# Pipe in space outputs gas - what is its angular momentum at $t$ and $t+dt$? [closed]

Given a pipe in space (neglect gravitational force):

• The speed of the gas is $v_0$ (in relative to the edge of the pipe)
• The length of the pipe is $l$
• The pipe rests (not moving) at $t=0$
• The gas spreads equally along the pipe

What would be the angular momentum at $t$ and at $(t+dt)$ during the process of outputting $(-dm)$ mass from the pipe?

I've tried using the definition: $L = \Sigma m_ir_i\times v_i$.
But at $t$ it is zero (which doesn't make sense): $L(t) = m \cdot 0 \times V_{cm}$. I calculated it in relative to the centre of mass of the pipe (i.e. $r_{cm} = \frac{1}{2}l$), therefore $r=0$.

What am I missing here?

## Edit #1:

At time $t$ it is before the mass leaves the pipe.
At time $(t+dt)$ it is after that the mass left the pipe.

• Why doesn't L=0 at t=0 make sense? Remember that it is a vector and you can have two contributions that cancel. Mar 2, 2015 at 11:50
• I maybe don't understand the figure. If the gas leaves the pipe through the opening that you show, the opening is at a distance $l/2$ from the center-of-mass of the pipe. Then $\vec r = \vec l/2$ and it is perpendicular to $v_0$. Mar 2, 2015 at 11:57
• @Eoin: L is zero at $t=0$. But at some time $t$ it can't be zero..
– Dor
Mar 2, 2015 at 12:21
• @Sofia: Indeed but at time $t$ it is before the mass leaves the pipe. At time $t+dt$ it is after that the mass left the pipe. Sorry for not specifying that, I thought it was a convention.
– Dor
Mar 2, 2015 at 12:23
• @Dor, yes it can. You sum up the contribution from the mas exiting the tube and the tube itself. Depending on your choice of axes, one is negative and one is positive Mar 2, 2015 at 12:32

After the mass exits the pipe, the tube will start to rotate from the recoil.

At $t=0$, there is zero angular momentum, $L=0$.

Let's take it that the pipe rotates about its centre-of-mass, and use that point as the origin from which to calculate the angular momentum.

At time $t+dt$, the puff of gas has angular momentum $L_{gas} = dm \times l/2 \times v0$.

So, through conservation of angular momentum, the pipe must have $L_{pipe}=-L_{gas}$.

The angular momentum of the pipe is from the relation

$L = \frac{m_{pipe}}{l}\int_{-l/2}^{l/2} r_i v_i dr_i$.

Since the pipe is a rigid body, $v_i = r_i \omega$,

$L = \frac{m_{pipe}}{l}\int_{-l/2}^{l/2} r_i^2 \omega dr_i$

$= \frac{m_{pipe}}{l} \frac{1}{3} \omega r_i^3 |_{-l/2}^{l/2}$

$= \frac{1}{12} m_{pipe} l^2 \omega$

At this point you might recognise the moment of inertia of a straight rod about its centre,

$I=\frac{1}{12} m l^2$, which would have been the other way to approach this.

Now all you have to do is equate the two expressions for the angular momentum to find the rotation rate

• (1) You showed the angular momentum at $t=0$ and at $t+dt$. What happens in $t$? (before the gas leaves the pipe). (2) Why did you divided the equation of the angular momentum with the length of the pipe $(l)$ ? Is it due to the additional $dr_i$?
– Dor
Mar 2, 2015 at 14:51
• (1) this depends on a lot of things, but to a first approximation, since I don't see that the gas imparts a force on the pipe (2) the quantity $\frac{m}{l} dr$ is the mass of an infinitesimal length of the pipe Mar 2, 2015 at 15:54
• (1) It seems that the sentence that you wrote is incomplete (??)
– Dor
Mar 2, 2015 at 19:51
• I don't think so... Mar 2, 2015 at 20:19
• Then I don't understand the sentence. You wrote "since I don't see that the gas imparts a force on the pipe". When writing "Since that X", the expression ", then Y" comes after it. But in your sentence there's only the first part - "Since X". Could you please write the same sentence in a different manner? Thanks :)
– Dor
Mar 3, 2015 at 14:21