why or how does gyroscopic precession work?


why or how does gyroscopic precession work?

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3 Answers

Anonymous 0 Comments

Lets imagine a spinning bicycle wheel. Ignore the mass of the hub and spokes. Only the rim and tire have mass here.

Each part of the wheel follows a path around around the hub in a circle. If you apply a sideways force to the wheel at one spot, the direction of motion of that part of the wheel changes, but its displacement has not yet changed (because no time has passed since the force was applied). That part of the wheel hasn’t moved yet, but the direction it is travelling has. As the wheel continues to rotate, this part of the wheel deflects from the path it would have followed had no force been applied, displacing it some distance from where it otherwise would have been during that part of the wheel’s revolution.

By the time the wheel rotates 90 degrees from where the force was applied, the part of the wheel that was acted upon has now displaced some distance from where it would otherwise be.

Now because the wheel is a solid circular ring, the force that was applied to one part of the wheel is effectively mirrored on the other side, but acts in the opposite direction (imagine you tilt a stationary wheel. One side goes up, and the other side goes down). So the deflection caused by changing the momentum of the wheel peaks about 90 degrees from its initiation, then decreases and reaches zero 180 degrees from initiation, then peaks again but in the opposite direction at 270 degrees, before returning to zero after completing a full revolution.

The reason precession causes the spinning wheel to tilt in a different direcrion than a stationary wheen is because the rim now has momentum, which carries the effect the deflection some distance from the initial preturbrance. The effect is always 90 degrees from where the force acts, because 90 drgrees is half way between nodes where the forces act in opposite directions to each other.

Anonymous 0 Comments

Conservation of angular momentum. Gravity pulling the gyroscope down will try and change the angular momentum, but since it has to be conserved, it has to go somewhere it ends up being made perpendicular to both the force and the axis of rotation. Feeling this force is really the best way to describe it. If you are holding a spinning bicycle wheel, if you try to rotate the whole thing perpendicular to the axis of rotation, you feel another force fighting you perpendicular to both your rotation and the wheel’s rotation.

When the gyroscope is spinning, it has angular momentum, just like regular momentum, but for spinning objects. L = p × r = I * ω. Angular momentum is linear momentum cross product with the distance to the origin, or the moment of inertia times angular velocity. Moment of inertia is basically angular mass, whereas in linear motion, mass is directly proportional to inertia, moment of inertia depends on how far that mass is from the origin. Ie something close to the center of rotation doesn’t need to move as much to have the same angular speed. Understanding these terms isn’t entirely necessary for what we are going to talk about.

If we have a gyroscope spinning, and it’s axis of rotation is at some angle φ from the vertical, we will see gyroscopic precession. Gravity will try and pull the gyroscope down, and surface the gyroscope is on will push it up. These forces balance, so we don’t get linear acceleration (the gyroscope doesn’t fall down) but the torques (angular force) they create are not balanced, so we get angular acceleration (the gyroscope begins to be rotated). That torque will be τ = F × r = ΔL/Δt (that second half is just the same relationship between force and linear momentum F = Δp/Δt)

This is where things get hard to explain without pictures. ΔL = Δt(F×r) now the way cross products work is when you multiply 2 vectors, you get a 3rd vector perpendicular to both initial vectors. Again, understanding it isn’t critical, but knowing the direction is all we are gonna care about for now. Gravity is acting down, and r is just the distance from the center of mass to where the gyroscope touches the ground, so it’s at an angle of φ. Our resulting change of angular momentum is going to be perpendicular to both of those, and since it has to be perpendicular to the downward direction, it must be horizontal. And since it has to be perpendicular to our axis of rotation, it can’t speed up or slowdown our gyroscope. The result is the entire gyroscope being pushed around seemingly by nothing.

[Video to hopefully show what I said](https://youtu.be/ty9QSiVC2g0)

Anonymous 0 Comments

A body in motion tends to stay in motion because of inertia. As a wheel spins inertia causes the wheel to continue to spin even after the spinning force is removed.

When a wheel is left to hang then the top of the wheel wants to pivot away from the point of contact while the bottom pivots towards the point of contact.

At the top of the wheel the inertia is traveling in one direction, and the tilting force is in 90⁰, a right angle to this. To a certain point the force of gravity is greater than the inertia and the wheel tilts a bit Dependant on the speed of rotation caused by inertia. At a certain point the forces of inertia and the force of gravity are in equality and so the wheel stops dropping, and other factors cone into affect.

The force of gravity is stored until such time as it exceeds the force of inertia. The bottom of the wheel also stores the force of gravity bit in the opposite direction.

At 90⁰ there are now 3 forces at play, the revolutionary momentum, the force of gravity pushing away from the point of contact from the top of the wheel, and the force of gravity pushing toward the contact from the bottom.

It is impossible to store both go away and come here at the same time so the inequality is resolved.

The top of the wheel has been pushing away since the far side, yet the moving toward the point of contact is just starting,bso the push away force is applied minus the small come here force, so the wheel at 90⁰ pushed away, causing gyroscopic progression.

On the far side of the wheel the opposite takes place with the bottom of the wheels come here force greater than the tops go away force (since it is just starting to break the plane of rotation) so it progresses the opposite way.