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Homework Help: Acceleration of the end of a hinged rod in a pulley system

  1. Oct 26, 2016 #1
    1. The problem statement, all variables and given/known data
    As shown in image.
    Screen Shot 2016-10-26 at 10.25.02 PM.png

    2. Relevant equations

    Moment of inertia of pulley = [tex] 1/2*M*R^2 [/tex]
    Moment of inertia of rod (about end) = [tex] 1/3*M*L^2 [/tex]
    Acceleration of end of rod in theta direction = [tex] L*α [/tex]
    Acceleration of end of rod in radial direction = [tex] L*ω^2 [/tex]

    3. The attempt at a solution
    Pretty sure this question requires solving a system of five simultaneous equations, but I cannot work out the final one.

    T1 = tension of rope attached to A
    T2 = tension of rope attached to B
    M = mass
    R = radius
    α = angular acceleration
    ω = angular velocity
    Positive movement defined upwards

    From free body diagram of block A:

    [tex] T1- M(A)*g = M(A)*a(A) [/tex]

    From taking the moment around the pulley:

    [tex] T1*R(pulley)-T2*R(pulley)=I(pulley)*α(pulley)=1/2*M(pulley)*R(pulley)^2*α(pulley) [/tex]

    From taking the moment around the rod:

    [tex] T2*L-M(rod)*g*L/2=I(rod)*α(rod)=1/3*M(rod)*L^2*α(rod) [/tex]

    From polar coordinates:

    [tex] a(A) = -R(pulley)*α(pulley) [/tex]

    One equation missing

    Once I work out the last equation and solve for α(rod), I should be able to use Pythagoras to work out the magnitude of the acceleration from:

    [tex] a=((α(rod)*L)^{2}+((L*ω^2)^{2}))^{1/2} [/tex]

    Can anyone see what I'm missing and/or if I'm going wrong in any of the other equations? Been trying this question for hours and can't get my head around it.
  2. jcsd
  3. Oct 26, 2016 #2
    I think there are only four variables so four equations should be enough .

    I guess you have considered anticlockwise positive , in which case you have a sign issue .

    Apart from that everything else looks fine :smile: . I hope I am not overlooking something.
  4. Oct 26, 2016 #3
    There's five in my equations, unless one I can work out from something else?:
    [tex] a(A), T1, T2,
    α(rod), α(pulley) ?
    Otherwise I think I still need a fifth equation.

    Thanks for the sign tip, though!
  5. Oct 26, 2016 #4
    OK .Fair enough .

    The tangential acceleration of point B is related to the acceleration of the block .
  6. Oct 27, 2016 #5
    Are you implying there's another equation I can get from this?
  7. Oct 27, 2016 #6
    Yes .

    Replace the rod attached to point B with a mass M .

    How would acceleration of block A and mass M be related ?
    Last edited: Oct 27, 2016
  8. Oct 27, 2016 #7
    Would it be that:
    [tex] a(A)=-a(B[tangential]) = -L*ω^{2} [/tex]

    As the displacement of A is the negative displacement of B (if considering it B as mass), so [tex]a(A)=-a(B)[/tex] in the tangential direction? This allows me to calculate a(A) without solving simultaneously, which doesn't seem right to me.
  9. Oct 27, 2016 #8
    Yes . The magnitude of tangential acceleration of B should be equal to that of A as they are connected by an inextensible string .

    This should give you the answer .

    What values are you getting for tangential and radial acceleration of B ?
  10. Oct 27, 2016 #9
    I am getting:
    [tex] a[radial]=7.677 m/s/s [/tex]
    [tex] a[tangential] = 6.728 m/s/s [/tex]

    (magnitudes only)

    These are not giving me the correct answer for overall acceleration.
    Last edited: Oct 27, 2016
  11. Oct 28, 2016 #10
    What is the given correct answer ?
  12. Oct 28, 2016 #11
    I won't know it until I get it right unfortunately, this is a question from an online task.
  13. Oct 28, 2016 #12
    I am getting 8.938 m/s2 as magnitude of tangential acceleration .
  14. Oct 28, 2016 #13
    I am just doing:
  15. Oct 28, 2016 #14
    This is radial acceleration .
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