h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

I was trying to explain the significance of a spinning top staying upright to my daughter last night, but suddenly realised I had no idea how it actually worked.

What exactly causes a spinning gyroscope to oppose any forces applied perpendicular to the plane of rotation, when a stationary one would just yield?

Angular momentum is the short answer.

For example, when a ball on string is swung around, basically the *speed* of the ball is constant, but the *velocity* (vector representing speed and direction) is continuously changing. So the change in velocity vector is just a change in direction. This change in direction is caused by the force applied by string on the ball. So basically, angular momentum is the normal, linear momentum of the ball, with a careully applied force which causes the ball to orbit.

For the top analogy, the outer parts of the top also want to go off in their separate, linear directions, but the the material properties hold the whole thing together, resulting in the spin.

So why when you apply a force to the axis of the top does it move at 90 degrees to the applied force?

We are talking about a perfectly upright top without precession here, Gnomon? Where 'it' equals the whole spinning top and not the angle of the central axis, yes?

The prosaic explanation is that you are not applying a constant force, accounting for friction, and that the top 'rolls' down the side of whatever it is you are applying the force with (let's say it's your finger). It then elastically rebounds at a normal angle from the _side_ of your finger because it is being impeded in that particular direction. This, of course, mucks about with the vector product that is angular momentum, and makes the top wheel off at right angles.

B

If your daughter is a child, there's a simple non-technical 'story' that can help with an initial 'grasp' of a gyroscope. I'll tell if you like...?

I'm up for hearing it, even though I'm neither a daughter nor a child

Yes please, let's here it.

hear it, even

I'm still not getting this though. Maybe I'm being thick this morning, but to me angular momentum is basically just saying "because it's spinning" isn't it?

An example: Years ago I once took a slight bend on a motorbike, at some speed that probably wasn't very sensible. I tried to lean the bike, but it decided it didn't want to lean and nearly put me in the barriers.

I can understand why momentum of the wheel (rims, say) want to keep going in a linear direction in the plane of rotation. But when I tried to lean, I applied a force to the left and the wheels applied one to the right in response. The wheels' mass hadn't changed, and their velocity - and therefore momentum - only acts in one plane, so the left-right forces should be completely independent of the fore-aft ones, shouldn't they? This is certainly the case with something at a constant velocity, so why is it different with something with a constant, albeit angular, acceleration? Where did the right-hand force come from?

That incident on your bike put me in mind of:

On a road. I hit an icy patch on a Right-hand bend (Left-hander to US & similar). The skid took me to (and over) the middle of the road.
perhaps not analogous, but similar to yours in effect, and not yet explained.
____

OK I'll try the 'story' & see if this one works:

Imagine (or mark) a bum towards the underside of your top (! a rounded 'W'?)

Set it spinning.

It gets tired & tries to sit down. Problem is, its bum's moved around, and around, and... so it's always trying but not quite making it.
____

It made me smile - I think it was at the initial induction talk (all departments together) for Sperry Gyro in Bracknell.

Precession, of course, is caused by the constant trying but only getting a little way, a bit like humanity really - it never learns.

Humanity's taken things a bit further, though - too clever by half but not half intelligent enough?


Dammit, why does my wireless router keep dropping out?

Bike's have a lot of mechanical compensation built in, I don't think you're going to get anywhere trying to understand angular momentum using a motorcycle.

angular momentum = linear momentum + constant, centrally applied force

Not aware of any compensation for gyroscoping wheels. In fact a bike relies on the rider's ability to apply sideways forces for it to work. I'm good with angular momentum, just don't understand how it can generate a force perpendicular to its own direction.

Rod - nice analogy, but same applies, what causes the upwards force that prevents the bum from sitting down?

Your incident sounds like understeer - maybe the back wheel slipped and caused the bike to lean into (and turn into) the bend? Odd nonetheless.

<>

'cos it's rubbish, stick to wires

Ah, Dave, your daughter's a bit more than a child, then. In that story/analogy, no upward force, just the downward force is effectively non-directional. And. If you accept that, you don't need an expert...

Saw this on telly once

sit on a swivel chair holding a wheel upright by the axle in both hands

get a helper to spin the wheel very fast

tilt the wheel over slightly

the chair will rotate as the wheel tries to get back to the vertical

try it

cheers

Taff
agent of kaos

"just don't understand how it can generate a force perpendicular to its own direction. "

Linear momentum does the same thing. It takes "force" to change the course of something travelling in a straight line, so it feels like the object is applying force in the direction perpendicular to which it is travelling.

Yes, but that force is not dependent on the object's speed or momentum. The force required to accelerate an object with mass M by a certain amount in direction Y does not vary with it's motion in direction X.

E.g. throw a ball horizontally and drop one from the same height and they'll both hit the ground at the same time. Consider the rotational equivalent, and the bum on the rim of a stationary gyro and the bum on a spinning one should hit the ground at the same time. But they don't.

good point, especially about the spinning vs. non-spinning gyro.

According to my classical dynamics book, in a frictionless environment the spinning gyro will not fall down. Even if you start the gyro inclined at 45 degrees, it will merely precess while spinning, and will not fall down.

Here's my hand waving explanation*

Imagine the spinning gyroscopy, inclined at 45 degrees. First, the force of gravity can be decomposed into 2 perpendicular forces - one along the axis of rotation of the gyro, and one along the plane of the gyro. We can ignore the one along the axis of the gyro.

Now imagine looking just at the plane of the gyro. In this 2D view, there is the spinning disc of the gyro, and the force of gravity is being applied from the "top" of the plane towards the bottom.

In the plane of the gyro, we have 4 quadrants. Assuming counter-clockwise motion:
quad 1 (+x, +y) : the torque applied by gravity is against the motion of the gyro.
quad 2 (-x, +y) : torque is with the motion
quad 3 (-x, -y) : torque is with the motion
quad 4 (+x, -y) : torque is against the motion

If we look at "half-cycles" of the gyro, we can determine the net effect of the force of gravity.

half cycle "A" = quad 1 and quad 2
result of quad 1 is cancelled by quad 2. Therefore, no net change in velocity going from start of quad 1 to end of quad 2.

half cycle "B" = quad 3 and quad 4
same as above.

The net effect of combining half cycle "A" and half cycle "B" is that there is *no* motion in the direction of the force of gravity. Very counter-intuitive to say the least.

half cycle "C" = quad 2 and quad 3
torque is with the gyro motion for both, hence going from beginning of quad 2 to the end of quad 3 there is a net increase in velocity

half cycle "D" = quad 4 and quad 1
torque is against the gyro motion for both, hence there is a net decrease in velocity (beg. quad 4 --> end quad 1)

This indicates that the velocity at the end of quad 3 is greater than the velocity at the end of quad 1. The gyro is a solid object, so the net result of this difference in velocities, is a translation of the gyro to the right in our 2D plane image. The result in 3D coordinates is a precession of the gyro.

This is truly very counter-intuitive to me. The angular momentum comes into play here in a more subtle way,




*There's similar problem, I think, in which a charged particle moves in constant electric and magnetic fields, but the electric and magnetic fields are perpendicular to each other. In this case the magnetic field is the centrally directed forces that causes the electron to orbit, and the surprising result is that the net motion of the system is perpendicular to the electric field - the electron spirals in the direction perpendicular to both the electric and magnetic fields.

Aha! I had to sketch it out but I've got you. So there is no upwards force to prevent the thing toppling over as such - rather it inherently moves to compensate for any force applied. A bit like a Segway. It's certainly not obvious.

Yeah, I believe that's the case!

I was sketching for a while trying to make sense of it...

Replying to post 11:

> angular momentum = linear momentum + constant, centrally applied force

Actually, angular momentum (about some point in space) = sum of (component of linear momentum not pointing towards / away from that point multiplied by distance to the closest point of approach). Roughly.

Or sigma (r ^ p) if you prefer

It is just 3D vector algebra.

That is probably what you meant, but the following statement is true:

"Turn off the forces binding the particles of the wheel together and its angular momentum will not change."

which would not follow from your definition.

It is the fact that these particles are bound together as a solid object with a moment of inertia that makes this an interesting question rather than the nature of forces between them. It is the fact that 3D vector algebra is required that makes it conceptually tricky!