What motor/payload mass ratio -> payload spins the motor

[Dukon]

Given an unconstrained (suspended without contact to anything) electric motor with its own power in total has mass m1

Given the motor shaft is connected to a payload of mass m2

If m1>m2 then: m1 no-spin + m2 spin

If m2>m1 then: m1 spin + m2 no-spin

What happens at the unit ratio (m1/m2 = 1 or m1=m2)?

Aside from the unit ratio are there any other non-trivial ratios which determine which mass spins the other?

 

[Merlin3189]

Given an unconstrained (suspended without contact to anything) electric motor with its own power in total has mass m1

Given the motor shaft is connected to a payload of mass m2

If m1>m2 then: m1 no-spin + m2 spin No

If m2>m1 then: m1 spin + m2 no-spin No

What happens at the unit ratio (m1/m2 = 1 or m1=m2)? Depends on mass distribution

Aside from the unit ratio are there any other non-trivial ratios which determine which mass spins the other? Think Newtons 3rd law perhaps?

I don't know how you can suspend this, but perhaps it's floating around in space, maybe an astronaut using his battery powered screwdriver or something.

Anyhow, your basic idea is wrong. Both objects spin. Whatever their relative size and mass.
If one is 'bigger' it rotates slower than the 'smaller' one.

Conservation of angular momentum is the key. So 'bigger' and 'smaller' refer to the moment of inertia. That depends on both mass and the way the mass is distributed

If you compare it to the non-rotational situation, think about two masses pushing linearly on each other - eg. a compressed spring between them. Which one moves and which one stays still?

Edit: For this last eg let's ignore relativity and assume we are watching from a fixed reference frame in which they start off stationary.

 

[Dukon]

Yes this is exactly a Newton's Third Law investigation.

The Problem

I want to know given the exact masses of $m_1$ and $m_2$ can I calculate the exact Newton's Third Law force operating on both masses so I can then select $m_2$ to exceed this so $m_2$ itself acts as a constraint (exceeding Newton's Third Law force) needed to hold $m_2$ so $m_1$ spins. In detail I need to exceed the Newton's Third Law force so I must calculate it exactly in terms of $m_1$ and $m_2$.

Is the calculation below getting me there?

The Preliminaries

Think outer space lightyears away from any large masses so there is identically zero gravity
Consider both $m_1$ and $m_2$ as uniformly distributed masses shaped as cylinders

Hypothesis: the split of motor "spin" between the two is

$M \equiv m_1 + m_2$
$f_1 \equiv m_1/M$ is the fraction of total motor "spin" $m_1$ receives
$f_2 \equiv m_2/M$ is the fraction of total motor "spin" $m_2$ receives

So for a ratio $\rho \equiv m2/m1 = 2$ then $f_1=\frac{1}{3}$ and $f_2=\frac{2}{3}$

Results for many values of $\rho$ would be

\begin{matrix}
\rho & f_1 & f_2 \\ \hline
1 & \frac{1}{2} & \frac{1}{2} \\
2 & \frac{1}{3} & \frac{2}{3} \\
3 & \frac{1}{4} & \frac{3}{4} \\
4 & \frac{1}{5} & \frac{4}{5} \\
\end{matrix}

And so on

So now as $m_1$ is fixed and $m_2$ increases, then $\rho$ increases, and $m_2$ receives a larger and larger fraction of the motor "spin" according to this tabulated ratio analysis (while $m_1$ receives a smaller and smaller fraction of the motor spin force).

The Concluding Questions

The Newton's Third Law question is can $m_2$ be selected big enough to exceed the Newton's Third Law force trying to spin it oppositely of the spin of $m_1$. Therefore, is there an $m_2$ large enough that by itself it will remain at rest while only $m_1$ spins even without any constraint actively constraining $m_2$ from moving, other than its own mass (inertia)?

Seems to me if $m_2$ is "larger the maximum motor force" (the motor has its own torque curve and cannot provide any acceleration or even constant speed above its characteristic limit) then the $f_2$ fraction of that motor force determines the Newton's Third Law force trying to move $m_2$. If I know the maximum motor force, can I not just pick $m_2$ to just slightly exceed it and thereby eliminate the Newton's Third Law force from operating on $m_2$?

Further Formalism of Newton's Third Law

If a constraint now holds either $m_1$ or $m_2$ but not both (don't worry about how this is realized in practice) then the constraint would have to be stronger than $f_i$ times the motor force $F$ trying to spin its shaft. So

\begin{align}
F_1 = m_1 a \nonumber \\
F_2 = m_2 a \nonumber
\end{align}

where $a$ could be specified in angles per second squared or

$\omega_\phi = \frac{\phi}{t}$
$a = \frac{\omega_\phi}{t} = \frac{\phi}{t^2}$

So now the motor force on

$m_1$ would be $f_1 F_1$
$m_2$ would be $f_2 F_2$

where $a$ would be the same for both but split by $f_i$ accordingly for each.

Introducing Newton's Third Law $F_3^i$ for the $i^{th}$ mass explicitly then derives:

$F_3^1 = - f_1 F_1$
$F_3^2 = - f_2 F_2$

The constraint would then have to exceed either of these to actually hold one of the cylinders fixed enabling the other to continue to spin

Further Questions

1) Can this be reformulated in terms of Moments of Inertia?

2) Is this ratio analysis above also wrong or is it correct and can be derived from the more accurate Moment of Inertia calculation?

3) how is that Moment of Inertia calculation started?

 

[Ibix]

Both masses must start to spin in opposite directions, otherwise the conservation of angular momentum is violated. Full stop (edit: or period. However you call that dot at the end of a sentence in your dialect of English). This applies even if one of the masses is the Earth and the other is a tiny strip of paper (or whatever extreme you wish to carry this to).

The speeds of rotation do vary - the thing that is conserved is the sum of $I_i\omega_i$, where $I_i$ is the moment of inertia of the $i$th body and $\omega_i$ is its angular velocity. Bodies that were initially at rest means that the total angular momentum must be zero, and hence for just two bodies $I_1\omega_1+I_2\omega_2=0$. Thus you can make the angular velocity of one body as small as you like by increasing its moment of inertia - but it can only be zero if the other body does not rotate either.

You can look up moments of inertia for common geometric solids of uniform density via Google easily enough.

 

[A.T.]

how is that Moment of Inertia calculation started?

https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum

 

[CWatters]

The Newton's Third Law question is can $m_2$ be selected big enough to exceed the Newton's Third Law force...

You are asking if a mass (m2) can be bigger than a force (or torque). They are different things. It's like asking if a colour can be smaller than a mouse.

 

[Merlin3189]

Wow! You have put a lot of effort into this. I hope it eventually brings you to a better understanding.

You can be quite sure you will not find any way around Newton's 3rd law, conservation of momentum and of angular momentum. This conservation of momentum principle is one of the most fundamental in Physics.

... can I calculate the exact Newton's Third Law force operating on both masses ...

Whether you can actually calculate the value of this force, depends on how you cause it, or on what measurements you can make. The force is exactly the same size and opposite direction on each body. That applies whether it is a linear push or a rotational torque.

so I can then select $m_2$ to exceed this so $m_2$ itself acts as a constraint (exceeding Newton's Third Law force) needed to hold $m_2$ so $m_1$ spins. In detail I need to exceed the Newton's Third Law force so I must calculate it exactly in terms of $m_1$ and $m_2$.

No way! You will never break Newtons 3rd law.

Hypothesis: the split of motor "spin" between the two is

$M \equiv m_1 + m_2$
$f_1 \equiv m_1/M$ is the fraction of total motor "spin" $m_1$ receives
$f_2 \equiv m_2/M$ is the fraction of total motor "spin" $m_2$ receives

So for a ratio $\rho \equiv m2/m1 = 2$ then $f_1=\frac{1}{3}$ and $f_2=\frac{2}{3}$

Good intuition, but not quite right.
Both objects acquire equal but opposite momentum (whether angular or linear.) So if "spin" means angular momentum, then they both get equal amounts irrespective of their mass.
BUT if "spin" means angular speed, then it is shared according to their moments of inertia and hence according to their masses, because the moment of inertia is proportional to mass. In that case, you are near but still no coconut, because it is the opposite way round. The smaller mass object must rotate faster and the larger mass rotate object slower, so that they both have the same size of angular momentum.

$| angular\ momentum_1 | = ω_1 I_1 = ω_1 k m_1$
$| angular\ momentum_2 | = ω_2 I_2 = ω_2 k m_2$

So $ω_1 k m_1 = ω_2 k m_2 \ \ or \ ω_1 m_1 = ω_2 m_2$

Then $\frac {ω_1} {ω_1 + ω_2} = \frac {ω_1} {ω_1 + w_1 \frac{m_1} {m_2}} = \frac {1} {1 + \frac{m_1} {m_2} }= \frac {m_2}{m_2 + m_1}$

BTW - Anyone - How do I stop Latex making the writing so small when the expressions get complicated?

So now as $m_1$ is fixed and $m_2$ increases, then $\rho$ increases, and $m_2$ receives a larger and larger fraction of the motor "spin" according to this tabulated ratio analysis (while $m_1$ receives a smaller and smaller fraction of the motor spin force).

Well, the opposite way round. Notice, one gets a smaller and smaller fraction, but never zero.

I hope that's a bit helpful. I'm giving up on the rest of your detailed work for now. You can find more info on the web about angular momentum, but please don't try to fight conservation of momentum and Newton's laws. That's just a waste of effort.

 

[CWatters]

+1

Notice, one gets a smaller and smaller fraction, but never zero.

Which, for example, means that if you spin a child's roundabout you really do cause the Earth to rotate in the opposite direction a very small amount. Of course that effect is reversed when you stop the roundabout or allow friction to stop it.

 

[Dukon]

BTW - Anyone - How do I stop Latex making the writing so small when the expressions get complicated?

Just use 1/2 instead of \frac{1}{2} in your denominators. Not the best solution but perhaps an acceptable temporary workaround...

 

[Ibix]

Then $\frac {ω_1} {ω_1 + ω_2} = \frac {ω_1} {ω_1 + w_1 \frac{m_1} {m_2}} = \frac {1} {1 + \frac{m_1} {m_2} }= \frac {m_2}{m_2 + m_1}$

BTW - Anyone - How do I stop Latex making the writing so small when the expressions get complicated?

The font size commands listed at https://texblog.org/2012/08/29/changing-the-font-size-in-latex/ seem to work. Using \Huge:
$\Huge{\frac {1} {1 + \frac{m_1} {m_2} }}$

 

[Ibix]

Also using the paragraph maths style (delimited by pairs of $ signs) instead of inline (delimited by pairs of # signs) helps.
Using #:
$\frac {ω_1} {ω_1 + ω_2} = \frac {ω_1} {ω_1 + w_1 \frac{m_1} {m_2}} = \frac {1} {1 + \frac{m_1} {m_2} }= \frac {m_2}{m_2 + m_1}$
Using $:
$$\frac {ω_1} {ω_1 + ω_2} = \frac {ω_1} {ω_1 + w_1 \frac{m_1} {m_2}} = \frac {1} {1 + \frac{m_1} {m_2} }= \frac {m_2}{m_2 + m_1}$$
This is probably the right thing to do. The inline mode is designed to be used for things like "$m_1$ is the mass", and tries to control the vertical extent to avoid mucking up line spacing in paragraphs. The paragraph mode just lays out the maths.

 

[Merlin3189]

Thanks both for the advice.
Mainly I think I need to practice using it more, so I remember details like these.