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Angular momentum to linear velocity

  1. Apr 7, 2006 #1
    Im stuck on this problem and Ive gotten to the point where Im just staring at my computer, so I figured Id give you guys a go at it.

    I got Part A which was fairly simple. But Ive tried everything but cant seem to get Part B. I must be missing something. This is one of those online assignments where you get as many attempts as you want. I keep getting Part A, but cant get Part B. Please explain how you did Part B if you get it. I have the correct answers too.

    In Fig. 11-52, a 7.3 g bullet is fired into a 0.46 kg block attached to the end of a 0.12 m nonuniform rod of mass 0.88 kg. The block-rod-bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertia of the rod alone about A is 0.038 kg·m2. Treat the block as a particle. (a) What then is the rotational inertia of the block-rod-bullet system about point A? (b) If the angular speed of the system about A just after impact is 7.2 rad/s, what is the bullet's speed just before impact?

    [​IMG]
     
  2. jcsd
  3. Apr 7, 2006 #2
    anybody have any hints? I know about conservation of momentum. And I know MV=Iw. Im pretty sure I need to convert units of w to m/s but I dont think Im doing it right.
     
  4. Apr 7, 2006 #3

    Curious3141

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    Homework Helper

    Remember the angular momentum of a point mass about an axis is simply the linear momentum times the distance from the axis, i.e. p*r = mvr.

    The block and the bullet are point masses. Can you proceed ?
     
  5. Apr 7, 2006 #4
    Nevermind I figured it out. It was a stupid error on my part. I simply had to divide the rad/s by the radius to get the correct units.
     
  6. Apr 7, 2006 #5
    Thanks for helping though, Curious
     
  7. Apr 8, 2006 #6
    Have a care. You may be able to get the right answer by converting units, but the equation MV=Iw is incorrect as it stands. You cannot equate linear and angular momentum. They have different units. What you have done in "converting units" is change your equation (to the correct one, if I am guessing correctly.) Make sure you know what this equation is and how to use it!

    -Dan
     
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