Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Euler's laws of motion for a distributed mass are:

$$F = \frac{d}{dt} MV_{cm},\ N = \frac{d}{dt} L$$

$F$ are the sum of the external forces, $M$ the total mass, $V_{cm}$ the velocity of the centre of mass. $N$ are the sum of the moments of the external forces about some given point, L the total angular momentum about the same point.

If a gyroscope is supported at its base with its axis horizontal, it precesses at a constant angular velocity. Using the above equations, how does one show this?

share|cite|improve this question

Start off ignoring gravity. The spin axis is horizontal? Well then, you have an L vector. Very simple, nothing happens.

Now turn on gravity. This pulls down on the centre of mass, which is elsewhere from the pivot. That gives you an N vector - cross product of force with location relative to the pivot. This is perpendicular to the axis - so perpendicular to L. This is the change in L, according to N=dL/dt. In an arbitrarily small change in time, dt, we find dL will not change the length of L but will change its directions. Repeat indefinitely for every dt in a finite time interval t_1 to t_2.

share|cite|improve this answer
+1 This answer is close to what I'm looking for, but you haven't said anything about Euler's first equation for the motion of the centre of mass in a circle. – Larry Harson May 17 '11 at 12:05

You have to expand out $L=I\,\omega$ and notice how both terms vary with time. In many books they expand upon this to come to the equation of $\rm{d}L/\rm{d}t = I\,\dot{\omega} + \omega\times I\,\omega$

So there is a solution with $\omega>0$ and $\dot\omega=0$ for a constant $N$.

PS. Is this related to a homework problem?

EDIT: Do a Free Body Diagram to balance gravity with the reaction force to notice there is a net moment of the center of gravity $N>0$. So the above has a stead state solution of $N=\omega\times I\,\omega$ (with $\dot\omega=0$, $L=I\,\omega$). If the angular momentum $L$ is not aligned with the rotation axis $\omega$ then there exists a non-zero vector $\omega>0$ to satisfy the above equation.

Example: if $\vec\omega = [\cos\varphi\,\dot\theta,\sin\varphi\,\dot\theta,\Omega]$ where $\Omega$ is the precession speed, $\varphi$ the orientation/precession angle and $\dot\theta$ the spin rate, then the angular momentum in an axis alinged with the object but not spining is $L_\rm{body}=[I_{xx}\,\dot\theta,0,\Omega\,(I_{zz}+m\,L^2)]$. Then

$$ N = [\dot\theta,0,\Omega] \times [I_{xx}\,\dot\theta,0,\Omega\,(I_{zz}+m\,L^2)] = [0,\Omega\dot\theta (I_{xx}-I_{zz}-m\,L^2),0]$$

all expressed in this intermediate coordinate frame. So a torque component about the $y$-axis supports the gyroscope precession.

share|cite|improve this answer
it's not homework and I appreciate your answer, but it didn't answer my question using Euler's laws of motion that I stated at the beginning. – Larry Harson May 16 '11 at 23:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.