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Torque required to rotate a circular sector inside a drum

  1. Mar 1, 2013 #1
    What I have ?
    • Have a 16’ x 72’ rotating drum
    • There is a solid circular sector inside the drum

    What I need ?
    Want to Calculate the torque required to move that solid circular sector from the intial position to the position shown in attached picture (90 Degrees rotation)


    Data :
    • RPM = 10
    • time = 15 seconds


    What I have caluted ?
    • Calculated the torque required to rotate the drum using Torque = Moment of Inertia X Alpha where Alpha = angular acceleration

    How do I proceed ?
    1. Do I have to calculate the moment of inertia of that solid circular sector and then use the same formula?
    2. But here its not a complete rotation. Its just 90 degrees. So how to arrive at torque to move a mass about an axis only to a certain degree?
    3. Or, Should it be just Torque = Force * Radius ? where force shall be its mass * acceleration due to gravity and radius shall be distance from the drum's rotation axis to the circular sector's center of gravity ?
    4. Does Perpendicular axes theorem or Parallel Axes theorem come into the picture ?
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2013 #2

    rock.freak667

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    Homework Helper

    You will need to get the total moment of inertia of the drum + circular section. For that you may need the parallel axis theorem.

    As for the rotation of the 90 degrees. What is the definition of angular acceleration?
     
  4. Mar 4, 2013 #3
    I am afraid , I didn't understand. My approach was to find the Inertia of both separately and then arrive at their individual torques. Did you mean the same ? Do we need a parallel axis theorem in that ? If yes, could you explain a little more elaborately ?

    Angle was taken to be ∏/2.
    Therefore, Angular Velocity = (∏/2)*rpm/60 = (∏/2)*10/60=.26 s^(-1)
    and Angular acceleration = .26/time = .26/15 =0.0174 s^(-2)

    Thanks for your help !
     
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