A question about why gyroscopes fall slowly

[jbriggs444]

This is, I think, only so because said gyroscope is supported on only one end in a fashion that it can "fall off".

As has been explained, it is because it does not need to fall at all. The Earth turns so that even if the gyroscope's orientation remains constant, the formerly vertical gyroscope ends up horizontal in less than 6 hours. It's not that the gyroscope tilted, it is that the ground did.

One might expect "equal to" 6 hours, but the sidereal day is less than 24 hours long.

 

[Shane Kennedy]

The Earth rotates by 90 degrees.

Irrelevant if the axis of the gyro is parallel to Earth's axis

 

[Shane Kennedy]

The Earth rotates by 90 degrees.

Irrelevant if the axis of the gyro is parallel to Earth's axis

 

[jbriggs444]

Irrelevant if the axis of the gyro is parallel to Earth's axis

True, but the context means that caveat does not apply.

If it starts vertical, on the equator of the Earth, it will take less than 6 hours to fall.

 

[GTrax]

As has been explained, it is because it does not need to fall at all. The Earth turns so that even if the gyroscope's orientation remains constant, the formerly vertical gyroscope ends up horizontal in less than 6 hours. It's not that the gyroscope tilted, it is that the ground did.

One might expect "equal to" 6 hours, but the sidereal day is less than 24 hours long.

Indeed you are right, and I get that. My apologies for perhaps not being complete enough. For example, the gyros made for Gravity Probe B were spinning spheres of niobium-coated fused quartz in orbit, with the instrument and indeed the rest of the spacecraft also in orbit "around" them, with only about 32 microns clearance in vacuum. After spin-up, they would remain pointed at their distant star. The levitation / suspension system allowed a safe "let go", and the spin-down time constant was around 15,000 years. Used to test Einstein's theory of General Relativity, the requirement was the spin axis not drift by more than one hundred billionth of a degree per hour.

Gyros in orbit are not experiencing the acceleration as would one on Earth, and if to be kept in place, need "support" somehow. I suppose if kept gimballed and with very low friction, in an evacuated space one could be left spinning for some days, but the "falling over" as I understood it from the OP, was of the typical child's toy gyro, with a support pin underneath, for a short while, able to hold itself from toppling. The overturning moment from gravity acceleration, of course results in the precession, as pretty well covered by other post in this thread. This is surely noticed by most of us who might spin up a bicycle wheel, or use an angle grinder.

 

[Andrew Mason]

A perfectly friction-free gyroscope that is fixed with a pivot along its axis does not so much 'resist' gravity as transfer any forces that act on it to the pivot.

The torque provided by gravity is tangential to the rotation of the centre of mass of the spinning wheel. Since the direction of "tangential" is constantly changing, the net change in angular momentum over a single revolution of 2π is zero IF there is no friction. But if there is even the slightest amount of friction, there is a net gravitational torque that will cause the gyroscope axis to tilt more (which will change the direction of the angular momentum spin vector which will lead to an increase the angular momentum precession vector to compensate). And eventually the spinning wheel/frame/top will touch the ground and stop.

AM

 

[hutchphd]

If the gyroscope is initially pointing at the Sun, directly overhead, and we wait six hours and it's still pointing at the Sun, which way is it pointing relative to the Earth's surface?

The gyroscope (mounted in gimbals), was named that by Foucault because it could show (scope) the rotation (gyro) of the Earth.

There is a famous incident involving the shutdown of an early SOYUZ launch in 1966 and subsequent gyro-triggered abort:
https://en.wikipedia.org/wiki/Soyuz_7K-OK_No.1
.

 

[AHaHaCiK]

A perfectly friction-free gyroscope that is fixed with a pivot along its axis does not so much 'resist' gravity as transfer any forces that act on it to the pivot.

So the answer is that a truly friction-free gyroscope will no more drop than would any stationary object resting on the pivot that such a gyroscope was attached to.

The table in my room has not dropped a mm in the last week, despite gravity acting on it all the time. The same for a 'perfect gyroscope', by translating its load to the pivot, the pivot provides the reaction force that keeps the gyroscope pointing in the same direction (which might look different to your moving frame, but it remains the same), just like the floor keeps my table at the same height (in this local gravity field) where it is now.

what happens with non-perfectly friction-free gyros, How is friction make the gyros fall?

 

[cmb]

what happens with non-perfectly friction-free gyros, How is friction make the gyros fall?

If you are asking me; AFAIU a torque from the bearing shaft of a gyroscope when combined with a straight-vector force on that shaft will cause a cross-product force and the gyroscope will precess around the pivot. As the gyroscope slows down from that torque, the precession would actually speed up (if the pivot itself was friction free) or reach some 'terminal presessional rotational speed' (friction at the pivot). That 'terminal rotation speed' will be a function of the strength of the cross-product force, which diminishes as the gyroscope slows down.

i.e. if the gyroscope is spinning on a friction axis, the axis will precess horizontally around the pivot due to the downward gravitational force, and both the gyroscope and precession will gradually slow down, whilst the angle of its axis drops downwards. The faster the gyropscope spins at the start of all this slowing-down, the faster the precession rate, and the longer it will take for the axis to dip downwards.

(AFAIU)

 

[hutchphd]

The faster the gyropscope spins at the start of all this slowing-down, the faster the precession rate

You should check this.

 

[cmb]

You should check this.

Yeah, not sure about that now you mention it. I was just assuming "a x b" would always be bigger, but it'd be translated back to the pivot point, so it might be the pivot point bears a higher load rather than a faster precession.

Sorry, I am not clear on the mechanics of that, happy to say I am not sure about that comment.

 

[snorkack]

Would a compass needle also resist toppling over and insist on precessing instead?
Remember, the aligned spins and orbital angular momenta of electrons confer on the compass needle an angular momentum even though the solid arrangement of iron nuclei is not rotating.

 

[Andrew Mason]

what happens with non-perfectly friction-free gyros, How is friction make the gyros fall?

See my post #36. Gravity provides a torque if the axis is not completely vertical. The torque vector is tangential to the direction of rotation of the centre of mass of the gyroscope.

If there is NO friction, then in one complete precession period, T, $\int_0^T \tau dt = 0$ so there is no change in angular momentum ($\tau = dL/dt$).

If there IS friction, then in one complete period of precession of the axis $\int_0^T \tau dt \ne 0$ (ie. the torque on the second half of the revolution is going to be applied for a slightly longer time than the torque on the first half of the revolution). This means that there is a change in angular momentum each precession period and it is cumulative.

AM