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An amusement park ride question, picture included

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data

    AmusementParkRide.jpg


    An amusement park ride consists of a rotating circular platform 8.07m in diameter from which 10kg seats are suspended at the end of 2.87m massless chains. When the system rotates, it makes an angle of 37.3° with the vertical.
    a) What is the speed of each seat?
    b) If a child of mass 45.3kg sits in a seat what is the tension in the chain for the same angle?


    m=55.3kg (mass of the seat + the mass of the child)
    Vt=5.56565m/s (calculated from part A, confirmed as correct)
    Ѳ=37.3° (listed)
    g=9.8m/s2 (listed)
    d=8.07m (listed)
    l=2.87m (listed)
    r=5.77419m (calculated from the length of the chain and the diameter of the circular platform, d/2+lsinѲ, which is its horizontal distance from the center of rotation, where d is the diameter of the platform and l is the length of the chain)


    2. Relevant equations

    ƩFx=0
    ƩFy=0
    Fc=[tex]\frac{Vt^2m}{r}[/tex]
    Fg=mg


    3. The attempt at a solution

    I have the answer to A, and that part is verified as correct already. B is what I'm wondering about.

    So, this swing is rotating in the z dimension, but in the x and y directions it is in equilibrium which means that the sum of the forces in the x direction and the y direction must be equal to 0.

    ƩFx=0
    ƩFy=0

    Then the forces in the x are the horizontal tension in the cable in the x and the centrifugal (equal and opposite to the centripetal), and the forces in the y are the vertical tension in the cable and the force of gravity.

    ƩFx=Tx-Fc
    ƩFy=Ty-Fg

    Since they're both equal to 0, they can be set equal to each other.
    Tx-Fc=Ty-Fg

    I rewrote these based on their definitions.

    Tsin(Ѳ)-[tex]\frac{Vt^2m}{r}[/tex]=Tcos(Ѳ)-mg

    Solve symbolically:
    T=[tex]\frac{m}{sin\Theta-cos\Theta}[/tex][tex]\frac{Vt^2}{r}[/tex] -g

    When I put all that together, I get T=681.28N. I have one more guess left for this problem so I need to be sure it's right beforehand. Can someone verify if this logic is correct please?
     
  2. jcsd
  3. Nov 24, 2009 #2

    Andrew Mason

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    You are making it more complicated than necessary. All you have to do is determine Ty. You know the angle, so you can determine T from that.

    AM
     
  4. Nov 24, 2009 #3
    It doesn't ask for the vertical component of the tension. It asks for the tension overall.
     
  5. Nov 24, 2009 #4

    Andrew Mason

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    Yes. But with the vertical component of T and the angle you can easily find T.

    AM
     
  6. Nov 24, 2009 #5
    So...like Ty/cos[tex]\Theta[/tex]?
     
  7. Nov 24, 2009 #6

    Andrew Mason

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    That's right.

    AM
     
  8. Nov 24, 2009 #7
    Wow. I can't believe I didn't think of that. And it equals the same thing I got but with so much less work. Thank you very much.
     
  9. Nov 24, 2009 #8

    Andrew Mason

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    You are welcome. And welcome to PF by the way.

    AM
     
  10. Nov 24, 2009 #9
    Wait...okay so I have another question.

    So Tcos[tex]\Theta[/tex]=mg. And when I solved for T, it gave me the same answer of 681.28N as the supercomplicated one. But my other equation was Tsin[tex]\Theta[/tex]=Vt^2m/r. When I solved that one, I got T=489.555N. It should be the same, right? So what's different about it? Am I missing a force in the x direction that I should take into consideration?
     
  11. Nov 25, 2009 #10

    Andrew Mason

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    Since [itex]T\cos\theta = mg[/itex] is always true, the tension is easily determined. It is simply a matter of relating the x component of tension to the centripetal force: [itex]T\sin\theta = mv^2/r = m\omega^2r[/itex]. It is a bit tricky because r is a function of [itex]\theta[/itex]. I am not sure what you were using for r but your formula is correct.

    AM
     
  12. Nov 26, 2009 #11

    Andrew Mason

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    As a follow up to this, I get 6.56 m/sec as a value for v (tangential speed of the chairs) not 5.65 m/sec.

    [tex]\frac{mg}{cos\theta} = \frac{mv^2}{r}[/tex]

    [tex]v = \sqrt{g\tan\theta(L\sin\theta + d/2)}[/tex]

    [tex]v = \sqrt{9.8*.762*(2.87*.606 + 4.035)[/tex]

    [tex]v = \sqrt{43.12} = 6.56 m/sec[/tex]

    AM
     
  13. Sep 3, 2011 #12
    hey. can i ask something about this question. how to get the V for the part (a). is it something to do with the horizontal and vertical forces which involve sin and cos? i am still not able to get it :(
     
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