Conservation of angular momentum and the gyroscopic force.

In the pictured path of the wheel, the wheel starts out rolling forward from our perspective. The rolling rotation has an angular momentum represented by a vector pointing out from the left face. By western convention, if you give a thumbs up with your right hand and orient your fingers to point in the direction of rotation, your thumb points in the direction of the angular momentum vector. That vector is what must be conserved. The thumb vector is useful to think about and keep track of, but it’s the curling fingers (rotation) that’s actually physically relevant.

When pushed over, the angular momentum vector tilts upwards. To keep angular momentum conserved, gyroscopic forces will add rotation with an angular momentum pointing downwards. To use the right-hand rule again, point your right thumb in the direction of the vector (down) and your fingers curl in the direction that the gyroscopic force will turn the wheel (to the right, as pictured in your path drawing).

Silly notes on the right hand rule:

Of course, the vector pointing out along either pole/axel/axis could be used as the angular momentum vector. Early mathematicians chose the right hand rule to define which was the positive vs negative vector. You can try with your left hand to show to yourself that that convention choice still gives the same result. You may have learned the “gang sign” method of the right hand rule, but it’s terrible and I recommend forgetting it in favour of the rotation/curl based definition I use above that is much more physically intuitive and quick to assess. The gang sign method is needlessly tied to the mechanical steps of calculating a cross product which is less intuitive than thinking about the underlying rotation that the cross product represents.

This may not be a clever answer, but even you push it over, the wheel’s weight rest more on the right side, increasing the friction on the right and decreasing the friction on the left. This alone would cause a wheel to turn.

I think because the vectors of the centripetal and the force from the side add and point to front-right direction.

I would not be able to explain it easily enough in words, but please try visualise

lets suppose i have a seconds hand going around the clock (assume a disk basically, clock numbers are for my help to define positions), and I apply a force perpendicular to close face around 12, now there is my force pushing this point outwards, and natural turning dragging it towards 1, if you imagine, in that plane, we have 2 perpendicular forces (there is technically no force turning the disk towards 1, it is the existing momentum of disk)

going a little bit technical, for rotating objects,

L = sum (I_i omega_i)

where L is the angular momentum, a and I_i is the moment of inertia (consider just it as mass only) and omega_i the angular velocity (consider it speed only). The i subscripts denote the particular direction of this velocity, and corresponding inertia along that direction (this may be a bit of perspective shift, but basically momentum inertia is not just mass but mass distribution, consider it like this, turning a wheel around it’s axis is easy, but rotating it about it’s side (about a axis passing through diameter or spork of that wheel, because there is just a lot mass that is away from it))

also, newton’s second law is force is rate of change of momentum. and rotation equivalent of force is torque.

getting back to our problem, basically there is a existing anggular momentum that the disk has (its axis is perpendicular to face of disk), now if we apply a force on 12 position, and assuming disk is rigid, this force is acting parallel to this momentum, and hence it can not effect it (consider a person pushing the central pole of merry go round downwards, nothing will happen (assume force is not strong enough to break stuff)). but there are other methods how disk can possible rotate according to this force. the disk can potentially fall about a axis passing through plane of disk, passing through 6, and parallel to 3-9 (for visualisation, assume a round disk, like a coin balanced on its edge on a flat surface, and slightly nudge it, it will fall flat, then this axis is passing through the point of contact of coin, and if you look from side, you will see this disk rotating about this.) There are 2 things to note, 1 is, this axis is from center, and has a high inertia (for now, i would not show calculation, but trust me here.) and the disk was already rotating about the other axis.

Now to understand the motion of any part of disk we have to take both of these into account. lets take 12, 3, 6, and 9 (here out means perpendicular to disk face, parallel to direction of applied force)

12 - moves towards 1, and out 3 - moves towards 4, and out 6 - moves towards 7 (not out, since it is on the axis, this point does not move) 9 - moves towards 10, and out

now imagine a line 12-6, tilted outwards, and rotating (as plane) - does it not look like it is rotating about a 12-6 which did not tilt. This is infact what happens. When you add the 2 angular momentum, you also get a component of angular momentum, which is parallel to 12-6 axis, outside the disk (at some finte distance). (this is the hardest part probably, but math works out this way)(imagine you are doing a cart wheel, maybe you have seen videos of other doing it. you would have seen the ones where they fit themselves in a ring, and rotate with the ring, this the exact same situation, there the perpendicular force is provided by them tilting in the beginning). I would really recommended you to try this out with coins. This the very effect you are looking for.

mathematically

we have L_total = L_original + L_push

now L original is along disk axis, and L push is along a axis in plane parallel to 3-6. Now, for toppling, there is a very important thing that is required, that is friction. If there is no friction, then disk would just slide. Since disk is not sliding, there is a friction force that acts at 6 position pushing inwards. This will not change L push (force passing the axis does not have an effect, like you trying to open a door by pressing hinges). But this sure can generate a seperate L about another axis. this axis is parallel to 3-6, but not passing at 6, this will provide us a balancing condition for torques (basically, will the disk topple or not).