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Angular acceleration of a rigid rod with a mass at both ends

  1. Mar 24, 2015 #1
    1. A rigid rod of mass M and length L rotates in a vertical plane about a frictionless pivot through its center of mass. Particles of masses m1 and m2 are attached at the ends of the rod. Determine the angular acceleration of the system when the rod makes an angle θ with the horizontal.
    In the accompanying picture, m2 is obviously larger and the rod is rotating clockwise.


    2. τ=Iα=Frsinθ

    3. The inertia I calculated for the system is (M/12 + (m1 + m2)/4)L2. I think that the angle between position vector r and the force F would be 90 - θ. So then the angular acceleration α would be m2g(L/2)sin(90 - θ) / I? I'm confused about whether or not the force of gravity on m1 also needs to be taken into account and if so, how.
     
  2. jcsd
  3. Mar 24, 2015 #2

    Orodruin

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    Yes, you need to take into account all of the forces acting on the system when you compute the torque. First order of business: How do you define the torque?
     
  4. Mar 24, 2015 #3
    It's the product of the rotational inertia and the angular acceleration. I know the angular acceleration is at/r, so would I need to subtract (m1g)/(L/2) from (m2g)/(L/2) to get this?
     
  5. Mar 24, 2015 #4

    Orodruin

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    Not in terms of the inertia and the acceleration, in terms of the forces acting on the object.
     
  6. Mar 24, 2015 #5
    Like τ = (r)(Fsinθ)?
     
  7. Mar 24, 2015 #6

    Orodruin

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    Yes. So the forces on each of the masses are going to provide a torque around the pivot. What is then the total torque?
     
  8. Mar 24, 2015 #7
    τ = (L/2)(m2gsin(90-θ) + (L/2)(m1gsin(90-θ) ?
     
  9. Mar 24, 2015 #8
    I guess I'm not sure if they need to be added or subtracted.
     
  10. Mar 24, 2015 #9

    Orodruin

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    Do the forces tend to accelerate the rod in the same or in opposite directions?
     
  11. Mar 24, 2015 #10
    Opposite, so they'd be subtracted?
     
  12. Mar 24, 2015 #11

    Orodruin

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    Correct. You have to define a direction for the torque, if the force acts to accelerate the object in that direction, the torque is positive. If it tries to spin it in the other, the torque is negative.
     
  13. Mar 24, 2015 #12
    So, α = [((m2g-m1g)Lsin(90 - θ))/2] / [(M/12 + (m1 + m2)/4)L2]? Was my calculation of the inertia correct?
     
  14. Mar 24, 2015 #13

    Orodruin

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    Yes, the inertia seems correct. Just also note that sin(90o - x) = cos(x).
     
  15. Mar 24, 2015 #14
    Right - thank you!
     
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