# Gyroscope precession torque

We're probably all familiar with the demonstration of gyroscopic precession where a man sits on a swivel chair holding a bicycle wheel (like this: https://www.youtube.com/watch?v=1fWl2LNpcak).

And the rate of precession is given by

$$\omega_{pr}=\frac{\tau}{L_s}$$

1. Suppose that we don't neglect the mass of the man on the chair, and instead, he had a relatively large mass (but still consider a friction-less swivel). What impact would this have on:

• the final rate of precession
• the rate of acceleration of the mass of the man.
2. What impact does the mass of the man have on the torque required to turn the wheel?

3. I expect that the angle the wheel is turned would have maximum effect (and therefore require maximum torque) at $90$ degrees to the vertical - beyond this the effect (and torque) would be reduced.

4. The above formula suggests that increasing the torque ($\tau$), or decreasing the angular momentum ($L_s$) will increase the rate of precession. Except that I expect there are limits (critical points) at which maximum torque will have no effect, or decreasing angular momentum (reduced spin velocity) will have no effect. - How do we consider these critical points mathematically?

• Welcome to Physics SE! Please use mathjax to format mathematical expressions. To learn more about mathjax, please read MathJax basic tutorial and quick reference. – Yashas Feb 24 '17 at 6:39
• Oh cool, I didn't know about that. Thanks – user33668 Feb 24 '17 at 6:58

## 1 Answer

This is not a rigorous explanation but hopefully shows how the questions can be answered in general terms.

If the bicycle wheel axle is horizontal then the spin angular momentum of the wheel is horizontal $\vec L_{\rm s}$ (cyan in the top diagram). The Professor now applies an anticlockwise torque to the axle of the wheel which causes the spin angular momentum of the wheel to change direction $\vec L_{\rm s}'$ (red in the bottom diagram).
The spin angular momentum now has au upward component of angular momentum of the wheel $\vec L_{\rm p}$ (red in the bottom diagram).
As there are no vertical torques applied to the Professor-wheel system (the swivel is friction-free) and there was no vertical angular momentum before the Professor started to tilt the axis, the Professor and the wheel must rotate about vertical axis as shown in the bottom digram $\omega$ in yellow as there is an angular momentum $\vec L_{\rm p}$ (green in the diagram) because the total vertical angular momentum must stay zero.

Using these ideas you should be able to answer your own questions.

For example, when the inclination of the axle of the wheel is greater the speed of precession will increase.

The mass distribution of the Professor and the wheel about the vertical axis through the centre of the swivel will determine the moment of inertia of the Professor and the wheel (and the turn table) and you can use $L = I\omega$ and the idea that the total vertical angular momentum is zero with conservation of angular momentum about that vertical axis to find out how the mass distribution of the Professor and the wheel affect the rate of precession.

• Thanks for that answer. It goes a long way towards clearing up my understanding. Let's now say that I want to figure out how much torque ($\tau$) the Professor needs to apply in order to rotate himself and the wheel at a rate of $\omega_{pr}$. Is it as simple as taking the value $\vec L_{\rm s}'$ and substituting it into the equation $L=I\omega$ where $\omega = \tau / L_s$ then solving for $\tau$? And if so, what do i need to do to find the relationship between velocity and time? Do I just derive it w.r.t. time? – user33668 Feb 26 '17 at 12:34