{m_2}{2}(x^2\dot \theta^2 + 2\dot x^2 + 2x \dot x \dot \theta \sin \theta - 2 \dot x^2 \cos \theta) + m_2g(x\cos \theta \dot \theta + \dot x \sin \theta)
the second term containing g has physical dimension of MLT^-2LT^-1=(ML^2T^-2)T^-1 but it should have dimension of ML^2T^-2, energy.
The first term ##~m_2gx\cos\theta\dot{\theta}~## also suffers from the condition of the same bad dimensions.
 
  • Love
Likes member 731016
I also have concerns about dimensions)

Your second term has ##g## in it along with generalized velocities. It seems to me that you’re mixing your Kinetic and Potential terms.

Why would a potential term have generalized velocities? Also, why would kinetic terms have acceleration ##g##?
 
  • Love
Likes member 731016
Back
Top