I have observed that, a spinning top, when toppled by knocking it with finger, becomes almost upright again (with negligible precessional motion), instead of undergoing precession while being heavily leaned.

I cannot find which force produces the required torque to make it upright.

(Please note that my question is not exactly why a top does not fall due to gravity, which has already been answered.)

EDIT: See this youtube video for an example of this "stabilization". Compare the lean of the top at the beginning of the video to the lean at time 1:30.


5 Answers 5


The torque that rotates a top upright, as happens in that youtube video, is due to sliding friction between the top and its supporting surface.

Crucial to this effect is the fact that the top in that youtube video has a rounded bottom, instead of coming to a sharp point at the bottom like some tops do. The effect is more pronounced and dramatic in tops that have a larger radius of curvature on their bottom, such as in the extreme case of a tippe top, which has a radius of curvature so large that it's possible for the top's center of mass to be at a height that's smaller than the radius of curvature. Indeed, the papers I've seen that show how sliding friction causes a top's center of mass to rise are specifically doing an analysis of a tippe top.

The analysis of a top in full generality, including the effects of friction, is quite complicated. To simplify the analysis enormously, I'll just look at the top at an instant in time at which the top has no linear momentum, and has a very large angular momentum that lies precisely along the top's axis of symmetry.

I'll also consider gravity to be negligible in this simple explanation. Gravity causes a purely horizontal torque on the top, but we're only interested in torque that has a vertical component, which will cause the top to become increasingly upright. In reality, if it weren’t for gravity holding the top and the table together, there wouldn’t be any sliding friction at the point of contact between the two, but we will simply assume that the sliding friction exists, without considering how the sliding friction is related to gravity.


The diagram above shows a vertical cross section through the top, that contains the top's axis of symmetry. The point $P$ lies on the axis of symmetry, as does the top's center of mass $O$. The top's angular momentum $\vec{L}$ points in the direction of the axis of symmetry.

Because the top has a rounded bottom instead of a pointed bottom, the top's point of contact isn't at $P$, but rather at some point $C$. From the assumptions stated above, at the instant of interest $P$ is stationary. In contrast, from the direction of $\vec{L}$, at $C$ the surface of the top is moving towards the viewer, straight up out of the plane of the diagram. The sliding friction is a force $\vec{F}_k$ (not shown) on the top at $C$, in the direction opposite to the top's motion at that point, i.e., straight down away from the viewer.

The position vector of $C$ from $O$ is $\vec{X}_C$. The force $\vec{F}_k$ on the top produces a torque on the top around the top’s center of mass,

$$\vec{\tau} = \vec{X}_C \times \vec{F}_k \,\, .$$

The torque $\vec{\tau}$ can be written as

$$\vec{\tau}=\vec{\tau}_{\parallel}+\vec{\tau}_{\perp} \,\, ,$$

where $\vec{\tau}_{\parallel}$ is parallel to $\vec{L}$, and $\vec{\tau}_{\perp}$ is perpendicular to $\vec{L}$.

The torque $\vec{\tau}$ is how the top's angular momentum $\vec{L}$ changes with time,

$$\frac{d\vec{L}}{dt} = \vec{\tau}=\vec{\tau}_{\parallel}+\vec{\tau}_{\perp} \,\, .$$

$\vec{\tau}_{\parallel}$ points in the opposite direction as $\vec{L}$, so the effect of $\vec{\tau}_{\parallel}$ is to reduce the magnitude of $\vec{L}$, i.e., to slow the top down.

If the top was in empty space, the effect of $\vec{\tau}_{\perp}$ would be to rotate the top around $O$ clockwise in the diagram. However, due to the constraint that the top remains in contact with the table, the effect of $\vec{\tau}_{\perp}$ is instead to raise $O$ away from the table, and to make $O$ closer to being above $C$.

For a much more detailed analysis of how sliding friction on a top's bottom causes the top's center of mass to rise, see pretty much any paper on the tippe top, such as this one.

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    $\begingroup$ Thanks for a clear explanation of what is, for me, a complex effect. Also, thank you for the links to the tippe top. I had seen these tops before, but never understood how they work. $\endgroup$
    – James
    Commented Sep 30, 2016 at 11:14

The top is a symmetric rigid body. The equations of motion of a rigid body around its center of mass are given by: (Please, see for example: Marsden and Ratiu , (page 6).

$$I_1\dot\Omega_1=(I_2-I_3)\Omega_2\Omega_3$$ $$I_2\dot\Omega_2=(I_3-I_1)\Omega_3\Omega_1$$ $$I_3\dot\Omega_3=(I_1-I_2)\Omega_1\Omega_2$$ Suppose that the rigid top is symmetric about one axis (let's say the third one), thus we have: $$I_1=I_2$$ and also that the third axis is slim: $$I_3<I_1(or I_2)$$ In this case the third equation of motion implies $$\Omega_3=\Omega = const.$$ and substituting into the other two equations, we get: $$I_1\dot\Omega_1=(I_2-I_3)\Omega\Omega_2$$ $$I_2\dot\Omega_2=(I_3-I_1)\Omega\Omega_1$$

Taking the first derivative of the second equation with respect to time and substituting the second equation, we obtain: $$I_1I_2\ddot\Omega_2= \Omega^2 (I_3-I_1)(I_2-I_3)\Omega_2$$

This is an equation of a harmonic oscillator: $$\ddot\Omega_2+k^2 \Omega_2 = 0$$

With $$k^2= - \frac{\Omega^2 (I_3-I_1)(I_2-I_3)}{I_1I_2}$$

Now, observe that k^2>0 since $I_3-I_1<0$ and $I_2-I_3>0$, thus the spring constant is real and the harmonic oscillator is stable.

When a small external force limited in time is applied to a harmonic oscillator, it returns to oscillate in its natural frequency around its equilibrium position. The same happens when the top is knocked, in this case a small limited in time torque is applied. If the angular momentum around the third axis is very large such that: $$I_3 \Omega_3>> \int T_3 dt$$ Where $T_3$ is the torque component along the third axis, thus the right hand side is the impulse due to the torque application. In this case the application of the toque will not change the top's angular velocity much and all our above assumptions guarantee the stability of the top.

  • $\begingroup$ I don't have the background to fully understand your answer, but I do have a question. Does your argument hold regardless of whether the tip location is fixed or whether it is allowed to wander about the table? $\endgroup$
    – James
    Commented Sep 28, 2016 at 14:09
  • $\begingroup$ @James If the table has no friction then nothing will prevent the top from sliding after it gets hit. However, in real situations there may be some friction at the contact point and if the hitting impulse of the force component at the contact point is small enough, then the friction will be enough to hold the top without sliding and the assumption that the top is rotating about one fixed point will still remain intact. $\endgroup$ Commented Sep 28, 2016 at 14:30
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    $\begingroup$ "In this case the application of the toque will not change the top's angular velocity much and all our above assumptions guarantee the stability of the top." However, when the torque has been knocked and it has slanted (this is not a very small angle), The direction of the top's angular momentum has already changed much. Am I missing something? $\endgroup$ Commented Sep 28, 2016 at 14:48
  • $\begingroup$ @ArchismanPanigrahi The change of direction that you observe is due to the oscillations in $\Omega_2$ for instance. When another component besides $\Omega_3$ is present, the top's total angular velocity vector is indeed changed, but this change is oscillatory about the angular velocity direction before the hit. $\endgroup$ Commented Sep 28, 2016 at 15:40
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    $\begingroup$ @DavidBarMoshe: Thank you for the clarification. Given that the top in the OP's video link does not have an initial vertical spinning direction but becomes vertical after some time, it seems to me that you are not answering the OP's question. $\endgroup$
    – James
    Commented Sep 29, 2016 at 13:06

This is called gyroscopic effect and it states that An object that is spinning has an angular momentum $\vec{L}$ Thus it tends to remain its axis of rotation, $\vec{L}=I\omega$ faster it spins (greater $\omega$) more it tends to remain its axis of rotation.

Consider an image bellow, A spinning top is rotating with an angular velocity $\omega$, Therefore it has an angular momentum $\vec{L}$, Faster it spins greater the $\vec{L}$ and more it tends to remain its rotational motion about a certain axis, Note that when it slows down (Due to frictional forces) it has smaller $\vec{L}$ Thus its precession increases due to its weight pulling it down. (A Gyroscopic effect demonstration video)

enter image description here


Notice that Force of gravity caused our spinning top to precess (precess its axis of rotation) Thus we can conclude from this observation that a force must be applied to change spinning tops axis of rotation, Greater the force applied more it will precess from an original axis of rotation (assuming $\vec{L}$ is constant). When this force is removed it will naturally return to the original state with no precession because of it having angular momentum.

Think of it as Newtons first law, but in rotational motion instead of translational motion.

Newtons first law states:

  • An object that is in motion tends to stay in motion moving in a straight line, unless acted by an unbalanced force.

We can recreate this law for rotational motion:

  • An object that is spinning tends to remain rotational motion about a certain axis unless acted by an unbalanced Force.

An object that has a translational motion with a velocity $v$ will require a force $\vec{F}$ to change its direction of motion similarly an Object that has angular velocity $\omega$ requires a Force to change its rotational axis.

Torque is defined as a tendency of a force to rotate an object about an axis and mathematically it is defined as a vector (cross) product of distance and force:


Where $r$ is a distance from the point of rotation and $\vec{F}$ is force applied.

Note that torque is a vector and this vector is represented in the image bellow:

enter image description here

spinning top becomes almost upright again because it has angular momentum and that means that if an object spins it is resisting its axis of rotation to be precessed and faster it spins more it is resisting this precession to occur so if I tipped it over if it still spins with same angular velocity it will return to zero precession, faster it spins faster it will return to its original state.

To explain this mathematically consider a spinning top on earth with angular velocity $\omega$ its angular velocity of precession $\omega_p$ and angle of precession as $\phi$

enter image description here

its angular momentum is defined as:

$$\vec{L}= \vec{\omega} I$$

Say spinning top rotated $\Delta \theta$ and its change in angular momentum is $\Delta L$.

Then we can express $\Delta \theta$ as following:

$$\Delta \theta \approx \frac{\Delta L}{L sin(\phi)}$$

Angular velocity of precession can be expressed as following:

$$\omega_p = \frac{\Delta \theta}{\Delta t}$$

now we can substitute first equation in this one.

$$\omega_p = \frac{\Delta L}{\Delta t L sin(\phi)}$$

Torque is defined as change in angular momentum:

$$\frac{\Delta L}{\Delta t}= I \vec{\alpha} = \tau $$

Now we substitute this in previous equation:

$$\omega_p = \frac{\tau}{L sin(\phi)}$$

and we get the following formula:

$$\omega_p = \frac{\vec{F}r}{L sin(\phi)}$$

From this equation we can see that if we apply a force on a spinning object its $\omega_p$ will increase because it is directly proportional to force applied. If force applied is zero, $\omega_p$ becomes zero as well, therefore it will no longer have precessional angular velocity thus it will stand upright again.

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    $\begingroup$ I don't see anything in this answer that explains why a spinning top will become almost upright again after being tipped over. I think the effect described by the OP only occurs if the tip is allowed to slide over the tabletop, but I couldn't find a reference describing exactly how it happens. $\endgroup$
    – James
    Commented Sep 20, 2016 at 18:21
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    $\begingroup$ @ArchismanPanigrahi: There are two distinct situations... 1) The tip of the spinning top stays in one location on the table (because the table is rough), and 2) The tip of the spinning top is allowed to wander about the table. I believe (by personal observation) that a top will only decrease $\phi$ in situation #2. $\endgroup$
    – James
    Commented Sep 21, 2016 at 11:08
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    $\begingroup$ In the last paragraph of this answer... The applied force causing precession is gravity. If gravity were removed, then precession would stop as you stated ($\omega_P=0$). However, it would NOT stand upright again. With no external force, it would maintain its CURRENT axis of rotation per your first paragraph. $\endgroup$
    – James
    Commented Sep 21, 2016 at 11:15
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    $\begingroup$ Maybe, in case 2) something like this happens : From centre of mass frame, friction gives the torque, and from the frame of the tip of the top, a pseudo force produces the top. Whatever, this answer explains why a top does not fall down, and not why it becomes upright. $\endgroup$ Commented Sep 21, 2016 at 12:32
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    $\begingroup$ I agree, the answer does not explain why the limit of $\omega_{p} \rightarrow 0$ implies that a torque will suddenly appear to tilt the top into an upright position. The precession occurs because of an external torque but that does not mean a lack of precession will result in a spontaneous torque. $\endgroup$ Commented Sep 22, 2016 at 16:31

This type of top tilts and uprights itself because the centrifugal force requires the spin pattern to be circular to the perpendicular of the gravitational force. When it is knocked to the side the pattern of the spin is oval to the perpendicular of gravity. So the top spirals back to the most efficient position for centrifugal force.

  • $\begingroup$ This isn't an answer. Centrifugal force doesn't "require" anything. It acts on the top, and the top responds. How does the combination of centrifugal force and friction against the table work to right the top? $\endgroup$ Commented Oct 21, 2016 at 8:06

To understand gyroscopic precession intuitively, without using moment vectors, consider this orbiting satellite analogy. Watch “TheHue’s SciTech” video from 25 seconds until you are comfortable with predicting which way a gyroscope will precess when a rotational moment is acting upon it. https://www.youtube.com/watch?v=n5bKzBZ7XuM . Now consider Red Acts’s diagram as shown in his answer, his top is spinning anti-clockwise when viewed from above, because the frictional force F is pointing away from the viewer, opposite to the surface of the top. This creates a rotational moment in the plane occupied by contact point C and the top’s centre of mass O. Resolve this moment into a plane parallel to the rotating components of the top, and a plane perpendicular to them. The parallel component only causes a slowing down of the top, without precession. The perpendicular component of this moment is the one that causes precession of the top back to vertical. This moment's direction of rotation causes a force that pushes on the edge of the top nearest the viewer towards the bottom end of the top, and upwards on the opposite side. But like the satellite analogy, those edges do not actually move up or down - it’s the edges to left and right of the viewer that actually move. The edge to the viewers right moves down, and the edge to the left moves up. And this is the precession that brings the top vertical. I realised a while ago that this is what’s happening to a Tippe Top, but there’s so much more going on with a Tippe due to fact that it’s a sphere with a large radius.


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