So, I remember this demo.
It was something that, back in grade school, the engineer father of one of my best friends really liked to put on -- did it on several occasions that I remember. It was one of these Physics is Cool demos.
Start with a room about 40-50 feet long, almost exactly like this one
, in fact if we can trust Google's scale marker
, that's pretty much what it was (yes, the room takes up the whole length of the building; yes, the Internet is really scary these days...). I remember the ceiling as being really high, but since I was smaller back then, that recollection is suspect. I also distinctly recall someone accidentally sticking the end of a flagpole into the ceiling (we had our Boy Scout troop meetings in that room), so it couldn't have been that
high. From the photo, 12 feet seems likely.
- There's a stepladder right at the back wall of the room.
- Somewhere towards the middle of the room, there's a hook in the ceiling.
- A piece of twine is tied to the hook, long enough to almost reach the floor; at the other end of the twine, a Rather Heavy Object of some sort, ... shotput or bowling ball or somesuch, I forget.
- Up at the front of the room is a chair with a 1-2 foot diameter basket on it.
- He pulls two volunteers out of the audience. Volunteer #1 is somebody's little sister. He positions her near the middle of the room, between the basket and the hook, and hands her this big-ass butcher's knife. "This is extremely sharp. I want you to hold this right here, just like that." Blade is pointed towards the back of the room, angled downward.
- Volunteer #2 takes the bowling ball up the stepladder. "Yeah, that's about high enough. Just let go of it."
- Ball swings forward, right down to the floor and up again, reaches the knife, the twine parts pretty much instantaneously, the ball sails through the air, and lands right in the middle of the basket.
And there was much rejoicing.
What's a bit frustrating now is that, since I was in 6th or 7th grade at the time, this was all before I'd had any Actual Physics. Meaning "Physics is Cool" was pretty much all
of the content that I got out of this. It was mentioned that the ball needed to be heavy and the knife needed to be sharp, but beyond that I had no idea which other details of the setup were crucial and which didn't matter at all. Nor was there really much of an explanation of why it worked, what principle was being demonstrated -- or maybe there was one, and I just didn't have the background to appreciate it and so it all just went kind of went whoosh.
Actually, now that I think about it, there may indeed have done some kind of brief "inertial foo gravitational mass mumble mumble", but I'm sure he knew up front that when you've got something that works mainly because it just drops out of the math and you've got an audience that doesn't know a whole lot of math, you're just going to lose them if you try to explain too much, so you just skip the boring part and get on with the show.
To be sure, it's fairly cool to be able to set up a scenario, calculate out in advance where things are going to go, then pull the trigger and have them Actually Go There
. Which perhaps describes pretty much every classroom physics demo ever
. And this may indeed be enough of a point for 7th & 8th graders (i.e., "Learn your math, kids. Keys to the universe!"
or maybe just, "It works, bitches
But still a bit frustrating, because, years later I learned some Actual Physics, and a few questions remain. Not that I'm particularly surprised that this worked, but,... what's the gimmick? Some particular sweet spot where to put the basket so that it doesn't matter quite so much where you put the knife or where you let the ball drop from? Or maybe it's a sweet spot in the knife placement or the original drop point? Was the scenario chosen the one that was easiest to calculate? Or the one that had the basket the furthest forward? Or the one had the longest flight time for the ball?
I suppose I could just call John up and ask, "Hey, that thing your dad did with the knife and the swinging shot put; what was the deal with that, anyway?" But this would be cheating. So I'm going to analyze this a bit. . .
There are two phases:
- Before the string is cut, we have a weight swinging on a string (duh). Once you know the initial conditions, the state of the system at any given time depends only on θ, the angle of the string from vertical. The force exerted by the string is always at right angles to the motion of the weight, thus never does any work. The only other force is gravity, which is happily constant (mg downward everywhere), and so, from conservation of energy (or a billion other different methods) we get mv²/2 = mgh
where h is the vertical distance the ball has fallen, or, since the mass m of the ball never changes (we're not doing relativity, and even if we were, people doing relativity generally don't think in those terms these days):
v²/2 = gh
We don't even have to measure any angles if we don't want to.
- Once the string has been cut, we now have a ball in free flight on the usual parabolic trajectory. Know the initial velocity (speed v, aimed at some angle θ from horizontal) and you know everything else: how high it goes, how far it goes. Horizontal velocity (vcosθ) is constant since there are no longer any horizontal forces. Vertical velocity is continually whittled down by gravity (vsinθ - gt); at some point (t = (vsinθ)/g) it gets to zero as the ball reaches its maximum height, an equal time later (t = (2vsinθ)/g) the ball has fallen back down to its original launch height, reaccumulated its original velocity, now directed downward, and the position is now (vcosθ)(2vsinθ)/g = (v²/g)(2sinθcosθ) = (v²/g) sin2θ feet farther down the room...
... which, just in case you were wondering, applies to any launching-crap-into-the-air scenario. E.g., you want to know what angle to throw a softball to get the best range, this tells you (sin() maxes out at 90 degrees, so with g fixed and v fixed (i.e., since your throwing arm is only so strong), you evidently need to throw at 45 degrees from horizontal, just like they always told you in gym class.
And now we see some gimmicks.
- the launch velocity depends solely on how much higher we dropped the ball from vs. how high the ball is at the moment we cut the string, as already noted
- knowing where the knife is also nails down the string angle θ at that point
- due to geometry, because the path of the ball up to that point is circular, the string angle from vertical is the launch angle from horizontal. The two θs are the same. Always.
Put everything together, we get something else interesting: the range from the point of release
(v²/g) sin2θ = (2gh/g)sin2θ = 2h sin2θ
does not even depend on g, the acceleration of gravity. Meaning this problem is now purely geometric
; we don't have to care what the release velocity actually is and the scenario works exactly the same way with all of the distances unchanged no matter if we're on Earth, the Moon or Mars or wherever (though it will
take different amounts of time
to play out in each case, but anyway...)
Though this is all independent of the mass of the ball, there are other factors we've been ignoring that could matter, particularly
- the weight of the string itself
- the possibility that whoever is releasing the ball might accidentally give it a little shove in one direction or another.
The heavier the ball is, the less we have to worry about either of these.
This still leaves any number of possibilities for h and θ. For h, it's fairly clear that the higher the better. But you still have to know h reliably; any mistake in measuring how high you're dropping the ball from gets multiplied by 2sin2θ. For now, let's suppose we're dropping all of the way from the ceiling -- keeps life simple; you know the length of the twine, r; then h = rcosθ.
What about θ? If we're going for ease of calculation θ = 45° is a pretty good candidate. sin2θ = 1. Basket is 2h in front of the knife.
Is that the best we can do? Actually, no. Look what happens if we decrease θ very slightly: h increases by some amount Δh and the release point moves back by roughly the same amount Δh, but the range increases by 2Δh
Solving for the best range, we want to maximize (wall o' math coming up...) rsinθ + 2hsin2θ = r(sinθ + 4sinθcos²θ) = rsinθ(5 - 4sin²θ)
d/dθ of that mess is rcosθ(5 - 12sin²θ) which zeroes out at sin²θ = 5/12 (just a little over 40 degrees) and a maximum range of h(40/12)sqrt(5/7) or 2.817h
or 33.8 feet forward from wherever the hook is if it's a 12-foot ceiling.
The other nice thing about going for maximum range, aside from the ooo-ahhh factor is that with (d/dθ)range being zero, mistakes in θ will only have a second-order effect on the range, which leaves that much more slack for whoever's little sister that was.