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Dynamics of Rotational Motion

  1. May 26, 2016 #1
    One end of a thin, uniform rod is connected to a frictionless hinge as shown in Figure 1. The rod has a length of 0.8 mand a mass of 2 kg. It is held up in the horizontal position (θ=90∘) and then released.
    1)Calculate the angular velocity of the rod at θ=90∘.
    2)Calculate the angular acceleration of the rod at θ=90∘.
    3)Calculate the angular velocity of the rod at θ=60∘
    4)Calculate the angular acceleration of the rod at θ=60∘
    5)Calculate the angular velocity of the rod at θ=0∘.
    6)Calculate the angular acceleration of the rod at θ=0∘.
    1067.PNG
    2. Relevant equations
    mgh=1/2 Iw^2
    a = ((mgcos)1/2)/I
    3. The attempt at a solution
    I got these answers and only 1 and 6 were correct
    1)0 rad/s
    2)74 rad/s^2
    3)8.6 rad/s
    4)64 rad/s^2
    5)12 rad/s
    6)0 rad/s^2

    Any help is appreciated thank you :)
     
  2. jcsd
  3. May 26, 2016 #2
    you are having problem with above - so check expression for angular acceleration.
    moment of inertia xangular acceleration = torque on the rod =forcex the the perpendicular distance from the axis of rotation
    and for angular velocity one can use the work- energy conversion
    work done by a torque is torquex angular displacement= rotational energy
     
  4. May 28, 2016 #3

    haruspex

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    We cannot tell where you are going wrong if you do not post your working.
     
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