1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A falling rod

  1. May 26, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider a uniform rod of mass 12kg and length 1.0m. At its 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.
    b) the angular speed of the rod as it passes through the vertical at C.

    2. Relevant equations
    PE = mgh
    KE = 1/2Iω2
    Irod = 1/3ml2
    ω2 = ωo2 + 2αθ

    3. The attempt at a solution
    a) PE = KE
    mgh = 1/2Iω2
    (12kg)(9.81m/s2)(1.0m) = (0.5)(1/3)(12kg)(1.0m)ω2
    ω = 7.668rad/s
    Can I use the length of the rod as its height?

    ω2 = ωo2 + 2αθ
    (7.668)2 = 2α90° (90° = 1.571rad)
    α = 18.7rad/s2
    is that right?

    b) ω2 = ωo2 + 2αθ
    ω2 = (7.668rad/s)2 + 2(18.7rad/s2)(1.571rad)
    assumptions: the α will stay the same and i'm trying to calculate from point B to C so θ = 90°
    ω = 10.84rad/s

    Thanks in advance!

    Attached Files:

  2. jcsd
  3. May 27, 2014 #2


    User Avatar

    Use the center of mass of the rod.
  4. May 27, 2014 #3
    This is incorrect .

    The relation in red is applicable when the angular acceleration is constant which is not the case in this problem.

    Write torque equation for the rod when it is horizontal.
    Last edited: May 27, 2014
  5. May 27, 2014 #4
    Ok, so:

    τ = Iα
    τ = 1/3ml2α

    τ = rF
    F = mg
    τ = rmg
    rmg = 1/3ml2α
    (1.0m)(12kg)(9.81m/s2) = (1/3)(12kg)(1m)2α
    α = 29.4rad/s2
    This seems rather high...

    would the radius of the rod be its length because it is pivoting on its end? I think i did something wrong... thanks for the help
  6. May 27, 2014 #5
    1) What is the point of application of the force mg which is producing the torque about the pivot ?
    2) What is the distance between this point of application and the pivot ?
  7. May 27, 2014 #6
    1) Would that be at the end of the rod, the opposite end of the pivot?
    2) if the point of application is the end, then it would be 1.0 m and the radius should be 1.0m, right?
  8. May 27, 2014 #7

    Where does weight of a body act ?
  9. May 27, 2014 #8
    Oh, the centre of mass? so the radius would be half of the length, 0.5m, right?
  10. May 27, 2014 #9
  11. May 27, 2014 #10
    Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted