Re: new problem

[Leigh Palmer <palmer@sfu.ca>]

> An ice-skater is travelling at 11 m/s when she grabs a 32 meter rope
> attached to a fixed vertical pole with a radius of R = 0.20 m. At the
> instant she grabs the rope it is stretched out from the pole,
> perpendicular to her path.

There is an ambiguity here that I suspect is important. Is the rope
stretched radially from the pole (which would make the initial radius
of the trajectory, the impact parameter, 32.2 meters) or tangentially
(which would make it D1 = 32.0 meters)? I'll assume the latter for
the time being, if only because the calculation is smoother that way.

> After she grabs the rope, her path is constrained by the length of the
> rope, and she spirals inward toward the pole as the rope winds around it.

I'm going to assume that she falls to the ice at this point and slides
frictionlessly the rest of the time. That is what you intend, isn't it?
The skates impose a difficult constraint otherwise and, given the
numbers, a physically difficult task for the skater.

> At the moment that the free end of the rope is 16 meters long, what is
> the speed of the skater?

Here I'll assume that you mean D2 = 16 meters from the point of
tangency of the rope to the pole.

I'm going to interchange the order of the last two parts.

> Is the skater's angular momentum conserved?  Why or why not?

Angular momentum of a system must be specified with respect to some
point. In this case the relevant angular momentum might be chosen
with respect to the polar axis (!) It is clear that the force which
the rope exerts on the skater is not a central force. The magnitude
of her angular momentum with respect to the polar axis will be
expected to decrease in this process.

> Is the skater's kinetic energy conserved?  Why or why not?

Once I knock her off her skates I see no frictional forces acting on
her. Given an inextensible rope and other idealities I will say yes,
her kinetic energy, and hence her speed, is conserved.

The answer to the question is that she is moving with a speed of 11 m/s
when the rope is 16 m long, and all my worries were for naught!

Having determined these answers let's analyze a little further.

The torque T exerted on the skater with respect to the polar axis:
                            2
                          mv
                   T = - ----- x R
                           D

Here the symbols are the usual suspects, D being the instantaneous
length of the rope from the point of tangency. We can now find the
expected change in angular momentum by integrating the equation of
motion:
                              t2
                             /
                   L2 - L1 = |  T dt
                             /
                              t1 = 0

Now we note that the rope must wrap around the pole at an angular
velocity w given by
                            v
                       w = ---
                            D

This means the rope's length at any time t is changing at a rate

                      dD              Rv
                     ---- = - Rw = - ----
                      dt              D

Solving for dt we obtain
                               DdD
                       dt = - -----
                               Rv
Our integral now becomes
                              t2        D2   2
                             /         /   mv          DdD
                   L2 - L1 = |  T dt = |  ----- x R x -----
                             /         /    D          Rv
                              t1 = 0    D1
Pretty easy!

                   L2 - L1 = mv (D2 - D1)

We note that this is exactly the result anticipated above. The lesson
here is that one should read the whole problem before starting (you've
seen my stream of consciousness analysis here). All the parts will
contribute to understanding.

Nice problem, Chip.

Leigh