Comments

1. #1 Harlan

   February 2, 2006

   Uh, letsee. No momentum is being transfered anywhere, so the instantaneous velocity will remain v. However, the rotational speed, in degrees/second or whatever, will increase by some amount I don’t know how to calculate.

   Is that true? I haven’t done any physics since High School, 15 years ago!

2. #2 Matt B.

   February 2, 2006

   The answer to this really depends on the number of distinct vacua. 10^499 vs 10^500 represents such a large difference to this problem. I am afraid that there is no distinct answer.

   So, 42 is my answer. I leave it as an excercise for the grader to find the valley in the landscape in which my answer is correct.

3. #3 Matt B.

   February 2, 2006

   Harlan,

   In order to conserve angular momentum, v will have to increase as L decreases. But the origin of L is shifting; I find that confusing to quantify.

   Good news. I have a solution for R=0. :)

   /C in classical mechanics

4. #4 Marcus Vitruvius

   February 2, 2006

   At the actual moment of impact, wouldn’t the speed be zero, since it is prevented from going forward and not yet begun the rebound?

5. #5 Yan

   February 2, 2006

   energy conservation
   v

6. #6 theophylact

   February 2, 2006

   Offhand, looks to me as though it’d be Lv/r, since momentum (mv) is conserved. But I’m just a chemist, and my last grade in college physics was a D.

7. #7 mo

   February 2, 2006

   Infinite (classical mechanics will fail, we need relativity here). Angular momentum is initially m*v(0)*L(0), and must be conserved.

   m*v(t)*L(t) = m*v(0)*L(0)
   v(t) = v(0)*L(0)/L(t) ==> infinity

   sucks to be a zero-size tetherball.

8. #8 David

   February 2, 2006

   The answer is v. I had to get this by calculus (shown below), what the heck is the intended solution?

   First of all, my physical intuition is that there is a significant torque on the center post which transfers angular momentum into the earth: if I imagine the post set loosely in the ground and surrounded by ball bearings, I think the post spins forever and the string doesn’t wrap. Since that is the case, I can’t use conservation of angular momentum. Curses.

   Let L(t) be the length of the string at time t and theta(t) the angle from the post center to the point of string tangency. Since the string can not stretch or flex, we have

   L(t)=L-R theta(t).

   Let’s use complex coordinates for the position z(t) of the ball. The post is at the origin and the string start out tangent at the point R, with the ball starting at R+iL. So its position is

   R e^{i theta(t)}+L(t) i e^{i theta(t)}=(R+iL-R theta(t)) e^{i theta(t)}

   The velocity is

   - (L-R theta(t)) theta’(t) e^{i theta}. (*)

   As a side note, this is entirely perpindicular to the string. That sort of makes sense — I can convince myself that if it were otherwise the string would break or go slack — but I can also see a counter argument that there should be a component along the string equal to the rate that the string is being taken up. Without calculus, I never would have known which argument was right.

   Its acceleration is

   ( some big mess in terms of theta, theta’ and theta”) e^{i theta}

   Now, we know that the acceleration must be along the string because that is the only force acting on the ball. So the real part of the mess must be zero. That gives

   R (theta’)^2+R theta theta” = L theta”

   Integrating both sides

   R theta theta’+C = L theta’ (**)

   where C is L theta’(0). I get that C=- vL^2/(L^2+R^2), but that’s not important so I’ll just keep calling it C.

   This can be integrated again to give an explicit solution for theta(t), but that isn’t important either. Instead, rewrite (**) to give

   (L-R theta(t)) theta’(t)=C.

   But (L-R theta(t)) theta’(t) is exactly the speed of the ball at time t! (Since e^{i theta(t)} has magnitude 1.) So we get that the speed of the ball is a constant. Since it starts at v, it must stay that way.

   As a side note, this predicts that the quantity of string wrapped per time becomes infinite as t approaches collision. I can’t decide whether that is reasonable or a sign that I have totally screwed up.

9. #9 Nicholas Condon

   February 2, 2006

   Umm… This may be the simplistic answer of a chemist yokel like me tryin’ to work one of dem “big city” physics problems, but I don’t see any energy entering or leaving the system here, so I’m pretty sure that the magnitude of the velocity remains v, so long as we assume that the post is anchored to a body with infinite mass.

10. #10 Matt B.

    February 2, 2006

    RE: Energy Conservation

    But there is a force acting on the ball: the tension in the string, right?

    It’s like gravity and orbits (where the string is gravity). The earth does speed up at its perigee. So this ball speeds up in a constant periapsis.

11. #11 Mark Paris

    February 2, 2006

    I vote for a “speed” of v purely on the basis of the wording of the problem.

12. #12 Clay Blankenship

    February 2, 2006

    Which way is up?

13. #13 John Casey

    February 2, 2006

    Nicholas,

    2 points,

    1. Clearly v is not constant, since it is a vector quantity, and the direction is changing, so there is some acceleration going on. V becomes constant only in the case the string severs.

    2. Having actually watched a tetherball wind around a pole with a starting non-zero speed, I can testify that speed increases as the string winds up. haven’t a clue as to the formula, tho.

    JC

14. #14 Dennis

    February 2, 2006

    There is a force on the ball (the tension of the string) but it’s always acting in a direction perpendicular to the direction of travel. The simplest solution is what Nicholas said – energy (or momentum, take your pick) conservation. The angular velocity will increase, but not the velocity.

15. #15 anthoanres

    February 2, 2006

    I vote for infinity if the tetherball has zero radius.

16. #16 David

    February 2, 2006

    Yeah, Nicholas and Dennis have the right approach. Once you know that the velocity is always perpindicular to the string, there is no work being done so the velocity has to be constant. I’m still thinking about how to best see this physically. The best I can come up with is to imagine marking a point on the string. At any moment, this point has some velocity along the string and some velocity perpndicular to the string and, to keep the string from tearing or going slack, the velocity along the string must be independent of the point we mark. Right next to the tether pole, it is pretty clear that there is no velocity along the string.

17. #17 Danil

    February 2, 2006

    You could also think about the case where R is really large compared to L, such that at the diagrammed point, the ball is only infinitesmally distant from the post. Thus there is no room left for the speed of the ball to change before it strikes.

18. #18 Ross Smith

    February 2, 2006

    You can also approach this with dimensional analysis and a bit of real-world intuition.

    The only parameters in the system are V, M, L, and R. M can’t play any part in the result, because there’s no mass anywhere else in the parameters to cancel it out. V is the only parameter with a time dimension in it, so the final speed must be V times a dimensionless number.

    The multiplier must be either a constant or a function of the ratio L/R. Anyone who has seen a real tetherball in action knows that the ball doesn’t speed up by a huge factor as it approaches the pole, so unless the function is surprisingly complicated, the f(L/R) option is ruled out by observation.

    That means the final speed must be V times a constant independent of the design of the system. At L=0 the initial and final speeds are the same, so the constant must be 1.

19. #19 Wheatdogg

    February 2, 2006

    The centripetal force F is parallel to L, so there is no torque on the ball. Thus, dL/dt = 0, and angular momentum L must be constant.
    Further, the orbital radius decreases by 2piR after each revolution.

    Initially, L0 = I0w0, where I = ML^2 and w0 = v0/L.
    After one rev, L = Iw, where I = M(L-2piR)^2 and w = v/(L-2piR).
    After n revs, L = Iw, where I = M(L-2npiR)^2 and w = v/(L-2npiR).

    Since I0w0 = Iw (dL/dt = 0), v-final = v0L/(L-2npiR). The value of n depends on the ratio of L/2piR (assuming of course that the ball is a point mass).

20. #20 Nikolai

    February 2, 2006

    Intuitively, this seems like an easier question if you look at the ball’s frame of reference. The post just coils up and approaches the ball. By conservation of momentum, the post would approach the ball with constant speed.

21. #21 Anonymous

    February 3, 2006

    John and Matt, I stand by my argument. There is a force acting on the ball, which is altering the direction of v, but not its magnitude. There are no external forces or energy sources or sinks in the problem, so the energy in the box Chad drew around the problem must remain constant. If the string is massless, then the only way that can happen is if |v|, the speed of the ball, remains constant.

    As to the real world, the period of the tetherball decreases, but I’d bet my hat (I don’t wear a hat, but still) that the period of the tetherball’s orbit decreases, but that the magnitude of its velocity does not.

22. #22 Nicholas Condon

    February 3, 2006

    John and Matt, I stand by my argument. There is a force acting on the ball, which is altering the direction of v, but not its magnitude. There are no external forces or energy sources or sinks in the problem, so the energy in the box Chad drew around the problem must remain constant. If the string is massless, then the only way that can happen is if |v|, the speed of the ball, remains constant.

    As to the real world, the period of the tetherball decreases, but I’d bet my hat (I don’t wear a hat, but still) that the period of the tetherball’s orbit decreases, but that the magnitude of its velocity does not.

23. #23 Andrew

    February 7, 2006

    Seems to me it’s just conservation of angular momentum. The initial angular momentum is mv(L+R). At the moment of contact, it’s mv’R (assuming the ball has zero radius). So mv(L+R) = mv’R which gives v’ = v(L+R)/R. I’ve assumed the string starts out 90 deg to the pole and at the point of contact, the string is effectively also perpendicular to the pole.

24. #24 Reef

    March 13, 2006

    I am a little curious as to why the word speed rather than velocity has been used throughout the definition of the problem, when the assumption “(which is, of course, massless, fricitonless, and non-elastic)” has also been used – displays that the poster does have at least a basic grasp of science (although I’m sure they know much more than that!) Is this a deliberate mistake in an attempt to send people off the scent?
    Also surely if a calculation were required, it would be as to the speed just before impact…

25. #25 Chad Orzel

    March 13, 2006

    I am a little curious as to why the word speed rather than velocity has been used throughout the definition of the problem, when the assumption “(which is, of course, massless, fricitonless, and non-elastic)” has also been used – displays that the poster does have at least a basic grasp of science (although I’m sure they know much more than that!) Is this a deliberate mistake in an attempt to send people off the scent?

    I’m not sure what you’re asking. The speed is the appropriate quantity, here, as I don’t actually care about the direction of motion at any given instant.

26. #26 Mondo

    December 13, 2006

    “Seems to me it’s just conservation of angular momentum.”

    Hmm. Seems wrong.
    The ball won’t speed up as it wraps around the pole so angular momentum must be changing. The ball and string aren’t doing any work on the pole (except for the teeniest of rotation of the earth (assuming it is planted solidly into earth)). So, it would seem like the ball shouldn’t lose or gain kinetic energy and will strike the pole with its initial speed of v.

    Would be interested to know what the solution really is.

27. #27 Clayton

    December 13, 2006

    The linear velocity (speed) represented as v remains constant upto the point of impact. So v as “linear” or “tangential” velocity is constant. If you were to cut the string at any moment during it rotation it would fly off at the same speed.

    The angular velocity defined v/r, increases as L becomes shorter, being that in this case L represents r in the v/r. L being the radius of the arc made by the tetherball.

    Not having a variable for the radius of the tetherball I suspect that determining the formula for angular velocity at the moment of impact was not intended. At least, I think, that variable would be needed to calculate the point of impact, and thus, the angular velocity at the point of impact…I think. Of course, even with it, I still wouldn’t know how to go about it :( That parts got me stumped.

    In the formula for angular velocity, ω = v/r, as the radius approches zero, the angular velocity approaches infinity. That would probably ruin most tetherball games, friction aside ;)

28. #28 Clayton

    December 13, 2006

    Actually I was thinking about the infinity thing while vacuming, and realized that this could be somewhat if a trick question.

    “What is the speed of the ball when it hits?”
    Rather than ‘What is the speed of the ball just before it hits?’

    So I think the answer you may be looking fo is zero.

29. #29 Chad Orzel

    December 13, 2006

    I actually posted the answer quite a while back. The first explanation is a little garbled, but it’s corrected in the comments.

    No peeking unless you’ve figured it out, though.