Direction of torque precession of a spinning wheel

Question

Consider a spinning wheel, which is held up by one end of it's axis like this:

To explain why the change of angular momentum is directed as shown in the figure above, one usually says that there is an applied torque $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ is in this case the radius vector from the point where the string is attached to the center of mass of the spinning wheel and $\vec{F}$ is just the gravitational force which acts on the center of mass of the wheel. So it seems to be clear that it must point into the direction as shown in the figure.

However is it possible to see this without applying the Formula for the torque above. I am thinking of something like this:

By applying the gravitational force one would move the free end of the wheel axis down a bit for a very small time, so we get the beginning of a rotation which angular velocity vector points into the same direction as the $\Delta L$ on the picture. Thus we get an additional angular momentum $\Delta \vec{L}$ into this direction, which changes the direction of the entire angular momentum vector $\vec{L}$.

However I don't see why the last "thus" must be true, since in general the angular velocity doesn't have to point in the same direction as the angular momentum vector.

This is true for the rotation about the wheel axis since this is a principal axis of the inertia tensor (which is needed to justify that the axis of rotation must turn, when the angular momentum vector turns), but the (beginning/infinitesimal) rotation above doesn't seem to be around a principal axis.

So it would be great if someone could make the argumentation I tried more clear and more rigorous and clarify the problem described above.

Answer 1

There is a paper written by Svilen Kostov and Daniel Hammer titled 'It has to go down a little, in order to go around' that addresses precisely the question that you ask here.

The idea for the paper came from a discussion of gyroscope motion in the Feynman Lectures on Physics.

Kostov and Hammer discuss that as you say the gravitational force moves the free end of the wheel axis down a bit. They also created an experimental setup to demonstrate the dip. They found very good agreement between theory and experimental result.

Answer 2

You are correct to observe that there is an often unstated assumption in the standard setup of this problem. When given this problem you are supposed to assume that the off-major-axis components of angular velocity make a contribution to L which is negligible compared to the on-axis angular velocity. Obviously this is a good assumption if the top is rotating fast enough, but it isn't exactly true.

If you start with the initial condition, that the top's angular velocity is completely aligned with its major axis at the time of release, you should find that the top's major axis does not really rotate uniformly in a circle but rather there is a very small sinusoidal variation about the uniform circular motion.

Answer 3

What I see here on the picture is that the formula

$$\tau = I \alpha$$

is probably not applicable for this situation. This is because $\omega$ is constant, i.e. $\alpha = 0$!

This situation is analogous to circular rotation with constant velocity. There you have velocity $\vec{v}$, whose magnitude is constant, but its direction is constantly changing. Acceleration $\vec{a}$ is always perpendicular to velocity.

Here you have angular momentum $\vec{L}$, whose magnitude is constant, but its direction is constantly changing. Torque $\vec{\tau}$ is always perpendicular to velocity.

Answer 4

The resultant moment on the body is the moment in x direction due to weight, so the change in angular momentum of the body must be only in x direction to achieve that from rotation equations (Euler's equations) we find that the body must be in steady precession ( nutation =0 , precesion = constant , spinning = constant ) and the resultant moment should be in this case (in x direction ):

M = I * spinning * precession = Moment of the weight

so the body modifies its precession to be a specific value

Answer 5

Angular momentum of wheel due to spin is in radial direction of wheel's axis, $\vec L=L\hat e_r$

Torque produced due to downward pull of centre of mass and perpendicular to both radial and vertical direction of cylindrical coordinates,$$\vec\tau=\vec r\times \vec f_g=-mgr(\hat e_r\times \hat z)=\tau(\hat z\times \hat e_r)$$ Torque cause change in angular momentum of wheel, but as torque is perpendicular to angular momentum, so there is only change in direction of angular momentum and not its direction. So,$$\vec\tau=\dot{\vec L}=L\dot{\hat e_r}\Rightarrow \dot{\hat e_r}=\frac{\tau}{L}(\hat z\times \hat e_r)$$ Now,$\quad \dot{\hat e_r}=\vec \omega_p\times \hat e_r.\quad$ Therefore,$$\vec\omega_p=\frac{\tau}{L}\hat z=\omega_p\hat z$$ $\omega_p$ is the rate wheel's axis moves in circular motion in horizontal plane and direction of plane is upward or against gravity and called as precessional angular speed of wheel.

Also, at the contact end of the wheel's axis, there is downward torque $\vec\tau_p=-\tau\hat z$, in direction of gravity is produced due to precessional motion which is equal in magnitude of torque that changes angular momentum of wheel and keeps free end of wheel's axis upward against gravity. The precessional angular velocity is constant $\omega_p=\sqrt g$, both in magnitude and direction. So increasing or decreasing wheel's spin changes torque only.