Top Precession

Why? Spin = 1 / precession?

That may be true but it does not explain why. I answered correctly because of my memory observing tops as a child.

Can you just give a detailed explanation why it is inversely proportional to the spin speed of the top. One thing is angular momentum of the top about the vertical axis is conserved as tvertical=0.

In order to understand it intuitively I think that it is crucial to realize that Torque ( $\tau$ ) is a vector, which in the picture above is pointing out of the screen straight into our eyes, but in general a vector tangential to the circumference of precession. Remember that $\tau =r \times F$ where $\times$ is the cross product and $r$ and $F$ are vectors. From the relation between Angular Momentum ( $L$ ) and Torque ( $\tau$ ) we can also write down:

$\Delta L= \Delta t . \tau$

This means that a torque $\tau$ after an infinitesimal amount of time will change the existing angular momentum of the top by $\Delta L$ , and $\Delta L$ will also be a vector pointing in the same direction of the torque $\tau$ . This $\Delta L$ vector added up to the top's current angular momentum $L_{n}$ determines the next iteration angular momentum $L_{n+1}$ . The top's angular momentum is another vector pointing up or down (depending in what direction the top is spinning) straight in the direction of the top's axis of rotation. Therefore $\Delta L$ is $\perp$ to $L_{n}$ .

Adding the vector $\Delta L$ to the vector $L_{n}$ will slightly change the angle of the latter by $\Delta \Theta$ , causing the top to precess. If $L_{n}$ is very large to start with, the same $\Delta L$ will cause a smaller $\Delta \Theta$ between $L_{n}$ and $L_{n+1}$ compared to a smaller initial $L_{n}$ , which will be perceived as a smaller precession angular velocity $\omega_{prec} = \frac{ \Delta \Theta }{ \Delta t}$ .

In other words the bigger it is the top's angular momentum to start with, the harder it is for the same torque to change it (change its direction).

As the top's angular velocity decreases, its angular momentum also decreases since $L = I. \omega$ , therefore, even if the torque $\tau$ would remain the same (it doesn't, it actually increases), $\Delta L$ will, by simple vector addition, cause a bigger $\Delta \Theta$ in the same $\Delta t$ , i.e. the precession angular velocity increases as the top's angular velocity decreases.

I thought that the axis passing through a spinning top was the vertical axis.