# Conservation of Angular Momentum, as related to a flywheel

Trying to work out some pesky flywheel dynamics for a project I'm working on, would love some for your assistance to better understand the underlying concepts.

For a given flywheel (thin-walled cylinder, assume a spoked bicycle wheel) rotating in the x-y plane, I'm trying to calculate the force generated in either direction along the z-axis.

It seems to me, in line with Newton's first law of motion extended to rotational dynamics, what forces are physically being generated that prevent a rolling wheel from falling over?

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Why will the flywheel fall over? Isn't it connected to a stick? And also, what is the direction of the gravitational field? – udiboy1209 Aug 5 '13 at 4:01
Hint: Ever heard of Euler's laws of rotational motion? – ja72 Oct 4 '13 at 2:34
Possible duplicate: physics.stackexchange.com/q/506/2451 – Qmechanic Oct 4 '13 at 6:49
– Brandon Enright Dec 3 '13 at 6:39
You'll need to give more details here. Does the wheel touch the ground? Why would it fall over? (Quick answer to that one, if wheel is on an horizontal axle: weight). At least provide a sketch. – André Neves Jul 5 '14 at 7:23

Conceptually, the reason the wheel does not fall over is because in order to change the direction of the angular momentum vector you require energy (more specifically, a torque), so the spinning wheel just wants to stay in the x-y plane.

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All you need is the following:

1. Sum of forces equals mass times acceleration at the center of gravity $$\sum \vec{F} = m \vec{a}_{cog}$$
2. Sum of torques at the center of gravity equals the rotated mass moment of inertia matrix times the angular acceleration plus the cross momentum terms $$\sum \vec{M}_{cog} = I_{cog} \vec{\alpha} + \vec{\omega} \times I_{cog} \vec{\omega}$$

where $\times$ is the vector cross product and the rotated inertia is $I_{cog} = E\,I_{body}\,E^\top$ derived from the 3x3 rotation matrix $E$ and the body aligned inertia matrix typically $$I_{body} = \begin{pmatrix} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \end{pmatrix}$$ derived from the principal moments of inertia. See here and here.

This is literally rocket science, since a lot of the problems in rockets is the dynamic stability as a result of the 3D motions.

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I'm not sure if this is your question, but, more than forces, I think it's stability that prevents a rolling wheel from falling over. To elaborate on this, I will say that I cannot argue the role friction plays in making sure the wheel doesn't just slip and fall, but I would say that a lot of it has to do with the center of gravity. Remember that, for an object to be stabe and not fall, the center of gravity must be perpendicularly above the bottom of the object. By this, I mean the following. Say you have an object that is on the floor. Now, at one moment in time, you mark the center of gravity of the oject. Now, you drop a perpendicular from the center of gravity to the floor. Is this in a region where part of the object lies or is out "out of range" of the object? If the former, it will not tip. If the latter, boom! Torques also play a role in this. If you have a torque, that's what'll move the center of gravity. But, at a given moment of time, it'll only fall if the center of gravity is not above the bottom of the object.

Hope this helped!

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Stability is a misleading term here. I would say that a big flywheel has a tremendously high moment of inertia (which we can easily calculate as $mr^2$. The high moment of inertia means that torque applied on the flywheel will produce a very small change in angular momentum, which alludes to the flywheel being very stable about its axis of rotation. – shortstheory Nov 3 '13 at 3:10