enter image description hereI recently bought a Gyroscope toy online and was keen to find out the Physics behind this fun little "gravity-defying" toy. I did some online research and found a force diagram of a gyroscope in precession. However, I am quite puzzled as to my understanding, there are two angular momentums: one is the spin angular momentum of the Gyro itself, and one is the angular momentum of precession about the origin O. Hence when gravity functions as a torque, which angular momentum is this torque changing, and is it vectorially or just in terms of magnitude?

  • $\begingroup$ The diagram is incorrect, $\vec{L}$ is not axial to the body due to the precession $\omega_p$ which contributes to angular momentum as well as spin. It is also missing the ground reaction which is what causes the torque at the CM and may not be equal to $Mg$ in all cases. $\endgroup$
    – JAlex
    Jun 15, 2021 at 15:20
  • $\begingroup$ About the physics: does this 2012 physics stackexchange discussion of gyroscopic precession answer your question? (That answer is written by me.). That discussion treats the most symmetrical case: a gimbal mounted gyroscope wheel so that all rotation axes go through the wheel's center of mass (with torque provided by an additional weight). Once you understand that most symmetrical case: the understanding transfers to the case here: the top's own weight gives rise to a torque. $\endgroup$
    – Cleonis
    Jun 23, 2021 at 18:56

3 Answers 3


I have added the "new" angular momentum although I have not moved the image of the top itself as this would make the diagram more difficult to interpret.

enter image description here

So you have $\vec L_{\rm new}= \vec L_{\rm old} + d\vec L$ with the change in angular momentum, $d\vec L$, caused by the torque, $\tau$, applied on the top about point $O$ due to the force of attraction of the Earth $mg$.

The tip of the angular momentum vector, $\vec L$, (and the axis of the top) moves around the circle shown dashed and labelled $\omega_{\rm p}$.


It is the torque $\vec\tau$ that gives rise to the precession. That is, it gives rise to the rotation of the axis $\vec L$ (as in your diagram) about the z-axis.

Torque is directly proportional to the timed rate of change of angular angular velocity (and identical to the timed rate of change of angular momentum) as is defined as $$\vec \tau=I\frac{d\vec \omega}{dt}=\frac{\vec {dL}}{dt} \rightarrow \vec {dL}=\vec \tau dt \ \ \langle \text{as in your diagram above}\rangle$$ where $I$ is the top's moment of inertia, and $\vec\omega$ is the angular velocity of the spinning top which is not $\vec\omega_p$ (as in your diagram), and $\vec\omega_p$ represents the rate at which the axis (pointing in the direction of $\vec L$ in your diagram) is rotating (precessing) about the vertical. The new vector $\vec {dL}$ is the infinitesimal change in the angular momentum $\vec L$ that is due to the torque $\vec \tau$.

It is the torque itself which acts on the center of mass of the top that causes the precession. You will notice that as soon as you start the top spinning, it is this torque which will cause a precession (in other words, the appearance of the second axis), and as the top begins to slow down, the precession angle $\theta$ gets larger and larger until the top falls.


Newton's Second Law for Rotation reads $\vec{\tau} = \frac{d\vec{L}}{dt},$ where $\vec{L}=\int \vec{r}\times\vec{p}$. Newton's Second Law holds in any inertial frame for all objects.

The reason why we think of two angular momenta separately in Gyroscope precession is that the expression for $\vec{L}$ can be simplified as $\vec{L}=\vec{L}_{p}+\vec{L}_{cm}$ due to the particular geometry of this problem. The remaining tasks are to (1) choose the inertial frame with which to solve the problem, and (2) solve the vector equation $\vec{\tau} = \frac{d\vec{L}}{dt}$.

Final question: which angular momentum is the torque changing? Strictly speaking, the total torque acts on the total angular momentum of the object. However, since $\frac{d\vec{L_{p}}}{dt} \perp \vec{\tau}$ in this problem, we can effectively say that the torque changes only the spin angular momentum of the gyro itself.


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