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Rotational Movement

  1. Jan 20, 2009 #1
    This is a problem that I am just having some trouble wrapping my head around. I don't know if I am doing it correctly.




    1. The problem statement, all variables and given/known data
    You are a member of a group designing an air filtration system for allergy sufferers. To optimize its operation you need to measure the mass of the common pollen in the air where the filter will be used. To measure the pollens mass you have designed a small rectangular box with a hole in one side to allow the pollen to enter. Once inside the pollen is given a positive electric charge and accelerated by an electrostatic force to a speed of 1.4m/s. The pollen then hits the end of a very small uniform bar which is hanging straight down from a pivot at its top. Since the bar has a negative charge at its tip, the pollen sticks to it as the bar swings up. measuring the angle that the bar swings up would give the particles mass. After the angle is measured, the charge of the bar is reversed, releasing that particle. It's a cool design but your friend insists it will never work. To prove it she asks you to calculate the length of the bar which would give you a reasonable angle of about 10 degrees for a typical pollen particle of 4 x 10^-9 grams. Your plan calls for a bar of 7 x 10^-4 grams with a moment of inertial 1/12 as much as if all of it's mass were concentrated at its end. is she correct?


    2. Relevant equations

    w = omega
    Angular momentum (L) = I(w) I= Mr^2
    Rotational Kinetic Energy = 1/2I(w)^2
    Potential energy = mgh

    I was given a formula by my prof though I don't know how he got it: rw = v, becomes v/r = w

    3. The attempt at a solution
    First I set up a conservation of angular momentum equation: Initial momentum of pollen + momentum of bar: Iw + Iw with the initial momentum of the bar = 0
    L(initial) = (mr^2)V/r = mrV = 5.6x10^-9R = L(initial)
    L(final) = (I of pollen + I of bar/12)w = (mr^2+mr^2/12)V/r
    Here I am confused how dividing the inertia of bar by twelve, I didn't know how do divide an unknown (r^2) by 12 so I only divided the mass by 12.
    Anyway the r in v/r cancels out one of the r's from inertia leaving the equation: 5.83x10^-5rv(final)
    conservation of momentum: 5.6x10^/9R = 5.83x10^/5rv(final), divide both sides by r, removing r. divide the right number from v to find v final.

    From here I do a conservation of angular energy and solve for height and from there I use h =L(1/cos(theta)) to find L.
    I am wondering if I am doing the first part right to get the info needed to solve the rest of the problem. I want to type out the rest of the equations but I have to get to work, I will finish entering in how I did the conservation of energy part when I get home.
     
  2. jcsd
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