One topic which was covered in university, but which I never understood, is how a spinning top "magically" resists the force of gravity. The conservation of energy explanations make sense, but I don't believe that they provide as much insight as a mechanical explanation would.

The hyperphysics link Cedric provided looks similar to a diagram that I saw in my physics textbook. This diagram illustrates precession nicely, but doesn't explain why the top doesn't fall. Since the angular acceleration is always tangential, I would expect that the top should spiral outwards until it falls to the ground. However, the diagram seems to indicate that the top should be precessing in a circle, not a spiral. Another reason I am not satisfied with this explanation is that the calculation is apparently limited to situations where: "the spin angular velocity $\omega$ is much greater than the precession angular velocity $\omega_P$". The calculation gives no explanation of why this is not the case.

  • $\begingroup$ There's a great debate about this. $\endgroup$ Nov 5, 2010 at 23:42
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    $\begingroup$ At some point you will have to do the math ... $\endgroup$
    – Cedric H.
    Nov 12, 2010 at 12:04
  • $\begingroup$ A correction. The floor/table resists the force of gravity (try using a free falling top to see what happens). The spinning resists the torque imposed by the gravity-reaction pair. It is the inertia of the part, not in the direction of spin that creates this reaction moment. Ever try to spin a pencil that has minimal inertia in the direction of spin? It won't work. $\endgroup$ Dec 6, 2010 at 18:53
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    $\begingroup$ This is a good edutainment video which covers pretty much exactly what you want. TL;DW the top does fall over in the end (unless it was perfectly straight initially which is impossible), but instead of just dropping it to the floor, it's axis precesses until it touches the ground. $\endgroup$ Dec 17, 2019 at 16:34

7 Answers 7


This is a rather old topic, but I felt I might have what you're looking for.

In response to some of the answers, you write:

Since the angular acceleration is always tangential, I would expect that the top should spiral outwards until it falls to the ground.

Absolutely, that is what you should expect to happen. And it does ... momentarily. The final solution is a little more involved than just being uniform rotation around the vertical axis.

In order to understand this, imagine that you take a spinning top which you have just set down at time $t = t_0 $ on the ground. Now, what happens in the next instant is exactly what you intuitively expect - the top begins to fall under gravity's influence and $ \phi $ (see figure for notation) starts to increase going from $ \phi \rightarrow \phi + \delta \phi $ at time $t_1$. Consequently the angular momentum $ \mathbf{L} $ of the top changes.

This is similar to what happens in the 2nd figure on the hyperphysics page, where $\delta \mathbf{L}$ is in the direction of $ \delta \theta $, only now $ \delta \mathbf{L} $ in the direction $ \delta \phi $ and lies in the plane containing the longitudinal axis $L_A$ of the top and the central vertical axis $V_A$.

Increasing $\phi$ lowers the center of mass of the top and thus its potential energy by an amount $ -\delta U $. Assuming energy conservation, this translates to an increase in the kinetic energy $\delta K$ of the top. Since the top is constrained to have zero linear momentum, this $ \delta K$ contributes entirely to the top's rotational energy.

Keep in mind, however, that the top is now rotating around two different axes. One component is the original spinning motion around its own longitudinal axes and the other is the rotation induced by gravity around the direction $N_A$ normal to the plane containing $L_A$ and $V_A$. Therefore, the $\delta K$ must be appropriately portioned between these two motions. Let's see how this happens.

The moment of inertia of the top ($I_A$) around the axis $L_A$ is clearly less than that ($I_V$) around the axis $N_A$. This is true for all but the most oddly shaped tops. Convince yourself that this is the case. In circuits more current flows through paths with lower resistance. Likewise in mechanics more energy is transferred to the component with lesser inertia. Thus the greater portion of $\delta K$ will go to increasing the angular momentum of the top around its longitudinal axis $L_A$ by some amount $\delta L'$

Now, conservation of angular momentum requires that there be a torque corresponding to this increase. The effect of this induced torque is to cause the falling top to start swinging back upwards. In this way, instead of a spiral, the tip of the top traces out something like a cycloid as it precesses around the central axis.

However, the diagram seems to indicate that the top should be precessing in a circle, not a spiral.

The circular trajectory is an idealization only achieved in the limit that $\omega_s / \omega_p \rightarrow \infty$, where $\omega_s$ is the spin angular velocity and $\omega_p$ is the precession angular velocity. Any top with realistic values of $\omega_s$ and $\omega_p$ will have finite "wobble".

I would not have known of this rather elaborate dynamics if not for one of Feynman's lecture volumes (Part I, I think) where this question is considered in great detail!

The above write-up is a little on the hand-wavy side and there probably are errors in my reasoning. For the full kahuna look up the Feynman lectures !

  • $\begingroup$ Your answer is extremely interesting. "Likewise in mechanics more energy is transferred to the component with lesser inertia" - I imagine that it would depend on how energy was transferred to the components. Surely it would be possible to increase one of the components without affecting the others? "Now, conservation of angular momentum requires that there be a torque corresponding to this increase." Are you saying that a change in angular momentum requires a torque to have changed it? If so, then this torque won't oppose itself? $\endgroup$
    – Casebash
    Dec 14, 2010 at 9:24
  • $\begingroup$ The rate of change of angular momentum is by definition torque. Anyway, the angular momentum in the case of a precessing spinning top is not conserved. $\endgroup$
    – Vagelford
    Dec 14, 2010 at 12:20
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    $\begingroup$ For clarity I recommend checking out Feynman's definitive exposition on this in Ch. 20, Sec. 5 of Vol I of his lectures. The name of the phenomenon I refer to "wobbling" is nutation. See nice video on youtube ! $\endgroup$
    – user346
    Dec 14, 2010 at 20:16
  • $\begingroup$ The difference between the moment of inertia around the two axes is not very relevant. The reason that the precession starts is that when the top falls, it causes the angular momentum to have a change $\delta \boldsymbol L$ pointing down. Since there is no torque along the vertical direction, this change of the angular momentum has to be compensated by an angular velocity pointing up, that is just the precession. $\endgroup$
    – Cosyn
    Jun 26, 2019 at 9:32
  • $\begingroup$ After the precession speed increases, some of the torque from gravity is used for maintaining the precession, so the falling acceleration decreases. Then the precession will overshoot the steady precessional velocity, causing the top to decelerate and then to move up. $\endgroup$
    – Cosyn
    Jun 26, 2019 at 9:37

When it is spinning its angular momentum is quite high. By conservation of angular momentum the spinning top is then more stable against small torques like the action of gravity on the top.

The angular momentum of the top is $J = I \omega$ where $I$ is the inertia tensor and $\omega$ is the Darboux vector, whose magnitude is proportional to the rotational speed.

You can find a detailed discussion on this page of Hyperphysics.

  • $\begingroup$ That's cool. Lots of cultures have their own terms for things in physics, it's hard to know which are international sometimes. $\endgroup$
    – Noldorin
    Nov 5, 2010 at 23:32
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    $\begingroup$ Conservation of angular momentum seems specious. If the top is tilted there's net external torque, but the top can precess without falling down. $\endgroup$ Nov 6, 2010 at 4:39
  • $\begingroup$ @Mark: that's my point... $\endgroup$
    – Cedric H.
    Nov 6, 2010 at 10:33
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    $\begingroup$ Oh, and for a low-level intuitive (sort of) explanation of why the top doesn't fall over: since the particles in the top are revolving around an axis, the downward acceleration induced by gravity gets "converted" into sideways acceleration as the particles move to a different point in their orbit. Of course this is not an easy view to understand, which is why we talk about angular momentum ;-) $\endgroup$
    – David Z
    Nov 7, 2010 at 8:04
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    $\begingroup$ So a frictionless spinning top of infinite hardness can hold any weight you throw at it, as if it were a rigid pillar. I'd pay anything for a house that stood on top of 4 spinning tops. =) $\endgroup$
    – Malabarba
    Dec 14, 2010 at 6:06

The point is that conservation principles are not generally intuitive. For example, why should energy be conserved? One must have a grip of the dynamics involved in order to understand them.

Anyway, the precession of the spinning top doesn't have to do with the conservation of angular momentum. It has to do with the strange nature of torque and its interaction with angular momentum. When a force acts on a spinning top, it excerpts a torque perpendicular to the plane defined by the axis of the top and the direction of gravity, which is a vertical plane. That direction is horizontal. On the other hand, the torque is the rate of change of angular momentum. That means that the direction of the torque is the direction towards which the vector of angular momentum changes. Thus, since the torque is horizontal and perpendicular to the angular momentum, it can only change the direction of angular momentum along the horizontal direction and not towards the ground. That means that the vector of angular momentum has its back on the ground, at the point that the tip of the top touches the ground, and its head is performing a circle on a plane that is parallel to the ground. That motion is the precession of the spinning top.

Finally, I think that the reason for assuming a much faster rotation than precession for the top, is to simplify the calculations and consider the top as a gyroscope.


when the mass is spinning it has an angular momentum pointing in a direction perpendicular to the plane it's spinning on.

The angular momentum has to be conservate: i.e. has to keep pointing in the plane-perpendicular direction. As cedric said, the gravity, works for the axis of the spinng mass to fall horizontally on the plane: if this happens also the angular momentum as to torque! and this is not convenient from an energy conservation point of view..

Then u can consider that the magnitude of the angular momentum is proportional to the spinning speed: so as the spinning velocity gets higher it gets, for lack of a better word, "easier" for the top to resist the gravity..

If u try to spin a top on an inclined plane you will need to spin it faster to obtain the same "resistance to gravity"!


From your linked article:

Spin a top on a flat surface, and you will see its top end slowly revolve about the vertical direction, a process called precession. As the spin of the top slows, you will see this precession get faster and faster. It then begins to bob up and down as it precesses, and finally falls over.

The drawing shows a circle instead of a spiral due to leaving out variables like friction and gravity.


The quick answer is that, for the top to fall over due to gravity, each fragment of the top that is moving around the spin axis has to change its individual direction of movement. They are already changing direction around the spin axis, due the rigidity of the top keeping them moving in a circle. But gravity is operating at 90º to their direction of movement, and its effect depends on the velocity or inertia of the fragment. For a fast rotating top, this slight change of direction is what causes the top’s precession. And as it slows down, the effect of gravity has more effect, and it falls over. It’s a similar situation to changing the orbit of a satellite with side thrusters.


This is a nice example which shows understanding does not come automatically after completing a calculation. But calculation still serves the (perhaps the most) important guide. Nobody in the above has mentioned the discussions given in \ittext{Landau & Lifshitz, Mechanics (BH, 3rd ed.), page 112}. I think these discussions have already elucidated the issue. Unfortunately, they proceeded using Euler angles. I have reformulated their discussions here. Hope that helps.


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