$\vec{\tau}$ = torque
$\vec{L}$ = angular momentum
$\vec{R}$ = radius
$\vec{F}$ = force
$M$ = mass
$\vec{A}$ = acceleration
$\vec{V}$ = velocity

I am familiar with the proof that $\vec{\tau} = \vec{R} \times \vec{F} = \vec{R} \times M\vec{A} = \vec{R} \times M\frac{\mathrm{d}}{\mathrm{dt}}\vec{V} = \frac{\mathrm{d}}{\mathrm{dt}} (\vec{R} \times M\vec{V}) = \frac{\mathrm{d}}{\mathrm{dt}}\vec{L}$ , and in my research, this is the proof that is cited to justify the claim that the torque on a gyroscope due to gravity is equal to the derivative of its spin angular momentum. But in the case of the gyroscope, the torque radius (extending from the pivot to the center of the gyro wheel) and the angular momentum radius (the radius of the gyro wheel) are not the same, that is, the $\vec{R}$ in the torque equation is not, according to my understanding, the same as the $\vec{R}$ in the angular momentum equation. So why does this proof apparently suffice to make this claim?

For example, the following diagram:


$\vec{d}$ = the arm extending from the pivot to the gyro wheel
$\vec{r}$ = radius of the gyro wheel
$\vec{g}$ = acceleration due to gravity
$\vec{V} = \vec{\omega} \times \vec{r}$ = spin velocity of gyro

What is the proof that $\vec{d} \times M\vec{g} = \frac{\mathrm{d}}{\mathrm{dt}}(\vec{r} \times M\vec{V})$

  • $\begingroup$ It's not clear what you have in mind when you refer to 'the angular momentum radius'. When a gyro wheel is in precessing motion the spin axis of the gyro wheel is sweeping out an angle; at some angular velocity. The rate of change of angular momentum correlates with that angular velocity (rate of change of angle). The angle being swept out is angle with respect to the pivot point. The torque radius is also with respect to the pivot point. $\endgroup$
    – Cleonis
    Nov 5, 2022 at 19:57
  • $\begingroup$ By 'angular momentum radius', I mean the radius of the gyro wheel. (I updated the question to hopefully clarify this.) Since the radius from the torque equation, which extends from the pivot to the gyro wheel center, is parallel to the angular momentum vector, it couldn't possibly be the same radius that is used in the cross product equation for angular momentum, right? So what is the correct formula for angular momentum in this case whose time derivative can be shown to be equal to the torque? $\endgroup$ Nov 6, 2022 at 4:58
  • $\begingroup$ One thing I find very odd here: in the case of a spinning gyroscop wheel the standard convention is to express the angular momentum of the gyro wheel in terms of the angular velocity of the gyro wheel, and the moment of inertia of the gyro wheel. $L=I\omega$ A gyro wheel cannot be treated as a point mass; the mass is distributed, symmetrical with respect to the intended spin axis. In the expression you ask about the angular momentum expression is of the form used for a point mass in circumnavigating motion. That mismatch does not make sense. It would appear that mismatch is throwing you off. $\endgroup$
    – Cleonis
    Nov 6, 2022 at 9:12
  • $\begingroup$ So does that mean that the proof at the top of this question is not sufficient to prove that $\vec{\tau} = \frac{\mathrm{d}}{\mathrm{dt}}\vec{L}$ for non-point masses? My textbook seems to use that proof as justification for claiming $\vec{\tau} = \frac{\mathrm{d}}{\mathrm{dt}}\vec{L}$ in all cases. Perhaps the more general proof for non-point masses is beyond the scope of my textbook and the authors failed to mention this? $\endgroup$ Nov 6, 2022 at 9:45
  • $\begingroup$ I tried writing a proof in terms of a summation over point masses. I got as far as showing that $\frac{\mathrm{d}}{\mathrm{dt}}L = I\alpha$, where $I\alpha$ is the wheel's moment of inertia times its spin angular acceleration. But I don't know how to prove that the torque around the center of mass, $\tau_0 = I\alpha$, and my textbook doesn't offer any proof of this either, it just says $\tau_0 = I\alpha$ because $\tau = \frac{\mathrm{d}}{\mathrm{dt}}L$ $\endgroup$ Nov 8, 2022 at 11:41

1 Answer 1


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What is the proof that $\vec{d} \times M\vec{g} = \frac{\mathrm{d}}{\mathrm{dt}}(\vec{r} \times M\vec{V})$

I don't think that this is correct ?.

with: $$\vec{d} \times M\vec{g}=\vec\tau=\frac{d}{dt}\vec L\tag 1$$

where $~\vec L~$ is the angular momentum of system.

but only where the components of the torque $~\vec\tau~$ are zero ,the components of the angular momentum $~\vec L~$ are conserved.


$$\vec L=\vec d\times M\vec v+\Theta\,\vec\omega$$ and

$$\vec d= d\left[ \begin {array}{c} -\sin \left( \psi \right) \cos \left( \varphi \right) \\ \cos \left( \psi \right) \cos \left( \varphi \right) \\ \sin \left( \varphi \right) \end {array} \right] $$

$$\vec v=d\, \left[ \begin {array}{c} \sin \left( \psi \right) \sin \left( \varphi \right) \dot\varphi -\cos \left( \psi \right) \cos \left( \varphi \right) \dot\psi \\ -\cos \left( \psi \right) \sin \left( \varphi \right) \dot\varphi -\sin \left( \psi \right) \cos \left( \varphi \right) p\psi \\ \cos \left( \varphi \right) \dot\varphi \end {array} \right] $$

$$\vec\omega= \left[ \begin {array}{c} \dot\varphi +\sin \left( \psi \right) \omega \\ \sin \left( \varphi \right) \dot\psi +\cos \left( \psi \right) \omega\,\cos \left( \varphi \right) \\ \cos \left( \varphi \right) \dot\psi -\cos \left( \psi \right) \omega\,\sin \left( \varphi \right) \end {array} \right] $$

$$\mathbf\Theta= \left[ \begin {array}{ccc} I_{{x}}&0&0\\ 0&I_{{y}}+ I_{{w}}&0\\ 0&0&I_{{z}}\end {array} \right] $$

thus equation (1)

$$\vec\tau=M\,g\,d\,\left[ \begin {array}{c} \cos \left( \psi \right) \cos \left( \varphi \right) \\ \sin \left( \psi \right) \cos \left( \varphi \right) \\ 0\end {array} \right] $$

from here

$$(\vec L)_z=\left(\vec d\times M\vec v+\Theta\,\vec\omega\right)_z=\left( M \left( \cos \left( \varphi \right) \right) ^{2}{d}^{2}+{ \it I_z}\,\cos \left( \varphi \right) \right) \dot\psi -{\it I_z}\,\cos \left( \psi \right) \omega\,\sin \left( \varphi \right) =\text{constant}$$

  • $\begingroup$ I'm a bit confused, I don't see where the proof that $\tau = \frac{\mathrm{d}}{\mathrm{dt}}L$ is in this answer. Am I missing it? I see a proof that the angular momentum about the z axis is constant. But ultimately what I'm looking for is a proof that the total external torque is equal to the derivative of the spin angular momentum of the gyro wheel. I agree that my original formulation of the equation to be proved is incorrect; the spin angular momentum can't be represented as a single point-mass cross-product. $\endgroup$ Nov 9, 2022 at 6:58
  • $\begingroup$ Angular momentum vector \begin{align*} \frac{d\vec p}{dt}&=\vec{F}\\ L&=(\vec{x}\times\vec{p})\quad\Rightarrow\\\\ \frac{d\vec{L}}{dt}& =\frac{d}{dt}\left(\vec{x}\times\vec{p}\right) =\underbrace{\dot{\vec{x}}\times\vec{p}}_{=\vec{0}}+\vec{x}\times\dot{\vec{p}}=\vec{x} \times\vec{F}=\vec\tau \end{align*} $\endgroup$
    – Eli
    Nov 9, 2022 at 13:26
  • $\begingroup$ I think this proof is equivalent to the one at the beginning of my question. Since $p=mv$, I believe this is only a proof for point masses. I am looking for a proof that deals with non-point masses, or else I am looking for an explanation for why the point-mass proof is sufficient to prove that $\tau = \frac{d}{dt}L$ for non-point masses as well. For example, can you show that $\tau = I_0\alpha$ where $\alpha$ is the spin angular acceleration of the gyro wheel? $\endgroup$ Nov 10, 2022 at 4:34

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