1
$\begingroup$

hero's engine

This is a simple hero's engine. The source of the gas and the rotating part is seperated by a frictionless bearing. So only the upper part rotates. My question is this:

Given ideal conditions ( no friction / air drag, unlimited gas ) is there an upper bound for the rotational speed of the upper part (in newtonian mechanics)?

I have done some calculations and it seems there indeed is an upper bound. Assuming the gas is released at a constant speed $v$ wrt the edge of the pipe the released gas produces a torque $\frac{\Delta m v r} {\Delta t} $ where $∆m$ is the mass of the released gas in small time $\Delta t$ and $r$ is the radius of the upper part. If the edge of the upper part was rotating at speed $V$ at that time wrt ground then to make the gas left inside the pipe travel to the edge (filing the void left by the released gas) $∆m V r$ amount of angular momentum is needed. Angular momentum is conserved here.

So,

-Change in angular momentum of the released gas = Change in angular momentum of the gas left inside the tube in the upper part + Change in angular momentum of the upper part (the metal part)

$$\implies\Delta m v r = \Delta m V r + I \Delta \omega$$ where I is the moment of inertial of the upper part (the metal part only)

$$ \implies\frac{dm}{dt}r v = \frac{dm}{dt} r V + I \frac{\frac{dV}{dt}}{r} $$

$$ \implies\frac{dV}{dt} = C (v - V) $$

here $C$ is a constant This means that $V$ eventually settles at $v$

Are my calculations right ?

Improve this question
$\endgroup$
3
  • $\begingroup$ I think you’re right. If the reservoir of gas were at the end of the arm, the speed would not be limited to v. But as it is, the input of gas from the center constitutes a torque which counteracts the one you get from spraying the gas tangentially. $\endgroup$ – Ben51 Feb 10 at 18:19
  • $\begingroup$ If the steam left the pipe at a higher velocity than it has traveling outward, you'd get a higher top speed, wouldn't your? And you could accomplish that by having a narrower nozzle toward the exhaust, right? $\endgroup$ – Kristoffer Sjöö Feb 10 at 21:37
  • $\begingroup$ I don't think the velocity travelling outwards is relevant, as the total change in angular momentum per unit mass during the trip outwards will still be Vr (regardless of both how long it takes for the gas to travel to the orifice and what path it takes to get there). A smaller nozzle will increase v, so the final velocity would be higher but still limited. A smaller nozzle may also reduce the mass flow, so constant C might be smaller (assuming the same internal gas pressure) - thus reducing acceleration to that final limited velocity. $\endgroup$ – Penguino Feb 10 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.