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Movement in relative motion

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data
    confession Olympic hammer competition, competitor turns mass 7.3 kg ball at the end of a metal cable length 1.2 meters. Throw the hammer was released some 1.3 meters high 21 ° angle above horizon.
    Last horizontal distance was 83 m hammer, what is the radial acceleration of the hammer just before liberation?


    2. Relevant equations
    Path equation - y=xtanTHETA - g / (2v0^2cos^2THETA) x x^2

    3. The attempt at a solution
    i set the details i know at the equation and i got a huge velocity 103.69 m/s
    i put it in the equation a=v^2 / r but this is an astronom acceleration
    what did i do wrong , i will glad for help and explanation...
    thanks!
     
  2. jcsd
  3. Mar 25, 2010 #2

    berkeman

    User Avatar

    Staff: Mentor

    It would be good to see more details of your calculations. Generally in problems like this, you write one equation for the x motion as a function of time and one for the y motion as a function of time, and solve for the time where the object hits the ground. This lets you get back to the initial velocity, which will give you the centripital acceleration in the circular throwing motion just before release...
     
  4. Mar 25, 2010 #3
    i will post mu calculiting :
    i talk the Y as the sin21 x 1.3
    sin21x1.3= 83 tan21 - ( ( 9.8 x 83^2) /(2Vo^2-cos^2 21)
    0.47 = 31.9- ( 67512.2)/(2Vo^2-0.88)
    0.94Vo^2= 0.141+(-28072)-67512.2
    -62.86Vo^2 = -28.072 - 67512.2
    -62.86Vo^2 =-67540.2
    V0=103/69 [m/s]
    Ar=v^2/r
    ------------------
    i didn't reference to any mass when do i need to refer it ?
    thanks
     
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