Two masses connected by Pulley - Lagrangian problem

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

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.

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!

 

[anuttarasammyak]

At a glance your Lagrangian does not have m1. Dimension of potential energy term has excess T^-1. Are they OK?

 

[member 731016]

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!

 

[kuruman]

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..

 

[anuttarasammyak]

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.

 

[kuruman]

$$\mathcal{L} = \frac{m_2}{2}(x^2\dot \theta^2 + 2\dot x^2 + 2x \dot x \dot \theta \sin \theta - 2 \dot x^2 \cos \theta) + m_2g(x\cos \theta \dot \theta + \dot x \sin \theta)$$
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.

 

[PhDeezNutz]

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 $g$?