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Dynamics of uniform circular motion

  1. Sep 28, 2004 #1
    In the Bohr model of the hydrogen atom, an electron (mass m = 9.1 x 10^-31 kg) orbits a proton at a distance of 5.3 x 10^-11 m. The proton pulls on the electron with an electric force of 9.2 x 10^-8 N. How many revolutions per second does the electron make?

    can someone help me solve this problem?
  2. jcsd
  3. Sep 28, 2004 #2
    first of all
    :mad: :grumpy: :mad: :grumpy: :mad: :grumpy:

    SHOW SOME WORK we're not here to do your work for you!

    Secondly F = m V^2 / R

    Secondly distance / velocity = time

    what is the distance travelled by this electron ? Is it a straight line or something else? If it something else , isn't htat something else defined by some formula?

    The time given in the above formula represents something, what does it represent?

    All the clues are here, show some work and i'd have been more helpful
  4. Sep 28, 2004 #3
    sorry i had done the problem but it was incorrect so i figured i would start from scratch. anyways i did complete some parts that are still useful. using F = mv^2/r
    i got the velocity = sqrt(F*r/m) so v = 1636802. for radius am i correct in assuming that the distance given is the diameter? if so, i got a time of 3.23 x 10^-17 s. the electron is orbiting a proton so do i use the time as the period and compute the velocity using v = 2*pi*r/T(period) ?
  5. Sep 28, 2004 #4
    ok i figured it out. thanks for your hints. basically i didn't realize that the distance given was the actual distance instead of my dividing it by two to get a radius. so i just used v = sqrt((distance * force)/ mass) to get the velocity and then to get the angular velocity i just did velocity / distance. then converted from radians to rev/s.
  6. Sep 28, 2004 #5

    period is given by 2 pi r / v or if you care for angular velocity w = v/ r then T = 2 pi / w and you're done like dinner
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