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Motion about a circular track

  1. Dec 21, 2008 #1
    1. The problem statement, all variables and given/known data
    A frictionless track with a loop of radius 36 cm sits on a table 0.9 meters above the ground. If a 0.15 kg object just makes the loop, how high above the table did the object start and how far from the table does it land?




    2. Relevant equations



    3. The attempt at a solution
    Can you check my work
    Mgh - Mg(2R) = 0.5MV^2
    gh - 2gR = 0.5V^2
    once i solve for V i can use a kinematics equation to figure our how far from the table it lands?
     
  2. jcsd
  3. Dec 22, 2008 #2
    I think I misunderstood question first time.
     
    Last edited: Dec 22, 2008
  4. Dec 22, 2008 #3
    Can you check my work
    Mgh - Mg(2R) = 0.5MV^2
    gh - 2gR = 0.5V^2
    gh-2gR = 0.5Rg
    h = 2.5R = 2.5(.36m)= 0.9
    for the second part to figure out how far from the table it lands:
    Y = Voyt + .5gt^2
    0.9 = 4.9t^2
    t = 0.428571s
    x = Voxt
    x =((Rg)^0.5)*t
     
  5. Dec 22, 2008 #4
    First part is correct. For the second part, you have put Y = 0.9 m and Vox = sqrt(Rg).
    But Vox = sqrt(Rg) at the top of the loop. At that time, the object's height from ground
    = table's height from ground + object's height from table
    = 0.9 m + 2R
    = 0.9 m + 2*0.36 m
    = 1.62 m
    So put Y = 1.62 m
    Rest is fine.
     
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