The Skater's Spiral.

[Rope secured and wrapped around a pole of radius R, free end extending 
straight.] A rope is tied securely and wrapped around a cylindrical pillar in the center of a skating rink, with a portion of the rope extending straight out from the surface of the cylinder. A skater on frictionless ice grabs onto the extended length of the rope. The skater's velocity is perpendicular to the straight portion of the rope when she grabs it. She retains her grip on the rope as she spirals around the pillar with continually decreasing radius. Assume that the rope segment she holds remains perfectly straight at all times. What is the skater's speed when she crashes into the pillar?

Answer

The skater crashes into the pillar with the same speed she had when she grabbed the rope. And her velocity at the time she contacts the pillar is perpendicular to the pillar's surface. The rope does no work on the skater. No work is done by or on the pillar. How can we know this without complicated analysis?

The rope has negligible mass compared to the skater, so we treat it as massless. The pillar, and the building it is part of, have huge mass, and may be treated as nearly motionless. So the force the pillar exerts on the rope does no work, for the portion of rope wrapped around the pillar doesn't have any component of motion along the rope's length. Therefore the other end of the rope does no work on the skater. Since no work is done on the skater, the skater's kinetic energy, and speed, remain constant.

[Tracing of the skater's path.] It follows that at all times the rope is perpendicular to the skater's velocity. That's true, though not obvious, and not particularly easy to show directly. It's probably a calculus problem. But you can get a lot of insight by taking a round bottle cap, fastening a string to it, and securing it on a sheet of paper. Put a loop in the free end, and with a pencil in that loop hold the string taut and draw the path on the paper as you swing around the round bottle cap. The path traced will be a smooth curve closing in on the cap, and the path will be perpendicular to the cap's rim where it contacts the cap. This picture was made in this way.

Consequences to ponder. The skater's speed remains constant. What about her angular velocity and angular momentum?

Answer: Since the speed and mass of the skater are both constant, the size of her linear momentum is constant also.

Now let's refocus our attention to a center of torques at the center of the pillar. The force of the rope on the skater is perpendicular to her velocity at any instant, as we noted. But this force does not pass through the center of torques, so it has a lever arm equal to the radius of the pillar. The sense of this produces a torque on the skater that opposes her velocity, thereby decreasing her angular momentum. At the moment of her impact with the pillar, her angular momentum has become zero and she impacts with a velocity that is directly toward the center of the pillar.

This has an interesting consequence. The angular momentum of the pillar must, by the momentum conservation law, be equal and opposite to that of the skater. The rope exerts a force on the pillar, with radius arm equal to the pillar radius, and the sense of this torque is such that the pillar's angular momentum increases. But torques do not cause acceleration (and do no work) unless there's motion. This is a subtle point. The end of the rope, where it first contacts the pillar, moving?

That point of contact progresses around the pillar's circumference. That end of the rope is steadily shortening. That "point" of contact is moving. Now we have learned how to deal with such situations.

If this seems like cheating, consider a motor driven winch with a rotating drum over which a rope is wrapped, lifting a load as it turns. The motor does work on the rope and the rope in turn does work on the load. The upper end of the rope is continually shortening and the work done on that end is the product of the tension there and the velocity with which the rope is wrapped up.

But wait. How can the angular momentum of that massive pillar be increasing when it isn't moving? It is attached to the building and to the earth, and all of that is so massive that we don't notice the infinitessimal change in its angular velocity. Nor do we noice anything of this sort when race cars go around a track, or an amusement park carousel accelerates to speed and later slows to a stop. Remember, we said earlier that the pillar may be considered so massive it is nearly motionless.

One reason people sometimes get off on the wrong track when analyzing this problem is their previous familiarity with problems of central force motion, where the force is always directed toward a fixed point. The force of tension on the skater with rope around the pillar is not directed toward a fixed point, so it is not a case of central force motion.

A similar problem is No. 9 in University of Chicago Graduate Problems in Physics with Solutions, Addison-Wesley, 1967. They give a solution using a Lagrangian method, and two other solutions as well.

  • Donald Simanek, 2017.

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