# Why can you assume that the angular momentum vector of a top will always track its axis of rotation?

My favorite physics 101 textbook (Giancoli) explains precession in terms of a spinning top whose axis is tilted from the vertical.

The way the book sets things up, $L$ (angular momentum) points along the top's axis, and $\operatorname{d}\!L$ is perpendicular and horizontal to $L$, pointing in the direction of precession (in their figure, $\operatorname{d}\!L$ ends up pointing to the right). It then goes on to state that since $\operatorname{d}\!L$ is always perpendicular to $L$, $L$'s magnitude is constant, and only its direction changes. Finally, it concludes:

Since $L$ points along the axis of the top, we see that this axis moves to the right... (Giancoli, 2008)

I just don't understand why you can assume that it will always be true that $L$ and the axis of the top will always lie on top of one another.

The book doesn't address this, but is there only one way in which the top can move in order to produce the required motion of the angular momentum vector? If that is the case, then I guess it would make more sense? I dunno. Help me out here folks.

(I know that there is a deviation in $L$ due to the precessional motion itself, but another one of the book's assumptions is that this is irrelevant as long as the top is spinning quickly enough.)

-

Essentially it is because the top is a rigid body with a well defined symmetry axis.

Examine the definition of angular momentum:

$$\vec{L} = \int_V dV \left( \vec{r} \times \vec{v} \right) \rho(\vec{r})$$

Choosing a (instantaneously) cylindrical coordinate system with $\vec{z}$ coincident with the axis of the top, we see that $\vec{v}$ is always in the $\phi$ direction so the angular momentum due to the spinning always points in the azimuthal direction.

$$\vec{L} = \int_V dV r^2 \omega \left( \hat{r} \times \hat{\phi} \right) \rho(\vec{r}) = -\omega \int_V dV r^2 \rho(\vec{r}) \hat{\theta}$$

by taking advantage of the symmetry of the top, we find that only the $\hat{z}$ component contributes.

$$\vec{L} = -\omega \hat{z} \int_V dV r^2 \sin^2(\theta) \rho(\vec{r})$$

You have already identified the first correction, and as the book says if the top spins fast enough (and precesses slowly enough) we can neglect that.

-
I like this answer, but the math part makes it look as if you're using $\hat{r}\times\hat{\phi} = \hat{z}$ rather than $\hat{r} \times \hat{\phi} = -\hat{\theta}$. I know the text identifies that the simplification can be made due to symmetry, but I think this step could be confusing for a beginning student. – Mark Eichenlaub Apr 14 '11 at 21:57
@MArk: You are absolutely right. I was being lazy. – dmckee Apr 14 '11 at 22:11