Re: Atwood's machine problem

[Ken Caviness <caviness@SOUTHERN.EDU>]

Herb,

One more try to see whether I've understood you, and then I'll leave you in
peace!

Quoting Herb Schulz <herbs@WIDEOPENWEST.COM>:

> Howdy,
>
> The L for the monkey about the Pivot Point of the Pulley is definitely NOT
> zero. The r vector points from the origin at the Pivot Point of the Pulley
> to the CM of the Monkey (the apparent point of application of the earth's
> gravitational force on the Monkey).

I thought that I understood you, but perhaps I didn't.  Are you not saying that
the r vector points straight downward, from the pulley to the monkey?

> That is NOT parallel to the velocity
> vector which is vertically upward.

Agreed, the velocity vector is straight upward.  Thus my musing that the r & v
vectors would be anti-parallel and so  | r x mv | = rmv sin(pi) = 0.

I don't like this answer.  When I read your response that we could use
conservation of angular momentum to solve the problem I was very happy.  But
either you're forgetting to multiply by sin(pi) or I still haven't correctly
understood what you mean.

> A similar thing is true about the Bananas
> except that the L vector for the Bananas is opposite in direction to the L
> vector for the Monkey so the two can add to give a net L vector of zero.

This I understand, but since I'm getting 0 + (-0) = 0 it doesn't help me much.

Ok, I've got it now.  You are right, we don't assume that r is straight down.
Its downward component is L, its horizontal component is R (the radius of the
pulley).  I'm picturing the monkey on the right side, the bananas on the left,
the center of the pulley at the origin.  Monkey's coordinates are (at some time
t):  (x,y) = (R,-L), so r = R i - L j.  Its velocity is v j, so

  L = (R i - L j) x (v j) = Rvk

Or, letting v_m be the speed of the monkey, that's L_m = R v_m k

The banana's angular momentum is clearly L_b = -R v_b k.

No external torque means angular momentum is conserved.  Before the monkey
starts pulling, we agree that

  L = L_m + L_b = 0 + 0 = 0.

Therefore L remains equal to 0, and even when the monkey's speed is non-zero,

  0 = R v_m k + (-R v_b k) = R (v_m - v_b) k.

The only way this can be true is if v_b = v_m, if they both ascend at the same
speed.

Nice!

Wait, what if we don't use a pulley, just a frictionless hook?  If R=0, the
angular momentum is already zero without putting any constraint on v.
Effectively that's what I was doing, ignoring the radius of the pulley.  I'm
inclined to think that in this case we'd have to consider the problem as the
limiting case of decreasing pulley radius.  For any non-zero radius we must
have v_b = v_m, so there's no dependence on the radius, and the result must
still hold for R = epsilon, a shade greater than zero, and so we take it to be
true for zero, too.  besides, even with a hook we'd have a non-zero thickness
rope, so there's still some radius.

Ok, I'm happy now.  Got to run to my next class.

Oh, but notice that if this problem really needs conservation of angular
momentum to do it correctly, it shouldn't be appear in the book in the section
dealing with linear momentum.  Perhaps the suggested analogy to two iceskaters
pulling on a rope is the easiest way to handle the problem after all.

Ken Caviness
Physics @ Southern