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Determine the angular acceleration and angular velocity

  1. Aug 7, 2017 #1
    1. The problem statement, all variables and given/known data

    A thin uniform rod (of mass 10.0 Kg and length of 1.20 m) is attached to a friction-free pivot. Initially, the rod is balanced vertically above the pivot (position A in the figure attached). If the rod falls from rest, calculate

    a. the angular acceleration at position B

    b. the angular velocity at position C

    2. Relevant equations

    m= 10.0 kg
    l= 1.20 m
    I=ml^2 / 3
    v initial = 0 m/2

    Torque (T) = moment of inertia (I) * angular acceleration (a)

    3. The attempt at a solution

    a.
    angular acceleration = T / I

    where I is known --> a = T / (ml^2 / 3) we know mass and l, therefore plug in to obtain:

    a = T / 4.8

    Net torque only comes from the weight, thus it is: mg * 1/2*l * sin (90) --> This is at position B

    Net torque = (10)(9.8)(1/2)(1.2)sin(90)
    Net torque = 58.8

    a = 58.8/4.8 --> 12.3 rad/s^2 clockwise (direction is given)

    Is this part correct? It was the only way I could think about doing this...

    b) I used energy to solve for the angular velocity at part c.

    I know: mgh = 1/2Iw^2 (after eliminating variable/etc. did I eliminate all the correct things and have the correct variables remaining?)

    w^2 = 2mgh/I where I is the moment of inertia, and was calculated to be 4.8

    Thus,

    w^2 = 2mgh/4.8 where m is known, g i known, and h is known

    w^2 = 2* 10 *9.8*1.2 / 4.8

    w^2 = 49

    w = 7.00 rad/s clockwise (direction is given)

    Is this correct? It is the only process I could think of how to solve...
     

    Attached Files:

  2. jcsd
  3. Aug 7, 2017 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me!
     
  4. Aug 7, 2017 #3
    Thank you so much!
     
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