# physics of a reaction wheel behind a gear chain

So as I understand it, due to angular momentum conservation, if you have two objects with inertia $I_1$ and $I_2$ and one of them applies a torque onto the other, they will start rotating in opposite directions.

Say you have a reaction wheel ($I_2$) mounted on a DC motor ($I_1$) hovering in space. If you apply a voltage to the motor, a current will flow through it and it will exert torque $T$ on the reaction wheel. Then the reaction wheel will start to accelerate with $a_2 = \frac{T}{I_2}$ and due to momentum conservation there will be an equal torque in the opposite direction on the DC motor which will cause it to accelerate with $a_1 = -\frac{T}{I_1}$.

Now my question is what happens if you put the reaction wheel behind a chain of gears. Say that there are a couple of them in a row (a gearbox), such that the overall reduction is $10$, i.e. $\frac{1}{10}$x RPM and $10$x $T$ on the reaction wheel.

I can't quite wrap my head around what will happen now. Obviously the torque (and therefore the acceleration) on the reaction wheel is 10 times greater now. But if you go back through the chain of sprockets, the torque applied by the DC motor on the motor shaft will still just be $T$. What does that mean for the magnitude of the opposing torque applied on the DC motor?

Is it $-T$ or is it $-10$x$T$?

I feel like the answer lies somewhere in the way the gearbox works. Obviously it is not just mounted in thin air, but is attached to the DC motor. But I can't seem to work out what that actually means for the torque/acceleration of the DC motor that applies torque $T$ on the motor shaft which will then go through the gearbox and result in a torque $10$x$T$ on a reaction wheel mounted on the output shaft of the gearbox.

Just like pulley systems which can be thought of as force multipliers so you can think of gears as torque multipliers.
A way of working out what happens in the ideal case is to say that work in (torque $\times$ angle) is equal to the work out.
So if $10$ revolutions at the input end produces $1$ revolution at the output end the torque increases by a factor of ten.

However that is not the end of the story because the gear wheels have themselves to be constrained to produce such a torque multiplication and in the isolated system that you are considering how do you do that?
What are the bearings attached to?

One also has to look at the motor itself without any reaction wheel attached.
When the motor is switched on what will happen?
The axle and armature will rotate one way and the rest of the motor will rotate the other way.
Joining the gear box to the rest of the motor will still result in such a rotation.

• Thanks for the input. I was thinking about sth like that: pololu.com/product/3036 . So the gearbox bearing the sprockets is attached to the DC motor. – hopfi Oct 14 '16 at 10:08
• I'm sorry, hit enter before I was done with the rest of the comment (and can't seem to edit?). If the system is set up like in the geared DC motor I linked to and the reduction rate is 10 and I mount the motor onto a plate. What would the torque be on that plate if by applying a voltage to that DC motor I exert a torque of 10T on the reaction wheel that is attached to the output shaft of the gearbox. – hopfi Oct 14 '16 at 10:13