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Uniform Circular Motion/Newton's 2nd Law?

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Question: A 3.45 kg object is attached to a vertical rod by two strings as in Figure P6.11. The object rotates in a horizontal circle at constant speed 5.65 m/s. What are the tensions of the two strings, individually?

    Variables:
    m = 3.45kg
    length of both strings (individually) = 2m
    vertical distance between two strings = 3m


    2. Relevant equations

    Sum(F) = m(v^2/r)
    Sum(F) = m(a)
    Basic trig.

    3. The attempt at a solution

    I assumed that the ball was revolving around the middle of the rod, thus we have a triangle with hypotenuse of 2m and side 1.5m and by solving we find the radius:
    r = sqrt(1.75) = 1.32m.
    We should be able to find the sum of the forces by using m(v^2/r):
    (3.45kg)((5.65m/s)^2/1.32m) = 83.25N.
    This is the centripetal force, right?
    a_c = v^2/r, thus a_c = (5.65m/s)^2/(√1.75m)=33.429m/s^2
    I then used Sum(F) = ma to find the x-component of T (tension of string):
    (3.45kg)(33.429m/s^2) = 115.33N
    We have a triangle with legs 1.5m and √1.75m and hypotenuse 2m, thus the angle made by T and r is θ = sin^-1(1.5m/2m) = 48.59°.
    Therefore, T = 115.33N/cos(48.59°) = 174.36N
    Both strings share the force equally and both have this value, correct?
    I just want to make sure I am doing this right...I have already lost points at wrong attempts already (online homework). Could someone look this over and make sure I did it correctly and notify me if there's anything I might have misunderstood? Thanks!
     
  2. jcsd
  3. Oct 1, 2011 #2

    rl.bhat

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    Hi Cryptologica, welcome to PF.
    In the equilibrium condition, weight of the object acts in the downward direction. In your attempt there in no mg term. Obviously the tensions in the two strings are not equal. Now try it again.
     
  4. Oct 1, 2011 #3
    So which one will have more tension? How do I account for mg?
     
  5. Oct 1, 2011 #4

    rl.bhat

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    Go through the thread " Newton's 2nd Law for a Particle in Uniform Circular motion" below.
     
  6. Oct 1, 2011 #5
    So would the y-comp(of T) = Tsin(θ) - mg
    and x = Tcosθ? I don't have to factor that into x because mg is a vertical force? Is that going in the right direction?
     
  7. Oct 1, 2011 #6
    Ah, sorry...I am trying to follow that other thread, but am having a hard time. I have been working on this for too long probably, but I think I am almost there? Any more hints you can give, I am still kind of confused?
     
  8. Oct 1, 2011 #7

    rl.bhat

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    Σ Fy = T1cosθ - (T2cosθ + mg) = 0
    ΣFx = (T1 + T2)sinθ - Fc = 0
    Here cos θ = 3/4
    Now solve for T1 and T2.
     
  9. Oct 1, 2011 #8
    θ = cos^-1(3/4) = 41.41
    t1cos(41.41)-(t2cos(41.41) + (3.45kg)(9.8m/s^2)) = 0
    (3/4)t1 - (3/4)t2 - 33.81N = 0
    t1 = 48.08N+t2
    substitute:
    (48.08N+2(t2))sin(41.41) = fc
    fc/2sin(41.41) + 24.04N = t2

    Did I do something wrong, or do I know fc? Then I could plug and solve for t2 and t1? fc = centripetal force, which is?
     
  10. Oct 1, 2011 #9
    Oh, is fc = m(v^2/r)?
    fc = (3.45kg)[(5.65m/s)^2/sqrt(1.75)] = 83.25N
    Then use that to solve, right?
     
  11. Oct 1, 2011 #10

    rl.bhat

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    t1 = 48.08N+t2
    check this step.
    fc/2sin(41.41) + 24.04N = t2
    Check this step.
     
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