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Inertia/angular momentum, not sure why this is wrong

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data

    In Fig. 11-54, a 8.49 g bullet is fired into a 0.200 kg block attached to the end of a 0.551 m nonuniform rod of mass 0.735 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 0.0349 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 5.73 rad/s, what is the bullet's speed just before impact?

    2. Relevant equations

    http://courses.wcupa.edu/mwaite/phy170/ch10to12gifs/tfg054.gif [Broken]

    3. The attempt at a solution

    I solved part A and got the correct answer by using mr^s for the mass of the bullet and block and the lenth of the rod as the radius and then adding this to the given Inertia of the rod. The answer is .098. I then used the formula v=wr, or veloctiy equals angular velocity times the radius, to find the velocity of the block just after impact. so v=(5.73)(.551)...this gave me 3.16. I then used conservation of momentum to find the velocity of the bullet. m1v1=m1v2+m2v2....or (.00849)v1=(.20849)(3.16) and then solved...this gave me 77.53 m/s, but the online thing where i turn this in says that is wrong....WHY???
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 12, 2009 #2


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    Homework Helper

    0.98 what? It's not a correct answer without the correct units :wink: But anyway, your procedure seems right to me. Unless you made a little math mistake, it seems like you should have the right answer.
  4. Jul 12, 2009 #3
    oh i figured it out now...i had to treat the problem as a conservation of ANGULAR momentum problem...Lrod2+Lbulletblock2=Lrod1+Lbullet1....
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