1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    Note:
    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) ?
    [/tex]
    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:
    [tex]0.8*2.9^2=6.728[/tex]
     
  15. Oct 28, 2016 #14
    This is radial acceleration .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Acceleration of the end of a hinged rod in a pulley system
  1. Pulley system question (Replies: 3)

Loading...