# Gyroscope design: does extra mass towards the axis of rotation help or harm the gyroscopic effect?

I am a hobbyist metalworker and I'm working on some gyroscopes for family for X-mas, and realized that there was something I did not understand completely with regard to gyroscopes, and while there is lots of information online, I'm struggling to find an answer to the following.

I generally get the sense that having absolutely as much mass as possible towards the outside edge of our spinning disk, with minimal mass towards the inside (center of the disk) is ideal, based on every gyroscope toy I've ever seen having some type of central mass reduction in place.

Why wouldn't a solid disk be better than say one with a thin center area(, often seen with cutouts like on the classic gyroscope toys)? Said another way, does having as much mass towards the outside, with extra mass towards the inside/center help or harm the gyroscopic effect?

Assuming you can get both to the same RPM having the extra mass would make for a better gyroscope, wouldn't it?

And if that is the case, then why are soo many gyroscopes thinner towards the center?

Goodness might well be equated to how long the gyroscope keeps going?

This will depend on the initial rotational kinetic energy of the rotating wheel $\frac 12 I \omega^2$ where $\omega$ is the angular speed of the rotating wheel and $I$ its moment of inertial about an axis through the centre of mass of the rotating wheel and at right angles to the plane if the wheel.

Individual point masses, mass $m$ and distance $r$ from the axis of rotation make a contribution of $m r^2$ to the moment of inertia.

For a given angular speed the larger is the moment of inertia then the larger the rotational kinetic energy.

The formula for the moment of inertia of a point mass $mr^2$ tells you that for a given mass to have a large moment of inertial requires $r$ to be as large as possible.
That is the mass needs to be concentrated near the periphery of the wheel.

Putting mass closer to the axis of rotation produces less of an increase in moment of inertia than having the mass further away from the axis of rotation.

• Good point on the goodness item, I never thought about what the end goal is before you pointed it out, I suppose holding momentum longer would be the goal, yes, longer spin time. | So that totally makes sense to me, more mass, or mass further from the central axis are both positive, so lets say we have designed with as much mass towards the periphery of the largest circumference we can manage, at that point, does extra mass closer to the axis of rotation help or harm? - I think help, based on mr2 still working for smaller r values, right? – Joshua West Nov 10 '16 at 23:05
• For a given mass have the largest distance from the centre to get the biggest moment of inertia. – Farcher Nov 10 '16 at 23:18

The question you need to ask yourself is "What is the same between two gyroscopes with different designs?" followed by "Knowing that, can I tell which is the most effective design?"

The answer to the first question (for toy gyroscopes pulled by a string) is "roughly the work done to the toy by the user and therefore the initial kinetic energy".

The answer to the second question depends on what you think make for good performance. When I do gyroscope demonstraction in class I'm looking for the maximum confounding of usual expectation which comes with maximum angular momentum.

OK. So let's assume that we We have some fixed input work $W$, and two gyroscopes with different moments of intertial $I_{1}$ < $I_{2}$. Which gives us the largest angular momentum.

The work done results in kinetic energy from which we can find the angular velocity $\omega$: \begin{align*} W &= \frac{1}{2}I \omega^2 \\ \omega &= \sqrt{\frac{2W}{I}} \;, \end{align*} then it is simple to compute the resulting angular momentum \begin{align*} L &= I \omega\\ &= I \sqrt{\frac{2W}{I}} \\ &= \sqrt{2 W I} \;. \end{align*} which makes it clear that the larges moment of inertia will give the largest angular momentum for fined work input.

And of course $I = \int_V r^2 \rho \mathrm{d}V$ is maximized by getting as much of your mass to large $r$ as possible.