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Rotation and Angular Momentum: Solving for thrust of a rocket

  1. Oct 17, 2010 #1
    1. A solid bar of length L = 0.5m and with W = 0.1m weighs 2kg. It also has two constant-thrust rockets attached on either end. Each rocket is small enough to be considered a point mass of 0.25kg. If the bar is initially at rest and in two seconds after the rockets are fired it achieves a rotation rate of 1000rpm, determine the thrust of each rocket.



    2. Relevant equations
    torque = I * angular acceleration
    I = (1/12)*(L^2+W^2)*M



    3. I found I, angular momentum, to be 0.0542. I am not sure what to do from there.
     
  2. jcsd
  3. Oct 17, 2010 #2

    diazona

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    I is moment of inertia, not angular momentum.

    What equation do you know that relates torque and angular momentum (which is usually written L)?
     
  4. Oct 17, 2010 #3
    Correct my apology, I is moment of inertia. An equation I know that relates torque with angular momentum 'L' is:

    torque = dL/dt = d(Iw)/dt

    The acceleration changed from 0 to 1000rpm in 2 seconds. and L=Iw, which is the angular velocity times the moment of inertia. I'm confused as to what to do
     
  5. Oct 17, 2010 #4

    diazona

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    It's not acceleration that changed from 0 to 1000rpm in 2 seconds :wink:

    Think about this: what else can you calculate from that 1000rpm figure?
     
  6. Oct 17, 2010 #5
    So, angular velocity changes from 0 to 1000rpm in 2 seconds, and we know I. I can therefore solve for angular momentum, right? Couldn't you also integrate the sum of the moments exerted by each rocket from time t to 0? I still am not sure how to translate all of this to the thrust of each rocket.
     
  7. Oct 17, 2010 #6

    diazona

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    Right, try that.

    By the way, you do know what kind of physical quantity thrust is, right?
     
  8. Oct 17, 2010 #7
    thrust must be in Newtowns (N), presumably? I was pretty certain of that.
     
  9. Oct 17, 2010 #8

    diazona

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    Yep, just checking.
     
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