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Uniform circular motion - finding velocity

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data
    klg2LjH.png

    2. Relevant equations
    radial acceleration = v^2/R

    3. The attempt at a solution
    I'm unsure if we have to mention the tension of the cable, but I'm guessing L2 will suffice? Correct me if I am wrong.
    Anyways, I drew the diagram from a side view (kind of like a circular pendulum) but L1 and the angle is throwing me off. I also drew a free body diagram of the seat at the 3 o'clock position: Tcable (tension or L2?) point towards the center, Tz pointing north, Tr pointing left, and Fg downwards. In the z direction, I have Tz + Fg = 0, so Tcable = mg/cos(theta). In the radial direction, I have Tr = ma(radial) > Tsin(theta) = mv^2/L1(?) so far.
    Help will be greatly appreciated.
     
  2. jcsd
  3. Feb 13, 2016 #2

    TSny

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    OK for the tension in the cable. However, you do not have the correct radius of the circle that the rider is moving along.
     
  4. Feb 13, 2016 #3
    Is it R = L1+L2x = L1 + L2sin(theta)?
     
  5. Feb 13, 2016 #4

    TSny

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    Yes it is. :oldsmile:
     
  6. Feb 13, 2016 #5
    Okay so after that part, I plugged in T (from the z direction) and R and I found v = sqrt( (L1+L2sin(theta))*g*tan(theta)). Does that sound about right? But the mass cancelled out, how come in the problem they asked for the expression to contain the mass?
     
  7. Feb 13, 2016 #6

    TSny

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    Yes.
    I don't know. Maybe they didn't want to give away that the mass doesn't matter.
     
  8. Feb 14, 2016 #7
    Okay thank you so much!
     
  9. Feb 14, 2016 #8
    Yeah I didn't think we would need to consider the tangent acceleration in this case.
     
  10. Feb 14, 2016 #9

    TSny

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    Right. There's no tangential acceleration in this case since the circular motion is "uniform".
     
  11. Feb 14, 2016 #10
    Consider the forces on the rider: there are three.
    1. g, acting downward; acceleration g = 9.8 m/s²
    2. centifugal force (call it C), acting outward; accleration C = v²/R
    3. The force on the cable (call it T), which MUST close the triangle at the hypotenuse, hence T = sqrt (g² + (v²/R)²)

    Then θ = arctan (C/g)

    Therefore θ = arctan ([v²/R]/g)
    tanθ = (v²/R)/g
    g tanθ = v²/R
    v² = R g tanθ
    v = sqrt ( R g tanθ)

    Now all you have to do is express r in the terms of the problem (L1, L2, etc) and plug it in ... I think you can figure that out ...
     
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