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Speed of a bullet (circular motion problem).

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data
    A bullet is shot through two cardboard disks attached a distance D apart to a shaft turning with a rotational period T, as shown.

    Derive a formula for the bullet speed v in terms of D, T, and a measured angle theta between the position of the hole in the first disk and that of the hole in the second. If required, use pi, not its numeric equivalent. Both of the holes lie at the same radial distance from the shaft. theta measures the angular displacement between the two holes; for instance, [itex]\theta = 0[/itex] means that the holes are in a line and [itex]\theta=\pi[/itex] means that when one hole is up, the other is down. Assume that the bullet must travel through the set of disks within a single revolution.


    2. Relevant equations
    [itex]\theta(radians) = \frac{s}{r}[/itex]
    [itex]v = \frac{2{\pi}r}{T}[/itex]
    where s is the arc length, r is the radius, and T is the period.

    3. The attempt at a solution
    I really do not know how to approach this problem. Nothing on circular motion has been covered in lecture and all I know about it is what I have read about it, which isn't much. Ideally, I assume the bullet will travel successfully through the disks in one revolution if both holes have the largest displacement (which would be if one is at its highest point and the other at its lowest, implying [itex]\theta = \pi[/itex].

    Then:
    [itex]\theta(radians) = \frac{s}{r}[/itex]
    [itex]\theta = \pi[/itex]
    [itex]\pi = \frac{s}{r} => r = \frac{\pi}{s}[/itex]
    Now, to substitute this into equation 2:
    [itex] v = \frac{2{\pi}r}{T} => \frac{2{\pi}(\frac{s}{{\pi}})}{T} => \frac{2s}{T}[/itex]

    But I can't go any further than that, and in fact, I am quite doubtful that I am even on the correct path to solving this problem thus far. Any help? Thanks in advance.
     
  2. jcsd
  3. Sep 27, 2011 #2

    PeterO

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    The speed of the bullet is given by D/t where t is the time taken to get from one card to the next.
    The time will be related to T, the time to make a full rotation.

    If the holes are opposite each other, the shaft has completed half a rotation while the bullet travelled from card to card - t = T/2.

    The angle theta will tell you what fraction of a rotation that the shaft made.
     
  4. Sep 28, 2011 #3
    How will T be related to t? I've yet to see an equation where the two variables have been expressed in an equation as separate quantities...

    So, the equations I was using, they don't apply at all to this problem?
     
  5. Sep 28, 2011 #4

    PeterO

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    I feel your equations are too involved.

    Can you solve this:
    A merry-go-Round at a fair rotates once every 12 seconds. A nearby small child fires a water pistol at the merry go round, leaving two spray marks.
    The angular separation of the two spray marks is 90 degrees - one quarter circle.
    What was the time interval between the two squirts.
     
  6. Sep 28, 2011 #5
    The time interval between them would be t = 12/4 = 3 seconds.
     
  7. Sep 28, 2011 #6

    PeterO

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    Perfect.

    Now these two discs parallel that sort of question. It is one bullet passing through, first, one disc, then continuing on to the next, rather than two sprays on the one merry-go-round, but the angular difference property still exists.

    The shaft turns in time T , rather than 12 seconds

    The angle travelled between strikes is "Theta" rather than 1/4 turn.
    and during the time interval, the bullet travels from one disc to the next - a distance D.
     
  8. Sep 28, 2011 #7
    So, then the speed of the bullet would then be:
    [itex]v = \frac{2D{\theta}}{T}[/itex]

    But then doesn't theta equal pi?
     
  9. Sep 28, 2011 #8

    PeterO

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    No. It is possible that theta = 17.4 degrees, or the radian equivalent. [I just chose 17.4 as an unusual number]

    2pi or360degrees comes in when you work out what fraction of a full turn this represents.

    Bullets travel pretty fast, so unless this shaft is spinning quite fast and/or is very long, the two holes will be close to lining up - perhaps only a degree or two off-set.
     
  10. Sep 28, 2011 #9
    Aside from the mistake from theta, the equation is correct thus far?

    Converting 17.4 to a radian value would yield a long numerical answer, but I assumed the solution would only composed primarily of variables and constants.

    I apologize if explaining this takes much time. I've never seen a question like this in the "suggested problems" and my book just barely touched on circular motion. Thanks for all the help thus far, though.
     
  11. Sep 28, 2011 #10

    PeterO

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    That is correct, your answer will be full of symbols, and perhaps a 2 for 2pi.
     
  12. Sep 16, 2012 #11
    Wrong....
    Almost everything
    just wrong...
    just look at that equation
    v=2Dθ/T
    D is in meters, θ is in radians, T is in seconds
    units on left side is meters/seconds
    units on right side is meters*radians/seconds

    time for one revolution = T, actual time that the cardboard wheels rotated t
    full revolution = 2pi, actual rotation = θ
    the relationship: t/T = θ/(2pi)
    v = D/t
    last step: algebra :)
     
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