Here’s the write up on the skater spiral for those interested. A little long, but an easy read.
[b]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. How can we know this without complicated analysis?
The rope has negligible mass compared to the skater, so we treat it as massless. 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 jar lid, fastening a string to it, and placing 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 jar lid. The path will end up with the curve closing in on the lid, and the path will be perpendicular to the lid’s rim where it hits. 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: Taking the fixed point at the center of the pillar as a center of torques, we see that the rope’s tension causes a torque on the pillar, giving the pillar and the Earth an angular momentum in the same sense as the skater’s angular velocity. Therefore the skater’s angular velocity and angular momentum decrease. (Her angular momentum reaches zero when she impacts the pillar.)
Another way to look at this: The rope exerts a torque on the skater. The lever arm of this force about the pillar’s center is just the radius of the pillar, R. At any instant the torque, R × F is a vector opposite in direction to the skater’s angular momentum, so this must decrease her angular momentum.
At the moment of her impact with the pillar, her angular momentum has become zero and she impacts with velocity that is directly toward the center of the pillar.
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. For an example of central force motion, consider a skater on a frictionless rink, holding a rope that is fastened at a fixed point. Now, as time goes on, the rope length remains constant, and the skater’s motion is constant speed around a circle. Now suppose, due to some clever mechanism, the rope is continually shortened, perhaps by having it pass over a pulley at the center of the rink, and then to a motor winch. But we take care that the rope at the center always passes through a fixed point there. By insisting that the rope pass through a fixed point, we ensure that there’s no torque on it about that point. Now the angular momentum must be constant, as it always is in central force motion. So as the radius decreases, the skater’s speed must increase as she spirals toward the center. This tells us that work is being done on the skater. The skater’s velocity is no longer perpendicular to the rope, but has a small radial component. This means there’s a radial displacement toward the center, and the radial force is responsible for that displacement. That’s where the work is done on the skater—work done by the motor that drives the winch that’s shortening the rope.[/b]
