Physics Malpractice

Via physics and physicists I see a story about how golf can be hazardous to your hearing. And the story botches the physics. (I don’t know if it’s the journalist or from the original journal article)

The coefficient of restitution (Cor) of a golf club is a measure of the efficiency of energy transfer between the golf club head and the golf ball. The upper Cor limit for a golf club in competition is 0.83, which means that a golf club head striking a golf ball at 100km per hour will cause the ball to travel at 83km/h.

Well, that’s just wrong. The Cor tells you about the kinetic energy, so it won’t be the same for the speed, because KE depends on v2. i.e. if a ball is dropped from 1 meter and bounces, returning to 0.83m, the impact speed is ~4.4 m/s and the return speed is ~4.0 m/s, which is 0.91 of the speed.

Another problem is that the mass of the clubhead is not the same as the mass of the ball. Even if the Cor applied to speed, the statement is incorrect. In the limiting case of Cor=1 and the ball’s mass being negligible, the ball would leave at twice the clubhead speed.

The actual equation is v = u*(1+e)/(1+m/M)

v is the ball’s speed, u is the clubhead speed, e is the Cor, m is the ball’s mass and M is the clubhead mass. (This is trivially derived using conservation of momentum and balancing the kinetic energy equation to account for the loss) Using e = 0.83, and assiming the M=4m, we see that v = 1.46u

Update — This is using a definition of Cor on terms of energy. I couldn’t find how the USGA was defining it when I was composing the post, but further research (and noted in the comments) indicates that it is indeed the fraction of the speed retained after the collision. That changes the details of the analysis, but the article’s numbers are still wrong — the ball’s speed is larger than the clubhead speed. I still haven’t found a mathematical definition of how the USGA applies this to a golf club

That makes e in the equation the square of the Cor (so e = 0.689), which means that the ball leaves the clubhead at v = 1.35u

Rolling, Rolling, Rolling

A few days ago I was relating the cans-in-a-blanket problem, and retelling the vacuum joke and story to someone who had not yet heard them. One of my colleagues commented on a problem he had been given during an interview, also involving cans of soda:

You have two cans, one filled with ice and the other with liquid, but otherwise identical. The cans are rolled down an incline. Which one reaches the bottom first?

Much like the previous problem, I think there is a common misconception at play here for some people who get the answer wrong, and I’ll get to the explanation below. One of the people in the conversation said his first impulse was the wrong answer, but when we discussed the physics, we all agreed on the solution.

I set up to do a demonstration, though my first attempt was thwarted — I filled up a can with water and popped it in the freezer, hoping the can would be strong enough to hold together and have the ice expand vertically. It wasn’t.

first-attempt

I think the problem being that since ice will freeze from the top down and outside-in, the ice adhered to the can too well to let it expand upward as much as I hoped. (BTW — Black Cherry Citrus? Blecch. I bought it by accident when they redesigned their color scheme and introduced the flavor)

So I did it again, adding a little bit of water and letting that freeze, repeating the process several times until it was full, and it worked. Here is the experiment to investigate the problem given above:

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For those who think that the liquid-filled can will roll more slowly, I think I know what the misconception is: most of us have seen or done the experiment with spinning an egg, and a hard-boiled egg spins readily while the unboiled egg doesn’t. So the intuition is that since liquids don’t spin readily, the liquid-filled can won’t want to roll very fast. And, as we can see, that’s wrong.

The reason the intuition is wrong is from a misinterpretation of the reason the unboiled egg doesn’t spin — it’s because it’s difficult to transfer energy and angular momentum to the liquid by spinning the container; the coupling between them is weak. And angular momentum tells you the tendency for something to spin — it only changes when you apply a torque. With the soda cans it means that the work being done, adding energy (gravity acts on it, and there is a torque from the friction of the treadmill causing rotation)but this energy isn’t being added to the liquid, so it must be going into the can itself, which isn’t very massive — almost all of the energy goes into translational kinetic energy. The frozen water, though, does rotate with the can, so the gravitational potential energy has to be shared between translation and rotation of the can + ice system, so the translational kinetic energy (and therefore speed) is smaller.