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Homework Help: What is the angular velocity of the shutter?

  1. Apr 3, 2009 #1
    You desire to photograph the dynamics of a bumblebee in flight and estimate that its wings flap somewhere about 10,000 times a second. The radius of the film drum is r = 0.501 m and you've loaded it with 35mm film.

    A) You desire to take the photographs at a rate of 20,000 pictures per second. What is the angular velocity of the shutter? (Note: shutter has one hole in it!)

    B) What must the angular velocity of the film drum be?

    When the drum has taken its pictures, the gas is turned off and the drum slows down uniformly, finally setting to a standstill after 2 minutes.
    C) What was the angular acceleration of the film drum after turning off the gas?

    D) How many times does the film drum rotate after the gas is turned off and before coming to rest? Give your answers in revolutions.

    I honestly dont know how to approach any of these problems any help is welcomed
  2. jcsd
  3. Apr 4, 2009 #2


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    Welcome to PF!

    Hi mattmannmf! Welcome to PF! :smile:

    (have a theta: θ and an omega: ω and an alpha: α :wink:)


    angle: θ

    angular velocity: ω = dθ/dt

    angular acceleration: α = dω/dt = d2θ/dt2

    distance round the rim: s = rθ

    ordinary (tangential) velocity: v = rω

    ordinary (tangential) acceleration: a = rα

    and the ordinary constant acceleration equations work for θ ω and α in the same way as they work for s v and a :wink:
  4. Apr 4, 2009 #3
    ok for the first one.. i have to find the angular velocity. i know it takes roughly 222 revolutions per second in order to get the 20,000 photos. Now i convert it to rad/s by multiplying it by 2pi..i got 1396.2. i checked it but it was wrong...dont know where i went wrong.
  5. Apr 4, 2009 #4
    ok, so i guess i found B actually.... haha, but than what am i doing wrong for A?
  6. Apr 4, 2009 #5
    i actually managed to solve for all but A... any help
  7. Apr 4, 2009 #6


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    What is the shutter? Is it a wheel that rotates over the exposure area, while the drum moves the film underneath? That's how I see it reading.

    If you have a picture that comes with the problem it would maybe be a help.

    But if it is a wheel with a hole in it, then it apparently needs to rotate at 20K rev/sec to expose the frame on the drum below. In which case it seems like it is just ω = 20k * 2π in radians/sec ?
  8. Apr 4, 2009 #7


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    Hi mattmannmf! :smile:
    I don't get it …

    you want 20,000 photos in one second, and the shutter only has one hole in it …

    how do you get 222 rps? :confused:
  9. Apr 4, 2009 #8
    ok, never mind for the 222 rps..90 photos for every revolution and you have 20,000 (20,000/90= 222)..that was for the film drum which i solved for already. Than im not sure where to take it from there
  10. Apr 4, 2009 #9


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    But for A, the shutter, it can only take a picture when the hole is opposite the bee … so isn't it just 20,000 rps for the shutter?
  11. Apr 4, 2009 #10
    i thought so too, but i tried that and its not correct
  12. Apr 4, 2009 #11
    here were 2 hints they gave me, but i dont know how that helps me:HELP:

    There are 2*p radians per revolution. You desire the solution in units of radians per second.

    HELP: angular frequency = 2 * p * frequency.
  13. Apr 4, 2009 #12
    ahhh, never mind. i got it
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