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A man on a swing

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data The rope of a swing is 2.9 m long. Calculate the angle from the vertical at which a 77-kg man must begin to swing in order to have the same kinetic energy at the bottom as a 1430-kg car moving at 1.13 m/s (2.53 mi/hr).
    2. Relevant equations
    ke=1/2 m v^2 w=f Δx cosθ



    3. The attempt at a solution i found the ke of the car to be 912.98 J and now iam trying to relate that to the man to find theta but am finding it hard to
     
  2. jcsd
  3. Oct 9, 2012 #2
    With a swing, potential energy is converted to kinetic energy. You want to match the KE of the car to the KE of the man. The man's KE can be related to his PE.
     
  4. Oct 9, 2012 #3
    so ke(912.98)=mgh where h=Lcos(theta) ??
     
  5. Oct 9, 2012 #4
    Setting KE equal to PE is correct. That provides you with the height the man must drop on the swing. But you have an error in your angle calculation.
     
  6. Oct 9, 2012 #5
    iam not sure where i went wrong on that???
     
  7. Oct 9, 2012 #6
    If you draw a picture of the swing at some angle from the vertical you will see where h is in relation to the triangle formed by the vertical, the rope position, and the horizontal line going from the vertical to the man's position. It should then be obvious what you should do to calculate theta.
     
  8. Oct 9, 2012 #7
    if h is the verticle line and L is the hypotenuse then doesnt that mean thath h=Lcos(theta) i know its wrong and this should be really easy but i dont understand y its not working.....
     
  9. Oct 9, 2012 #8
    h is the distance the man must drop so that his KE equals the KE of the car. h is the vertical line below the triangle where you are to compute the angle. To compute theta, you need the triangle leg which is vertical and lies on the vertical radius.
     
  10. Oct 9, 2012 #9
    The KE match is at to bottom of the arc of the swing.
     
  11. Oct 9, 2012 #10
    ok sorry iam just really confused...so in the triangle h is the bottum and is the length that theman is swining...L is the vertical component or the hypotenuse??
     
  12. Oct 9, 2012 #11
    No. Draw the picture so that the swing is off from the vertical by about 50 degrees - right of vertical. Draw a horizontal line from the swing to the vertical. The h you have computed is the distance from where that line intersects the vertical radius and the bottom of the arc.

    The hypotenuse of the triangle is the radius, 2.9 m. The horizontal leg of it is unknown and is the line you drew from the swing to the vertical radius. The vertical leg of the triangle is the difference between the radius and h.
     
  13. Oct 9, 2012 #12
    ok i think i understand that so the verticle commonent is r-h? so that mean h= r-rcos(theta)??
     
  14. Oct 9, 2012 #13
    That's it. You've got it. Bye....
     
  15. Oct 9, 2012 #14
    ok thanks so much!!!!!
     
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