# Why is only one component of spin angular momentum considered in the precession of a top?

This was how the orbital angular velocity/ angular velocity of precession was shown to be calculated in my textbook ($$\phi$$ is angle made by top with vertical)

$$dL=L_\text {spin}\sin\phi d\theta$$

But , $$\omega_ \text{precs} =\frac{d\theta}{dt}$$

$$\omega_ \text {precs}= \left|\frac{d\vec L}{dt}\right| \left[ \frac{1}{L_ \text {spin}\sin\phi} \right]$$

Now $$\frac{d\vec L}{dt}$$ is equal to torque from gravity $$=mgr \sin\phi$$

So, $$\omega_\text {precs}= \frac{mgr}{L_\text {spin}}$$

My doubt is : Why is the torque changing only the horizontal component of the angular momentum vector, if its also perpendicular to the vertical component (since its direction is out of the plane)?

What I am thinking is that the torque will pull down on the vertical component and slowly align it with the direction of the torque itself. While this is happening it will of course also be shifting around the horizontal component but remain always perpendicular to it. As the previously vertical component will also now be in the same direction as torque this means that the top is no longer vertical at all? And is simply going around circles on the ground? I don't know, I am really confused.

The torque is always horizontal (in the direction of precession). Since $$\vec{\tau}=d\vec{L}/dt$$, the vertical component of $$\vec{L}$$ does not change.
Here, the direction of the spin (figure) axis of the top is visualized for three different initial conditions. The nutation amplitude ($$\theta_1 - \theta_2$$) is approximately inversely proportional to the square of the spin angular velocity: the faster the top spins, the less it nutates. The nutation frequency is approximately proportional to the spin angular velocity. For a very fast top, nutation can be imperceptible. Since the figure axis closely tracks the angular momentum, your equation for the precession angular velocity is still valid.
• Why are we first measuring the change ($dL/dt$) and then equalling it to the torque? Shouldn't we instead be defining torque to be $mgrsin\phi$ and then calculating $dL/dt$ from that? – l1mbo Jun 18 at 8:29
• But what is the problem if it is perpendicular to the vertical component? If at any one point of time we take the horizontal surface on which the top is rotating to be the $x$ and $y$ plane, with the $y$ axis pointing into the plane of the paper, and the $z$ axis to be the vertical axis, the torque is acting in the $y$ direction, the horizontal component of angular momentum is in the $x$ direction and the vertical component in the $y$ direction. Each of them is mutually perpendicular so if torque is changing the horizontal component why won't it change the vertical one? – l1mbo Jun 18 at 8:35
• You are still calculating how $\vec{L}$ changes from the torque. $d\vec{L}/dt$ is the torque. You just use the observation that $\phi$ is approximately constant for a fast top (so the torque magnitude is constant), and that $\vec{L}$ precesses about the $z$-axis so that we can speak of precession in the first place. – Puk Jun 18 at 8:44
• Horizontal in this context means in the $xy$-plane. The torque is in the $y$ direction, so it only changes $L_y$. As the top axis moves accordingly, the angular momentum vector picks up a $y$-component, hence the torque picks up an $x$-component, and starts changing $L_x$. The point is $\tau$ remains horizontal. At no point in time does it point in the vertical direction, so $L_z$ cannot change. – Puk Jun 18 at 8:49