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Rotational Motion Question, lever arm with two masses

  1. Jul 2, 2017 #1
    1. The problem statement, all variables and given/known data

    A uniform rigid rod with mass Mr = 2.7 kg, length L = 3.1 m rotates in the vertical xy plane about a frictionless pivot through its center. Two point-like particles m1 and m2, with masses m1 = 6.7 kg and m2 = 1.6 kg, are attached at the ends of the rod. What is the magnitude of the angular acceleration of the system when the rod makes an angle of 51.1 degrees with the horizontal? (m2 is 51.1 degrees above the horizontal, and m1 is 51.1 degrees below the horizontal).

    2. Relevant equations

    Torque=Inertia*Angular Acceleration

    Inertia of uniform rigid rod = mL^2/12

    3. The attempt at a solution

    I began by calculating inertia, like this:

    I = mL^2/12 = (2.7kg)(3.1m)2/12=2.16225 kgm2

    Then I calculated the gravitational forces from m1 and m2:

    m1g=(6.7kg)(9.8m/s2)=65.66N

    m2g=(1.6kg)(9.8m/s2)=15.68N

    Then I determined the component of these forces which act perpendicular to the lever arm:

    65.66N*cos51.1=41.23N

    15.68N*cos51.1=9.84N

    Then I calculated the net torque/moment (clockwise negative, counterclockwise positive):

    Radius = 1/2(3.1m)=1.55m

    Torque = (1.55m41.23N)-(1.55m9.84N) = 48.65Nm

    Then I calculated the angular acceleration:

    Angular acceleration = Torque/Inertia

    Angular acceleration = 48.65Nm/2.16kg*m2

    Angular acceleration = 22.52 rad/s2

    I have been told my answer is incorrect through an online system where I am able to check my answers, however I'm not sure where I went wrong. If anyone would be able to spot an error, or guide me in the right direction, it would be greatly appreciated!!
     
  2. jcsd
  3. Jul 2, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi shmoop,

    Welcome to Physics Forums!

    Take another look at your moment of inertia for the system. What masses will contribute to the moment of inertia?
     
  4. Jul 2, 2017 #3
    Thank you SO much! I figured it out by adding the inertias of the two masses at the ends to the total inertia.
     
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