Be Master of the Midway—With Physics\!
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Come on. Some of those carnival games are just plain silly. There’s one ( check out the video ) with a bowling pin and a ball suspended from a string above it. You’re supposed to swing the ball around and hit the pin from behind. Spoiler alert: It can’t be done. This is a sucker game. How do I know? Physics, my friend.
If you swing the ball out to the side, it will circle around the pin like a planet orbiting a star, and the center of the orbit will be right below the spot on the ceiling where the string is attached. If the ball doesn’t pass through the center when you release it, it can’t do so on the return. No possible way.
In physics, both the swinging ball and the orbiting planet are called central force problems. What that means is that there is a net force on an object that pulls toward a central point. For a planet, it’s the gravitational pull of the star. For a ball on a string, it might not be so obvious. But let’s figure this out!
What Goes Around
Here's a diagram of a ball on a string, as seen from the side:
There are actually two forces acting on this ball: the downward gravitational force ( mg ) and the tension in the string ( T ). (Quantifying T is not trivial, because it's a “force of constraint.” Basically, the magnitude is whatever it takes to keep the ball from flying away. That's just how strings work.)
As you can see, the T force points up on a diagonal toward the pivot point of the string. But, as usual, we can decompose this force into horizontal and vertical components.
Now, to simplify, let’s assume the ball swings around in a horizontal plane, parallel to the table (mostly true). If we view it from above, the orbital path would look something like this:
Here, F r is the horizontal component of tension, and you can see that this is a central force. Wherever the ball is in its elliptical path, F r will point toward that center point, which is right below the pivot point of the string.
I also showed two other things above. One is an arrow representing the linear momentum of the ball ( p ) at a given instant, which is the product of its mass and velocity. Linear momentum is always tangent to the orbital path. (Why p for momentum? I guess m was already taken for mass.)
Second, I’m describing the position of the ball relative to the center point with an arrow labeled r , for radius. Note that r points away from the center; you'll see why that matters later. With these I can calculate the angular momentum of the ball, which is the whole key to this carny game.
What Is Angular Momentum?
Angular momentum is a measure of rotational motion. We can calculate it as the vector product of an object's position and its linear motion. (And for angular momentum we use the symbol L , because … to be honest, I have no idea.) That gives us the first equation below:
The arrows show that these are vector variables, meaning they have more than one dimension. Specifically, three: for the x, y, and z axes of the 3D space we live in. This lets them describe direction and location. An example would just look like this: (1, 5, 2). Not too scary, right?
Multiplying vectors is complicated, but in our case we can skip the work, because we really only need the magnitude of angular momentum, which is a scalar. And we can get that from the magnitudes of the p and r vectors, along with the sine of the angle θ between them. (Yes, I’ve used θ twice—sorry about that.) This gives us the equation on the right above.
Now that is pretty slick, because if you look at the orbital diagram again, you’ll see that the r and p vectors are always perpendicular, and the sine of a 90 degree angle is 1. So L = r × p . No arrows, nice and simple!
Let's Talk Torque
You know about torque, right? You use it every time you push on something to rotate it. For instance, when you open a door, the amount of torque you create depends on three things: (1) the force ( F ) that you apply (i.e., how hard you push), (2) the distance ( r ) from the door’s axis of rotation (the hinges) to the spot you push on, and (3) the angle (θ) between those force and distance vectors.
Mathematically, torque is also a vector, represented by the Greek letter tau (τ). But there is a scalar version of this equation too:
Think about it. It’s easier to open a door by pushing on a spot farther away from the hinges. Why? Because it gives you a bigger r . That's why doorknobs are where they are. And it’s easier if you push perpendicular to the door instead of at some off-kilter angle. Why? With a θ of 90 degrees, you get a maximum sin θ of 1. Result: small push ( F ), big torque (τ).
And what does torque do? It changes the angular momentum of an object. This is called the angular momentum principle, and it looks like this:
Now we can put it all together for the case of our carnival game. Since the horizontal component of the tension force ( F r ) points toward the center, and the vector r always points away from the center, the angle between them is always 180 degrees. The sine of 180 is zero, which means there is no net torque on the ball. No net torque means no change in angular momentum. Yes, the angular momentum of the ball is always the same .
A Simple Simulation
This is huge. But it might be difficult to grok, so let me make a numerical model. Here is an animation that is something like the motion of the ball (as seen from above). Oh, I made this in Python— here is the code .
Notice that at some points the ball gets closer to the center of the "orbit," which means the magnitude of the position vector r decreases. What happens to the ball's speed (and thus its momentum)? Here is a plot of momentum vs. position.
Right, the closer the ball gets to the center, the faster it goes and the more linear momentum it has. But what about angular momentum? Since that depends on both r and p , it's not so clear. One of these values goes down while the other goes up.
But wait! Don’t forget about the angle between r and p . As the ball moves around its “orbit,” the combination of these three quantities gives a constant angular momentum. So, really there are two reasons that the angular momentum is constant. First, because there is zero torque, and second, because of the mathematical relationship between r , p , and θ.
Save Your Money
So that carnival trick? Here's what we learned: If the ball passes through the center, then r = 0, and because L = r × p , the angular momentum is zero—which means it travels in a straight line, forward and back. That can't work, because the ball needs to swing around the pin and hit it from behind.
And if the ball passes on the side of the pin, then r > 0, and the angular momentum will be nonzero. And with nonzero angular momentum, it's impossible to pass through the center. The angular momentum would have to change, and we’ve already seen that it’s constant.
Which means: The only way to beat this game is to set the pin off-center. And that’s exactly what the carny guy does when he demonstrates it to people, as you can see at the end of the video. That way the ball can pass just to the side of the center on the way out. On the way back, it passes the same distance from the center to conserve angular momentum, but now there is a pin in its path.
Bottom line: This is a terrible game—but a great example of central force. And as a bonus prize, you can use your understanding of angular momentum to figure your odds before getting on those crazy, gyrating carnival rides.
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