RobertGC said:
Yes. Because the length of the rotating string is shrinking the path of the smaller weight can not remain circular.
The length of the rotating portion of the string may be shrinking initially, but in the absence of frictional losses it will reach a minimum and then grow again.
RobertGC said:
I have a question about the calculation in post #6 however. It would seem there should be a solution for r depending on time. But the equation provided only has 2 or only 1 solution.
I was solving for the maximum ##r## and the minimum ##r##. Apoapsis and periapsis. Naturally there are two solutions, one for each. Or just the single solution in the case of a perfectly circular orbit.
As I mentioned in post #2 above, the same idea can be used to compute ##v_r## and ##v_t## for any permissible value of ##r##.
You know ##v_t## immediately from angular momentum conservation. Divide angular momentum by ##r## and out pops ##v_t##.
With ##v_t## in hand, you can then compute ##v_r## from energy conservation. You know the original total energy. You know potential energy from ##r##. You know tangential kinetic energy from ##v_t##. Whatever is left over is radial kinetic energy. Solve ##E_r = \frac{1}{2}(M+m){v_r}^2## for ##v_r##.
Note that you cannot calculate the sign of ##v_r## from energy conservation alone. Only its absolute value. Similarly, you can only determine the absolute value of the trajectory's angle from the horizontal in this way. You cannot tell whether the angle is upward or downward. The fact that the trajectory is smooth helps you out in this respect. The trajectory angle can only change sign at periapsis and apoapsis.
While the equations do not give you position as a function of time, they do give you a differential equation. You could attempt to solve that. Or you could just iterate forward in time with a
computer program. You can even use
Excel for this.
Note also that you only have to simulate one half of one orbit. The second half will be the mirror image of the first. Then the next half will be the same as the original half. And so on.
If you turn friction on, you get a new term in the equations. There will be an energy drain from the system at a non-constant rate and a possibly lop-sided in-spiral. While Excel could be used to model this, it lacks the graphic capability to display a proper
x-y plot in polar coordinates.
Interesting question... With friction, is the resulting inspiral finite or infinite? e.g. an Archimedean spiral is finite but an exponential spiral is infinite.
Assume a smooth in-spiral. The tangential velocity is responsible for centripetal force approximately equal to ##Mg##. The required velocity and, accordingly, the energy loss rate will scale with the square root of ##r##. The orbital energy will scale directly with ##r##.
So we have a pool of energy depleting at a rate proportional to its square root. ##\frac{dE}{dt} = -c\sqrt{E}##. What function has a first derivative that is its own square root? Let us try something of the form ##t^k##. Like ##t^2##. Oooh, that guess works out well. So we would expect energy and orbital radius to follow a parabolic arc to the bottom while tangential velocity follows a straight line graph to zero in finite time.
Or more directly without the differential equation, if the descending weight is approaching the rope limit, we have an orbitting weight with velocity ##v## subject to a retarding acceleration of magnitude ##\mu g## so the orbitting weight slows to zero speed in finite time given by ##\frac{v}{\mu g}##.