Understanding the Concept of a Gyroscope: How Rotation Defies Gravity

[Felix83]

Consider the concept of a gyroscope. You tie a string to one end and attach the other end to the ceiling so only one end is supported, If you let go when it is not spinning, the obvious occurs - it falls from its horizontal position to a vertical position with the other end pointing towards the ground. However, if you spin the wheel fast enough, start it in a horizontal position and let go, it doesn't fall. It remains in the horizontal position and slowly rotates around the axis of the string. This is explained in most physics books by stating that when an object is rotated fast enough, any force applied to a point on the wheel tends to act at a 90 degree angle - a force down on a point on the bottom of the wheel when spinning (gravity) will act horizontally. In the case of gravity, the force causes it to rotate slowly about the axis of the string.

If there is no string and the spinning gyroscope is simply dropped, why does this same concept not cause the gyroscope to defy gravity and instead of fall down, 'fall' horizontally?

If one end is attached to the string and it is rotating around the axis of the string, if the force is constant (gravitational force is constant), why does the wheel spin at a constant speed rather than accelerate?

 

[Andrew Mason]

It remains in the horizontal position and slowly rotates around the axis of the string. This is explained in most physics books by stating that when an object is rotated fast enough, any force applied to a point on the wheel tends to act at a 90 degree angle - a force down on a point on the bottom of the wheel when spinning (gravity) will act horizontally.

I doubt that there are any real physics text that would explain it that way without mentioning the conservation of angular momentum. The angular momentum of the spinning gyroscope is in the direction of its axis (ie. horizontal).

Because one end of the gyroscope is fixed by the string, the force of gravity on the centre of mass of the gyroscope creates a torque about the fixed end and is adding angular momentum to the system. The torque is:$\vec\tau = m\vec{g}\times\vec{r}$ where $\vec{r}$ is the vector representing the displacement of the centre of mass of the gyroscope from the fixed end.

This causes the gyroscope to move in a horizontal circle about the string (see explanation below). It seems counter-intuitive, and it is, but it is quite understandable in terms of conservation of angular momentum.

If there is no string and the spinning gyroscope is simply dropped, why does this same concept not cause the gyroscope to defy gravity and instead of fall down, 'fall' horizontally?

When you remove the pivot point, gravity does not exert a torque and so cannot create a change in angular momentum. It just accelerates the whole gyroscope downward. So the gyroscope does not rotate horizontally at all. It just falls.

If one end is attached to the string and it is rotating around the axis of the string, if the force is constant (gravitational force is constant), why does the wheel spin at a constant speed rather than accelerate?

The system is accelerating. The system is experiencing a constant change in the direction of the angular momentum as the gyroscope rotates, so there is a constant angular acceleration. However, the change is always perpendicular to the direction of the angular momentum of the gyroscope. This is analogous to a body moving in a circle - it is accelerating (centripetal acceleration) but its acceleration is always perpendicular to its direction of motion. So while the gyroscope is rotating about the string at constant angular speed, it is undergoing constant angular acceleration (in a direction perpendicular to its angular momentum vector).

AM

Explanation of the horizontal circle prescribed by the gyroscope:
In small interval of time $\Delta t$, the torque $\vec\tau = m\vec{g}\times\vec{r}$ creates a change in the angular momentum of the gyroscope system, $\Delta \vec L = \vec\tau\Delta t = m\vec{g}\times\vec{r}\Delta t$.

The direction of this change is perpendicular to the angular momentum of the gyroscope. Assuming the gyroscope angular momentum is pointing to the right (at 3 o'clock) and the free end of the axis is to the right from your position, the change in momentum is directly away from you (ie. at 12 o'clock).

This means that the resulting angular momentum of the gyroscope system is now pointing right and a tiny bit forward of you (ie. a little before 3 o'clock, say 2:59) so the gyroscope has rotated a tiny amount counterclockwise around the axis made by the string. This changes the direction of the torque $\vec\tau$ so the change in angular momentum in the next small interval $\Delta t$ is perpendicular to the new angular momentum vector and the gyroscope rotates a little more. This keeps repeating. The result is that the gyroscope prescribes circular motion about the string.

Gyroscopic precession also causes a boomerang to move in a circle (except that the torque is not gravity but a rotational torque caused by greater lift on the uppermost arm).

AM