Rocket engine, perpendicular to the end of stick, located in cosmos, without atmosphere and gravitation. As in picture below: enter image description here

In what direction will the stick move? In V or another direction?

  • $\begingroup$ You have to clarify if it is really a jet engine or a rocket. The difference is that a jet engine needs oxygen and without atmosphere it will not work. A rocket moves with the impulse given by the exhaust of the fuel being burnt. $\endgroup$ – anna v Apr 25 '11 at 17:51
  • $\begingroup$ It's a rocket engine, with a impulse physics. $\endgroup$ – Squ Apr 25 '11 at 17:53
  • $\begingroup$ Foe an exact answer one needs the mass of the "stick" and the mass of the engine. $\endgroup$ – Georg Apr 25 '11 at 19:07


The rocket is "magic" - it has no weight and the fuel it burns doesn't decrease the mass of the system. This can also be thought of as working in the limit where the time of burning a more realistic rocket is small enough that the mass of fuel burned is negligible compared to the mass of the stick.

The force supplied by the rocket is constant and always perpendicular to the orientation of the stick.


Initially, the stick will move in the direction of $\vec{V}$, but it will also start rotating, and as it does, its path will curve.

The stick spins faster and faster. Call a rotation of the stick the time during which $\theta$, the angle the stick makes with horizontal, starts at a multiple of $2\pi$ and increases to the next multiple of $2\pi$. In each rotation the stick spins a full circle, and the rocket points in all directions once. Thus, the forces at different times in the same rotation mostly cancel, and the stick doesn't simply blast away.

However, the stick spins a bit more slowly at the beginning of each rotation than at the end. At the beginning, the thrust is pointing up and to the right, so there is a bit more momentum transfer in those directions. The net result is that the stick picks up a little bit of momentum up and to the right with each spin.

As the rocket spins faster and faster, the percentage change in its spin rate during each spin goes down, and the time-averaged acceleration of the center of mass decreases.


The stick is subjected to a constant torque $Fl$, with $F$ the force exerted on the stick by the rocket and $l$ the half-length of the stick.

The moment of inertia of the stick about its center is $ml^2/3$, with $m$ the mass of the stick.

The angular acceleration of the stick is the torque divided by the moment of inertia, $Fl/(ml^2/3) = 3F/ml$. Integrating twice with respect to time, we get the angle the stick makes with horizontal

$$\theta(t) = \frac{3F}{2ml}t^2$$

If we choose units correctly we can set $3F/2ml = F = 1$.

The motion of the center of mass comes from $F = ma$. At any given moment, the force on the stick has components

$$F_x = \sin\theta = \sin(t^2)$$ $$F_y = \cos\theta = \cos(t^2)$$

Integrating these gives the motion of the center of mass.

The integrals for velocities are called Fresnel Integrals. They are mostly used in optics. The intriguing result is that a plot of the velocity as a function of time is the Cornu Spiral or Euler Spiral.

enter image description here

The picture is copied from Wikipedia. It's a little bit backwards because $x$ is usually taken to be the integral of $\cos t^2$, whereas we've taken it as the integral of $\sin t^2$. If you imagine rotating the picture 90 degrees clockwise, then mirror-flipping left and right, you get something that matches up with the original drawing.

The left half of the (transformed) picture shows the velocity integrating backwards into the past. The right half shows it integrating into the future. The velocity is asymptotic to

$$v_x = v_y = \frac{\sqrt{\pi/2}}{2} \approx .627$$

  • $\begingroup$ Ah Klotho, the spinstress (or spinner?) $\endgroup$ – Georg Apr 25 '11 at 20:17
  • 1
    $\begingroup$ spinster , the etymology of the old maid, always spinning for her dowry. $\endgroup$ – anna v Apr 26 '11 at 3:42
  • $\begingroup$ @Mark: Your solution actually ignores the principle of the rocket, based on the conservation of momentum. The mass of the fuel may be negligible compared to the mass of the stick but the momentum of the burned fuel must be equal to the momentum of the center of mass of the stick at each moment. $\endgroup$ – Martin Gales Apr 26 '11 at 5:49
  • $\begingroup$ @Martin You're right that the momenta must be equal. This answer simply assumes the fuel is ejected at "infinite" speed. It is working in the domain where the ejection speed of the fuel is very high compared to the speed of the stick. $\endgroup$ – Mark Eichenlaub Apr 26 '11 at 6:02
  • $\begingroup$ @Mark: Yes, but in that case it does not make sense to talk about trajectory of the center of mass and involve integrals. $\endgroup$ – Martin Gales Apr 26 '11 at 6:19

The obvious answer is: "In a different direction." Or, more correctly, the direction of vector V will change continuously."

The engine will force the stick to rotate, continuously gaining angular velocity. In addition, assuming the stick begins at rest, there will be a linear component to the AVERAGE OVERALL velocity of the center of mass of the contraption. But it will not be in the direction of vector 'V' as you have it drawn. Somewhere slightly clockwise of 'V' for most reasonable choices of (impulse/mass) and (impulse/rotational_inertia)

Assuming duration of thrust is long enough, vector 'V' will eventually point in EVERY direction in the 2-D plane of the page in which it is drawn, multiple times. Any more significant detail than that requires more information on the mass, center of mass, moment of internia, initial angular velocity, thrust magnitude, and thrust duration.


Without atmosphere you have the simple case of no movement due to no atmosphere for the jet action to work upon, and the not so simple case of the stick rotating in a plane perpendicular to v due to conservation of momentum when its turbine blades rotate.

  • $\begingroup$ This is the jet rocket engine, without blades or something mechanical else $\endgroup$ – Squ Apr 25 '11 at 17:42
  • $\begingroup$ well by common definition there is a jet engine or a rocket engine, perhaps editing your question would be advisable $\endgroup$ – Nic Apr 25 '11 at 17:45
  • $\begingroup$ Sorry. Rocket engine. Fixed. $\endgroup$ – Squ Apr 25 '11 at 17:48

If the force of the engine is constant and equal to $F$, while the length of the stick is $L$, and it's mass moment of inertia $I=m\,L^2/12$ then there is a constant torque applied resulting in a constant rotational acceleration

$$ \ddot{\theta} = \mbox{-}\dfrac{F\,L}{2 I} $$

where $\theta$ is the orientation of the stick relative some arbitrary fixed axis. Positive $theta$ is counter-clockwise from horizontal in the picture.

$$ \theta(t) = \theta_0 + \omega_0\,t-\dfrac{F\,L}{4\,I}\,t^2 $$

while the center of gravity is accelerating linear in whatever direction the engine in point at

$$ \vec{a}_G = \begin{pmatrix} \mbox{-}\frac{F}{m}\sin\theta & \frac{F}{m}\cos\theta \end{pmatrix} $$

The integration of the acceleration vector over time escapes me, but it does look like it is related to Fresnel integrals.


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