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Rotational speed versus linear motion problem

  1. Oct 19, 2004 #1
    Here is the question:

    The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance 0.921 m apart on the same axle. From the angular displacement 27.4 degrees of the two bullet holes in the disks and the rotational speed 1050 rev/min of the disks, we can determine the speed v of the bullet. What is the speed of the bullet in units of m/s?

    Okay to start the problem, I wrote out my known values:

    d = o.921 m
    r = 0.4605 m
    initial angular velocity = 1050 rev/min
    initial angular velocity = 1050 (rev/min) (1 min/ 60 sec) (2 x 3.14 rad/rev)
    initial angular velocity = 109.9557429 rad/s
    Theta = 27.4 degrees X ( 1 rev/ 360 degrees) = 0.0761111111

    Those are my known values and their derivations. To solve the problem, I figured the speed of the bullet was reliant upon where the hole would be in the second disk. Thus I set up the equation:

    theta = (1/2) alpha (t^2) + (initial angular velocity X t)
    - Here I assumed alpha was 0, since I assumed the bullet was moving at a constant speed. Therefore:

    theta = (initial angular velocity X t)
    t = (0.0761111111/109.9557429)
    t = 0.000692197688 seconds

    With the time value found for the second disk, I figured that the bullet had to trave the diameter at that time. In other words, velocity is equal to diameter/time.

    V = d/t
    V = 0.921 m / 0.0006921976888 s
    V = 1330.544749 m/s

    This was my final answer, yet it is incorrect. Can anyone point out what I have done wrong and how to fix it? Sorry for the inconvienince.
  2. jcsd
  3. Oct 19, 2004 #2


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    I have'nt checked your arithemtic, but I'd have marked it wrong just because of the excessive number of digits you used. Read up on significant digits.
  4. Oct 19, 2004 #3


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    If you convert to radians do it everywhere, you did not express the angle between the holes in radians, you have simply expressed them as a fraction of the full 360deg circle. Multiply your angle(.07...) by 2 [itex] \pi [/itex] that should help your answer.

    As I said before, save yourself time and effort, do not even write down non significant digits.
  5. Oct 19, 2004 #4
    The answer is supposed to be in 6 significant digits, which I answered the problem with. I just posted the numbers in their entirety here, in case I needed to use them to find the answer eventually. There is something else wrong with the problem unfortunately.
  6. Oct 19, 2004 #5
    I'll do that and get back to you guys on whether or not it worked ;).
  7. Oct 19, 2004 #6
    Wow, all that work, and I just missed such a little thing. Thanks Integral ;). It oftentimes takes somone else to point out the careless mistakes. I really appreciate the help.
  8. Oct 19, 2004 #7


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    I think you over complicated the problem a bit, doing the check I went straight to deg/s then used that, with the 27.4 deg, to get the time of flight.
  9. Oct 19, 2004 #8
    Ah...No need to convert to radians then? I just remember my prof. hammering into us to work with radians :D.

    The changes this causes, for those who look at this problem later are as follows:

    theta = 0.47822 radians
    time = 0.0043492 seconds
    Velocity = 211.763 m/s
    Last edited: Oct 19, 2004
  10. Oct 20, 2004 #9


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    Converting to radians is fine, in fact a very good habit.
  11. Oct 20, 2004 #10


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    I dpn't understand why you converted to radians, since the only angle is given in degrees.

    Here's how I would have done it:

    The papers are rotating at 1050 revolutions per minute which is 1050*360= 378000 degrees per minute and therefore 378000/60= 6300 degrees per second.

    The total rotation in T seconds is 6300T degrees. Since we are told that the papers rotated 27.4 degrees in the time the bullet was moving between the two papers, we must have 6300T= 27.4 or T= 27.4/6300= 0.003921 seconds.

    Taking v to be the speed of the bullet, v= .921m/0.003921 seconds= 234.9 m/s

    I also don't understand why you say "The answer is supposed to be in 6 significant digits" when the distance is given to only 3 significant digits. I would give the answer as 235 meters (or, better, 2.35x102 m.).
  12. Oct 20, 2004 #11
    The place I turn in the homework to always wants the answers in 6 significant digits....not the correct amount ;).
  13. Oct 20, 2004 #12
    wow, that is horrible that any professor would have their students do that...such horrible practice....
  14. Oct 21, 2004 #13
    Its an online homework thing that our professor uses. For our classwork, we actually have to use correct significant digits, but the homework agency doesn't "believe" in significant digits. Therefore, they assigned an arbitrary number...6...for all answers to come in. It makes things easier for the homework, but you have to watch yourself in the exams!

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