Angular momentum and torque in gyroscope

Question

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

Answer 1

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

Additional resource:
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.

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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.

Answer 2

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.

Answer 3

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).

Answer 4

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.