How Does a Variable Mass Affect a Simple Pendulum?

[DannyJ108]

Homework Statement

Find the solutions to the equations of motion of a simple pendulum with a mass which varies in time

Relevant Equations

$L= T -V =\frac 1 2 m(t) l^2 {\dot \theta}^2 -m(t)gl\cos\theta$

Hello,

I've got to rationally analice the form of the solutions for the equations of motion of a simple pendulum with a varying mass hanging from its thread of length $l$ (being this length constant).

I approached this with lagrangian mechanics, asumming the positive $y$ direction is pointing down, I get that:

$L= T -V =\frac 1 2 m(t) l^2 {\dot \theta}^2 -m(t)gl\cos\theta$

I work out my Euler-Lagrange equations with respect to $\theta$ and get:

$\dot m(t) l \dot \theta + m(t)l \ddot \theta - m(t)g\sin \theta = 0$

If i assume the mass varies slowly with time and use small angle approximation ($\sin\theta \approx \theta$) I thought to proceed like this:

$\frac d {dt} (m(t) \dot \theta) = m(t) \frac g l \theta$

I don't know how to proceed from here. I think I don't have to give a explicit solution to this equation, but rather just interpret how it'll be. Either way if you could help me out finding a solution it would be great.

Thank you!

 

[etotheipi]

Note that if the mass varies with time, you need to think about how the pendulum is exchanging mass with the environment. If you're saying that mass is just a variable parameter, then it's fine to use the product rule. But if the mass is being ejected (i.e. sand falling out), then there are some other considerations to take into account and life gets more complicated.

Also, shouldn't it be $L= T -V =\frac 1 2 m(t) l^2 {\dot \theta}^2 + m(t)gl\cos\theta$

 

[DannyJ108]

Note that if the mass varies with time, you need to think about how the pendulum is exchanging mass with the environment. If you're saying that mass is just a variable parameter, then it's fine to use the product rule. But if the mass is being ejected (i.e. sand falling out), then there are some other considerations to take into account and life gets more complicated.

Also, shouldn't it be $L= T -V =\frac 1 2 m(t) l^2 {\dot \theta}^2 + m(t)gl\cos\theta$

It doesn't specify anything else regarding the variable mass, just that it varies slowly in time and to consider small angles, so I think I wouldn't need to assume anything else, just that it's a variable parameter like you said, so I could use the product rule. In that case, it's correct how I wrote it, right?

$\frac d {dt} (m(t) \dot \theta) = m(t) \frac g l \theta$

But then what kind of explanation would I give for the solution form to this? I have no idea whatsoever. Or maybe I should've taken a different approach to this, I'm not sure how to proceed

Regarding the Lagrangian, I'm assuming the positive $y$ axis direction is down, so the potential energy would be positive, thus substracting it for the formulation of the Lagrangian it would end up as a positive term

 

[etotheipi]

Regarding the Lagrangian, I'm assuming the positive $y$ axis direction is down, so the potential energy would be positive, thus substracting it for the formulation of the Lagrangian it would end up as a positive term

How you setup your coordinate system is irrelevant. The potential energy is $mgh$ where $h$ is the vertical distance from the reference position in the opposite direction to the gravitational field.

With your construction, the pendulum gains GPE as it swings down.

 

[DannyJ108]

How you setup your coordinate system is irrelevant. The potential energy is $mgh$ where $h$ is the vertical distance from the reference position in the opposite direction to the gravitational field.

With your construction, the pendulum gains GPE as it swings down.

Okay, thanks. I'll change my formula.

Then again, can I get some sort of guidance on the solution form for the equation of motion?
If mass varies slowly can I make the approximation that $\frac {dm} {dt} \approx 0$?

Or maybe assume that mass variation is constant, then:

$\frac {dm} {dt} = \alpha \Rightarrow m(t) \approx m_0 +\alpha t$ and assume that $m_0 \gg \alpha t$ (i.e. mass loss is negligible)?

 

[etotheipi]

If you set $\frac{dm}{dt} = 0$, then you wind up back with $\ddot{\theta} = -\frac{g}{l}\theta$, which is I suppose reassuring.

You might consider the case where $\dot{m} = -k$ so that $m = m_0 - kt$ which would give you an equation like $$(m_0 - kt)\frac{d^2 \theta}{d\theta^2} - k\frac{d\theta}{dt} + \frac{(m_0 - kt)g}{l} \theta = 0$$ I admit I have no idea whether an analytical solution is possible. I would be inclined to try solving it numerically. Perhaps someone else will have a better idea!

 

[haruspex]

It doesn't specify anything else regarding the variable mass, just that it varies slowly in time

Then you are in the ridiculous position of having to guess what is in the question setter's mind. What can you rely on to stay constant?
If we suppose the bob is subliming, won't the equation just be the normal one?

 

[etotheipi]

I'm not actually sure how we deal with variable mass systems within Lagrangian mechanics. In the Newtonian domain, NII doesn't apply to systems of variable mass and instead they're governed by the slightly modified equation $$\mathbf{F}_{ext} + \mathbf{v}_{rel}\frac{dm}{dt} = m\frac{d\mathbf{v}}{dt}$$ You could try deriving the an equation of motion from that. But I wonder if anyone else knows an equivalent variable mass formulation for Lagrangian mechanics?

 

[haruspex]

I'm not actually sure how we deal with variable mass systems within Lagrangian mechanics. In the Newtonian domain, NII doesn't apply to systems of variable mass and instead they're governed by the slightly modified equation $$\mathbf{F}_{ext} + \mathbf{v}_{rel}\frac{dm}{dt} = m\frac{d\mathbf{v}}{dt}$$

how are F_ext and v_rel defined there?

 

[etotheipi]

how are F_ext and v_rel defined there?

I think $\mathbf{v}_{rel}$ is the velocity of the mass leaving/entering the system w.r.t. the centre of mass and $\mathbf{F}_{ext}$ is the external force on the system (e.g. not to do with the thrust of the ejected mass etc.). This page has some nice derivations.