The h Bar

4:49 AM

@tpg2114 I cycle every day (if it isn't raining). Not to race but for exercise. I find it's a lot kinder on the knees than running, though yes it can be hard on your rear :-)

user434058

5:52 AM

Hi

user434058

user434058

Funny thing, my flag on Jeff Atwood's post wasn't right out declined, rather it was disputed :))

Q: Rolling equations from a non Inertial frame

Disputed means a user who reviewed the flag disagreed

Declined means a moderator declined it

user434058

O_o I thought it meant that the people who reviewed the flag weren't able to come to a common decision.....

user434058

Anyways that was expected!

user434058

6:14 AM

user434058

Can anyone help me in finding the Lagrangian of the above system? The string is connected to the ceiling at the left end and the ball is release at t=0

user434058

7:00 AM

Released*

user434058

In one of the suggested edits, I typo-ed out "MethJaz" in the summary, looks like a nice alternative to MathJax :P

user434058

7:27 AM

0

In particle dynamics, I had used "pseudo forces" frequently, and i had come to the conclusion that pseudo forces ensure that analysis from a non inertial frame yield the same conclusion(s) as the ground/inertial frame. I wanted to extend this thought further, to rigid body dynamics. Consider a d...

user434058

This isn't a close-worthy HW question, right? Or am I being gullible?

user434058

7:55 AM

Hi! @ACuriousMind Did you delete the comments on this question? I know you aren't obliged to be accountable for mod actions, but still I am just wondering....

user434058

5

Q: Given two vectors (with no common point), is a dot product between them possible?

Most of the descriptions I have come across related to dot products between two vectors start off by stating/showing the vectors have a common point* to begin with - makes the notion of an angle between the vectors very easy to deal with in the final formula for dot products. I am showing below...

vectors coordinate-systems geometry

8:46 AM

morning

Morning.

Hi :-)

user434058

Hi!

9:13 AM

Hi, @Yuvraj.

@PM2Ring hi sir :-) is it possible for is to launch a missile towards the sun.

?

@JohnRennie "the Parker Solar Probe that is going to use seven slingshots to reduce its speed enough to approach the Sun." Reduce it's speed? That doesn't sound right. Surely it needs to reduce its perihelion and increase its speed, doesn't it? physics.stackexchange.com/a/440258

@YuvrajSingh... Funny you should ask that right now. :) See the question I just linked & en.wikipedia.org/wiki/Parker_Solar_Probe

@PM2Ring Getting it into a Hohman orbit that intersects the required orbit round the Sun is easy, but the perihelion velocity would be far too large to shed using the thrusters on the satellite.

So you need to use a series of transfers to increasingly smaller orbits using flybys to shed the excess perihelion speed.

Also this article about a very important technique for traveling through space with (just about) the lowest fuel requirements:

Hohmann transfer orbit

In orbital mechanics, the Hohmann transfer orbit () is an elliptical orbit used to transfer between two circular orbits of different radii around a central body in the same plane. The Hohmann transfer often uses the lowest possible amount of propellant in traveling between these orbits, but bi-elliptic transfers can beat it in some cases. The orbital maneuver to perform the Hohmann transfer uses two engine impulses, one to move a spacecraft onto the transfer orbit and a second to move off it. This maneuver was named after Walter Hohmann, the German scientist who published a description of it in...

@JohnRennie Ah, that makes sense. I'm still waking up after an afternoon nap. :)

That pesky potential energy :-)

9:22 AM

I should re-read that Hohmann transfer article myself. :)

Would we have to reverse the velocity to change the orbit of our satellite?

I should know this stuff inside out, I've written enough sims that do Hohmann orbits...

user434058

It's nice that I find first posts before the review queue finds them. And thus when I go to review, I already have done the stuff that I wanted to and so I just need to click "Reviewed".

@YuvrajSingh... Sending a missile into the Sun is a little easier than putting something into orbit very close to the Sun because if you're crashing into the Sun you don't need to worry about shedding excess speed. But it still takes a lot of energy to get it into an orbit that intersects the Sun. IIRC, it takes less fuel to send something to Neptune, which is a lot further away.

user434058

1 min ago, by FakeMod

It's nice that I find first posts before the review queue finds them. And thus when I go to review, I already have done the stuff that I wanted to and so I just need to click "Reviewed".

9:28 AM

Wouldn't the sun gravity attract the satellite towards it, as the satellite is near the sun.

user434058

I meant the "I'm done" button!

9:41 AM

@PM2Ring do you watch YouTube channel, for some physics related videos?

@YuvrajSingh... Very rarely. I prefer to read physics, although videos are good for demos. I mostly use YouTube to watch music clips.

Lol

I remember you shared

An video on political

Tedeschi Trucks Band feat Jerry Douglas - These Walls - Beacon Theater 9-19-14

@YuvrajSingh... Yes, but you need to put the satellite into some kind of orbit to get it near the Sun. And when you launch that satellite it has some velocity due to the rocket engine, but it has a lot of energy due to the circular orbital motion of the Earth. Please read the Hohmann article, and look at the diagrams there. It will be helpful.

@YuvrajSingh... That song Politician is pretty heavy. Here's a mellow song by one of my favourite modern American bands. Derek Trucks is a wonderful slide guitar player, he also studied sarod under an Indian guru. This melancholy song has an Indian feel to it, especially in the introduction. I call this style "Country and Eastern". ;)

►

The woman singing is Susan Tedeschi, she's married to Derek. And she is also a very good guitar player, and song writer.

user434058

10:07 AM

4 hours ago, by FakeMod

user434058

Anyone?

@FakeMod hi.

What you want to find?

are you sure l>d?

Susan Tedeschi - Just Won't Burn (Live at Farm Aid 1999)

Here's Susan playing guitar & singing a song she wrote, when she was younger, before she married Derek.

►

Wow

data electromagnetism radiation

I'm glad you like it, Yuvraj. I think @ZeroTheHero would also enjoy it too, since he likes the blues.

user434058

10:21 AM

@YuvrajSingh... Yes, that is why the rope is slack. I want to find the equations of motion for the mass m.

user434058

@ACuriousMind if you haven't seen it before, then you would be glad to know that people have written books to capture your greatness :P

user434058

Son of Stackoverflow, William Street, New York, NY, USA

101 1

user434058

This user's about and user name is quite funny....:)

@FakeMod Sorry, I don't know how to do that. The string starts out as a catenary, and I think it continues to be a catenary as the ball starts falling, before the string becomes taut, but I'm not certain. Can we assume the string is inextensible, with negligible mass?

user434058

10:28 AM

@PM2Ring Yes the string is a massles, inextensible iseal strin.

Hm

user434058

@PM2Ring Unti the string becomes taut, the mass will be under free fall.

@FakeMod That's nothing, everything I say is written down and bundled for people to read! (Communications of the ACM)

user434058

I am able to find the equations of motion using Newtonian mechanics but I want to use Lagrangian mechanics to solve this @PM2Ring

How to show that $$(\gamma^* \omega)_t (X) = \omega_{\gamma(t)}[\dot{\gamma}] X$$

Sans coordinates

10:30 AM

"by definition"

Well no

user434058

@ACuriousMind How do I subscribe to that magazine?

Seriously, if my guess at that what notation means is right that's just the definition of the pullback

(if my guess is incorrect you have to explain yourself)

By definition it is

\begin{equation}
(\gamma^* \omega)_t (X) = \omega_{\gamma(t)} [d\gamma_t (X)]
\end{equation}

@FakeMod acm.org/membership/membership-options

user434058

10:34 AM

@ACuriousMind Shit! That joke was for real? Nice....

With coordinates, that would be
\begin{eqnarray}
\omega_{\gamma(t)} [d\gamma_t (X)] &=& \omega_\mu \theta^\mu [(d\gamma)^\mu_t \theta_t \otimes e_\mu (X^t e_t)]\\
&=& \omega_\mu \theta^\mu [(d\gamma)^\mu_t e_\mu X^t ]\\
&=& \omega_\mu \delta^\mu_{\nu} (d\gamma)^\mu_t X^t\\
\end{eqnarray}

which is indeed the correct thing

@Slereah $\mathrm{d}\gamma(\cdot) = g(\dot{\gamma},\cdot)$, no?

I think it only works because it is one-dimensional

@ACuriousMind I mean yes, but I'm not sure from what that would stem, without coordinates!

Yes, what I just wrote is special to 1d

@Slereah There's nothing wrong with proving such a local statement in coordinates

I know

But still

It feels WRONG

like a proof by brute force

Or by using Zorn's lemma

user434058

10:38 AM

@ACuriousMind any ideas about this question?

user434058

31 mins ago, by FakeMod

4 hours ago, by FakeMod

Oh well

it will have to do for now

@FakeMod No, I'm afraid I don't find mechanics exercises interesting

user434058

@ACuriousMind okay, np

@FakeMod why would you use Lagrangian mechanics for that problem?

The mass just falls straight down until the string becomes taut, then it moves in a circle with radius equal to the length of the string.

10:48 AM

@FakeMod Of course. But I don't think that means the string stops being a catenary... but I'm certainly happy to hear any arguments that prove it's not. :)

user434058

@JohnRennie Yes! That's exactly what happens. But if I were to solve it using Lag. Mech., how do I even write it's Lagrangian? Also, this example is a good one, in the sense that the total energy is not conserved. When the string gets taut, the angular momentum is conserved and after that we can safely conserve energy, but some energy is lost when the tension increases impulsively.

@FakeMod I am too badly out of practice with Lagrangian mechanics to comment.

It is generally a bad idea to try to apply Lagrangian mechanics when non-conservative forces are involved.

Sometimes it works, but it's never worth the trouble

user434058

@ACuriousMind But there are no apparent non conservative forces, which makes me think that Lag Mech should not fail here.

user434058

@JohnRennie No problem :)

10:58 AM

@FakeMod the string is dissipating some of the KE of the object. Presumably converting it to heat.

user434058

@ACuriousMind AFAI have thought, we'll need to use a discontinuous Lagrangian, the kind of one whose value suddenly jumps when the string gets taut. This way the acceleration at the taut point will become undefined because L is not differentiable there.

user434058

@JohnRennie Yes, that seems the most probable sorce of dissipation....

@FakeMod if you made the string perfectly elastic then that would be more interesting because then energy would be conserved and the Lagrangian approach would work fine.

You'd get a complicated mixture of rotation and oscillation.

user434058

@JohnRennie looks interesting! Moreover, in that case we won't be able to find a solution using Newton's laws of motion.

user434058

Then the string would essentially become a spring.....

11:03 AM

You can always find a solution from Newton's laws. It's just that it can become computationally impenetrable.

user434058

@JohnRennie I suppose the system will become chaotic if we take a spring instead of a string, what do you think?

user434058

11:22 AM

user434058

FWIW This is the Lagrangian ☝️

user434058

A bit of cleverness, tweaking and signum functions make it possible to solve it using Lag Mech. That sgn is signum fuction

11:46 AM

Argh

Trying to define the wilson loops properly is a huge pain :V

Aaron Stevens

@FakeMod I still think the question should be closed :p