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Circular Motion at an angle, finding angles and tension

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data
    At the royal show there is a ride for younger children where they sit in a seat and it spins in a circle. The seats weigh 10.0kg and by comparison the mass of the chains holding the seats is negligible. When the system rotates the chains make an angle of 28 degrees to the vertical if there is nobody sitting in them. The radius of the system supporting the chains is 4m.The length of each chain supporting a chair is 2.50m long. a) what is the speed of each seat? c) what angle would the chain make with the vertical with the child in a seat? d) what is the tension in the chain during the ride with the child in the seat?

    2. Relevant equations
    Fc = mv^2/r
    W=mg
    T=ma+mg

    3. The attempt at a solution
    Part a)
    W=mg=10*9.8 = 98N
    I found the horizontal component of the weight at the end of the chain = 98tan28=52.11N
    As Fc=Horizontal Weight to keep it in position, i figured that 52.11=mv^2/r
    R=2.5sin28 = 1.17m+the 4m off the base of the support = 5.17m
    52.11=mv^2/5.17 = 10v^2/5.17. V=*square root*(52.11*5.17)/10=5.19ms-1

    Part c)
    In this i wasnt sure how to find the two unknown variables, as i dont know both V and theta so i assumed V was the same as part a), 5.19ms-1
    Fc=mv^2/r = (10+40)*5.19^2/(2.5sin*theta)+4
    As Fc=Horizontal component of Weight, Fc=mgtan theta.
    = [(10+40)*5.19^2/(2.5sin*theta)+4]=490tan*theta
    And so rearranging gave me [(50*5.19^2)/(2.5*490+4)]=sin*theta*tan*theta.
    from this i solved in solver of my calculator and got an angle of 53.68 degrees. I am pretty certain to say this is wrong as the mass has increased, and so the angle should decrease no? how would you solve this question for both V and theta?

    Part D)
    I only know the equation that i think is right for tension, i have no idea what to do for this part of the question. Ive tried using the weight down of 490N (50*9.8) and using the angle gathered in part C and using trig to solve, in which i got a tension of 827.3N but im not certain about that at all.

    If a picture is needed for the question, please ask and ill try upload one
     
    Last edited: Sep 14, 2010
  2. jcsd
  3. Sep 14, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    Hi etern1ty, welcome to PF.

    If the velocity remains constant, θ is independent of mass in the chair. But tension will increase.
     
    Last edited: Sep 14, 2010
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