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Circular Motion using polar coordinates - Mechanics

  1. Nov 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A block of mass M is on a frictionless table that has a hole a distance S from the block. The block is attached to a massless string that goes through the hole. A force F is applied to the string and the block is given an angular velocity w0 , with the hole as the origin, so that it goes in a circle of radius S. The force F is increased, starting at time t = 0, so that the distance between the block and the hole decreases according to S-c1t2 . Here c1 is a known, positive constant. Assuming the string stays straight and can only pull along its length, find the torque about the hole exerted by each force acting on the block and the angular acceleration of the block as a function of time.

    Here is the picture.

    2. Relevant equations
    acceleration vector = [d2r/dt2 - rw2] ir + [2(dr/dt)(w) + r α] iθ

    3. The attempt at a solution
    Looking at the system from the top, the torques are:
    TN = -mg(S-c1t2) iθ
    Tmg = mg(S-c1t2) iθ
    TF = 0

    given that
    r = S-c1*t2
    dr/dt=-2c1t
    d2r/dt2 = -2c1

    Then using F=ma
    Fr = m[-2c1 - (S-c1t2)w02]
    Fθ = m[(S-c1t2)α - 4(c1t)(w02)]

    Assuming that the Fr and Fθ components are correct, I still don't know where to go, especially because it doesn't tell us about this force, which is not a constant.
    I am working out review problems for a test, so I know that the final answer is:
    α = (4S2w0c1t) / (S - c1t2)3
    I am trying to understand how to get the solution, so I will be able to do similar problems on the test. Also, are the torques I wrote above correct?
     
  2. jcsd
  3. Nov 22, 2014 #2

    haruspex

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    I don't see that any of those torques are interesting.
    You seem to be assuming that angular velocity is constant (w0). What conservation law are you basing that on?
     
  4. Nov 22, 2014 #3
    I only listed the torques because it was part of the question. I don't think they are of any real significance.

    I guess that is wrong. Would I plug in (w0 + αt) instead?
     
  5. Nov 22, 2014 #4

    haruspex

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    That will be true if α is constant. Is it?
     
  6. Nov 22, 2014 #5
    No, it isn't a constant. However, if I am understanding the problem, α is what I am trying to solve for. And if that is correct, I am confused as to what I can set the components of force equal to, so that I can solve for alpha.

    I also believe I had a mistake in my Fθ component. Where I have (w0)2, which we determined was already incorrect, but it should not be squared. So:
    Fr = m[-2c1 - (S-c1t2)(w0+αt)2]
    Fθ = m[(S-c1t2)α - 4(c1t)(w0+αt)]

    If I set Fθ = 0, because there are no forces acting in the iθ direction (I think). I was able to solve for α, but it came out to an incorrect answer of α = 4c1t*w0 / (S-5c1t2)
     
  7. Nov 22, 2014 #6

    haruspex

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    No, you've not understood what I told you in the previous post. Because α is not constant you cannot write w =
    w0+αt.
    If you write that equation correctly you have a differential equation in terms of w. However, solving it is effectively the same as applying a certain conservation law.
    Actually, I was wrong to say that none of those torque equations are interesting. What I should have pointed out instead is that a torque is meaningless unless you identify an axis for the torque. Just look at the last one, TF = 0. You seem to be taking a vertical through the hole as the axis, so let's go with that. If there is no torque about that axis, what quantity is conserved about that axis?
     
  8. Nov 23, 2014 #7
    Angular momentum is conserved.
    L0=Lt

    L0=r x p = rm(rw)
    L0=S2m*w0

    Lt= (S-c1t2)(m)(V)
    Vθ = r*dθ/dt
    Vθ = (S-c1t2)(w)​
    S2m*w0=(S-c1t2)(m)((S-c1t2)(w))
    S2w0=(S-c1t2)((S-c1t2)(w))

    w = S2w0 / (S-c1t2)2


    α=dw/dt
    α= S2w0 * d/dt[ 1 / (S-c1t2)2 ]
    α= -2*S2w0 / (S-c1t2)3

    This would be the answer if it were multiplied by -2c1t, which is dr/dt. I assume this comes from the chain rule when taking the derivative of dw/dt, but I don't quite see why. Also, I just want to take a second to thank you for all your help. I really appreciate it!
     
  9. Nov 23, 2014 #8

    haruspex

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    Quite so. Do you not understand how to apply the chain rule here?
     
  10. Nov 23, 2014 #9
    Haha, yes. I am confused at what point the dr/dt is entering into the equation. If I had to guess, it comes from the 'r' term that we plugged in. (S-c1t)

    However, since this is only in terms of 't' and the other values are all constants, I didn't think chain rule would apply.
    Would you mind explaining where the dr/dt comes from?
     
  11. Nov 23, 2014 #10

    haruspex

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    No, it's just a matter of executing this correctly:
    This takes the form (d/dt)(f(g(t)), where g(t) = S-c1t2 and f(x) = x-2. Can you quote the chain rule in terms of f and g?
     
  12. Nov 23, 2014 #11
    Wow, I don't know how i missed that! Thanks!
    (d/dt)(f(g(t)) = f'(g(x))*g'(x)
    (d/dt)(f(g(t)) = -2(S-c1t2)-3*(-2c1t)

    So α= S2w0 * d/dt[ 1 / (S-c1t2)2 ]
    α= (-2S2w0 * (-2c1t)) / (S-c1t2)3
    α=4S2w0c1t/ (S-c1t2)3


    You've been really helpful! Thanks!
     
  13. Nov 23, 2014 #12

    haruspex

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    You're welcome.
     
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