# 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 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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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