Torque required to prevent precession

[rcgldr]

TL;DR Summary

A torque is applied to a spinning gyro, and a second torque is applied to prevent precession. How to quantify the second torque?

A torque is applied to a spinning gyro, and a second torque is applied to prevent precession. How to quantify the second torque?

Example, a gyro spinning about the z axis is connected to a frame that can only rotate about the x axis. A torque about the x-axis is applied to the frame. What is the torque about the y-axis exerted onto the frame by the gyro and its Newton third law counterpart, the torque the frame exerts onto the gyro about the y axis, that prevents the gyro from precessing? The rate of change in angular momentum about the x-axis would be due to the combined external torques.

 

[Andrew Mason]

Summary: A torque is applied to a spinning gyro, and a second torque is applied to prevent precession. How to quantify the second torque?

A torque is applied to a spinning gyro, and a second torque is applied to prevent precession. How to quantify the second torque?

Example, a gyro spinning about the z axis is connected to a frame that can only rotate about the x axis. A torque about the x-axis is applied to the frame. What is the torque about the y-axis exerted onto the frame by the gyro and its Newton third law counterpart, the torque the frame exerts onto the gyro about the y axis, that prevents the gyro from precessing? The rate of change in angular momentum about the x-axis would be due to the combined external torques.

Are we to assume that this gyro is not affected by gravity?

Since there is a constant torque about the x axis, the torque vector is a constant vector along the x axis. As a result, if the frame was free to move about all axes, the change in the angular momentum vector would be: ##d\vec{L}=\vec{\tau}dt## ie the change in angular momentum is in the direction of the x axis.

The resultant of ##\vec{L}+d\vec{L}## would be a new ##\vec{L'}## in which the axis of spin has rotated about the y-axis at an angle to the z axis ##\alpha## such that ##\alpha=\frac{\vec{\tau}dt}{\vec{L}}##.

But that cannot occur, because rotation about the y-axis is not allowed. So it seems to me that the frame must apply a torque to the x-axis that is equal and opposite to the applied torque along the x-axis to negate that applied torque.

AM

 

[rcgldr]

Are we to assume that this gyro is not affected by gravity?

But that cannot occur, because rotation about the y-axis is not allowed. So it seems to me that the frame must apply a torque to the x-axis that is equal and opposite to the applied torque along the x-axis to negate that applied torque.

The gyro in question is not affected by gravity.

The frame is free to rotate about the x-axis, so how does it generate a torque about the x-axis?

I recall some video where with the same or similar configuration: if rotation about y-axis is blocked, the reaction is the gyro accelerates about the x-axis as if the gyro was not spinning, no angular momentum, just angular inertia.

I haven't been able to find that video again, but I did find this old (1974) example claiming that there is no angular momentum related to precession. I set the link time to where an 8 lb gyro supported at one end is spun up to several thousand rpm and is precessing (with assist to get it near level) due to torque from gravity, then a small peg is placed in a hole that blocks the other end, stopping the precession, and the gyro just drops down (and bounces back a bit). The support structure is then vertically oriented, but there's too much play in the hole, so the peg moves instead of breaking. In another part of the video, a false claim is made about assisting the precession affects the downwards force at the support, but any movement in the supporting spring is probably nutation.