A question about why gyroscopes fall slowly

[AHaHaCiK]

why when the gyroscope work for a short time, it will fall down slowly.

I know that the system loses the energy by work done by friction between the wheel and its axis, but I don't understand how this makes the gyro drops. I think the friction makes the wheel spin slowly, and precession velocity increases, but this is not related to the phenomenon I mentioning.

I need your help, thanks for your attention and have a good day

 

[Andrew Mason]

If I understand you correctly, you are asking why a spinning gyroscope resists falling.

The spinning gives it angular momentum. In order to fall over, that angular momentum needs to change. A torque is needed to change that angular momentum. Torque x time = change in angular momentum. Torque is supplied by gravity and by friction. Since it is almost perfectly vertical initially (it can never be perfect) the gravitational torque is initially small. It takes a sustained torque (caused by gravity) over a sufficient period of time to change the angular momentum of the gyroscope enough for it to fall over.

AM

 

[vanhees71]

In the limit of no friction the gravity makes the top precess (more precisely the angular momentum precesses) but not letting it fall down as is the case when it's not spinning. The reason is indeed the torque. It's visualized nicely, e.g., here

http://hyperphysics.phy-astr.gsu.edu/hbase/rotv2.html#rvec4

 

[Delta2]

So @vanhees71 if there is no friction that reduces the magnitude of the angular momentum (by reducing the angular velocity) a gyroscope will never fall?

 

[Baluncore]

So @vanhees71 if there is no friction that reduces the magnitude of the angular momentum (by reducing the angular velocity) a gyroscope will never fall?

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

 

[Delta2]

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

You implying that even in the absence of friction , gravity will make it fall, it would just be very very slowly?

 

[Ibix]

You implying that even in the absence of friction , gravity will make it fall, it would just be very very slowly?

No - remember that gyroscopes resist changing orientation, and think about what happens in six hours on the Earth.

 

[Delta2]

No - remember that gyroscopes resist changing orientation

ok fine I can understand that.

and think about what happens in six hours on the Earth.

I am completely clueless, what happens every six hours on Earth?

 

[Baluncore]

The Earth rotates by 90 degrees.

 

[Delta2]

The Earth rotates by 90 degrees.

I could imagine that it has something to do with Earth's spin but why rotation by 90 degrees makes the gyroscope fall? Sorry I am a bit slow on this lol.

 

[Baluncore]

The gyroscope keeps pointing at the same point in the stars, while the Earth rotates under it. The axis of the gyroscope will gradually move to be horizontal, pointing West, after 6 hours.

 

[Ibix]

I could imagine that it has something to do with Earth's spin but why rotation by 90 degrees makes the gyroscope fall? Sorry I am a bit slow on this lol.

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?

 

[Delta2]

Ok I got it now thanks.

 

[Delta2]

So the correct statements is "If Earth had no spin and there wasn't any friction too, then the gyroscope would never fall"..

 

[Ibix]

So the correct statements is "If Earth had no spin and there wasn't any friction too, then the gyroscope would never fall"..

As long as you mean no spin with respect to distant stars, rather than something like tidal locking, yes. More simply, put your frictionless gyroscope at the north or south pole.

 

[AHaHaCiK]

If I understand you correctly, you are asking why a spinning gyroscope resists falling.

The spinning gives it angular momentum. In order to fall over, that angular momentum needs to change. A torque is needed to change that angular momentum. Torque x time = change in angular momentum. Torque is supplied by gravity and by friction. Since it is almost perfectly vertical initially (it can never be perfect) the gravitational torque is initially small. It takes a sustained torque (caused by gravity) over a sufficient period of time to change the angular momentum of the gyroscope enough for it to fall over.

AM

I'm not sure if adding a link is okay. If not, please remind me.



at the end of this video, the wheel is falling, though it continuously spins. I want a formula showing how to friction affect the gyroscope

 

[DaveC426913]

As long as you mean no spin with respect to distant stars, rather than something like tidal locking, yes. More simply, put your frictionless gyroscope at the north or south pole.

I was about to ask this question myself.
A friction-free gyroscope, placed at the pole, will not fall?

 

[sophiecentaur]

I was about to ask this question myself.
A friction-free gyroscope, placed at the pole, will not fall?

That doesn't take into account the Earth's gravity. A perfectly spherical Earth, spinning on an unchanging axis would still put a vertical gyroscope in unstable equilibrium - just like a pencil on its point. Any displacement will cause the gyro to droop and precess by a finite amount. For a slow rotation rate, the axis of the gyro would precess around an axis nearly vertically downward.
Rather than discussing 'letting go' a spinning gyro, it would probably be best to discuss the steady state condition or the lack of friction would allow a constantly changing situation (bouncing around the stable situation).

 

[ergospherical]

You can work all of this out; it's quite a tractable problem. Take the configuration linked by @vanhees71, for instance, and assume that the angle between the local vertical and the line joining the Earth's centre to the North Pole is $\gamma$ [you may set $\gamma = 0$ if the apparatus is located at the North Pole]. Consider an Earth fixed frame $\{ \hat{\mathbf{r}}, \hat{\mathbf{x}}, \hat{\mathbf{y}} \}$ and a gyroscope fixed frame $\{ \hat{\mathbf{r}}, \hat{\mathbf{x}}', \hat{\mathbf{y}}' \}$ which are related by a rotation $\mathsf{R}(\varphi)$ around the $\hat{\mathbf{r}}$ axis$$\begin{pmatrix} \hat{\mathbf{x}}' \\ \hat{\mathbf{y}}' \end{pmatrix} = \begin{pmatrix} \cos{\varphi} & \sin{\varphi} \\ -\sin{\varphi} & \cos{\varphi} \end{pmatrix} \begin{pmatrix} \hat{\mathbf{x}} \\ \hat{\mathbf{y}} \end{pmatrix}$$Write the angular velocity at which the Earth rotates about its axis as $\boldsymbol{\Omega} = \Omega (\cos{\gamma} \hat{\mathbf{r}} + \sin{\gamma} \hat{\mathbf{x}})$.

Let $\psi$ be the angle through which the gyroscope has rotated about its axis. Relative to an inertial (space fixed) frame, then, the angular velocity of the gyroscope takes the form \begin{align*}
\boldsymbol{\omega} &= \boldsymbol{\Omega} + (\dot{\psi} \hat{\mathbf{x}}' + \dot{\varphi} \hat{\mathbf{r}}) \\

&= (\Omega \sin{\gamma} \cos{\varphi} + \dot{\psi}) \hat{\mathbf{x}}' - \Omega \sin{\gamma} \sin{\varphi} \hat{\mathbf{y}}' + (\Omega \cos{\gamma} + \dot{\varphi}) \hat{\mathbf{r}}
\end{align*}where we have chosen $\hat{\mathbf{x}}'$ such that it lies along the axial direction of the gyroscope; the inertia tensor relative to the centre of mass of the gyroscope is hence also reduced to diagonal form $(I_{ij}) = \mathrm{diag}(a,b,b)$. Then the angular momentum about the centre of mass relative to the inertial (space fixed) frame is nothing but $\mathbf{L} = a (\Omega \sin{\gamma} \cos{\varphi} + \dot{\psi}) \hat{\mathbf{x}}' - b\Omega \sin{\gamma} \sin{\varphi} \hat{\mathbf{y}}' + b (\Omega \cos{\gamma} + \dot{\varphi})\hat{\mathbf{r}}$.

This should be sufficient information for you to determine the equations of motion, after writing $\dot{\mathbf{L}} = \mathbf{G} = \alpha \hat{\mathbf{y}}'$ for some $\alpha \in \mathbb{R}$. It might also help to know that each vector in the gyroscope fixed basis satisfies the equation $\dot{\mathbf{e}} = (\dot{\varphi} \hat{\mathbf{r}} + \boldsymbol{\Omega}) \times \mathbf{e}$.

 

[vanhees71]

No - remember that gyroscopes resist changing orientation, and think about what happens in six hours on the Earth.

Well, ok, I was not considering the rotation of the Earth ;-)). The gyro compass is, btw. also a very nice exercise for advanced mechanics courses (the use of Lagrange is highly recommended ;-)).

Of course, what happens in the case of a "fast top" is of course that there's both precession and oscillating nutation. It's also known as "pseudo-regular precession".

Maybe I should translate the chapter on the theory of the top of my mechanics manuscript to English. The German version is here

https://itp.uni-frankfurt.de/~hees/publ/theo1-l3.pdf

Sect. 4.2 Rigid-body kinematics
Sect. 4.3 Theory of the spinning top
Sect. 4.4 Gyro compass

 

[AHaHaCiK]

seem everyone is off-topic. What about my question

 

[Delta2]

seem everyone is off-topic. What about my question

So you want a detailed answer on how the slowing by friction is making the gyroscope drop. Maybe @vanhees71 has the answer in his notes linked in #20, I ll take a look at them though i don't know very well German.

 

[vanhees71]

I've no friction in this section on the spinning top.

Qualitatively it's clear that with friction, the angular momentum around the direction of gravity is no longer conserved but gets smaller, and that's why the top is "falling" if this spin components gets smaller and smaller due to the friction.

 

[AHaHaCiK]

I've no friction in this section on the spinning top.

Qualitatively it's clear that with friction, the angular momentum around the direction of gravity is no longer conserved but gets smaller, and that's why the top is "falling" if this spin components gets smaller and smaller due to the friction.

I think that when this spin components get smaller, the precession angular speed Ω=wr/Iω become larger, but not related to the falling of wheel (with w: weight of the wheel, r: distance of wheel from the pivot, I: angular momentum, ω: angular speed of the wheel)

 

[AHaHaCiK]

So you want a detailed answer on how the slowing by friction is making the gyroscope drop. Maybe @vanhees71 has the answer in his notes linked in #20, I ll take a look at them though i don't know very well German.

I even don't know german

I think paying attention to the rotation of the Earth is an interesting idea. I've not noticed that before. Thank you a lot

 

[Baluncore]

I think paying attention to the rotation of the Earth is an interesting idea. I've not noticed that before. Thank you a lot

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

In a gyrocompass, the gyroscope is constrained to remain with the axis horizontal, but it is free to rotate in azimuth. Rotation of the Earth results in it aligning with the meridian.

 

[AHaHaCiK]

my teacher explains that the reason for the falling of gyroscope is the effect of Coriolis force

Thanks again for everyone's answers!

 

[Andrew Mason]

my teacher explains that the reason for the falling of gyroscope is the effect of Coriolis force

Thanks again for everyone's answers!

The Coriolis "force" is an apparent change in direction of motion of a body relative to a rotating surface due to motion of the body perpendicular to the direction of rotation. I don't see any connection with a stationary gyroscope.

AM

 

[cmb]

why when the gyroscope work for a short time, it will fall down slowly.
A
I know that the system loses the energy by work done by friction between the wheel and its axis, but I don't understand how this makes the gyro drops. I think the friction makes the wheel spin slowly, and precession velocity increases, but this is not related to the phenomenon I mentioning.

I need your help, thanks for your attention and have a good day

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.

 

[GTrax]

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

This is, I think, only so because said gyroscope is supported on only one end in a fashion that it can "fall off". A fully gimballed gyroscope in partly evacuated casing keeps going for longer than 6 hours as everything around it, planet included, changes orientation.

 

[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.