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Angular momentum of a bullet problem

  1. Dec 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A 4 g bullet traveling at 500 m/s strikes a disk of mass 1 kg and
    radius 10 cm that is free to rotate around an axis passing through its
    center. The bullet’s incoming path is 5 cm above the rotation axis and
    the bullet comes to rest in the position shown in the figure. At how
    many revolutions per second is the disk is rotating afterwards?
    (Ignore the mass of the bullet after it hits the disk.)

    https://gyazo.com/7ba8193726c76f57aa6c3ffa9e0c2930

    the answer is 3.18 but i can't seem to figure out how to do this.

    2. Relevant equations

    L(initial) = L (final)

    LDisk = 1/2MR^2


    3. The attempt at a solution

    Here is what i got so far using conservation of angular momentum

    R* Mass of Bullet * Velocity of bullet * sin([PLAIN]http://physics-help.info/physicsguide/appendices/si_units_images/image002.gif[/I]) [Broken] = IDisk * ωFinal
    0.1m * (0.004kg)(500)sin([PLAIN]http://physics-help.info/physicsguide/appendices/si_units_images/image002.gif) [Broken] = (1/2) (1kg) (0.1)^2 * ωFinal

    But i can't seem to figure what the theta is in this case and how to incorporate 5cm above the rotation axis into the problem.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Dec 18, 2016 #2

    TSny

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    Hello, and welcome to PF!

    You have the right approach using conservation of angular momentum.

    For a refresher on the meaning of θ, look at second paragraph here:
    https://cluster31-files.instructure...iles/apb11o/resources/guides/G11-4.angmom.htm

    This discussion shows that you don't really need to worry about finding θ in this problem.

    Why did you not include the contribution of the bullet to the final moment of inertia ##I##?
     
  4. Dec 18, 2016 #3
    Thanks.

    I did not include the bullet's contribution to the inertia because the problem said to Ignore the mass of the bullet after it hits the disk. I am still confused about how the 5cm above the rotation axis should be used in this problem.
     
  5. Dec 18, 2016 #4

    CWatters

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    It says to ignore the mass of the bullet after impact so no effect on the moment of inertia.
     
  6. Dec 18, 2016 #5

    TSny

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    OK, great. I somehow overlooked that part of the statement of the problem. If you had included the bullet, you would have found that it does not change the answer to 3 significant figures.

    Do you understand why you can write ##L = p r_{\perp}## for the bullet before the collision?
     
  7. Dec 18, 2016 #6
    Because the disk is stationary before the bullet hits so right before the moment of impact, the only thing angular momentum that should exist should be for the bullet which is the cross product of r and p.
     
  8. Dec 18, 2016 #7

    TSny

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    OK. If you understand the meaning of a cross product of two vectors, then you should understand the meaning of θ in the formula
    L = r p sinθ.

    From the link I gave, you see the following diagram (which I slightly modified)
    upload_2016-12-18_19-28-6.png

    This shows a particle moving with constant speed along a straight line. Note how θ is indicated for the three positions A, B, and C. Hopefully, that agrees with your understanding of θ. Using trig, can you relate rA, θA, and rB. Note rB is the perpendicular distance from the origin O to the line of travel of the particle.
     
  9. Dec 18, 2016 #8
    sin(180-θA) = RB / RA
     
  10. Dec 18, 2016 #9

    TSny

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    OK. Can you simplify sin(180-θA) using a trig identity? Then solve your equation for RB in terms of RA and sin θA.

    (Note, there is a formatting toolbar that you can use for subscripts, etc.)
     
  11. Dec 18, 2016 #10
    RB = sin(θA) * RA
     
  12. Dec 18, 2016 #11

    TSny

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    Good. So, how can you express the angular momentum of the particle at point A in terms of RB?
     
  13. Dec 18, 2016 #12
    L = P * RB
     
  14. Dec 18, 2016 #13

    TSny

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    Yes. Note that RB is the perpendicular distance from the origin to the line along which the particle is traveling. So, you can see now that no matter where the particle is along the line, the angular momentum may be calculated using L = R##_\perp## p.

    What is the value of R##_\perp## in your problem?
     
  15. Dec 18, 2016 #14
    is it 10cm?
     
  16. Dec 18, 2016 #15

    TSny

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    No. Draw a line representing the line along which the bullet is moving. You can extend that line as far as you like. R##_{\perp}## is the shortest distance from the origin (center of disk) to this line. See RB in the previously posted figure.
     
  17. Dec 18, 2016 #16
    Oh it's 5cm
     
  18. Dec 18, 2016 #17

    TSny

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    Yes. Good.
     
  19. Dec 18, 2016 #18
    So the left hand side of the equation should be
    R * P = IDisk * ωFinal

    But is there something else i am missing here?
     
  20. Dec 18, 2016 #19

    TSny

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    ##r
    Yes, that's it.
    I don't think so. Why do you feel unsure?
     
  21. Dec 18, 2016 #20

    TSny

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    Note, however, that the question asks for the number of revolutions per second. So, make sure you express your answer in those units.
     
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