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Conservation laws in rotational movement

  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider a uniform rod of mass 12kg and length 1.0m. At it's end the rod is attached to a fixed, friction free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine,
    a) the angular acceleration of the rod as it passes through the horizontal at B. (rod is horizontal with left end at pivot point)
    b) the angular speed of the rod as it passes through the vertical at C. (rod is vertical with upper end at pivot point)

    For the diagram picture a clock with hand pointing to 12 (for position A, initial position), 3 for position B, 6 for position C.

    Given:
    m=12kg
    L= 1.0m

    2. Relevant equations
    τ = Iα
    PE = mgh
    KE (rotational) = 1/2Iω^2
    Irod = 1/3ml^2

    3. The attempt at a solution
    a) α=?
    τ = Iα
    τ = rF, F = mg, τ = rmg, r=L/2 since center of mass will be at centre of rod
    τ=L/2*mg
    Iα=1/3ml^
    L/2*mg =1/3ml^
    Solve for α
    m cancels out
    3L/2L^2 = α
    L cancels out
    3/2L=α
    3/2(1)= α = 1.5 rad/s^2

    b) ω=?
    mgh=KE
    mgL/2=1/2(1/3ML^2)ω^2
    mass cancels out
    6gL/2L^2 = ω^2
    3g/2L = ω^2
    ω= sqrt (3*9.8/2*1)
    ω = 5.42 rad/s

    I'm just not sure if I got the answer correct. Is someone able to check? Was a little unsure of what numbers I was using as radius and center of mass.

    Thanks

    Miral
     
  2. jcsd
  3. Sep 27, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    You lost a factor g somewhere in the calculation of the acceleration. This would have been obvious if you would have included the units, because the left side is a length and the right side is an angular acceleration - they cannot be equal.

    How did you find h in (b)?


    Concerning the notation: Things like 1/2L are problematic: is it 1/(2L) or (1/2)L? Usually the latter - if the L is supposed to be in the denominator, you should write 1/(2L).
     
  4. Sep 27, 2016 #3

    Whoops! You're totally right. I lost a g in there.

    L/2*mg =1/3ml^
    Solve for α
    m cancels out
    3Lg/2L^2 = α
    L cancels out
    (3g)/(2L)=α
    3(9.8)/2(1)= α = 14.7 rad/s^2


    Found the h or change in h by taking half of the length of the rod. L=1.0m, H = L/2
    If the rod is starting at position B with rod horizontal, gravity acts on the center of mass, the change in height from position B to C is therefore 0.5m or L/2.

    Does that look better?
     
  5. Sep 27, 2016 #4

    mfb

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    2016 Award

    Staff: Mentor

    I don't think it is. I don't see the diagram, but the starting position should be "12" at the clock.

    (a) looks better now.
     
  6. Sep 27, 2016 #5

    Ohhh...I think I see your point. If my starting point was B then the rod would have already gained some KE from A to B. So if I use conservation of energy and start at position A, the height would be 1.0m. mgh (point A) = KE (point C).
     
  7. Sep 28, 2016 #6

    mfb

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    2016 Award

    Staff: Mentor

    Right.
     
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