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Spinning Mass

  1. Jun 3, 2014 #1
    1. The problem statement, all variables and given/known data
    A mass of 3.50 kg is suspended from a 1.59 m long string. It revolves in a horizontal circle as shown in the figure.
    The tangential speed of the mass is 2.99 m/s. Calculate the angle between the string and the vertical.

    2. Relevant equations
    a_c = v^2/r
    F = m*a_c
    tanθ=sinθ/cosθ
    sin^2θ=1-cos^2θ

    3. The attempt at a solution
    Vertical: T*cos(θ)-m*g=0 → T=(m*g)/cos(θ)
    Horizontal: T*sin(θ) = m*a_c
    Combined the two: ((m*g)/cos(θ))sin(θ)=m*a_c → m*g*tan(θ)=m*a_c → g*tan(θ)=a_c
    we know that a_c=v^2/r, plug that in to get g*tan(θ)=v^2/r
    r=L*sin(θ) → g*tan(θ)=v^2/(L*sin(θ))

    I cannot get past this point and the notes on the problem say that a quadratic equation needs to be solved at some point.
     

    Attached Files:

  2. jcsd
  3. Jun 3, 2014 #2

    Nathanael

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    You wrote r in terms of θ and proceeded to try to solve for θ. It left to a bit of a dead end, so now try to solve for r

    See if you can write tan(θ) in terms of known quantites and r (it can be done) then you'll be able to solve it
     
  4. Jun 3, 2014 #3

    TSny

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    Good work so far. Continue from your last equation. Try isolating the trig functions on one side of the equation and all the constants on the other side. Let x = cosθ and see if you can get a quadratic equation in x.
     
  5. Jun 3, 2014 #4

    Nathanael

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    I don't think that will work. The equation becomes:

    [itex]\frac{v^{2}}{gL}=\frac{sin^{2}(θ)}{cos(θ)}[/itex]


    There is another way to solve for θ (involving the quadratic equation) if you solve for r and then use r to find θ (once you know r, θ is trivial)


    EDIT: actually you're right you could solve it that way... (solution removed) sorry about that you are correct
     
    Last edited: Jun 3, 2014
  6. Jun 4, 2014 #5
    I understand where this equation comes from, but I am not sure how you would go about solving that for θ.

    I don't understand at all what I would use to solve for r or how this would result in a quadratic equation.
    I am struggling with every aspect of this problem.
     
  7. Jun 4, 2014 #6
    Nevermind!!!! I figured it out, thank you all so much for you guidance in this problem.
     
  8. Jun 4, 2014 #7

    Nathanael

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    I know you said you figured it out now, but I'll still show you the other methods.

    You could rewrite [itex]sin^2(θ)[/itex] as [itex]1-cos^2(θ)[/itex] (from pythagorean theorem) and then the equation becomes:
    [itex]\frac{v^2}{gL}=\frac{1-cos^2(θ)}{cos(θ)}=\frac{1}{cos(θ)}-cos(θ)[/itex]
    Multiply both sides by cos(θ) and rearrange and you'll get a quadratic equation (with x=cos(θ)) which can be solved for cos(θ):
    [itex]cos^2(θ)+\frac{v^2}{gL}cos(θ)-1=0[/itex]
    There will only be one root with an absolute value of less than one, so simply take the arccosine of it and you'll have θ
    You had written the equation [itex]g*tan(θ)=\frac{v^2}{r}[/itex]
    Since [itex]L^2=r^2+y^2[/itex] (where y is the unknown side of the triangle) you could use the equation [itex]L^2=r^2+(r*cot(θ))^2[/itex]
    (... r*cot(θ) is the unknown side ...) to rewrite [itex]tan(θ)[/itex] as [itex]\sqrt{\frac{r^2}{L^2-r^2}}[/itex]
    Plug that expression into the "tan(θ)" part of your expression (and then square both sides and rearrange) and you'll get a quadratic equation (with [itex]x=r^2[/itex]) which you can solve for [itex]r^2[/itex]
    The quadratic equation would be:
    [itex]g^2r^4+v^4r^2-v^4L^2=0[/itex]
    There will be only 1 positive root, so take the square root of that to find r, then take the arcsin of (r/L) to find θ


    Using either method, you'll find θ=41.1° (I double checked)
     
    Last edited: Jun 4, 2014
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