Why doesn't a spinning top fall over instantly?

[ashishsinghal]

This is not a homework question but could not find a better place to post it.

Why doesn't a spinning top fall over instantly? Also why does it fall later? What makes spinning so special? Are the answers different for top spinning with its axis fixed and the top with its axis moving conically?
Please answer...

 

[Andrew Mason]

This is not a homework question but could not find a better place to post it.

Why doesn't a spinning top fall over instantly? .

Essentially the top keeps spinning because it has angular momentum and the gravitational force on the top cannot exert sufficient torque to change that angular momentum. To understand the physics of a spinning top requires a good understanding of mechanics. You will have to understand moment of inertia, angular momentum, and torque.

Also why does it fall later?

As the rotational speed slows, the angular momentum decreases. Gravity can now more easily change the angular momentum by exerting torque on the top. The top starts to wobble and gravity can now exert more torque on it. Once gravity starts exerting a torque on the top, the angular momentum changes and keeps changing. The top starts to precess and eventually falls down.

What makes spinning so special?

Angular momentum. No one really knows why angular momentum or inertia exists.

AM

 

[supratim1]

initially the angular momentum is high enough to defeat gravity. slowly, due to friction and air resistance, it slows down, and gravity wins.

 

[ashishsinghal]

Essentially the top keeps spinning because it has angular momentum and the gravitational force on the top cannot exert sufficient torque to change that angular momentum.

What do you mean by sufficient torque? I mean there is a torque, there is an angular momentum, the torque has to change the angular momentum.

Also according to supratim due to friction and air resistance, it slows down, and gravity wins.
This means that if friction and air resistance were not present the top would not stop spinning? But even if the torque by gravity is small, it should bring a change in angular momentum after some time. I am confused! Please Help!

 

[Andrew Mason]

What do you mean by sufficient torque? I mean there is a torque, there is an angular momentum, the torque has to change the angular momentum.

Theoretically, if you had a perfectly still surface and surroundings and a perfectly balanced top one could place the top on its point perfectly vertically and it would stay there. Gravity would be straight down on a line from the centre of mass through the tip of the top. The normal force on the top would be straight up and the top would not fall. Why? Because the only way the top can tip over is if gravity can exert a torque about the fulcrum (the tip) and pull it down. This is the only way to create an unbalanced downward force on the centre of mass.

In the perfect scenario I have given there is no gravitational torque about the fulcrum (the tip) at all and the top does not tip over.

\tau = m\vec{g} \times \vec{r} = Fr\sin{0} = 0.

In an imperfect world, the slightest horizontal movement of the centre of mass (even by one atomic diameter, created by random motion of the atoms in the surface in contact with the tip) means that \sin{\theta} \ne 0 meaning there is a slight torque on the centre of mass about the fulcrum. This causes the centre of mass to move even further increasing the angle, thereby increasing the gravitational torque. This increases the movement of the centre of mass even further, thereby increasing torque and so on. And the result is that the top falls over.

The torque acting on the centre of mass is equal to the rate of change of angular momentum of the top:

\vec{F} \times \vec{r} = Fr\sin{\theta}\hat z= \vec{\tau} = \frac{d\vec{L}}{dt}

If the top is not spinning then L = 0 initially. The torque on the centre of mass creates angular momentum by toppling the centre of mass about the fulcrum (the direction of angular momentum being horizontal, perpendicular to the initial vertical axis). The change in angular momentum needed is just enough to make the top fall over..

The situation is very different when the top is spinning. \vec{L} \ne 0. If the top is to tip over, the direction of angular momentum has to change. If L is large, this requires a large change in angular momentum (\tau = dL/dt -> \Delta L = \tau \Delta t).

Initially the gravitational torque is not very large (small angle) so the change in angular momentum is small relative to the angular momentum of the top. So the torque has to act for a long time before the change of angular momentum is enough to change the direction of L enough to topple the top. (\Delta L = \int \tau dt \approx \tau\Delta t)

Also according to supratim due to friction and air resistance, it slows down, and gravity wins.
This means that if friction and air resistance were not present the top would not stop spinning? But even if the torque by gravity is small, it should bring a change in angular momentum after some time.

It is not a matter of stopping the top's spin. The top can keep spinning at the same speed and gravity will still eventually topple it. Friction and air resistance are not the cause of the top falling over. Gravitational torque acting for sufficient time is the reason it falls over.

AM

 

[ashishsinghal]

torque = r vector cross force vector
http://en.wikipedia.org/wiki/Torque

 

[supratim1]

It is not a matter of stopping the top's spin. The top can keep spinning at the same speed and gravity will still eventually topple it. Friction and air resistance are not the cause of the top falling over. Gravitational torque acting for sufficient time is the reason it falls over.

AM

How? if we consider ideal case, the gravitational force will pass through the COM always, and never produce any torque, so it would never topple. this is how inertial navigation of satellites using gyroscopes in evacuated chambers and nearly frictionless contacts.

 

[Andrew Mason]

torque = r vector cross force vector
http://en.wikipedia.org/wiki/Torque

Thanks for pointing that out. I have corrected it above.

AM

 

[Andrew Mason]

How? if we consider ideal case, the gravitational force will pass through the COM always, and never produce any torque, so it would never topple.

Correct. That would be the perfect situation that I described except now with the top spinning. The top will keep perfectly vertical forever.

But that is not the real world. In the real world, the top is not placed in a perfectly vertical position on a perfectly smooth surface having no vibrating molecules and no currents in the surrounding air. The top will eventually accumulate a large \Delta L in the form of a precession that keeps increasing until the top falls over.

this is how inertial navigation of satellites using gyroscopes in evacuated chambers and nearly frictionless contacts.

That is not done to keep the gyroscope from slowing down. In fact, the spin of the gyroscopes are maintained with motors. It is done to eliminate forces that can create a torque on the gyroscope. It is also be done to minimize wear on the bearings in order to ensure it keeps working.

AM

 

[ashishsinghal]

Thanks for all of this. Now can we move to a top which spins conically.

 

[Cleonis]

The situation is very different when the top is spinning. \vec{L} \ne 0. If the top is to tip over, the direction of angular momentum has to change. If L is large, this requires a large change in angular momentum (\tau = dL/dt -> \Delta L = \tau \Delta t).

Initially the gravitational torque is not very large (small angle) so the change in angular momentum is small relative to the angular momentum of the top. So the torque has to act for a long time before the change of angular momentum is enough to change the direction of L enough to topple the top. (\Delta L = \int \tau dt \approx \tau\Delta t)

It is not a matter of stopping the top's spin. The top can keep spinning at the same speed and gravity will still eventually topple it. Friction and air resistance are not the cause of the top falling over. Gravitational torque acting for sufficient time is the reason it falls over.

Some additional remarks:

We have that the Earth is subject to a torque, arising from the Sun's gravitation. Because of it's rotation the Earth is flattened; the equator is 20 km further away the geometrical center than the poles. Because of the flattening the Earth's center of of mass and the center of gravitational attraction to the Sun do not coincide.

The tendency of the torque from the Sun is to bring the Earth's equatorial plane in alignment with the plane of the Earth's orbit around the Sun. As we know, those two planes are at an angle to each other, the angle is about 23 degrees.

The combination of the Earth's rotation, together with the torque from the Sun, results in gyroscopic precession of the Earth. The Earth's axis sweeps out a cone. This gives rise to the precession of the equinoxes, which has a cycle of 26,000 years.

The Earth is 4 billion years old. This illustrates that in the absence of friction a state of gyroscopic precession will persist forever.


It's probably not feasible to engineer a setup with a spinning top where on one hand friction is zero, while at the same time a torque is exerted upon the spinning top. Still, we can say what the laws of mechanics predict for that case. Once a state of gyroscopic precession has set in this state will persist forever.

Gyroscopic precession is a cyclic process, whereas a spinning top falling over is an irreversible event.
In the absence of friction a torque cannot topple a spinning top all the way down.

Presence of a torque (exerted on a spinning top) can lead to a state of gyroscopic precession, but it cannot lead to the top toppling over all the way to the ground.


Starting at 10 degree angle
Let me discuss the following case: a spinning top is released with the axis at an initial angle of, say, 10 degrees. Obviously, a non-spinning top released that way will fall all the way down very rapidly. But what will happen to a spinning top (when there is zero friction)?
The spinning top will keel over a bit, and as it keels over the top goes into a pattern of gyroscopic precession. The gyroscopic precession then stops the spinning top from keeling over further. In the (theoretical) case of zero friction that is the final state.

How much will the spinning top keel over? That depends on the rotation rate. The faster the spin of the spinning top, the slower the rate of gyroscopic precession in the final state. Usually the top spins fast enough so that the initial keeling over is not visible to the naked eye.

Enter friction. When there is friction (say, air friction on the spinning top) then the rotation rate will gradually go down, reducing the capacity of the gyroscopic precession to prevent keeling over. With friction present the spinning top will keep sagging down.

 

[Andrew Mason]

Some additional remarks:

We have that the Earth is subject to a torque, arising from the Sun's gravitation. Because of it's rotation the Earth is flattened; the equator is 20 km further away the geometrical center than the poles. Because of the flattening the Earth's center of of mass and the center of gravitational attraction to the Sun do not coincide.

I'm not sure that's quite right. The centre of mass and the centre of gravity of the Earth relative to the sun do not coincide because the sun's gravitational field is not uniform across the diameter of the earth. This creates a slight torque due to the Earth rotation about its centre of mass.

However, because the Earth is an oblate spheroid rather than a sphere, and I think also because the plane of the moon-earth orbit is not the same as the plane of the earth-sun orbit, the centre of gravity of the Earth relative to the sun and the centre of gravity of the Earth relative to the moon are in slightly different locations. So there is a moon-sun torque on the Earth as well the solar torque.

The Earth is 4 billion years old. This illustrates that in the absence of friction a state of gyroscopic precession will persist forever.

Well, at least for a very long time. Does the gravitational torque not cause a steadily increasing precession?

It's probably not feasible to engineer a setup with a spinning top where on one hand friction is zero, while at the same time a torque is exerted upon the spinning top. Still, we can say what the laws of mechanics predict for that case. Once a state of gyroscopic precession has set in this state will persist forever.

Only if it is a torque-free precession. If the torque persists, angular momentum must change with time. The change in angular momentum is always increasing with time: \Delta L = \int \tau dt

Gyroscopic precession is a cyclic process, whereas a spinning top falling over is an irreversible event.
In the absence of friction a torque cannot topple a spinning top all the way down.

Presence of a torque (exerted on a spinning top) can lead to a state of gyroscopic precession, but it cannot lead to the top toppling over all the way to the ground.

I disagree. Angular momentum is a very non-intuitive and confusing phenomenon. One has to explain it in terms of mathematics.

The toppling of a top does not depend on friction. Friction may cause it to occur sooner because it reduces the magnitude of the spin angular momentum, making it easier for gravitational torque to change the direction of that angular momentum vector.

AM

 

[Cleonis]

Well, at least for a very long time. Does the gravitational torque not cause a steadily increasing precession?

Well, as you point out: in the particular case of the Earth-Moon system orbiting the Sun there are multiple factors, so that system is not a good candidate for detailed discussion. (The tilt of the Earth equatorial plane varies somewhat, but for as long as the Earth exists it has been around that 23 degrees.)

Instead we can take the simplest possible case: a sun with a single planet, with the planet's equatorial plane tilted relative to the orbiting plane. Then the precession rate will not change over time. No such thing as 'steadily increasing precession'.Comparison
Let me make the following comparison: let there be two celestial bodies, in tidal lock with each other, orbiting in circular motion. Then their state of motion is permanent: they will keep orbiting each other. At all times the centripetal force is at right angles to the instantaneous velocity, so the centripetal force is not doing work. Since there is no friction there is no opportunity for dissipation of energy; the principle of conservation of energy implies that they will keep orbiting each other for eternity.

A state of gyroscopic precession is analogous to that. The torque does affect the angular momentum, but the change in angular momentum is at right angles to the existing motion. (As you know effect of torque on a spinning body is calculated with the vector cross product.) The torque is not doing work. Also, there is no friction, hence no opportunity for dissipation of energy.As a matter of general physicsforums conduct:
Discussing mechanics in general is not the purpose this section of physicsforums, this is the 'Homework & Coursework Questions' section. (And the original poster apologized in advance for maybe posting the question in an inappropriate section.)

I propose we move this discussion (which is between the two of us) to the Classical Mechanics section here on physicsforums.Here is what we disagree about:
I have described that the laws of mechanics imply the following: that if you have a frictionless spinning top, subject to torque from gravity, then the top won't keel over all the way down. Instead it will settle in a cyclic pattern of gyroscopic precession.

As I understand it you claim that even in the absence of any friction gravity will still make a spinning top go all the way down.

 

[Andrew Mason]

Instead we can take the simplest possible case: a sun with a single planet, with the planet's equatorial plane tilted relative to the orbiting plane. Then the precession rate will not change over time. No such thing as 'steadily increasing precession'.

And that can be shown mathematically: for a system with no external torque acting on it, <br /> \tau = dL/dt = 0.

Comparison
Let me make the following comparison: let there be two celestial bodies, in tidal lock with each other, orbiting in circular motion. Then their state of motion is permanent: they will keep orbiting each other. At all times the centripetal force is at right angles to the instantaneous velocity, so the centripetal force is not doing work. Since there is no friction there is no opportunity for dissipation of energy; the principle of conservation of energy implies that they will keep orbiting each other for eternity.

As does the principle of conservation of angular momentum.

A state of gyroscopic precession is analogous to that. The torque does affect the angular momentum, but the change in angular momentum is at right angles to the existing motion. (As you know effect of torque on a spinning body is calculated with the vector cross product.) The torque is not doing work. Also, there is no friction, hence no opportunity for dissipation of energy.

You seem to be saying that torque can do no work. A torque \tauapplied through an angle \theta does work W = \tau \cdot \theta

AM

 

[Cleonis]

You seem to be saying that torque can do no work.

You must be joking.
Obviously, in the general case a torque will do work.


Now to the the specific case of gyroscopic precession.
When a spinning top is in a state of gyroscopic precession the spin axis sweeps out a cone.
After one precession cycle is completed the spin axis is right back at the starting orientation.
During the precession cycle both the gravitational potential energy and the kinetic energy remain the same.


This is the analogy with circular motion in the presence of a centripetal force:

In general a force will do work, but the specific case of circular motion is an exception to that. In the case of circular motion the change of velocity is at right angles to the existing instantaneous velocity. A force will do no work if and only if the change in velocity is at right angles to the existing velocity.

In the specific case of spinning objects there is a counter-intuitive property: the change of angular velocity in response to a torque being exerted is motion at right angles compared to the exerted torque. The right angles effect makes it into an exceptional case. In general a torque will do work, but in this specfic case there is a right angles effect.


The spinning tops that we see with our own eyes are subject to friction. In the precence of friction we see that the cone that is swept out is slowly opening up; in the presence of friction the process of gyroscopic precession is almost cyclic, but not quite.

This is analogous to the case of circular motion with small but non-zero friction present. With non-zero friction the circular motion will gradually spiral inwards. That is, with non-zero friction the centripetal force has opportunity to do work.

You have stated a belief that the opening up of the cone will happen anyway, even in the absence of friction. You have stated a belief that the the effect of friction is that the cone opens up faster than would be the case without friction.

If I hazard a guess I think you do not quite accept the right angles effect. A spinning object responds differently to an exerted torque than a non-spinning object. The laws of mechanics imply that this difference is exactly at right angles, but to actually see that is far from straightforward.

 

[Cleonis]

we can take the simplest possible case: a sun with a single planet, with the planet's equatorial plane tilted relative to the orbiting plane. Then the precession rate will not change over time. No such thing as 'steadily increasing precession'.

And that can be shown mathematically: for a system with no external torque acting on it, <br /> \tau = dL/dt = 0.

I notice there is something here that must be emhasized separately:

The case:
A sun with a single planet, the planet is spinning, and has a corresponding equatorial bulge. The planet's equatorial plane is tilted relative to the orbiting plane.

Then there is an torque exerted upon the planet.
That torque is exerted by that sun upon that planet.
That torque causes continuous change of the planet's angular momentum.

Gyroscopic precession is a process of continuous change of the direction of the angular momentum. In the specific case of a torque exerted upon a spinning object only the direction of the angular momentum is changed, while the magnitude of the angular momentum remains the same.Let me repeat:
In the case of gyroscopic precession the angular momentum of the precessing object is changing all the time. The point is that the change is a cyclic process. After a cycle has been completed the angular momentum has returned to its direction at the start of the cycle. The fact that in gyroscopic precession there is no dissipation of energy relates to the fact that the process is a cyclic process. The energy of the object in gyroscopic precession remains the same; the (direction of the) angular momentum is changing all the time.It is crucial that you distinguish between the angular momentum of the sun + planet system, and the angular momenta of the sun and the planet separately.
A statement about the system as a whole is distinct from a statement about the angular momentum of the planet separately.

 

[Cleonis]

The top can keep spinning at the same speed and gravity will still eventually topple it. Friction and air resistance are not the cause of the top falling over. Gravitational torque acting for sufficient time is the reason it falls over.

It occurred to me that there is another way of addressing this issue.

What happens if you use a motor to maintain a constant angular velocity of the object in gyroscopic precession?
The motor is then used only to sustain an existing angular velocity. The motor replenishes energy lost due to friction. For the physics taking place that setup is equivalent to a frictionless setup.The site physics-animations.com has a page about http://physics-animations.com/Physics/English/gyro_txt.htm"
The movieclip shows a single precession cycle of a precessing assembly. The torque is huge, but so is the angular velocity of the rotor.

According to Andrew Mason that type of setup is not sustainable. According to Andrew Mason the weight will always sag down further and further, even if you use a motor to replenish energy lost to friction.

Well, that experiment can be done. Clearly it has been done; presumably the movie shows just one cycle to reduce the kilobyte size of the movie-file.

Andrew, if it turns out that the device shown in the movie can go through, say, 10 cycles of its precession cycle, and still not sag down, would you be willing to reconsider? If you want more than 10 cycles, how many more?

[Later edit]
I came across a Youtube video showing a http://www.youtube.com/watch?v=MlCZJwV_iXk". The rotor is driven. Also, an electronic device is present to monitor the angular velocity.
As the laws of motion predict: you can obtain a precessing motion, but as long as the angular velocity of the rotor remains the same the precession remains the same: no sagging.
[/Later edit]

 

[James_Frogan]

I'm particularly interested in the motion of tops, from what I understand, I would have to go with Cleonis on the perpetual motion of the top. Assuming that there is no external torque or work done (i.e. the Lagrangian equations of motion simply consists of the derivatives of the kinetic and potential energy), the top will keep performing a nodding motion (nutation).

That being said, I am not an expert and have a question myself. In practical spinning tops, we find that the tops have a tendency to correct themselves to a vertical sleeping position. If we have a fixed bottom point of contact, can the top still stand up (under the force of friction) or must we allow an extra 2 degrees of motion across the table? Has anyone derived equations of motion for these cases?

 

[D H]

It is not a matter of stopping the top's spin. The top can keep spinning at the same speed and gravity will still eventually topple it. Friction and air resistance are not the cause of the top falling over. Gravitational torque acting for sufficient time is the reason it falls over.

No, it is not, at least not in general. Time for you to revisit your classical mechanics class, either at the undergrad or graduate level. Both address the topic of a heavy symmetric top with one point fixed.

A heavy symmetric top with one point fixed will undergo precession, nutation, and rotation. Without friction, the rotation rate will remain constant. The nutation will make the top wobble between some (fixed) minimum and maximum can't angles -- assuming of course that this maximum can't angle is small enough such that the body of the top never touch the ground.

 

[Andrew Mason]

No, it is not, at least not in general. Time for you to revisit your classical mechanics class, either at the undergrad or graduate level. Both address the topic of a heavy symmetric top with one point fixed.

A heavy symmetric top with one point fixed will undergo precession, nutation, and rotation. Without friction, the rotation rate will remain constant. The nutation will make the top wobble between some (fixed) minimum and maximum can't angles -- assuming of course that this maximum can't angle is small enough such that the body of the top never touch the ground.

You are right. It is me who is confused.

The gravitational torque causes a change in the angular momentum vector that is always perpendicular to the vertical plane. This causes the axis of spin to precess at constant speed. The additional angular momentum that this creates (a small vector in the vertical direction) means that the angular momentum vector is not in the same direction as the axis of spin and so the top nutates. But the nutation would be constant as well.

Sorry for the confusion I have caused.

AM

 

[James_Frogan]

A heavy symmetric top with one point fixed will undergo precession, nutation, and rotation. Without friction, the rotation rate will remain constant. The nutation will make the top wobble between some (fixed) minimum and maximum can't angles -- assuming of course that this maximum can't angle is small enough such that the body of the top never touch the ground.

D H, if you take into account the nutation, the spin rate will not be constant but will have a slight sinusoidal increase/decrease in speed as the kinetic energy is passed between nutation and precessing.

If however the case were constant precession (no nutation) then the spin rate would be constant as well, I think.

Just highlighting this ;)

 

[D H]

A heavy symmetric top with one point fixed will undergo precession, nutation, and rotation. Without friction, the rotation rate will remain constant.

D H, if you take into account the nutation, the spin rate will not be constant but will have a slight sinusoidal increase/decrease in speed as the kinetic energy is passed between nutation and precessing.

Precession, nutation, and rotation are an Euler rotation about the z, x' and z'' axes^1[/color]. Precession is a rotation about the canonical space z axis (i.e., upward), nutation is a rotation about the (rotated) x axis, and rotation is a rotation about the (doubly rotated) z axis, which now coincides with the canonical body z axis (the axis from the fixed point through the center of mass).

The final rotation, the rotation about the canonical body z axis, is a constant of motion in the case of a heavy top with one point fixed. The precession and nutation rates are not constant.


^1[/color]Aside: This is *the* Euler rotation sequence. All other eleven sets of three rotations (zyz, yxy, yzy, xyz, xzy, ...) are sometimes also called Euler rotations, but the zxz sequence is the one that Euler developed. When astronomers and physicists talk about Euler angles they are most likely talking about a zxz sequence.

 

[James_Frogan]

I refer to the derived motions of equations here:
http://www.maths.surrey.ac.uk/explore/michaelspages/Spin.htm

Don't the derivations show that the rotation'' (ie psi'') is not equal to zero? Or am I confused with something else?

 

[D H]

The angular velocity in body frame coordinates in terms of the time derivatives of the Euler angles \phi, \theta, and \psi are

\aligned<br /> \omega_1 &amp;= \dot{\phi}\sin\theta\sin\psi + \dot{\theta}\cos\psi \\<br /> \omega_2 &amp;= \dot{\phi}\sin\theta\cos\psi - \dot{\theta}\sin\psi \\<br /> \omega_3 &amp;= \dot{\psi} + \dot{\phi}\cos\theta<br /> \endaligned

It is this final component, the angular velocity about the body z axis, that is a constant of motion.

 

[James_Frogan]

Thanks for the explanation. I merely assumed that "the angular velocity about the body z axis" was just the top's spin rate. Why does it stay as a constant of motion, but not \\omega_1 and \\omega_2?