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Rotational Motion of a thin rod about a pivot

  1. Nov 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A uniform thin rod of Length L and mass M is pivoted at one end is held horizontally and then released from rest. Assuming the pivot to be frictionless, find

    a) Angular velocity of the rod when it reaches its vertical position
    b) The force exerted by the pivot at this time
    c) Now we want to swing the rod 270° clockwise so that it will reach a vertical position at the top of its swing. What initial angular velocity is needed for the rod to achieve that?

    2. Relevant equations
    Part a)
    Parallel-axis theorem
    Conservation of Energy

    Part b)
    ∑Fcm = mω2r

    Part c)
    Conservation of Energy?

    3. The attempt at a solution
    I actually have no issues with parts A and B but only with C.

    My initial concept for C was to use the principle of COE
    Ugrav, 1 + Krotational, 1 = Ugrav, 2
    but this was wrong.

    May I know what my concept for part c should be and why I was wrong in the first place?

    P.s. Apologies if this post was not well done, it's my first post and feedback is appreciated. Thanks!
  2. jcsd
  3. Nov 11, 2016 #2
    Why do you think it is wrong ?

    Please show your calculations and result you obtained .
  4. Nov 11, 2016 #3


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    Having answered a, there is a very quick route to c.
  5. Nov 11, 2016 #4
    Damn, apologies everyone, my initial assumptions were right. I got confused because of some grading error. Sorry for taking up your time!

    In any case it was:

    Ugrav1 + K rot 1 = Ugrav2
    1/2Iω2 + mgL = mg(3/2)L
    where I = 1/3mL2 with parallel axis theorem, which leads to:
    ω = (3g/L)1/2 rads/s
  6. Nov 11, 2016 #5


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    Right... and do you see the connection with the answer to a?
  7. Nov 12, 2016 #6
    Apart from them being the same, I can't. Could you enlighten me?
  8. Nov 12, 2016 #7


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    In both cases the potential difference relates to a rotation from horizontal to vertical, and in both cases it is stationary at one end of the motion, so the speeds at the other end must be the same.
  9. Nov 12, 2016 #8
    I never thought of it this way before, thank you for the insight.
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