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Linear Momentum

  1. Sep 29, 2011 #1
    (Q) a 108 cm, 0.73kg golf club is swung for 0.5s with a constant acceleartion of 10 rad/s (squ). What os the linear momentum of the club head when it impacts the ball?

    Known:
    H= 108cm = 1.8m
    m = 0.73 kg =7.16N
    t = 0.5s
    [itex]\alpha[/itex]= 10 rad/s (squ)

    I know momentum = mass x velcoity

    So velcotiy = d/t.... (so i need to find the dispalcement)

    But Im thorwn but the angular accelertion o 10 rad/s(squ)
    The only thing I can think of is using 2(pi)r with r being the 1.8m?

    Any sugegstion where to go from here ??
     
  2. jcsd
  3. Sep 29, 2011 #2

    BobG

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    If you were talking about linear motion, you could find your linear velocity using your linear acceleration.

    [tex]v_f=v_i + at[/tex]

    Same principles apply to rotational motion, except now you're looking at:

    [tex]\omega_f = \omega_i + \alpha t[/tex]

    with omega your angular velocity
    and alpha your angular acceleration
     
  4. Sep 29, 2011 #3
    Hi.. thanks what you have said makes sense...

    But is I use:

    wf=wi + at

    wf= 0 + 10 X 1.5
    wf = 5 rad/s (sqR)

    The linear momentum = mv = 0.73 x 5 = 3.65 kg. m/s

    But the answer is 3.9 kg. m/s, I know im only .25 off but is the book wrong or my calculations wrong ??
     
  5. Sep 30, 2011 #4

    BobG

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    Your angular velocity is 5 radians/sec.

    You need to covert that to linear velocity.

    If your unit of measure is your radius (radians is the plural of radius) then you have to multiply by the radius to get your linear velocity. Your final answer was only close because your radius (the club shaft) was close to 1 - it was 1.08 meters.
     
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