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Bullet speed problem

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A device consist of two rotating disks, separated by a distance of d = 0.85 m, and rotating with an angular speed of 95 rad/s. The bullet first passes through the left disk and then through the right disk. It is found that the angular displacement between the two bullet holes is [tex]\theta[/tex] = 0.24 rad. From these data, determine the speed of the bullet.


    2. Relevant equations
    w=[tex]\theta[/tex]/Delta t
    *v=vo+at ~ w=wo+[tex]\omega[/tex]t
    *v2=vo2+2ax ~ w2=wo2+2[tex]\omega[/tex][tex]\theta[/tex]
    *x=vot+(1/2)at2 ~ [tex]\theta[/tex]=wot+(1/2)at2

    the ones with * means i changed it to what the problem is about. means the same thing just different letters so you wont get confused i guess.


    3. The attempt at a solution
    well i know the formulas but when the bullet hits the first disk wont it loose some speed and same with the second thing? does it mean speed after its done hitting or what im confused please help.

    thank you!
    sweedeljoseph
     
  2. jcsd
  3. Nov 17, 2008 #2
    It doesn't seem that speed lost in each impact is considered. After all you don't know anything about the material or the bullet.

    You do have the amount of time that passes between impacts however. It is the angular displacement divided by the rate of angular displacement.

    After calculating the time, you will have time and distance...
     
  4. Nov 17, 2008 #3
    so the angular displacement is .85 and the rate of the angular displacement is 95 rad/s. if i divide those two i get .009. thats the time? i just plug that into one of the equations to find distance?
     
  5. Nov 17, 2008 #4
    No, in your post you state that the angular displacement is .24 rad. Is that correct?
     
  6. Nov 17, 2008 #5
    oh yeah. haha i didnt see that but thats what i meant! so the answer is .002. and im guessing thats time. just plug that into the equation to find distance? is that it?
     
  7. Nov 17, 2008 #6
    Well, think about it like this...

    If it took the bullet .002 sec to travel .85m, what is the speed of the bullet in meters/sec?
     
  8. Nov 17, 2008 #7
    .85/.002? that would equal 425. that would be the speed of the bullet?
     
  9. Nov 17, 2008 #8
    Close, but you need to keep more precision in the intermediate step.

    .24/95 = ?

    The rounding error is very large.
     
  10. Nov 17, 2008 #9
    wait why are you using those numbers to find the speed? i thought we just solved for something else.
     
  11. Nov 17, 2008 #10
    Yes, you are correct. I didn't check your division before and find the rounding error right away. My apologies.

    If you keep more precision until the end.

    You would get time= .24/95 = .002526316 sec

    Then .85m/time = 336.46 m/sec

    sorry to confuse you...
     
  12. Nov 17, 2008 #11
    oh wow ok i know what youre talking about now. its ok for confusing me im always confused in physics its great. so the problem was that easy haha i feel so stupid. i get it now.

    thank you so much!
     
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