Does rod in free space rotate around CoM if continuous force act at one end?

This site has lots of similar questions like mine, there is force just instant strike after that object moves freely, so my question is not like these.

Rocket engine produce continuous force at one end of rod in free space. Initial condition: rod is at rest, not rotate, not translate in relation to space station that is near him.(Neglect tiny gravitation due to mass of station and neglect rocket fuel mass loss)

Will rod when RPM increase enough end up in pure rotation around CoM or CoM will orbit aorund center of smallest circle?

Is this left trajectory of Euler spiral, path of rod CoM?

Fersnel integral

Like this?

• I suppose if pivot point/COM is not unmovable, then in addition to rotation, translation of COM to some degree must also exist, because part of force will be transferred to com. Commented Oct 11, 2023 at 17:57
• That is quite a nose there ... :-) Commented Oct 12, 2023 at 14:34
• @JohnAlexiou :) What is your opinion for this case? Commented Oct 12, 2023 at 14:55
• If the rotational speed approaches infinity, the rod will just perform pure rotation around the CoM, if that's what you're asking. Commented Oct 12, 2023 at 15:37
• Hi user628075. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. Commented Oct 15, 2023 at 3:18

The rod can be characterized by its position $$\vec r = (x,y)$$, which is the location of the center of mass (CoM), and its orientation $$\theta$$, which is the angle that the end of the rod with the rocket makes with the $$x$$ axis.

The torque provided by the rocket is a constant $$\tau=FL$$ where $$F=|\vec F|$$ is the magnitude of the thrust and $$L$$ is the distance from the CoM of the rod to the rocket. So we have $$\tau=I\alpha$$$$FL=I\ddot \theta$$ where $$I$$ is the moment of inertia of the rod about the CoM.

Then the force is given by $$\vec F = F (-\sin(\theta),\cos(\theta))$$. So we have $$\vec F = m a$$$$F(-\sin(\theta),\cos(\theta))=m \ddot{\vec r}$$

To get the trajectory of the CoM solve the above equations for $$\vec r$$. They don't have any easily described shape.

• Is it possible that parts of CoM trajectory is curve? Commented Oct 11, 2023 at 18:54
• Yes. It is guaranteed. It is a rather complicated curve.
– Dale
Commented Oct 11, 2023 at 20:53

Following Dale's excellent solution, I shall provide some additional insights based on numerical simulations.

Solving for the centre of mass $$\boldsymbol{r}_{\mathrm{COM}} = \left(x(t), y(t)\right)$$, we get the differential equations for $$\boldsymbol{r}(t)$$ and $$\theta(t)$$: $$\ddot{\boldsymbol{r}}(t) = \left( \ddot{x}(t), \ddot{y}(t) \right) = \frac{F}{M}\left(-\sin\theta(t), \;\cos\theta(t)\right),$$ $$\theta(t) = \beta t^2, \quad \beta = \frac{FL}{2I} \,.$$

There are no elementary solutions for $$x(t)$$ and $$y(t)$$, so we express them in integral form

$$x(t) = -\frac{F}{M}\int^t_0 \mathrm{d}\tau\int^\tau_0\mathrm{d}\sigma\cos\left(\beta\sigma^2\right)\,,$$ $$y(t) = \frac{F}{M}\int^t_0 \mathrm{d}\tau\int^\tau_0\mathrm{d}\sigma\sin\left(\beta\sigma^2\right)\,,$$

and numerically find the trajectory (we can wlog set $$F=M=1$$):

Surprisingly, the trajectory of the rod COM is not a spiral as the question suggests, but an oscillation that asymptotically approaches a straight line $$45^{\circ}$$ anticlockwise to the direction of the initial force. This derives from asymptotic expansions of the Fresnel integrals, since $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = -\frac{ C\left(\sqrt{\beta}t\right) }{ S\left(\sqrt{\beta}t\right) }\, ,$$ so $$\frac{\mathrm{d}y}{\mathrm{d}x} \sim - \frac{ \sqrt{\frac{\pi}{8}} \; \mathrm{sgn}\left(\sqrt{\beta}t\right) + O\left(t^{-1}\right) }{ \sqrt{\frac{\pi}{8}} \; \mathrm{sgn}\left(\sqrt{\beta}t\right) + O\left(t^{-1}\right) } \sim -1 \, .$$

In fact, as the following graph of various $$\beta$$ value shows, the asymptote seems to be given by $$y = -x + 1/(2\beta)$$, where $$1/(2\beta)$$ happens to be the distance between the COM and the instantaneous centre of rotation in units where $$F=M=1$$, but I don't have a proof of this fact yet.

• Excellent answer! Thanks for going to all of the effort for doing the numerical solution and plotting the results.
– Dale
Commented Oct 18, 2023 at 14:46
• @user628075 I would recommend accepting this answer.
– Dale
Commented Oct 18, 2023 at 14:49
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– Buzz
Commented Oct 22, 2023 at 4:51
• @jono94 Did you try calculate in case we have air drag? Commented May 15 at 10:25
• @22flower I don't think adding in drag would affect the motion too greatly, but I may as well be wrong about this. Besides, I'm afraid the new equation of motion seems quite complicated. Commented May 30 at 15:14

It will never do a pure rotation about the COM as long as there is a force applied, because this force imparts an impulse which will accelerate the center of mass.

The geometry of the problem is as such

When a force is applied a distance $$d$$ from the center of mass, the body will have an instantaneous rotational acceleration (Eulerian acceleration) about a point at a distance $$c$$ from the other side of the center of mass. The distance is

$$c = \frac{I_{\rm CM}}{m d} = \frac{ \kappa^2}{d} \tag{1}$$

where $$\kappa$$ is the radius of gyration of the body about the axis of rotation at the center of mass, and $$I_{\rm CM}$$ the mass moment of inertia at the center of mass.

The above relationship is purely geometrical and since the application point does not change, and the radius of gyration is assumed to be fixed, the resulting rotation point is also fixed relative to the center of mass and the orientation of the body.

Only when the distance $$d$$ approaches infinity, does the rotation center $$c$$ near the center of mass. On the other hand, if the force-distance nears zero, the rotation center goes to infinity and the body purely translates (in an accelerating fashion).

Edit The statement below is incorrect.

In deciphering the tracks of the CoM shown, place the pivot point at the center of the circles, and the CoM rotates about the pivot point at a radius $$c$$ as shown above.

I think what might be going on is when the tracks are near circular, the rod is actually tangent to the track and the force pointing inwards. The radius of orbit $$r$$ is not the same as the pivot distance $$c$$. The radius of orbit is such that the force is $$F \approx m \omega^2 r$$ (when near-circular orbits).

To derive (1) apply an impulse $$F {\rm d}t$$ at a distance $$d$$ from the center of mass. Use the equations of motion summed at the center of mass to get the response

\begin{aligned} F {\rm d}t & = m {\rm d}v \\ d\; F{\rm d}t & = I {\rm d}\omega \end{aligned}

and note that if the body is pivoted about a point located $$c$$ distance from the center of mass, then the velocity at the CoM is $${\rm d}v = c \, {\rm d}\omega$$

If you divide the top expression from the bottom expression you have

$$\frac{d \, F {\rm d}t}{F {\rm d}t} = \frac{ I {\rm d}\omega}{m c {\rm d}\omega}$$ which is solved for $$\boxed{c = \frac{I}{m \,d} = \frac{\kappa^2}{d} }$$ when $$I = m \kappa^2$$.

• The CoM track goes to a circle of radius $c$ when the body rotates about a fixed point in space. You can see from the spirals that the CoM is never fixed at a point in space and thus the body never rotates purely about the CoM. Commented Oct 12, 2023 at 15:52
• I see your point. In reality, what you need is a force perpendicular to velocity to curve and this is what happens. I don't know how the trial graphs were made and so I cannot comment further. Commented Oct 13, 2023 at 13:13
• The rod cannot be perpendicular to the track at all times. Only initially. The force vector is split into parts, one perpendicular to the track to curve it, and another part parallel to the track to change the speed. In general, the rod might be parallel to the track rather than perpendicular, unlike what you have shown. [More edits happened] Commented Oct 13, 2023 at 13:38
• @user628075 - yes, that looks more correct. Commented Oct 13, 2023 at 21:17
• @Jono94 - I edited the answer to show the derivation. Radius of gyration is just a geometric way to interpret the mass moment of inertia of an object. The definition is $I = m \kappa^2$. So no, the orbital radius is not equal to the radius of gyration. Those two are independent of each other. Commented Oct 14, 2023 at 18:16