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Newton's 2nd Law for a Particle in Uniform Circular motion

  1. Jun 24, 2008 #1
    1. A 4.0kg object is attached to a vertical rod by 2 strings. Object roatates in a horizontal circle at a constand speed 6.00 m/s. Find the tention in the upper and lower string. The strings are 2 meters long. the distance between the attachment points for each string is 3 meters.



    2. F=ma=m(v^2/r)



    3. I know that you must sum the forces in the x and y direction but cant seem to get rolling on this
     
  2. jcsd
  3. Jun 24, 2008 #2

    G01

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    Yes, you must sum the forces from both strings. The sum of the forces is a net centripetal force. What does this mean it has to be equal to?

    HINT: What are all centripetal forces equal to?
     
  4. Jun 24, 2008 #3
    i don't know.
     
  5. Jun 24, 2008 #4

    rock.freak667

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    Draw a vector diagram...then split the tension into it's components.
     
  6. Jun 24, 2008 #5
    Y direction
    Fupper= T*cos(theta)+mg=0
    Flower= T*cos(theta)-mg=0

    yes no?
     
  7. Jun 24, 2008 #6

    G01

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    Yes, you do. You said it in your first post.

    F_c=mv^2/r

    So, if:

    [tex]\Sigma F_{radial direction} = F_c[/tex] and [tex]F_c=\frac{mv^2}{r}[/tex] then...

    Also,

    You need to sum the radial components (x components) of the forces, since only those will contribute to the net centripetal force. To find these it would help to draw a free body diagram as clintau said. How far can you get now?
     
  8. Jun 25, 2008 #7
    so Fc=4*6^2/r=109.09

    That gives the force in the x direction....What about the Y?
     
  9. Jun 25, 2008 #8
    how would you find the tensions in the strings?
     
  10. Jun 25, 2008 #9

    rock.freak667

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    [tex]\Sigma F_{vertical}=0[/tex]
     
  11. Jun 25, 2008 #10
    Fcx is the T in the upper string. How do you get the T in the lower?

    Tupper-mg? gives app 69N. Answer should be 56.2

    what is wrong?
     
  12. Jun 26, 2008 #11

    rl.bhat

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    Instead of taking componets of tention on Y direction, take the component of mg along the tensions.
    So T(upper) = T1 + mgcos(theta)
    T(lower) = T2 - mgsin(theta)
    Now T(upper), T(lower) and Fc are in equilibrium. Apply the Lami's theorem.
    [T1 + mgcos(theta)]/sin(pi/2 + theta) = [T2 - mgsin(theta)]/sin(pi/2 + theta) = [mV^2/r]/sin(pi -2*theta).
    Now solve for T1 and T2.
     
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