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Change in angular momentum

  1. Nov 14, 2006 #1

    fro

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    I have no clue as to how to solve this. Any hints/suggestions will be helpful.

    Problem: An object's angular momentum changes by 20kg*m^2/s in 4 seconds. What magnitude of torque acted on the object?
     
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  3. Nov 14, 2006 #2

    Hootenanny

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    Do you know of any expressions which relate a change in momentum to a torque? Or perhaps a torque to an angular acceleration?
     
  4. Nov 14, 2006 #3

    fro

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    Maybe I could use L = I*w and T = I*angular acceleration? But I don't know what to do after that.
     
  5. Nov 14, 2006 #4

    Hootenanny

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    It may be useful to note that for a constant acceleration;

    [tex]\alpha = \frac{\Delta\omega}{\Delta t}[/tex]
     
  6. Nov 14, 2006 #5

    fro

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    Sorry, I'm really confused about this.

    By your equation, angular acceleration should be 5kg*m^2. Not sure how and where to plug it in to get the answer.
     
  7. Nov 14, 2006 #6

    Hootenanny

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    Another useful observation; from your equation (I is constant);

    [tex]\Delta L = I \Delta\omega \Leftrightarrow I = \frac{\Delta L}{\Delta \omega}[/tex]
     
  8. Nov 14, 2006 #7

    fro

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    [tex]\Delta L = 20{kg} \cdot{m^2}[/tex]
    [tex]\alpha = \frac{20kg\cdotm^2}{4s} = {5kg} \cdot {m^2}[/tex]
    [tex]\tau = \frac{\Delta L}{\Delta \omega} \times \alpha[/tex]
    [tex]\tau = \frac{20kg\cdot m^2}{\Delta \omega} \times {5kg} \cdot {m^2}[/tex]

    So, how would I find [tex]\Delta \omega[/tex]?
     
    Last edited: Nov 14, 2006
  9. Nov 14, 2006 #8

    Hootenanny

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    From my previous posts;

    Using those to facts and the formula;

    [tex]\tau = I\alpha \stackrel{(1)\;\&\;(2)}{\Rightarrow} \tau = \frac{\Delta L}{\Delta \not{\omega}} \cdot \frac{\Delta\not{\omega}}{\Delta t}[/tex]

    [tex]\therefore \boxed{\tau = \frac{\Delta I}{\Delta t}}[/tex]

    This is a good formula to remember :wink:
     
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