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Don't know where to start!

  1. Nov 11, 2004 #1
    Hey, any ideas on this double starred, hard question :(

    A golf ball passing trhough a windmill at a miniature golf course. The windmill has 8 blades and rotates at an angular speed of 1.25 rad/s clockwise. The opening between successive blades is equal to the width of a blade. A golf ball of diameter 4.50x10^-2 m is just passing by one of the rotating blades. What must be the minimum speed of the ball so that it will not be hit by the next blade?

    Ok, my thought is, I need to find time for a blade to rotate theta, and I need the arc length for that. With arc length I have some way to compare the diameter and the space opening between the ball and blade. Lost from that point... :yuck:

    Any ideas, Great!
     
  2. jcsd
  3. Nov 11, 2004 #2

    NateTG

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    How far does the ball have to travel?
    How big (as an angle) are the gaps in the windmill?
     
  4. Nov 11, 2004 #3
    Ok, knowing w, I think I can use theta= w(avg) x time . I don't have time. I have 1.25rad/s. Hold it is this approach right. Ok I have 1.25 rad, 2pie is a full revolution or 360 degrees. There is a 5.02 factor between 1.25 rad and 2 pie. Ok, 1.25 rad is equal to 71.61 degrees. I multiply that by 5.026 and divide by 16 (total equal proportions including space and blade) and I get roughly 22.5 degrees for each gap?

    As for how far does the ball have to travel, is that relevant? I mean it just has to travel pass the width of the blade.
     
  5. Nov 11, 2004 #4

    NateTG

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    You should be looking at things like: There are 8 blades, so there are 8 gaps. That means that each gap is going to be 1/16 of a revolution, or an angle of [tex]\frac{2\pi}{16}=\frac{\pi}{8}[/tex].

    Then you can determine the time that the windmill takes to turn that far and so on.
     
  6. Nov 11, 2004 #5
    ok, so my way was also corrrect than ... 22.5 degrees is equal 0.39 rad which is equal to pie/8. Now for the next part, time. I'm thinking we can use theta= 1/2 (w avg )t. Since the mill will spin always at 1.25 rad/s the w avg is still 1.25rad/s. Theta = 0.39 rad, I substitute and solve for t.

    t = 0.628 s.
    Thinking...
     
  7. Nov 11, 2004 #6
    ok I figure: need the angular velocity of ball (w) then we could use this value and substitute it into V=rw? so diameter we know is 0.0450, d/2 = r. So, w = theta/time. So, 0.39/0.628 gives 0.621 rad/s. Substitute it into V=rw, V=0.01397m/s?
     
  8. Nov 11, 2004 #7

    NateTG

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    [tex]\theta=\omega_{avg}t [/tex]
    (There's no [tex]\frac{1}{2}[/tex] you may be thinking of [tex]\theta=\alpha_{avg}t[/tex].)
    Which will change your time.

    Then the ball needs to travel it's diameter in the time. so [tex]v=\frac{d}{t}[/tex]

    (This would be a lot easier to explain with drawings.)
     
  9. Nov 11, 2004 #8
    Everything sounds good, except I'm not sure I understand the concept, the ball needs to travel it's diameter?
     
  10. Nov 11, 2004 #9

    NateTG

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    I can try to do ascii art:
    Imagine you're looking at the windmill from the top
    Then the beginning of the ball crossing will look like this
    Code (Text):

     |  Blade (moving down on screen)



    o|--> (Ball about to cross)
     |
     | Blade (moving down)
     
    And the end of the ball crossing will look like
    Code (Text):

     |
     |Blade (moving down on screen)
     |o--> (Ball having crossed)



     |
     
    So, if the ball barely scrapes the tail end of the blade before it starts crossing, and barely gets scraped by the blade at the end of the crossing, then it's traveled it's diameter in distance.
     
  11. Nov 11, 2004 #10
    Well, the diagrams make some sense, excpet I don't see how both parts of the mill can move down, shouldn't it be rotating so that one side goes up while the other comes down. But yes, I get it now, I thought it was the travel distance before and reaches the mill plus the distance to get pass it.

    Thanks!
     
  12. Nov 11, 2004 #11

    NateTG

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    I was only drawing the bottom parts of the mill that interact with the ball.
     
  13. Nov 11, 2004 #12
    If I could remember correctly, this is a question from "Cutnell & Johnson Physics", right? For the golf ball to pass through the gap of the windmill with minimum speed, it must be at the beginning of the open gap before going through it as suggested by the diagram. The distance travelled is the diameter of the ball because this is the minimum distance the ball must travel in order to get itself pass through the gap. From the question, we know the angular speed and can find the angular displacement easily. From these infos, we can work out the time needed for the blade to sweep out the open gap. The time needed here is equal to the time needed for the ball to travel through the gap. Knowing the displament of the ball which is the diameter and the time, we can then calculate the minimum speed of the ball.
     
    Last edited: Nov 11, 2004
  14. Nov 11, 2004 #13
    Yup Cutnell and Johnson, and thanks for making it even more clear lol, esp the distance/diameter travelled by ball, was a tad fuzzy.
     
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