Not long ago, a friend of mine asked the following question:

*“Theoretically how fast would you have to be going on one of these things to be able to shoot a gun and then hit yourself with the bullet?”*

A good question indeed.

One could immediately answer “much faster than the bullet”, but we will soon see that the result actually depends on what you really mean when you say “speed of the bullet”.

We’ll make several assumptions that will make our lives easier. I’m sure my friend won’t mind being treated as a zero dimensional point in space. The bullet will be treated as point object as well. We’ll assume the shooter is an excellent sniper and has perfect aim, having no trouble calculating where to shoot in order to intersect the bullet in the future. Further, the bullet travels in a straight line parallel to ground level, and does not lose velocity due to drag (this is accurate enough for carousels that aren’t stupendously large, say, dozens to hundreds of meters in diameter, because then the bullet flight time will be very short).

Here is the carousel as viewed from above:

In this picture, the marksman fires the gun at point A during time t = 0. The bullet cuts through the air in a straight line, heading towards point B, eventually reaching it at time t_{1} (the straight purple line). In the meantime, the marksman goes in a curved circular arc, and is supposed to meet the bullet at point B at time t_{1} as well (the curved blue line). We’ll mark the half-angle of the arc as θ, as shown in the picture.

We immediately note that the path the marksman has to take is indeed longer than the path the bullet takes, so his linear speed should be larger than that of the bullet. Further, the larger θ is, the faster he has to go. For small angles, the curved path is nearly linear, so there shouldn’t be a big difference. But for an angle of θ = π/2, the bullet has to cross the carousel – it travels along the diameter. However, the marksman is carried on the edge – a half circle – so he goes a full π/2 greater distance. If we keep increasing θ beyond that, we see that the distance the bullet travels is actually shortened, while the distance the marksman has to go continually grows.

Now for some simple calculations:

Let’s denote the bullet velocity by V_{bullet}. This is the velocity of the bullet along the line AB, as if someone from outside the carousel were to fire a gun along the direction AB. As we will see later, this is not equal to the maw velocity of the gun. But for now let’s assume we know what it is. The distance it travels is:

Hence, the bullet flight time is:

Similarly, let’s denote the carousel angular velocity as ω. The angle the shooter on the edge of the carousel has to go through is 2θ. Hence, the carousel rotating time is:

Equating between the two, we can find the relation between ω and V_{bullet.}

It’s ok that we have R in there: we found out an equation for the angular velocity. At very large radii, even a small angular velocity translates into a very high linear velocity. The carousel’s linear velocity at the edge can easily be found out by V_{carousel} = ωR.

Let’s test some edge cases to see if we are ok. For small angles, θ ≈ sinθ, and we get that the speeds are pretty much the same. For θ = π/2, sinθ = 1, so we get the π/2 factor we considered earlier. Lastly, increasing θ to beyond π/2, sinθ only gets smaller, yielding a larger V_{carousel}, as expected.

It seems as if our question is finally answered. But in fact, there is a physical consideration which we have not yet dealt with that I would like to discuss. We have looked at this problem from an outside reference frame, meaning that we looked “from above” at the carousel and saw it spin. The bullet was fired with velocity V_{bullet} in the desired direction. However, we did not take into account the contribution of the carousel’s spin. The carousel’s spin changes the ratio between the velocity V_{bullet}, which is always in the right direction in order to intersect our suicidal marksman, and the gun’s maw velocity, which is the velocity with which the bullet leaves the gun’s barrel.

The carousel, spinning with angular velocity ω, has linear velocity V_{carousel} = ωR. This linear velocity **is added** to the gun’s maw velocity. So even if our gun did nothing, and just “dropped” the bullet out of the barrel, it will still go flying off at speed ωR to the side, ejecting out the carousel in a straight line.

Let’s consider the case where we want to hit ourselves after a half circle, meaning θ = π/2.

If we fire straight up, out bullet will **not** go straight up and meet us at point B. The carousel’s spin will give it some horizontal component, causing it to intersect the carousel’s edge at point P instead (the green line). If we want to hit point B, we are going to have to aim to the right, at such an angle α, so that our shot’s horizontal component will completely eliminate the velocity added by the carousel’s linear velocity (orange line).

We denote the gun’s maw velocity by V_{maw}. In order to compensate for the horizontal linear velocity, we have:

The remaining vertical component is:

Replacing V_{bullet} by 2V_{carousel}/π (as we got earlier for θ = π/2):

We now have a relation between the carousels linear velocity, and the gun’s maw velocity. The factor is larger than 1, so our carousel actually needs to travel slower than the maw velocity. In fact, we have:

So the simple quick answer “much faster than the bullet” turns out to be false in two ways: first, for small θ angles, it’s not “much faster”, and the two speeds are very similar. Second, you only need to go faster than the actual component along the bullet’s trajectory, and for some angles θ (but not all of them), the maw velocity has to “fight against” the spin of the carousel.

Homework for the reader: find out the relation between V_{maw} and V_{carousel} for any θ.