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Centripetal forces

  1. Mar 17, 2014 #1
    1. The problem statement, all variables and given/known data

    A 1.5 kg mass tied to the roof rotates with constant speed in a horizontal
    circle. The string makes an angle of 30 deg to the vertical.

    a) determine the velocity , the centripetal acceleration and the centripetal
    force on the mass.

    b) determine the tension in the string.

    c) say the mass is given a push so that now the mass rotates with a constant
    velocity of 9.4 m/s . Determine the angle the string makes with the vertical
    , the centripetal force on the mass and the tension in the string.


    2. Relevant equations

    ∑Fradial = TsinΘ = mv2/R

    ∑Fy = TcosΘ - mg = 0


    3. The attempt at a solution

    To solve part a:

    First I found my R by,

    R = LsinΘ = 1.5sin30° = .75m

    Second my T by,

    T= mg/cosΘ = 1.5*9.8/cos30° = 16.97 N

    Now I find my velocity by,

    V= sqrt((R*T*sinΘ)/m)) = sqrt((.75*16.97*sin30°)/1.5) = 2.06 m/s

    the centripetal acceleration,

    v2/R = a

    a = TsinΘ/m = 16.97*sin30°/1.5 = 5.66 m/s2

    for part b I already found my tension to be 17N.

    Part C is where I am having trouble, how can i find the angle Θ with the new constant velocity if im not given a radius or tension?
     
  2. jcsd
  3. Mar 17, 2014 #2

    TSny

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    Hello, mpittma1. Welcome to PF!

    Note that you have 3 "unkowns": θ, T, and R. But you also have 3 equations: your two "Relevant equations" and your equation for R in terms of L and sinθ.

    Can you see how to do the algebra to get to an equation where the only unknown is θ?
     
  4. Mar 17, 2014 #3
    Hey TSny,

    So my three equations are:

    R = LsinΘ (1)

    TcosΘ - mg =0 (2)

    &

    TsinΘ = mv2/R (3)

    I see how you could plug equation 1 into equation 2 and come up with

    TsinΘ = mv2/LsinΘ which is closer because I have my m, v, and L but Im still not seeing how to get rid of the T by using equation 2.
     
  5. Mar 17, 2014 #4

    TSny

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    Use (2) to find T in terms of θ. [EDIT: Or divide (3) by (2) after rearranging (2)]
     
  6. Mar 17, 2014 #5
    So I am coming up with:

    TanΘ = v2/LsinΘg

    after dividing eqn 3 by eqn 2 and plugging in eqn 1 for R.

    After doing some more algebra I come up with tanΘsinΘ = v2/Lg

    something doesnt seem right there though....... how could i solve for tanΘsinΘ?
     
  7. Mar 17, 2014 #6

    TSny

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    Looks good so far! Can you express the left side of the equation solely in terms of sinθ?
     
  8. Mar 17, 2014 #7
    well I could easily express the left side as sin2Θ/cosΘ

    im not seeing how to express it solely as sinΘ though
     
  9. Mar 17, 2014 #8

    TSny

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    Can you think of a trig identity that relates sinΘ and cosθ?
     
  10. Mar 17, 2014 #9
    so heres what im doing:

    sin2Θ/cosΘ → using a half angle formula (1-cos2Θ)/2 * 1/cosΘ →

    (1-cos2Θ)/2cosΘ

    my eqn thus far is (1-cos2Θ)/cosΘ = 2v2/Lg

    how could you simplify the left side further?
     
    Last edited: Mar 17, 2014
  11. Mar 17, 2014 #10

    TSny

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    I'm not sure the half-angle formula will help. Can you make use of sin2Θ + cos2θ = 1?
    [Edit: I said earlier to express everything in terms of sinθ, which will work. But it might be a little easier to express sin2Θ/cosΘ in terms of cosθ.]
     
    Last edited: Mar 17, 2014
  12. Mar 17, 2014 #11
    I was trying to find a way to get a +cos2Θ but its just not there
     
  13. Mar 17, 2014 #12

    TSny

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    So, use the trig identity to express sin2θ in terms of cos2θ.
     
  14. Mar 17, 2014 #13
    (1-cos2Θ)/cosΘ

    or secΘ-cosΘ so im almost there
     
  15. Mar 18, 2014 #14

    TSny

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    Good

    I don't recommend writing it this way. Use (1-cos2Θ)/cosΘ for tanθsinθ in your equation from post #5 and see if you can proceed from there.
     
  16. Mar 18, 2014 #15
    After plugging in my values for v L and g i get:

    (1-cos2Θ)/cosΘ = 6.011

    any suggestions any solving for Θ?
     
  17. Mar 18, 2014 #16

    TSny

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    Let x = cosθ and write your equation in terms of x. Can you solve for x?
     
  18. Mar 18, 2014 #17
    ok so what i did:

    let x = cosΘ

    so, 1-x2/x = 6.011

    1-x2 = 6.011x

    0 = x2 + 6.011x -1

    then after solving for the roots of x

    x = -6.173 and .162

    plugging x = .162 into x = cosΘ

    gives me Θ = 80.7° which is the correct answer!!!

    thank you for all your help!
     
  19. Mar 18, 2014 #18

    TSny

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    Good work!
     
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