# Angular momentum and torque in gyroscope

In my textbook (Kleppner), the principle of a gyrocompass is given to be

"A ﬂywheel free to rotate about two perpendicular axes tends to orient its spin axis parallel to the axis of rotation of the system."

While explaining the working, they do a step that I don't understand.

This is the first part of explanation which I understand. I get that (moment of inertia)*(angular acceleration) will make a contribution to the rate of change of angular momentum along AB.

Now this is the second part of their explanation. They explain that the
spin angular momentum
that is rotating with omega is also
trying to have a component in the total angular momentum along AB

This is where I get confused. In my mind, the rotating spin angular momentum can never have a component on AB. It will always stay perpendicular to AB and will not contribute in the change in total angular momentum along AB.

I think I am missing something here. All I know is that if
TORQUE ALONG A DIRECTION IS ZERO, ANGULAR MOMENTUM WILL NOT CHANGE ALONG THAT DIRECTION
WHAT I DON'T KNOW IS THAT WHAT WILL HAPPEN IF THE DIRECTION ITSELF IS MOVING.

I would highly appreciate answers that are not extremely advanced. I know RIGID BODY DYNAMICS till EULER'S EQUATIONS

It seems to me that in the schematic diagrams in the screenshots that you present something essential is missing.

The idea of a mechanical gyrocompass is that the spin axis of the gyroscope wheel eventually becomes aligned with the externally imposed rotation.

For a gyrocompass that externally imposed rotation is of course the Earth's rotation.

Let me introduce a naming scheme for the axes.

I define three axes:

• Roll axis - the gyroscope wheel spins around the roll axis.
• Pitch axis - motion of the red frame.
• Swivel axis - motion of the yellow frame.

The following youtube video, Gyrocompass, shows a demonstration on table top scale.

The gyroscope used in that video has friction in all the bearings.

By contrast: in the idealized case we have that all parts move without any frictions. Without any friction the spin axis of the gyroscope wheel would never become aligned wih the Earth's axis. Instead the externally imposed rotation (Earth rotation) would cause the spin axis of the gyroscope wheel to sweep out a cone, never becoming aligned with the Earth axis.

In the video the wheel spin axis does become aligned, thanks to the friction in the bearings. Due to that friction the cone that the spin axis of the gyroscope sweeps out shrinks, so that eventually the spin axis becomes aligned with the externally applied rotation.

The mechanical gyrocompass is a design that is obsolete now, superseded by instruments that perform the same function, but that internally operate with fiber optic rotation measurement or rotation measurement with MEMS technology

I can hardly find any quality information about gyrocompasses. The Encyclopedia Britannica article about Gyrocompass is the best I have encountered so far

On the website of the San Francisco Maritime National Park Association:
the Service manual for the Sperry Mark XIV, Mod. 1, 17-1400D Gyrocompass, which generously also contains an appendix with extended discussion of the fundamental principles of the Gyro-compass

The Sperry design includes finely tuned damping (involving mercury) so that when the gyrocompass is started from a zero spin state it settles on the geometric north in the shortest time possible.

Coming back to the treatment of gyrocompasses in the Kleppner textbook.

As far as I can tell the statements in the parts of the textbook that you copied are at odds with how gyrocompasses actually work.

That is, as far as I can tell the statements about gyrocompasses in that textbook are erroneous.

• There is one thing that I was previously unaware of. All of the two or three different gyrocompass designs that have been in wide use have in common that the suspension is such that the spin axis of the gyroscope wheel is at right angles to the local gravity. The design supports some pendulosity (which is an essential feature), but other than that: a right angle. That means that a gyrocompass will perform best at the Equator. At the equator the precession induced by the Earth's rotation will naturally turn the wheel spin axis into alignment with the Earth rotation axis. Commented Oct 13, 2019 at 4:50

In my mind, the rotating spin angular momentum can never have a component on AB. It will always stay perpendicular to AB and will not contribute in the change in total angular momentum along AB.

I think I am missing something here. All I know is that if TORQUE ALONG A DIRECTION IS ZERO, ANGULAR MOMENTUM WILL NOT CHANGE ALONG THAT DIRECTION WHAT I DON'T KNOW IS THAT WHAT WILL HAPPEN IF THE DIRECTION ITSELF IS MOVING.

You have the right doubt, and when in doubt, best to revert to the fundamentals to seek resolution. Let us recall the basic principles:

1. Rate of change of angular momentum in an inertial frame is equal to the torque of external forces (assuming torque of internal forces is zero).

2. A vector can be changed by changing the magnitude or the direction.

Keeping the above two principles in mind, we first choose the inertial frame to be the lab frame (in which the entire assembly is rotating about the vertical with angular speed $$\Omega$$). Let us station the origin of the lab frame at the center of the gyrocompass which is clearly stationary in the lab frame.

Next, although the direction $$AB$$ is changing in space, imagine a fixed moment in time $$\ t = t_{o}$$. At $$\ t=t_o$$, $$AB \$$ will point in a fixed direction in space. The torque equation (principle 1 above) then tells us that the torque in direction $$AB$$ at $$\ t = t_o$$ is equal to the instantaneous rate of change of angular momentum in the direction of $$AB$$. Mathematically, this means calculating $$\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$$ at $$t=t_o$$ and taking its projection along $$AB$$.

As already explained by Kleppner-Kolenkow, the component of torque on the system along axis $$AB$$ in the lab frame about our chosen origin is zero (they are assuming that the center of mass of the gyroscope is at its geometrical center and no friction on the axle $$AB$$). So, the only task at hand is to calculate $$\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$$ at $$t=t_o$$ along $$AB$$.

Now, to do the calculation for $$\big(\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}\big)_{t=t_o}$$, note that the "spin angular momentum" has a vertical as well as a horizontal component. But the horizontal component is precessing about the vertical with angular velocity $$\Omega$$ (because the entire assembly is rotating about the vertical)! This implies that the "direction" of the horizontal component of "spin angular momentum" in the lab frame is constantly changing. By principle 2 (stated above), this precession leads to a contribution in the expression for $$\big(\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}\big)_{t=t_o}$$.

I'd now recommend (re-)reading the previous section in the same chapter of this book (probably goes by the name "gyroscope precession"; also check out this for visualization). The essence of that section is that in case of a purely precessional motion $$-$$ imagine a vector, $$\vec{V}$$, of fixed length, spinning about a fixed axis with instantaneous angular velocity $$\vec{\omega} \ -$$ we have $$\frac{\mathrm{d}\vec{V}}{\mathrm{d}t} = \vec{\omega} \times \vec{V}$$

In this particular case, $$\vec{\omega} = \Omega \ \hat{k}$$, and $$\vec{V}$$ is the horizontal component of the "spin angular momentum" (because remember the entire assembly is spinning about the vertical and so the horizontal component of the "spin angular momentum" is precessing too). The only minor caveat here is that $$\vec{V}$$ might change in magnitude $$-$$ however, this contributes nothing in the direction $$AB$$ because ($$\vec{V}$$ is directed perpendicular to $$AB$$). Clearly then, the precessional contribution in the direction $$AB \$$ is given by $$\Omega L_s \sin \theta$$, and happens to be the only other contribution to $$\big(\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}\big)_{t=t_o}$$ along $$AB$$ apart from the usual $$I_{\perp}\ddot{\theta} \ -$$ and this is exactly what Kleppner-Kolenkow are claiming.

Thus, we have, $$\boxed{ \bigg(\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}\bigg)_{t=t_o}\cdot\vec{e}_{AB} = I_{\perp}\ddot{\theta} + \Omega L_s \sin \theta = 0 }$$ where $$\vec{e}_{AB}$$ is a unit vector in the direction $$AB$$.

While this heuristically proves the torque equation, I'd still suggest using Euler's equations or explicitly writing out the components of $$\vec{L}$$ in the lab frame and taking time derivatives in order to not miss other contributions in more complex setups.

Besides this, as explained by others, friction damps this (pendulum like) oscillatory motion in $$\theta$$, eventually aligning the axis of the gyroscope with the axis about which the platform is spinning ($$\theta = 0$$).

Note: This problem just illustrates the principle of a gyrocompass $$-$$ for an actual gyrocompass device the spinning platform is the earth.

Hope this helps.

The rotation of the turntable will exert a torque on the axle AB. That torque vector is directed upward, is transmitted to the axle of the wheel, and will cause the angular momentum vector of the wheel to swing upward and eventually stabilize in the vertical position. Your equations predict the rate at which this swing will occur. In swinging up, the wheel develops an angular momentum about the AB axle, (it vector directed along the axle). This momentum will carry it beyond the vertical position and lead to oscillations (perhaps damped by friction).

Although Cleonis’s answer provides a bunch of excellent detail, perhaps a physical model will help understand what’s going on:

• Gyroscopic precession in its usual form goes on forever.

• Friction that opposes the precession can/will provide a torque.

• That torque opposes the precession by tending to align the momentum axis, essentially reducing the size of the precessional circle

• Once that process runs its course, the gyroscope stabilizes with its axis aligned and can be used as a compass.

To understand those middle two steps, imagine you’re looking parallel to Earth’s axis. You’ll see the gyroscope tip rotating in a circle around that, let’s say counter-clockwise. Friction opposing that at every point is a torque toward you, pulling the gyroscope’s L vector to point toward you, hence make a smaller precession circle. That repeats until it’s aligned.

This is Example 8.11 Gyrocompass Motion from Daniel Kleppner and Robert Kolenkow's An Introduction to Mechanics, 2nd edition, page 307. This is a badly written example with misleading wording. I will try to give an explanation in a few steps below. Without formulas and calculations since they are redundant here.

Preliminaries

There is a good explanation of precession on Youtube. Gyroscopic precession -- An intuitive explanation Still, in my view the difficulty in this example is not in precession but in understanding why there is no change in angular momentum along AB, $$dL_h/dt=0$$.

Warm up

First, we have to assume that the system on the picture is an ISOLATED system. That is, we rule out all external forces on it. And consequently all EXTERNAL torques. Gravitation plays no role as we have assumed the system is well balanced in the first sentence. Yes, the system is somehow rotating with angular speed $$\omega$$ relative to inertial coordinate system, but for this example we have to assume that it is what it is without going into engineering intricacies. Rotation $$\omega$$ will be INTERNAL for this system.

Second, the book mentions the three axes of rotation. One axis goes through the rotor and points to the angular momentum for this spin. Second is along AB-axis. Angular momentum for this sideways rotation of the rotor is along AB and orthogonal to the first. The third axis of rotation point vertically and it is orthogonal to the first two axes of rotation. In the general case there can be no rotation about all or some of these axes.

Third, in this Example 8.11 quantity $$L_h$$ means specifically the absolute value of the angular momentum for the vector pointing along AB (or BA, does not matter). On previous pages of this book $$L_h$$ could mean absolute value of angular momentum vectors projected on the horizontal plane. In example 8.11 quantity $$L_h$$ related exclusively to the direction along AB.

Fourthly, the book does not qualify its arguments. This is the most important thing. The system is ISOLATED. All kinds of tilts for the base are ruled out. All kinds of things like spinning up rotations with human hands along the three axes are completely forbidden. The rotation of the base at the angular speed of $$\omega$$ happens at exacly $$\omega$$ at all times. It cannot be faster or slower than $$\omega$$ even for a infinitesimal amount of time.

In view of these four points the phrase of the book 'There can be no torque along the horizontal A-B axis because that axle is pivoted' is WRONG. The book does not quantify angular velocities along the three axes of rotation, technically they can have any numerical value. The fact that the rotor is fixed on the rectangular frame along AB plays no role. There CAN be torque along AB if we stop all rotations in this system and simply spin up the disk sideways on axis AB! What matters in this case is that we do not interfere with this system with EXTERNAL manipulations: external forces and torques are zero. Example 8.11 mutes this point so the whole idea become more difficult to understand.

Mental massage for rotations

For each of the rotations along the three axes we need to understand clearly the change direction and absolute value of angular momentum. Again, we do our analysis only along axis AB. If the base is rotating, we are mentally rotating with axis AB and calculate all changes in angular momentum (i.e. torques) moving along with it.

Changes in quantity $$I_\perp \ddot{\theta}$$ means the rotor is speeding up (slowing down) its rotation on axis AB. In this case the rotor may already be rotating with constant angular speed along AB (with changes in $$\theta$$, but for the change in the torque along AB there must be angular acceleration.

Changes in quantity $$\omega$$ means torque along the vertical axes on one hand. On the other hand, it will mean higher absolute value of the change of the spin vector that goes through the rotor. That is, quantity $$\omega L_s sin\theta$$. The angular momentum arising from the spin $$w_s$$ changes direction while the base is rotating with $$\omega$$. This change is torque pointing orthogonaly to the spin axis and is along AB. It is constant as long as the base is rotating at constant $$\omega$$.

Next, there may be precession effect from tilting of the rotor to the side. In this example if the spinning rotor is tilted to the side on which there is marking $$w_s$$, then the base will be rotating faster in the direction of $$\omega$$. But the book seems to rule out this effect by stating (implicitly) that rotation $$\omega$$ is always constant.

Conclusions

After thinking about this system for a while, analyzing all possible changes in angular speeds of rotations, one eventually comes to a conclusion that in order to make a change in the angular momentum along axis AB there have to be EXTERNAL changes to the system. The book stated that angular momentum is conserved for an ISOLATED system. In Example 8.11 we have to forbid all changes in angular speeds for all axes due to external events (human interfering), imperfections of the device (bad ball bearings and friction force).

The statement $$dL_h/dt=0$$ means there are no external changes to the system, and angular momentum along axis AB in particular.