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Circular Motion of a golf club

  1. Apr 12, 2004 #1
    High-speed movies reveal that the time interval during which a golf club is in contact with a golf ball is typically 1.0 x 10^-3 s and the speed of the ball when it leaves the club is about 65 m/s. The mass of a golf ball is 0.045 kg.

    a) Determine the magnitude of the average force exerted by the club on the ball. (Ignore the foce of gravity)
    b) Why is it reasonable to neglect the force of gravity in this case?
    c) What kinds of materials (for the club and the ball) help increase this average force? How does the material maximize the average force?

    ---------------------------------

    a) I tried to draw a FBD. The only force seems to be working on the golf ball is applied force (since no gravity, and I am assuming no air friction). Does the action and reaction force have anything to do with it?

    So far, I have m = 0.045 g ; v1 = 65 m/s ; t = 1.0 x 10^-3 s . What formulas do I use? I can only think of v = root of Rg, but the question said ignore gravity. Anyone has any idea how to do this question?

    b) I dont know :(, cause I dont understand how this works

    c) If I knew what force is really acting on it, may be I could give it a try, but I am hopeless now :(.


    I am always used to do my homework, but this time, looks like I am really stuck. I came here as my last resort :(. I am trying to do this question for the last 3 days, but its useless, and its due tommorow :(. Can anyone help me with this? I would really really appreciate it :)..

    And mods, admins, the owner may be, if you want I can design a skin for this forum...freelance :). I've been working with vBulletin for over 2 years now, and with vBulletin 3 ever since the first beta came out. Two of my sites are powered by vBulletin. I have experience designing skins for both vBulletin 2 and 3, I will present my portfolio if you are interested :).
     
  2. jcsd
  3. Apr 12, 2004 #2

    Doc Al

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    Staff: Mentor

    Impulse

    The average force can be found by considering the impulse given to the ball. The impulse, which is a force acting for a period of time, equals the change in momentum of the object. Impulse = FΔT = Δ(mv).
     
  4. Apr 12, 2004 #3
    The problem is we havent got to momentum and impulse yet. We are still doing basic grade 12 motion in 2d, and recently started circular motion for the first time. So theres got to be an easier way out :(
     
  5. Apr 12, 2004 #4
    There's no "easier" way to do it. Doc Al's method is the easiest way and, honestly, pretty much the only way. I'm not sure what any of this has to do with circular motion, though.

    He's given you the equation, just plug in the numbers.

    For (b), this answer will be obvious once you see the answer to part (a). Just compare values.

    (c) Here's a hint: You want as elastic a collision as possible. Now do you think a harder material will result in a more elastic collision (like two pool balls) or a softer material will result in a more elastic collision (like two balls of putty)?

    cookiemonster
     
  6. Apr 12, 2004 #5
    A) I thought there'd be a different method of doing it because we havent learned about impulse forces or that formula yet. We are doing circular motion now, so I thought this releates to that. Beside, the golf club is swang at a semi circular motion so I thought that'd be it.

    b) You neglect Fg because its not necessary for the equation? (Impulse = FΔT = Δ(mv))

    c) I thought of the same thing, just after posting that :). I was thinking, like the longer the time interval between the ball hits the club and the ball leaves the club is, the larger the force will be. So I was thinking, would a hard type material would cause the time interval to become longer, or would a rubber type material would make the interval longer :). I thought the softer material would result in a more elastic collision...am I right?

    Thank you very much both of you guyz, for helping me out with this, I really appreciate all the help :).
     
  7. Apr 12, 2004 #6
    1. I guess it's a review problem. Or maybe a preview problem.

    2. No, that's not the reason. Solve part 1 first.

    3. What happens when you throw two balls of putty at each other? How about what happens when you throw a ball of putty at a wall? What happens when you throw one of those magic bounce balls at a wall? Which is harder to squeeze, a ball of putty or a magic bounce ball?

    cookiemonster
     
  8. Apr 13, 2004 #7
    I gave it some more thought, and heres what I got (some help from another friend).

    a) calculated acceleration with a = v/t formula. Then used that a to find the netforce Fnet = ma. The accleration was 65000 m/s^2 (whoa!) and the netforce was 2925 N.

    b) The golf ball leaves the club almost horizontally. A very small vertical component of the force (compared to the very large average force considering the mass of the object) applied to the ball balances the weight of the ball. The weight of the ball is very small, because of the small mass, and also the time period is very short, so it is reasonable to neglect the force of gravity.

    c) I was being really dumb up there :p. Oviously a magic bounce ball is harder to squize, and it bounces more. So the material has to be hard? Like hard plastic?
     
  9. Apr 13, 2004 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    1) Force= mass times acceleration! You are given the mass of the golf ball and you know it goes from 0 to 65 m/s in 10-3 seconds so you can calculate its acceleration.

    2) What would the force of gravity be on a golfball of mass 0.045 kg? How does that compare with the force you got in part 1?

    3) In a "non-perfectly elastic" collision, some of the energy is lost because on or the other (or both) parties to the collision (in this case, ball and club) are deformed and energy is used up there. What kind of materials do NOT deform easily?
     
  10. Apr 13, 2004 #9

    Doc Al

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    Staff: Mentor

    Yes, there is an "easier" way. HallsofIvy explained it, but it looks like you already figured it out yourself: Just use Newton's 2nd law (F = ma).

    I put quotes around "easier" because as you will no doubt learn, this turns out to be exactly equivalent to using the "impulse = Δmomentum" method. :wink:

    And this doesn't seem to have much to do with circular motion.
     
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