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Homework Help: Maximum angular velocity from a swing carousel

  1. Mar 23, 2014 #1
    The problem is to find out the maximum angular velocity that the swing carousel can spin at before the chain holding the passenger breaks. The maximum tension that the chain can withstand is known (T). The radius from the centre of the ride to the chain when stationary is known (r). The maximum mass that a passenger can be is known (m). The length of the chain is known (l). Given this information, derive a formula that will tell us the maximum angular velocity before the chain breaks.

    2. Relevant equations

    At first, I thought this was easy where T = mgcosθ + mω2rsinθ . But then I realised there were several problems with this firstly that we don't know theta and so can't work out ω when θ is also unknown. But also that using r is incorrect as the true radius that the passenger is revolving round is larger due to the centripetal force making them swing out.

    I derived this formula for the necesarry condintions for the passenger being in equilibrium mgsinθ = mω2cosθ(r + lcosθ) . However it doesn't help me much at this stage. The problem is I know that θ and ω are dependant in that if I determine an angular velocity, I will get a corresponding value for θ when the system is in equilibrium.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 23, 2014 #2


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    T is in fact ##\vec T ##. You can't just add the components.

    Draw a free body diagram for the person in the seat and realize that the sum of ##\vec T ## and ##m\vec g## should do nothing else than cause a circular trajectory.
  4. Mar 23, 2014 #3
    Can I ask why you are taking the sine of the centrifugal force?
  5. Mar 23, 2014 #4
    Does anyone how I can draw up and post free body diagrams? It would help me respond to the last two replies with much better clarity.
  6. Mar 23, 2014 #5


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    When you go advanced you get the opportunity to upload a graphics file (under attachments). What kind depends on what you have available to make one.
  7. Mar 23, 2014 #6
    Ok well for now I will try to get through this with plain algebra. After correcting and refining my equations, I've come to the conclusion that the angle theta can be no larger than such a value that mg/cosθ could exceed T. In other words the maximum value for theta is θ=〖cos〗^(-1) mg/T .
    However, the maximum angle is not what I want, I want to know the maximum angular velocity I can use. A formula I was able to derive is ω = √(gsinθ/(cosθ(r+lsinθ))) . If I substitute θ for 〖cos〗^(-1) mg/T I can determine what the angular velocity should be using all known values (as I don't know what theta is).

    I hope this is right though!
  8. Mar 23, 2014 #7
    I think your first attempt was better.

    You can solve this in two ways:

    1) set up the equations of motion and solve a differential equation
    2) use conservation of energy to get the maximum ω
  9. Mar 24, 2014 #8
    Ok here's my attempt at setting up an equation of motion with time.
    I going to define the variable x, as the distance across the circumference of the quarter circle the passenger travels as the centripetal force throws them ourwards across this path. The other variable is time t. The angular velocity ω is constant as is r and the length of the chain l.

    So, as the carousel swing starts to spin we get a force mω2rcosθ pushing the passenger up. However the force is increasing with time because the radius is increasing as θ increases (but is also decreasing because of cosθ (whether on the whole it is increasing or decreasing I don't know)). So the force now looks like mω2(r+lsinθ)cosθ. However I don't want theta here as it is a third variable, but I can define it terms of x. θ (in radians) = x/l . So it now looks like this mω2(r+lsin(x/l))cos(x/l) . However, there is also another force acting directly against this which is mgsin(x/l) and it increases with time however I don't know at which stage this force is less than the opposing force and then is greater and then is equal, however I do know that these changes happen.

    So from here, I have no idea what to do nor can I do anything with conservation of energy.
  10. Mar 24, 2014 #9
    Some notes:
    1) You have 2 forces acting on your mass m: T and mg
    2) The mω^2r is the centrifugal force which is simply the resultant of T and mgcosθ
    3) The radius r does not change with time
    4) θ=x/l is an approximation that is valid for small values of θ only ≈60° or less for ≈5% error.
    5) Set up the 3 equations of motion: ∑Fx=max, ∑Fy=may, ∑M=Iα. The problem can be simplified if you take the y-direction to be in the same direction as T
    6) Your formula looks wrong. It should come out to be a simple looking differential equation at the end. You can solve the differential equation to get ω.
  11. Mar 24, 2014 #10
    Sorry I appreciate your time but I can't understand how 3) (x is the circumference and l is a radius) and 4) are true and how my formula is wrong and I can't understand points 2) and 5).
    Sorry again.
  12. Mar 24, 2014 #11
    That's why a diagram of the situation would help explain better.

    I assumed we were talking about a simple pendulum that is free to swing back and forth. I see know you seem to be talking about something else.
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