Re: Bullet penetration of spinning disc

To do this right, you would have to run a full simulation -- which may or may not then provide any insight as to what was really going on. So I'll instead tackle it from a simplistic physics perspective, and see what we can glean from this approach.

First of all, the reason that bullets can penetrate the disk is that a fast moving object can collide with (and push forward) the atoms in the disk. Molecular bonds also need to be broken, adding a lot of complexity, so we'll ignore that part. Each collision slows down the bullet a little bit due to conservation of momentum... so even if disk isn't spinning, the bullet is still strongly affected by the object. Each collision pushes the bullet in the opposite direction that the bullet is moving. The bullet will therefore come out of the disk moving slower than it went in, thanks to all the mini-collisions it experienced in the disk itself. (Make the disk thick enough, and it might not come out at all!)

So lets now spin the disk. Now there are two sets of collisions. One set, along the axis that the bullet is moving, still proceeds as described in the previous paragraph. But now there is another set of collisions happening along a perpendicular axis, where the atoms in the disk next to the bullet are colliding with the side of the bullet itself.

Now, it's an interesting thing about basic mechanics, that momentum in one axis is independent of momentum in a perpendicular axis. In other words, as far as these new collisions are concerned, it doesn't matter *how* fast the bullet is moving. In the axis along which these new collisions take place, all that matters is that the bullet has zero velocity (and zero momentum). This means that it will be relatively easy for the atoms in the disk to noticeably change the transverse momentum of the bullet.

As for how much the bullet will be deflected by this -- well, that depends on the numbers. If a 1cm-long bullet is moving at 1000 meters per second, it would only take 10^-5 seconds for it to pass through a thin disk (let's say the disk is 1mm thick). Lets also suppose the disk is moving at 100 meters/second (pretty fast!), and the bullet is 5mm in diameter. Then the distance that the disk moves during the collision is 100 m/s x 10 ^-5 s = 1mm. The total volume of atoms that collide with the side of the bullet is therefore 1mm x 5mm (bullet height) x 1mm (width of disk), or 5 mm³. Keep in mind these atoms are moving at 100 m/s.

Meanwhile, the atoms that the bullet hits head-on have a volume of the area of the bullet (pi x 2.5 mm x 2.5 mm) times the width of the disk (1mm), or almost 20mm³. Also these atoms presumably are knocked up to a speed of ~1000 m/s, so the total momentum change of these atoms is about 40 times more than the total momentum change of the atoms hitting the bullet on the side. (20 x 1000 vs. 5 x 100) However much the disk slows down the bullet, then, (say 200m/s) needs to be divided by 40 before figuring out how much transverse velocity the bullet gets (in this case 200/40 = 5 m/s.) So with these numbers the bullet would come out at 800m/s, with a transverse velocity of 5 m/s. The deflection angle can then be calculated with trigonometry, yielding about a third of a degree. (Not much!)

Of course, if you spin the disk much faster, you might even get to the point where there would be more energy in the disk than the bullet!

Hope that was helpful enough that you can play with the numbers on your own... And of course, a full treatement would require all sorts of material properties -- stress, strain, elasticity, plasticity, shock waves, relative densities, etc.