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Homework Help: Friction and spinning mass

  1. Sep 25, 2009 #1
    Spinning mass

    1. The problem statement, all variables and given/known data
    1) A mass of 7.10 kg is suspended from a 1.21 m long string. It revolves in a horizontal circle as shown in the figure. The tangential speed of the mass is 2.90 m/s. Calculate the angle between the string and the vertical.

    Picture 1) http://psblnx03.bd.psu.edu/res/msu/...Force_Motion_Adv/graphics/prob03_pendulum.gif

    3. The attempt at a solution

    For 1) I have the equation down to:

    [tex] \frac{gL}{v^2} - \frac{gL}{v^2}cos^2 \theta - cos \theta = 0 [/tex] How would I put this into the quadratic formula? (L = length of string)
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Sep 25, 2009 #2
    You should show what you did. For both of them you just need to draw an fbd and find the net force.
  4. Sep 25, 2009 #3
    I have shown what I did? If I were to show everything that would be a hell of a lot of typing. I've already done FBD, all I need is what has been asked.

    Problem 1:

    [tex] T_y - mg = 0 [/tex] X-Direction: [tex] T sin \theta = \frac{mv^2}{r} [/tex]

    [tex] T cos\theta = mg [/tex] X-Direction: [tex] T sin \theta = \frac{mv^2}{Lsin \theta} [/tex]

    [tex] T = \frac{mg}{cos \theta} [/tex] X-Direction: [tex] T = \frac{mv^2}{Lsin^2 \theta} [/tex]

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

    [tex] mgL sin^2 \theta = mv^2 cos \theta [/tex]

    [tex] gL(1-cos^2 \theta) = v^2 cos \theta [/tex] Thus:

    \frac{gL}{v^2} - \frac{gL}{v^2}cos^2 \theta - cos\theta = 0
    Last edited: Sep 25, 2009
  5. Sep 25, 2009 #4
    I have no idea what you did for (1). What were you x and y equations for force?
  6. Sep 25, 2009 #5
    I edited above for #1. The x-direction is on the right side, y on the left.
  7. Sep 25, 2009 #6
    Alright, when I did it I had the radius as L >>
    In your final equation just set cos[tex]\theta[/tex] = x.
  8. Sep 25, 2009 #7
    I did that but wouldn't you get something like

    [tex] \frac{1 \pm \sqrt{1^2 - 4(1.41)(1.41)} }{2(1.41)} [/tex]

    Which would simplify to:

    [tex] \frac{1 \pm \sqrt {-6.9524}}{2.82} [/tex]


    Edit: For reference the equation would be [tex] 1.41x^2 - x + 1.41 [/tex] Edited out 2nd question.
    Last edited: Sep 25, 2009
  9. Sep 25, 2009 #8

    \frac{gL}{v^2} - \frac{gL}{v^2}cos^2 \theta - cos\theta = 0

    So the quadratic equation would be
    -1.41x^2 - x + 1.41
    You had a positive when it is actually negative.
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