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Spinning Mass and Finding Omega

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data
    http://imageshack.us/f/41/physicsc.png/
    My professor gave the following hints:
    Write down what you know about how r (position of mass from hole)
    varies with time. Then write Newton's 2nd law in polar coordinates.
    For part (a), the differential equation that you are looking for comes
    from the F_theta = ma_theta equation. Part (b) involves integrating
    the result from part (a), and Part (c) uses the F_r = ma_r equation.

    2. Relevant equations
    F(θ)=m[rθ"+r'θ']


    3. The attempt at a solution
    For part a, I labeled the force of tension pointing towards the hole. And that's the only force I had.

    For part B, I tried doing this:
    r=vt (same v of the rope)
    r'=v
    r'=0

    θ=wt
    θ'=w
    θ"=0

    Then, I plugged the above results into the equation:
    F(theta)=[rθ"+2r'θ']
    F=0+2(-v)w

    I then got:
    θ''=θ'
    m(-v)w=-bw(given)
    bw(given)/m=w

    Would this be right? I do not thinkso...


    Thanks all!
     
  2. jcsd
  3. Sep 5, 2011 #2
    You know that angular momentum will be constant.

    angular momentum = m*omega_0*(r_0)^2 = m*omega(t)*(r(t))^2

    but r(t) = [r_0 - v*t]

    so omega(t) = omega_0*(r_0)^2/[r_0 - v*t]^2

    Also the centripetal force = m*v^2/r = m*omega^2*r

    We know omega and r so we should know the centripetal force.

    ?
     
    Last edited: Sep 5, 2011
  4. Sep 5, 2011 #3
    Oh, that's right. I misrepresented my r(t) function.

    We had to use Diffy Q's to solve the equation, so I did manage to do:
    rΘ"+2rΘ'=0

    Then ended up with:
    m*omega_0*(r_0)^2 = m*omega(t)*(r(t))^2
    (Which I thought was wicked cool equation because I completely forgot about conversation of angular momentum).

    Thank you for your help! It's good to know that I could have made that assumption from the start, and thank you for fixing my r(t) function!
     
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