Shoot the Hostage

(Because, as anybody knows, that’s the answer to “Pop Quiz, Hotshot”…)

The answer to the pop quiz posted below is “v.” That is, the speed is unchanged between the start of the problem and the collision between the ball and the pole. There are several ways to see this– conservation of energy is my usual approach (the only energy at the start of the problem is the kinetic energy of the ball’s motion, and nothing else in the problem takes up any energy, so you’ve got to have the same kinetic energy at the end)., but I really like Ross Smith’s dimensional analysis argument. If I were giving prizes, he’d get one.

A couple of other comments on things that were said by various people:


  • Angular momentum is a red herring. If you look carefully at the situation, you’ll find that the string is not quite perpendicular to the ball’s velocity (it might help to draw a better picture than mine), so there’s a small component of force in the backwards direction. This can be thought of as a torque that keeps the angular momentum from increasing as the ball is pulled in.
  • If the force isn’t perpendicular, why doesn’t the magnitude of the velocity change? It would change, if the force was perpendicular– the work done moving the ball in to a smaller orbit would be converted to kinetic energy, and the ball’s speed would increase (the inward force is slightly greater than the required centripetal force). The fact that it doesn’t increase is due to the torque thing mentioned above.
  • Doesn’t a tetherball speed up as it gets closer to the post? The linear speed doesn’t change, but the angular speed does. As the orbits get smaller, the ball takes less time to complete each revolution, so it appears to be moving faster.

Credit where due: I heard this problem from somebody on Usenet years ago, long enough that I’ve forgotten who it was. It was introduced as a problem from a test given to prospective TA’s at one of the Ivies, consisting of problems that could be solved using freshman physics principles that would induce the same sort of confusion in first-year grad students that the classic frictionless-block-sliding-on-an-inclined-plane problems induce in actual freshmen.

These sorts of problems are sort of hard to come by, so I doubt I’ll be making this a regular feature. If you know any good ones (I’m stealing the three-cylinders one from BioCurious for the next time I teach something having to do with thermodynamics), I’d love to hear them.

Comments

  1. #1 Ben V-L
    February 3, 2006

    Ummm. Sorry, but that’s wrong.

    The string is perpendicular to the ball’s instantaneous velocity. Which is a good thing, because that is why the string does no work on the ball, which is why the speed doesn’t change.

    The angular momentum issue is that the string is not oriented in the radial direction, so r x F gives a nonzero torque with respect to the origin and angular momentum isn’t conserved.

  2. #2 Ben V-L
    February 3, 2006

    For what it’s worth, I’m not as sold on Ross Smith’s argument either. I love these kinds of arguments, and he was fine up to v_f = v*f(L/R) and even f(0)=1. But the observation that “the ball doesn’t speed up by a huge factor” doesn’t just the extension to f(x)=1 for all x.

  3. #3 Ben V-L
    February 3, 2006

    I don’t want to come across as a font of negativity, so let me add a positive comparison. There is a related problem of a hole in a table. A ball on the frictionless table is attached to string, the other end of the string passes through the hole in the table where it’s attached to something, and the ball is circling the hole. So it looks like Chad’s problem, only this time the string is radial, so there is no torque and angular momentum is conserved.

    Now start pulling the string from below the table so that the distance from the ball to the hole is decreased from L to R. The string remains radial, so angular momentum is conserved, but the radius of the ball’s orbit is decreasing. From ang mom cons: v*L = v_f*R, so v_f = (L/R)*v (an example like Ross Smith’s where the function isn’t 1).

    So how does this square with work? At one level it seems obvious: to pull the string one has to do work, and this ends up in the increased kinetic energy of the ball. But how can the radially directed string do work on the ball? The answer is that the inward spiraling ball has a component of velocity in the radial direction. That is, in this case the string is not perpendicular to the velocity of the ball.

    I think these problems work well in tandem as an application for work and angular momentum concepts. When can you use which principle.

  4. #4 Bryan Bischof
    February 3, 2006

    Wow i feel sheepish for overlooking conservation of energy… I was looking at the angular momentum and i had arrived at the fact that the counter torque would take care of it but i hadn’t concluded its effect.

    Thanks, this was fun!

  5. #5 Chad Orzel
    February 3, 2006

    The string is perpendicular to the ball’s instantaneous velocity. Which is a good thing, because that is why the string does no work on the ball, which is why the speed doesn’t change.

    The angular momentum issue is that the string is not oriented in the radial direction, so r x F gives a nonzero torque with respect to the origin and angular momentum isn’t conserved.

    You’re absolutely right, and this is what I get for posting at 7:00 in the morning…

    What a week…

  6. #6 PhilipJ
    February 3, 2006

    The Tether-ball problem as I’m now going to call it will definitely end up in a problem set or tutorial I have to give in a future classical mechanics course. That one is great.

    As for more three-cylinders type problems, see the UWashington group‘s paper, Am. J. Phys. 73, 1055 (2005) for more! There are three or four really excellent problems that have all been used in the thermo course I’m TAing now, either on problem sets or in quizzes.

  7. #7 kyle
    February 3, 2006

    I learned one that was exceptionally good to explain why we need to resort to thermodynamics to calculate the behavior of a macroscopic object. It has to do with isolation.

    Assume you have a macroscopic, classical system a liter of hard sphere gas molecules. Now assume that you have isolated the system from the enviroment as well as you can. This means that you haven’t isolated it gravitationally, since there is no known way to do this even in principle.

    Now if someone living near Proxima Centauri were to move 1 electron by 1 centimeter, how long would it take for the tidal force changes this causes to change the trajectories of the gas molecules enough so that they begin to overlap.

    I’ll give you a hint. Geometric series are awesome. And it’s not a long time. Okay that ws two hints.

    The reason this problem is interesting is that is point out to the futility of attempting to think of a macroscopic object as the time evolution of 10^23 ballistic objects. Even if you assume you had a computer that could calculate the time dependent positions of these particles for all time to some great precision (a rather silly assumption i know but bear with me) then if the external conditions change by truly miniscule… nay picoscule ammounts, your calculations are incorrect.

  8. #8 Kirchner
    September 10, 2008

    Is gravity ignored for this problem? In that case the mass has kinetic and potential energy, and it will speed up as it falls.

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