Reaction wheel: angular momentum conservation or action-reaction?

Question

Reaction wheel are commonly used in spacecraft to change its attitude: an onboard inertia wheel is accelerated or decelerated along an axis to make the spacecraft rotate around the same axis.

While books explain how reaction wheel works with angular momentum, I cannot convince myself that it is entirely true and an important point seems to be missing: the spacecraft and the reaction wheel are necessarily connected through the motor that drive the reaction wheel angular velocity. And for me it is the key point of how a reaction wheel works.

Consider rotations of the reaction wheel (of moment of inertia $I_{RW}$) and the spacecraft (of moment of inertia $I_S$) around the same axis. Without the motor, the reaction wheel and the spacecraft are not connected at all. If the wheel angular velocity $\omega_{RW}$ change because of an external torque $\Gamma$, the spacecraft angular velocity $\omega_S$ should not change as this torque does not apply on it. Consequently, the whole system angular momentum $L$ evolves but the spacecraft angular momentum remains unchanged, which is not how a reaction wheel works :

\begin{eqnarray} L &=& I_{RW}\omega_{RW} + I_{S}\omega_{S}\\ \dfrac{d(I_{RW}\omega_{RW})}{dt} &=& \Gamma\\ \dfrac{d(I_{S}\omega_{S})}{dt} &=& 0\\ \implies \dfrac{dL}{dt}&=& \Gamma \end{eqnarray}

Now, consider a motor linking the reaction wheel and the spacecraft. As the motor apply a torque $\Gamma$ on the reaction wheel, it also apply a torque $-\Gamma$ (of the same intensity and opposite direction) on the spacecraft through action-reaction principle (same as when I screw a screw). So now we have

\begin{eqnarray} \dfrac{d(I_{RW}\omega_{RW})}{dt} &=& \Gamma\\ \dfrac{d(I_{S}\omega_{S})}{dt} &=& -\Gamma\\ \implies \dfrac{dL}{dt}&=& 0 \implies L \text{ is constant} \end{eqnarray}

We get the same explanation as the books: reaction wheel principle is based on the fact that the total angular momentum is constant. We get the same conclusion, but not considering the motor is for me a shortcut.

So I have two questions :

• Is my reasoning correct ?
• If so: How does the action-reaction principle work with a DC motor ? Even though I read about it, I can't find any answer (it does not seem to be the the counter electromotive force).

Thank you by advance for your answers.

Answer 1

Let me put this in a wider perspective, discussing the action-reaction theorem for the case of linear mechanics (the reasoning transfers to rotational dynamics, of course.)

I propose the following two axioms:

• space is uniform
• $F = ma$ (Newton's second law)

Let there be two unequal masses $m_1$ and $m_2$ (To help thinking about them you can set them to, say, $m_1$ ten times the mass of $m_2$)

Let's say the two masses are spacecrafts, and $m_1$ is using some form of traction to reel in $m_2$. The acceleration of $m_2$ is described by Newton's second law: $F=ma$

But then, we have of course that $m_1$ is not immovable. By contrast: on Earth if you want to reel in an object you dig in your heels; you secure yourself to make yourself immovable with respect to the Earth. But in space nothing is immovable. Still, $m_1$ does have more leverage than $m_2$ simply because $m_1$ has more bulk. To have more bulk is to have more inertia.

The amount of leverage that you have available to reel in some other object is described by Newton's second law: $F=ma$

(Let me elaborate on the concept of 'available leverage'. Compare the case of hammering in a nail. If you would try to push in that nail slowly you need a very large force, far more force than you are physically capable of. So how is the hammer capable of exerting that amount of force? The hammer exerts its peak force during the change of velocity as it strikes the nail. That is, at the instant of the impact the hammer head does have the leverage to push the nail a bit deeper. That leverage is described by Newton's second law: $F=ma$)

Back to the spacecraft example:
The change of velocity of the other spacecraft is described by $F=ma$, and your own spacecraft's change of velocity is also described by $F=ma$. So the case is symmetric; you can reverse the perspective of the description and the logic remains the same.

For motion in space the following two assertions are logically equivalent:

• action and reaction are equal and opposite
• in space nothing is immovable; the only leverage you have is your own inertia

I want to point out specifically that it isn't necessary to consider the actual mechanism of reeling in; this reasoning is valid for any actual mechanism.

The above reasoning has the following implication:

$$m_1 \vec{a_1} = - m_2 \vec{a_2}$$

Which implies that at every point in time the following holds:

$$m_1 \vec{v_1} = - m_2 \vec{v_2}$$

So we have arrived at conservation of momentum.

Now to your question:
Momentum conservation or action-reaction?

Logically the point is moot. Momentum conservation and action-reaction follow logically from the two axioms that I started with.

Still, I have a preference for $F=ma$ as the axiom, and thinking of momentum conservation as a consequence of that.