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Homework Help: Bullet hits door, find angular velocity

  1. May 6, 2007 #1
    Unfortunately I was absent when a similar example was done in class.....

    1. The problem statement, all variables and given/known data

    A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.

    What is the angular velocity of the door just after impact?

    2. Relevant equations

    v = ωr

    3. The attempt at a solution

    Attempt #1:

    I considered the bullet as a particle in linear motion. After collision I considered the mass of the door to be (M = Mdoor+Mbullet). I set up the problem like so:

    0.5mv^2 = 0.5Iω^2

    the 0.5 is a constant so it cancels..

    mv^2 = Iω^2
    *note* small 'm' is mass of the bullet, big 'M' is for the door, same goes for 'v' and 'V'

    Then I plugged in the moment of inertia I, which is given in a table as 1/3ML^2

    the width of the door is 1m, so the L^2 is simply 1

    so putting this into the equation, I get:

    mv^2 = (1/3M)ω^2

    and rearranging and solving for ω, I get:

    ω = √[(mv^2)/(1/3M)]|
    (square root)

    this did not work....

    Attempt #2:

    I tried to find the linear velocity by using just conservation of linear energy

    0.5mV^2 = 0.5MV^2

    0.5 cancel, so it is mv^2 = MV^2 and find the linear velocity as so:

    V = √((mv^2)/M)|

    and then using V = ωr, I solved for ω

    but still incorrect.....
    Last edited: May 6, 2007
  2. jcsd
  3. May 6, 2007 #2

    Here is an image I made on paint, this is the way I see it, bird's eye view...

    http://photo.ringo.com/210/210049338O350481129.jpg [Broken]

    here is the link if it doesn't come up for some reason

    http://photo.ringo.com/210/210049338O350481129.jpg" [Broken]

    http://photo.ringo.com/210/210049338O350481129.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  4. May 6, 2007 #3
    when you right 1/3M is that M/3? In any event, I think momenta is what should be conserved.
    Last edited: May 6, 2007
  5. May 6, 2007 #4
    yes, this is the moment of inertia: 1/3 ML^2

    but L is 1, so L^2 = 1 and therefore it is 1/3 M, or as you stated, M/3
  6. May 6, 2007 #5


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    Staff Emeritus
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    Gold Member

    conservation of momentum would be the way to go here.

    The alternate answer is zero, because the door is locked and the latch holds
  7. May 6, 2007 #6

    door locked, nice one haha

    ok ill try conservation of momentum

    so do i do the bullet as linear momentum, and the door as rotational

    so it would be:

    mv = rMω

  8. May 6, 2007 #7
    WOOT WOOT!!!!!

    Got it!!!

    i used conservation of momentum:

    mv = Iω (found this part in the txtbook)

    and I is given as 1/3ML^2 for a 'thin rod about side'

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