# Angular momentum and pinwheel rockets

I was a physics major a long time ago, but haven’t kept up with it. Recently, a problem’s been bugging me, and my old freshman textbook doesn’t seem to answer it.

How in the world is angular momentum supposed to be conserved in the case of a pinwheel rocket?

John W. Kennedy
“Compact is becoming contract; man only earns and pays.”
– Charles Williams

I think you need to elaborate a little. Angular momentum would be conserved just as it is on any spinning object once the rocket motors are spent. It will be turned into heat from hub and air friction and eventually stop.

I think you are asking “How can momentum be conserved when I light the fuse and the thing spins up?” - correct me if I’m wrong.

If I understand what you mean by this correctly, then it is the same question as “How is momentum conserved in my car when I step on the gas?”

Momentum is only conserved in the absence of external forces, or when you consider ALL of the participants in the motion. So, in the case of the car accelerating, it looks like it is not conserved, but really the earth is accelerating the other direction, conserving momentum.

Likewise, for the pinwheel, the exhaust shooting out of the rockets is going in the other direction, for a brief instant conserving angular momentum. The exhaust’s momentum is quickly lost due to friction with other gases in the air. If you imagine the pinwheel in a vacuum, the exhaust goes off to infinity, carrying momentum with it.

I understand about the linear momentum. But the rocket exhaust never has a rotational component.

John W. Kennedy
“Compact is becoming contract; man only earns and pays.”
– Charles Williams

Sure it does. Let me build up from basic blocks here:[ul][li]Mass(m) is a scalar.[]Velocity(v) is a vector.[/li]Thus, linear momentum (p = m * v) is a vector.[li]For purposes of calculating Angular Momentum(L) we need to define the center of rotation. This is a completely arbitrary definition, so we could choose Omaha Nebraska, but we’re going to choose the center of the pinwheel to simplify calculations. Let us then define a radius vector from the center of rotation to the object whose L we wish to calculate and call it r.[/li][li]L is r X p, a vector cross product. This has the property that when r & p are paralell, r X p is zero. When r & p are perpendicular, the magnitude of r X p is equal to |r||p|. The exact value of a cross product is a vector of magnitude |r|*|p|*sin <font face=“symbol”>q</font>, where <font face=“symbol”>q</font> is the angle between them. The direction of the resultant vector is perpendicular to both r and p, and the direction is set by convention to be defined by your right hand (point fingers in direction of r, curl fingers to point in direction of p, and stick your thumb straight out.)[/ul][/li]
Now that we have that, any object moving with respect to a point in space has L with respect to that point. Imagine a traffic light stuck in the middle of a road, and a car approaching it at a steady speed. Define r to be the vector from the traffic light to the car. At large distances, p is nearly parallel to r (r X p small compared to r & p). At small distances, |r| is very small and that exactly makes up for the fact that r is nearly perpendicular to p. If you do the math you’ll find that in the absence of external forces (friction, engine power, etc.) L is constant.

Back to the rocket pinwheel. Take a rocket at the edge of the pinwheel, assume a vacuum. The exhaust nozzle of the rocket is pointed perpendicular to r. Imagine a chunk of rocket exhaust that is ejected and now travels away from the pinwheel. Since its r vector grows larger, the fact that it is moving linearly is OK - it is carrying angular momentum and balances out the angular momentum that the rocket has acheived by accelerating itself around the pinwheel.

Did that help at all or am I just confusing people?

Yow! That’s almost as much fun as Special Relativity. I guess I can accept that (though part of me wants to stomp downstairs into my official Mad Scientist’s Lair grumbling, “Hmmppphhhh! Smoke and mirrors!”).

John W. Kennedy
“Compact is becoming contract; man only earns and pays.”
– Charles Williams