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Kinematics of Rigid Bodies question

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    The uniform 3.6m pole is hinged to the truck bed and released from the vertical position as the truck starts from rest with than acceleration of 0.9m/s^2. If the acceleration remains constant during the motion of the pole, calculate the angular velocity of the pole as it reaches the horizontal position.
    Diagram attached.


    2. Relevant equations
    I believe these equations are relevant, however, I am not given a mass for the pole, so i'm not entirely sure.
    a(tangential)=mrθ''
    a(normal) = mrω^2
    ƩMo=Iθ''+Ʃma(vector)d
    I=k^2m
    ω=2Vx
    ω=(ωo^2+2aθ)^(1/2)

    3. The attempt at a solution
    So I eventually want to realise the angular velocity ω.
    I have done a similar question that utilised energy and momentum methods, however, with a negligible mass, i'm not sure whether that will affect the equations, since the example question used mass.

    So essentially, I believe it may be using the last formula I provided, since it is not determined by time or mass.
    So, all I need to find is the acceleration, since I already know that ωo is at rest, and θ=90. Finding the acceleration, from the given positive direction of acceleration of the truck body, is something that I can't figure out.

    Thankyou for any feedback!
     

    Attached Files:

  2. jcsd
  3. May 13, 2012 #2
    your last formula is valid for constant angular acceleration,so from vertical to horizontal position does it remain the same.
     
  4. May 13, 2012 #3
    @ Andrien
    Well, the pole accelerates to the left (opposite the direction of the car) at 0.9m/s^2, and accelerates downward at 9.8m/s^2.
    These are both constants, but I don't know which to use in the formula.
    It can't possibly as simple as angular vel=(0^2+2*0.9*90)^1/2
     
  5. May 14, 2012 #4
    consider taking moment about hinge point, and write ma as a pseudo force along with gravity which will act towards left i.e. opposite to acceleration of truck.integrate from 0 to pi/2,you will get
    w=[3(g+a)/l]^.5, where l= length of rod.assuming k^2=l^2/12,
    w is around 3.
     
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