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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?

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2 Answers 2

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.

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+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.

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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

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