The spolling action of Euler's disk is similar to what happens when you spin a coin on a flat surface, but it lasts much longer!

Click on the link below to see a video of Euler's disk in action.

http://www.youtube.com/watch?v=nEF7fBvX_4Y (opens in new window)

It's very interesting how the disk spins faster and faster as time goes on. The reason for this will be investigated.

To analyze the physics of Euler's disk let's start off by drawing a schematic (shown below) which illustrates the disk spolling on a flat surface.

Where:

To analyze the physics of Euler's disk as simply as possible, we shall treat

Further simplifying assumptions include:

• The disk rolls on the flat surface without slipping

• The disk is thin relative to its radius

• The disk rolls such that

With simplifying assumptions in mind, let's now redraw the schematic. The disk is treated as "thin" such that the thickness can be neglected.

Where:

A global

A local

Lastly, we shall define

The angular velocity of the disk, with respect to ground, is

The angular acceleration of the disk, with respect to ground, is

Looking at the first term:

Looking at the second term:

Therefore, for the disk

Note that the terms

For the general case of slipping between disk and surface, the velocity of point

where

For rolling without slipping,

and

Now, for

This expression conveniently relates

By Newton’s Second Law:

Where:

If

Next, apply the Euler equations of motion for a rigid body, given that

We have,

Now,

This is the angular velocity of the disk (with respect to ground) resolved along the local

Furthermore,

This is the angular acceleration of the disk (with respect to ground) resolved along the local

Thus, the second and third equations (of Euler's equations) are equal to zero. Therefore Σ

Therefore, we only need to consider the first equation:

Where:

Σ

By symmetry (treat the disk as a thin circular disk),

and

Now,

This is the sum of the moments in the

Substitute this into equation (2) and we get

Now, substitute

Thus,

This informative result tells us that as

However, the above formula for

As Euler's disk loses energy due to friction and rolling resistance losses, it also loses gravitational potential energy by an equal amount. This means that the center of mass

Euler's disk works best when friction and rolling resistance is minimized. This is achieved on hard, low friction surfaces, such as glass.

Although the model here neglects friction, Euler's disk cannot be modeled as a conservation of angular momentum problem. This is because there is a net moment (torque) acting on the disk about its center of mass

In closing, it is worth mentioning that the physics of Euler's disk is similar to the physics of a gyroscope. Their analysis is in fact very similar.

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