Comments for Steven Landsburg | The Big Questions: Tackling the Problems of Philosophy with Ideas from Mathematics, Economics, and Physics

Comment on About That Boxcar by db

db — Thu, 23 May 2013 22:51:40 +0000

Sorry I have a bracket in the wrong place, but hopefully point is clear that the two problems live in a space together.

I’m sure Bernard is off solving this chap as we speak (it’s a little trickier…)

Comment on About That Boxcar by db

db — Thu, 23 May 2013 22:39:50 +0000

@KenB (passim)

Indulge me for a moment and imagine that we have a straw, but it’s one that we can choose the angle back or forth if we please.

If the straw points straight out then we have the problem that Bernard’s equations describe. If it points a little rightwards then we have a bit of a rocket added on top (and conversely if it is pointed leftwards).

Indeed I could allow you a straw pointing rightwards and still have the boxcar deviate initially left and then roll off to infinity.

Bernard’s equation becomes:-

0 = (m+w)x – int[0-t] {x + l + (t-u)(x’-CSw’)w’ du}

Where
– 1/C is the density of the fluid x cross-section of the hole
– S is sine(angle of straw)

Comment on About That Boxcar by Bernard

Bernard — Thu, 23 May 2013 20:42:51 +0000

@kenB,@db
Yes I am assuming zero l-r momentum of the water relative to the boxcar (‘steady flow’).
Not sure why my latex code didn’t go through,
the 2 results are:

$$x’(0)=\frac{l w’(0)}{m+w(0)}$$

and

$$x’(0)=\frac{l w’(t)}{m+w(t)} + \int_0^t \frac{l w’(u)^2}{(m+w(u))^2} du$$

Comment on About That Boxcar by db

db — Thu, 23 May 2013 20:25:57 +0000

@Bernard — the inclusion of the x’(t-u) term in the integral suggests that the water exits the boxcar with no relative l-r momentum.

I’m fine with that, but you probably need to state it as an assumption, assert it as a universal fact, or appeal to the framing of the question.

Comment on About That Boxcar by Ken B

Ken B — Thu, 23 May 2013 20:02:18 +0000

@Bernard:
I’d like to see your derivation of the equation. I cannot frankly make it out but it looks like you are doingf something with the com, and so are assuming the the expelled water has no x momentum. Which is the whole question.

Comment on About That Boxcar by neil

neil — Thu, 23 May 2013 20:00:28 +0000

I do not think it is a rocket. That answer can be crossed out. The jet leaving the boxcar perpendicular to the allowed direction of motion cannot accelerate the boxcar, we all agree on that I believe. I do not think that the water moving within the box car can accelerate either, ignoring turbulence. It is no different than if the water jetted out from a hole in the center or bottom of the boxcar. Water flowing to the right has to “stop” and turn to exit the hole, which offsets reaction of the right flow action.

Comment on About That Boxcar by Ken B

Ken B — Thu, 23 May 2013 19:58:01 +0000

@Scott H 66:
db is assuming that but saying otherwise. Why do I say he says otherwise? Because db wrote this: ” it isn’t the whole debate about whether the exiting water carries rightwards momentum relative to the car”. But since momentum is conserved where it goes is the whole issue.

IF there is no momentum leak THEN there is no rocket. There will be oscillations I think.
IF there is a momentum leak THEN there will be a rocket. Maybe a really slow pathetic shuddering one that accelerates and decelerates but one that at the end rolls off forever.

Lets rig up a hose under the hole. It slopes down and to the right. All the water exits via this hose. So with the hose in place all the exiting water, all of it, has rightward momentum relative to the car. And it is clearly a rocket.

Comment on About That Boxcar by Scott H.

Scott H. — Thu, 23 May 2013 19:42:47 +0000

@Ken B 63 & 61. I think what db is saying is that the water could be leaving with zero horizontal momentum relative to the car, but — for a while at least — the car/water would have imparted momentum relative to a stationary observer. This imparted momentum leak of the falling water is part of the debate.

btw… thanks a lot Steve. I needed to work today.

Comment on About That Boxcar by Bernard

Bernard — Thu, 23 May 2013 17:49:12 +0000

looks like the comment section didn’t like latex notations. And since I don’t want to write it again, I’ll just give the method:
(1) center of mass doesn’t move
(2) differentiate once when time=0
(3) differentiate once and integrate by parts once with position of the box car on the left and position of the water on the right.
(4) assume that the flow intensity decreases and gets sufficiently close to 0

Comment on About That Boxcar by Ken B

Ken B — Thu, 23 May 2013 17:45:58 +0000

@db 62: I think that depends on the hole. As Ron notes, a plane is usual assumption in idealized problems. But just make the hole large compared to the ball. Like say a mouse hole compared to a water molecule!

You are simply wrong about what the whole debate is about. If the water on net *leaves* with a rightward momentum then the car ends with leftward momentum. If the water does not on net carry away rightward momentum then there is no rocket, there is the shimmying that some predict. If I am in a boxcar with balls and no holes I cannot make a rocket. If I am on a flat car with balls and no walls I can make a rocket.

The “on net” means doing a sum. You can do this in any inertial frame or over any set of inertial frames.