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Using Centripetal Forces to find the radius of a circle

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data

    A 100g (0.1kg) rock is attatched to a 1.0m rope and spun around in a circle with a period of rotation of 1.0s. What is the Radius of the circle that it forms?

    2. Relevant equations


    Fc = (mV^2) / r
    V= (2∏r/T)
    LCosθ = r


    3. The attempt at a solution

    Im quite stick here, as Im not sure how I could ever find r without knowing the θ of the circle.

    First thing I tried was making a triangle like so:

    Illustration in 2D

    Im going to start with the simple equation TCosθ = Fc. We also know that (mV^2)/r = Fc. If you put those 2 equations together, you get TCosθ = (mV^2)/r. This leaves us with the equation:

    TCosθ = (mV^2)/ r

    Furthurmore, we know that V = 2πr/P (Where P is period, should normally be T but theres Tension in this equation, so ill just do P)

    TCosθ = m (2πr/P)^2 / r

    The period of rotation is 1, so the equation simplifies again to:

    TCosθ = m (2πr)^2 / r

    We can also sub r in to the formula we just moved in.

    TCosθ = m (2π (Cosθ)^2) / Cosθ

    From here, we can make the equation:

    T = (m (2π) ^2 * (Cosθ)^2) / (Cosθ)^2 (Correct? Im not sure if thats logical)

    Now the (Cosθ)^2 can cancel out, making:

    T = m (2π)^2

    Sub in the values, we have T = (0.1)(2π)^2
    = 0.4π^2

    We now know Tension, Which we can use in the triangle.

    Triangle illustration

    If I use Sin, I can find that the angle is 14.37321*.

    We can use this θ in the equation LCosθ = r
    (1)(Cos14.37321) = r
    r = 0.97 m.

    Is that correct?
     
    Last edited: Apr 17, 2013
  2. jcsd
  3. Apr 17, 2013 #2

    gneill

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    Staff: Mentor

    Your method and result look fine.

    Note that you might consider the problem in terms of similar triangles (I know, geometry was a looong time ago :smile:).

    attachment.php?attachmentid=57981&stc=1&d=1366247405.gif

    Consider the ratio ##g/a_c##...

    You can use the rotational motion form for centripetal acceleration, ##a_c = ω^2 r##, where ##ω = 2\pi/T##. A little algebra and you can find an expression for r that does not require sines or cosines.
     

    Attached Files:

  4. Apr 17, 2013 #3
    Do you mind explaining that a little? I'd like to check my answer, as I think 0.97m is a little large for the radius concidering you only make a full turn per minute. Or if you could do the math for this perticular situation and make sure it comes out correct, that would be awesome. Its for marks, so I just dont want to get it wrong.

    Thanks :D
     
    Last edited: Apr 17, 2013
  5. Apr 17, 2013 #4

    gneill

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    The problem says that the period of rotation is 1 second, not 1 minute. So ω is ##2\pi/1sec##.

    L is the length of the rope. r is the resulting radius of the circle that the mass follows at one revolution per second. ##a_c## is the centripetal acceleration acting on the rotating mass. g is the gravitational acceleration. You should be able to pick out similar triangles in the diagram. Equate the side-length ratios.
     
  6. Apr 17, 2013 #5
    Yea, sorry you're right. 1 second.

    Ohh, so the Theta in both Triangle Ac/g and Lr are the same, correct? So you can equate the two, like this?

    Tanθ = g/ac
    Tanθ = √1^2 - r^2

    g/ac = √ 1-r^2

    g/ (2π)^2 = √ 1-r^2

    (9.8 / (2π^2)) ^ 2 = 1-r^2

    r^2 = 1 - (9.8 / (2π^2)) ^ 2

    r^2 = .938

    r= .969

    Thats what I get when I do similar triangles, which Im 99% sure isnt right. Is the g (.1)(9.8) or mg, or is it just 9.8?
     
    Last edited: Apr 18, 2013
  7. Apr 18, 2013 #6

    haruspex

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    It's g, not mg. The mass cancelled out.
     
  8. Apr 18, 2013 #7
    Alright, so that makes both of them the same answer :). I end up with 0.97. Can someone do it and verify that that is right?

    Thanks :)
     
  9. Apr 18, 2013 #8

    haruspex

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    I agree with .97
     
  10. Apr 18, 2013 #9
    Great, thanks :D
     
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