Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum rotational vs linear velocity of thrown ball

  1. Aug 5, 2003 #1
    For a thrown ball of constant density, what is the maximum possible rotational/linear velocity ratio?
  2. jcsd
  3. Aug 5, 2003 #2
    Well one could assume the ball is thrown vertically. So it will reach a point where v=0 but it is still rotating. So the answer would be infinite.

    I'm sure you're thinking of something else, so please be more specific.
  4. Aug 5, 2003 #3
    I'm thinking of a potentially infinite coefficient of friction and moment of inertia characteristic to a perfect sphere.
  5. Aug 5, 2003 #4


    User Avatar
    Staff Emeritus
    Gold Member

    "potentially infinite coefficient of friction"?

    What do you mean with that?
    Why "potentially"?

    You need to give more context. Where would the limit come from? Are you thinking relativity? How should friction be considered?
  6. Aug 5, 2003 #5


    User Avatar

    Staff: Mentor

    Ratio of rotational to linear velocity? Infinite in both directions. You can spin the ball in place or throw it with no spin.
  7. Aug 6, 2003 #6
    I thought that applying torque to rotate the free ball during a throw would cause a minimum requisite linear velocity due to the moment of inertia of the ball. Remember the somewhat analogous problem of an object (a sphere, a tube, a cylinder, etc.) with given moment of inertia which begins to slide, rather than just rotate, down a plane at a certain critical angle. Keep to classical mechanics, and take the coefficient of friction as 1 (sorry about the infinity). Again, for a [free] thrown ball of constant density [and surface friction coefficient of one], what is the maximum possible rotational/linear velocity ratio?
  8. Aug 6, 2003 #7


    User Avatar

    Staff: Mentor

    Where are you getting this torque? Ordinarily when you throw a ball (like a pitch in baseball) it kinda rolls off your hand, which makes it spin. But haven't you heard of a knuckleball? You literally hold the ball with your knuckles and flick it at release, giving it virtually no spin at all.

    Forward speed and rotational speed of a thrown ball are completely independent of each other.
  9. Aug 6, 2003 #8
    are you maybe thinking of a ball being accelerated by just one force that acts tangentially upon its perimeter, so moving & (at the same time) spinning it? Without taking into account air friction or gravity? I think, this may be easily answered.

    Could this help, if we use 'sphere' instead of 'cylinder':
    Last edited: Aug 6, 2003
  10. Aug 7, 2003 #9
    arcnets, you have stated the problem fairly well as I see it; I propose one must apply both tangential and radial acceleration to the ball when spun off, say, from one point on its surface. Yes, russ, one can throw with forward velocity a ball practically without imparting spin, but here I have asked the converse. First, is it possible to spin an otherwise freely "thrown" ball (released at one point on its surface) without causing it to move linearly, and secondly, is its proportion of rotational to linear velocity limited in classical theory?
  11. Aug 7, 2003 #10


    User Avatar

    Staff: Mentor

    Ok, thats much clearer. So the answer to the first part is no. It is not possible to apply a tangential force at one point and impart only spin. The ball will spin and move forward. How much spin and how much forward motion you get can still vary infinitely but with practical limitations. The key is moment of inertia. An object with high mass and low moment of inertia (dense center, less dense outside) will spin but won't move forward much. A ball the opposite (hollow) will move forward with less spin.

    There is a theoretical minimum spin:velocity ratio, but not a theoretical maximum for this case.
  12. Aug 7, 2003 #11
    Probably not. Because when spinning the ball by a single force, you will always have a tangential component of the accelerating force, thus making the ball move.

    Remember Apollo days: Spaceships had always 2 jets located at opposite positions wrt. the center of mass (and in opposite directions), which were fired simultaneously to make the ship rotate. Yes you could rotate the ship with just one jet, but that would cause linear acceleration, too. I think they knew why they had to install 2 jets, even though that was more expensive.
  13. Aug 7, 2003 #12
    Can you prove the latter assertion?
  14. Aug 8, 2003 #13


    User Avatar

    Staff: Mentor

    Yeah, but frankly it would take me more effort than I feel like right now. If you've already looked at the equations though, you can easily figure it out for yourself. I'll get you started:

    The lower the moment of inertia for a given mass, the faster something will spin, right? Thats kinda the definition of moment of inertia. Moment of inertia depends on mass distribution and there is no theoretical limit on how you distribute the mass. You can never quite get to zero moment of inertia, but you can get as close as you are willing to take the effort to get. Which means infinite rotational rate. And the linear speed for a given mass and radius will always be the same (assuming you mean IMPULSE, not just force).

    So its a limit problem - derive or find the equation relating moment of inertia to rotational rate and take the limit as I -> 0.
    Last edited: Aug 8, 2003
  15. Aug 8, 2003 #14
    Yes sure, but in the initial post, Loren said 'ball of constant density'.

    I think this can be done if we treat it as an elastic collision problem.

    Let's assume that a particle of mass M and velocity V is fired away tangentially from the ball's surface (much like in my Apollo example, for the 1-jet case). Assume that M << m where m is the ball's mass.

    Initial energy: Ei = 1/2 M V2.
    Initial momentum: pi = MV.

    Final energy: Ef = 1/2 m v2 + 1/2 I w2
    Final momentum: pf = mv.

    For a full sphere, I = 2/5 m r2,
    Ef = 1/2 m v2 + 1/5 m r2w2.

    If I am not mistaken, there are 2 unknowns: v, w.
    And 2 Equations:

    So it should be possible, right?
  16. Aug 8, 2003 #15
    Good progress, arcnets. But doesn't pf=mv(cos[the])+Iw(sin[the]), where [the] is the radial angle from the ball's center to the point of initial trajectory ? The rest seems correct.
  17. Aug 9, 2003 #16
    I think I found out how to do this. Let's consider the same process (particle fired away tangentially). I think it should be treated as inelastic. Linear and angular momenta are conserved, we don't care about energy.

    pi = 0
    pf = MV - mv

    Ji = 0
    Jf = MVr - Iw

    Let's assume I = Cmr2 with some constant C.

    I. MV - mv = 0
    II. MVr - Cmrw = 0

    I. V = mv/M (into II)
    II. mvr - Cmr2w = 0 (m cancels. Good!)

    II. v = Crw, or

    <drumroll> rw/v = 1/C.
    That is the desired ratio, isn't it?

    Now, for a full sphere, C = 2/5, so rw/v = 2.5
    The maximum C value is C = 1 (for a ring). So the minumum ratio is 1.
    There is no minimum C value, so there is no maximum for the ratio. As russ_watters already said.
    Last edited: Aug 9, 2003
  18. Aug 9, 2003 #17


    User Avatar

    Staff: Mentor

    Oops. In that case, there is only ONE ratio I think.
  19. Aug 9, 2003 #18
    Great work, a. n. Do you agree with his derivation for the most part, r. w.? I have some questions.

    1. Is there no minimum C (or moment of inertia) value in general? Perhaps all I[>=]mr2/12.

    2. Aren't we talking here about the constant density sphere with C=2/5?

    3. With fixed r0, doesn't the ratio r0w/v=2.5 then always hold, where r0w is the rotational velocity and v the linear velocity of the ball?
  20. Aug 11, 2003 #19
    1. No there isn't. You could concentrate most of the mass near the object's center and so make I as small as you like.

    2. I think, yes. A ball of constant density, as you said.

    3. I holds when the accelerating force acts tangentially. You could of course hit the ball more 'head-on', thus causing less spin and more linear motion.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook