User:Art Carlson/Gyroscopic exercise tool

This is trying to figure out the physics of the Power Ball. If I can make it coherent, and it isn't too much OR, I'll add it to the article later. --Art Carlson (talk) 13:30, 23 February 2010 (UTC)

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The fundamental equation describing the behavior of the gyroscope is:

${\boldsymbol {\tau }}={{d\mathbf {L} } \over {dt}}={{d(I{\boldsymbol {\omega }})} \over {dt}}=I{\boldsymbol {\alpha }}$

where the vectors τ and L are, respectively, the torque on the gyroscope and its angular momentum, the scalar I is its moment of inertia, the vector ω is its angular velocity, and the vector α is its angular acceleration.

Under a constant torque of magnitude τ, the gyroscope's speed of precession Ω_P is inversely proportional to L, the magnitude of its angular momentum (assuming the applied torque is a right angles to the angular momentum):

$\tau ={\mathit {\Omega }}_{\mathrm {P} }L={\mathit {\Omega }}_{\mathrm {P} }I\omega \!$

The axle will roll along the side of the groove when speed of the surface of the axle is equal to the speed with which the axle moves along the groove:

$\omega r_{\mathrm {axle} }={\mathit {\Omega }}_{\mathrm {P} }R_{\mathrm {groove} }$

For a given rate of rotation of the central wheel, the geometry determines what the rate of precession must be, and this is achieved by applying a particular torque, given by

$\tau =\left(\omega r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)I\omega =\left(r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)I\omega ^{2}\!$

Note that the required torque increases quadratically with ω, and the frequency of precession (which is the speed at which you must swirl your hand) is directly proportional to ω.

(That the magnitude and direction of the roll at both ends of the axle is correct is an important but not immediately evident fact that must be explained here!)

These conditions provide for a smoothly moving motion, but where does the acceleration (or deceleration) come from?

You want to stay in synch with the precession of the gyroscope, so you can't produce acceleration by moving faster that the rate given above. It also doesn't help to introduce a phase shift, since torque in the orthogonal direction does not affect the gyroscope. What you can do is slightly increase the torque above the level given by the rolling condition. If the axle starts to slip, the dynamic friction will act in the direction to accelerate the gyroscope. If the axis does not slip, then the static friction supplies an additional torque, which pushes the axle in a direction ...

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Without loss of generality, one can assume that the axis of rotation is vertical and the direction of rotation is counter-clockwise as seen from above, that the groove goes around the ball from the front over the top to the back and then under the bottom, and that the applied torque is clockwise. When the clockwise torque is applied, the axis of rotation must shift such that the rotation acquires a component also rotating clockwise, that is, the top of the axle will move away and the bottom will move closer. Quantitatively, the torque is equal to the cross product of the angular velocity of precession and the angular momentum. Since the applied torque is clockwise, the top of the axle will move along the left side of the groove. This motion will reduce the relative velocity of the surface of the axle relative to the surface of the groove if the torque is small, and will even reverse the direction of the relative velocity if the torque is large enough. In the former case, friction will slow down the rotation, in the latter case, friction will accelerate the rotation. Therefore the relative velocity is a critical quantity. Without precession the relative velocity is equal to $\omega r_{\mathrm {axle} }$, where $r_{\mathrm {axle} }$ is the radius of the axle. The precession reduces this velocity by ${\mathit {\Omega }}_{\mathrm {P} }R_{\mathrm {groove} }$, where $R_{\mathrm {groove} }$ is the radius of the groove (half the length of the axle). Therefore the net relative velocity is equal given by

$v_{\mathrm {rel} }=\omega r_{\mathrm {axle} }-{\mathit {\Omega }}_{\mathrm {P} }R_{\mathrm {groove} }$.

The condition for acceleration of the rotation is $v_{\mathrm {rel} }<0$ or ${\mathit {\Omega }}_{\mathrm {P} }>\omega \left(r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)$. Since the relation between precession frequency and torque is $\tau ={\mathit {\Omega }}_{\mathrm {P} }I\omega$, the condition for acceleration of the rotation can also be written

$\tau >I\omega ^{2}\left(r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)$.

The direction of the force of sliding friction is opposed to the direction of the velocity but independent of the magnitude of the velocity. The magnitude:$F_{\mathrm {f} }$ is given by

$F_{\mathrm {f} }=\mu _{\mathrm {k} }F_{\mathrm {n} }$

where $\mu _{\mathrm {k} }$ is the kinetic coefficient of friction, and $F_{\mathrm {n} }$ is the normal force.

torque on the gyroscope is r_axle*(sum of F_f)

(sum of F_f) = mu_k*(sum of F_n)

(sum of F_n)*R_groove = tau

so torque on the gyroscope is mu_k*tau*(r_axle/R_groove)

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In the center of the device is a massive wheel fixed on an axle, which, when it is spinning, acts as a gyroscope. Surrounding the wheel is a shell with an equitorial groove around the inside. The ends of the axle are fitted into the groove so that the wheel has no direct contact with the shell, but the wheel can tilt by sliding the ends of the axle along the groove. The question is, How can the gyroscope be made to spin faster, simply by moving the shell?

Since the wheel is balanced, the only possibility to speed up the rotation is for the sides of the groove to exert forces on the ends of the axle. Furthermore, the normal and axial forces will have no effect, so tangential force must be provided by friction. Simply turning the shell parallel to the groove is not sufficient for two reasons. For one, the tangential forces on the ends of the axle will generally not both act to speed up the rotation. The second reason is that the shell cannot be rapidly turned for many revolutions by hand.

The solution to both problems is to apply a torque when the wheel is already spinning (at least slowly). This can be accomplished by tilting the shell in any direction except exactly in the plane of the groove, and results in a shift of the axle ends along the groove. The direction and speed of the shift can be found from the formula for the precession of a gyroscope: the applied torque is equal to the cross product of the angular velocity of precession and the angular momentum of the wheel. The most important observation here is that the direction is such that, if the torque is large enough, the friction between the axle and the surface of the groove will speed up the rotation.

This may seem odd. After all, if the axle were shifting in a horizontal groove, the friction on one end that acts to speed up the rotation should be cancelled by the friction at the other end, operating in the opposite direction. The difference is that a torque is being applied, so one end of the axle is pushing against one side of the groove, while the other end is pushing against the other side. Likewise, it doesn't matter in which direction the torque is applied. If the torque is reversed, each end of the axle will then be pressing against the opposite side of the groove, but the direction of precession is also reversed. The only restriction is that the relative speed of the surface of the axle and the side of the groove due to precession, $\omega r_{\mathrm {axle} }$, must exceed the relative speed due to the rotation of the wheel, ${\mathit {\Omega }}_{\mathrm {P} }R_{\mathrm {groove} }$. The minimum torque required to meet this condition is $I\omega ^{2}\left(r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)$, where I is the moment of inertia of the wheel, and ω is its angular velocity.

Since an acceleration of the rotation will occur regardless of the direction of the applied torque, as long as it is large enough, the device will function without any fine-tuning of the driving motion. The tilting of the shell does not have to have a particular phase relationship with the precession or even to have the same frequency. Since sliding (kinetic) friction is usually nearly as strong as static (sticking) friction, it is also not necessary to apply precisely the value of torque which will result in the axle rolling without slipping along the side of the groove. By applying the proportionality of the force of friction to the normal force, $F_{\mathrm {f} }=\mu _{\mathrm {k} }F_{\mathrm {n} }$, where $\mu _{\mathrm {k} }$ is the kinetic coefficient of friction, it can be shown that the torque spinning up the wheel is a factor of $\mu _{\mathrm {k} }\left(r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)$ smaller than the torque applied to the shell.

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How it works

A DynaBee gyroscopic wrist exerciser.

The device essentially consists of a spinning mass inside an outer shell. The shell almost completely covers the mass inside, with only a small round opening allowing the gyroscope to be manually started. The spinning mass is fixed to a thin metal axle, each end of which is trapped in a circular, equatorial groove in the outer shell. A lightweight ring with two notches in it for the ends of the axle rests in the groove. This ring can slip in the groove; it holds the spinning gyroscope centered in the shell, preventing the two from coming into contact (which would slow the gyro down), but still allowing the orientation of the axle to change.

Since the spinning mass is balanced, the only possibility to speed up the rotation is for the sides of the groove to exert forces on the ends of the axle. Furthermore, the normal and axial forces will have no effect, so tangential force must be provided by friction. If the axle is stationary, the friction will only act to slow down the rotation, but the situation is very different if the axle is turned by applying a torque.

This can be accomplished by tilting the shell in any direction except exactly in the plane of the groove, and results in a shift of the axle ends along the groove. The direction and speed of the shift can be found from the formula for the precession of a gyroscope: the applied torque is equal to the cross product of the angular velocity of precession and the angular momentum of the spinning mass. The most important observation here is that the direction is such that, if the torque is large enough, the friction between the axle and the surface of the groove will speed up the rotation.

This may seem odd. After all, if the axle were shifting in a horizontal groove, the friction on one end that acts to speed up the rotation would be canceled by the friction at the other end, operating in the opposite direction. The difference is that a torque is being applied, so one end of the axle is pushing against one side of the groove, while the other end is pushing against the other side. Likewise, it doesn't matter in which direction the torque is applied. If the torque is reversed, each end of the axle will then be pressing against the opposite side of the groove, but the direction of precession is also reversed. The only restriction is that the relative speed of the surface of the axle and the side of the groove due to precession, ${\mathit {\Omega }}_{\mathrm {P} }R_{\mathrm {groove} }$, must exceed the relative speed due to the rotation of the spinning mass, $\omega r_{\mathrm {axle} }$. The minimum torque required to meet this condition is $I\omega ^{2}\left(r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)$, where I is the moment of inertia of the spinning mass, and ω is its angular velocity.

Since an acceleration of the rotation will occur regardless of the direction of the applied torque, as long as it is large enough, the device will function without any fine-tuning of the driving motion. The tilting of the shell does not have to have a particular phase relationship with the precession or even to have the same frequency. Since sliding (kinetic) friction is usually nearly as strong as static (sticking) friction, it is also not necessary to apply precisely the value of torque which will result in the axle rolling without slipping along the side of the groove. These factors allow beginners to learn to speed up the rotation after only a few minutes of practice.

By applying the proportionality of the force of friction to the normal force, $F_{\mathrm {f} }=\mu _{\mathrm {k} }F_{\mathrm {n} }$, where $\mu _{\mathrm {k} }$ is the kinetic coefficient of friction, it can be shown that the torque spinning up the mass is a factor of $\mu _{\mathrm {k} }\left(r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)$ smaller than the torque applied to the shell. Since frictional force is essential for the device's operation, the groove must not be lubricated.

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Keeping with kinetic friction/slipping axle for now, how does the rotation of the flywheel respond to the torque around the y-axis?

$I{\dot {\omega }}=\mu _{\mathrm {k} }\left(r_{\mathrm {axle} }/R_{\mathrm {groove} }\right)\tau _{y}$

$\tau _{y}/I=\mu _{\mathrm {k} }^{-1}(R/r){\dot {\omega }}$

How do I make a connection between the slipping and the rolling cases? The equations look very different! Do I perhaps have to apply the ordering ${\dot {\omega }}\ll \omega ^{2}$?

For a constant torque, the rotation will change linearly with time, increasing or decreasing until it reaches the value $\omega ={\sqrt {(R/r)(\tau _{y}/I)}}$. With rolling, suddenly applying a constant torque to a previously constant omega will result in a quadratic increase in omega, which will overshoot the equilibrium value of $\omega ={\sqrt {(R/r)(\tau _{y}/I)}}$, and then oscillate around that value. What is the char4acteristic time in each case?

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The case of static friction, i.e., when the axle rolls rather than slips along the groove, is a more difficult problem. It turns out to be necessary to add an inertial term to the precession term in the equation for rotation around the y-axis:

$\tau _{y}=I{\ddot {\Theta }}_{y}-I\omega {\dot {\Theta }}_{x}$

The last term can be reduced (as for the slipping case) by the kinematic rolling constraint:

${\dot {\Theta }}_{x}=-(r/R)\omega$

The rotational acceleration around the y-axis in the first term on the right side can be found by differentiating the rotational velocity around that axis, in turn found from the equation for rotation around the x-axis (ignoring the inertia term, which can be shown to be small):

$\tau _{x}=I\omega {\dot {\Theta }}_{y}$

Since the torque around the x-axis and that around the z-axis are both the result of tangential forces (friction) on the contact surfaces, they can be related as:$\tau _{x}=(R/r)\tau _{z}$:

$\tau _{x}=(R/r)\tau _{z}$

Finally, the torque around the z-axis spins up the flywheel:

$\tau _{z}=I{\dot {\omega }}$

We are left with this expression:

${\dot {\Theta }}_{y}=(R/r){\dot {\omega }}/\omega =(R/r)(\mathrm {d} /\mathrm {d} t)\ln \omega$

Putting that all together, we have:

$\tau _{y}/I=(R/r)(\mathrm {d} /\mathrm {d} t)^{2}\ln \omega -(r/R)\omega ^{2}$

We recognize the previous formula for the critical torque $(r/R)I\omega ^{2}$, which is the condition for steady-state rolling and marks the boundary between acceleration and deceleration. If we fix the torque and consider small deviations, we see that we should expect (nearly) harmonic oscillations in the rotation of the flywheel.

Of course, we cannot in fact just assume that the surfaces will not slip. There is a maximum value of the ratio of the tangential force to the normal force, namely the coefficient of static friction, mu, for the materials involved. This can be written as a lower limit on the torque around the y-axis:

$\tau _{y}/I\geq \mu ^{-1}(R/r)\omega ^{2}$

The presence of the first derivative of omega in this expression makes it difficult to perceive the nature of the solutions.

If I stick to rolling, how much faster can I spin up the flywheel than if I use only slipping?

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