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Acceleration of particle along the curve

  1. May 27, 2015 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=30b910ee9f21323c40cfd9e699f77c32.png


    2. Relevant equations


    3. The attempt at a solution

    Since the tangent moves at uniform angular speed ,the particle also moves with constant angular speed . Is it correct ?

    If it is right then the particle has only radial acceleration given by ##ω^2r## ,where 'r' is the radius of curvature .

    r = sec2x

    r ( x= πa/4) = 1/√2 .

    Hence resultant acceleration at x= πa/4 = 2√2 .

    I don't have the answer key . Have I interpreted the question correctly and done it right ?

    I would be grateful if somebody could verify my work .
     

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  3. May 27, 2015 #2

    SammyS

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    It may be true that the particle moves with constant angular speed (not sure how you would define that), but not for the reason you give.

    I think that the problem statement is saying that the angle the tangent line makes with the x-axis increases at a constant rate.
     
  4. May 27, 2015 #3
    Ok . So what should be the correct approach ?
     
  5. May 27, 2015 #4

    SammyS

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    Get dy/dx.

    Slope of tangent line = tan(θ), where θ is the angle the tangent line makes w.r.t the x-axis. Right ?

    Start there.
     
  6. May 27, 2015 #5

    D H

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    No, that isn't correct.

    Angular velocity is not the way to solve this problem. As Sammy mentioned, determining dy/dx is the first step.
     
  7. May 27, 2015 #6

    D H

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    The problem statement says the tangent to the curve rotates with a constant angular velocity. I would take that as meaning the unit tangent.

    HINT: The magnitude of the time derivative of any unit vector is the rate at which that unit vector is rotating.
     
  8. May 27, 2015 #7
    Ok .

    dy/dx = tan2x i.e tanθ = tan2x .

    θ = 2x .

    dx/dt = 1

    Is it correct ? If yes , what next ?
     
  9. May 27, 2015 #8

    SammyS

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    Where did the ##\ a\ ## go ?
     
  10. May 27, 2015 #9
    Sorry .

    dy/dx = tan(x/a) i.e tanθ = tan(x/a)

    θ = x/a

    dθ/dt = (1/a)dx/dt

    dx/dt=2a

    Is it correct now ?
     
  11. May 27, 2015 #10

    SammyS

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    Looks good!
     
  12. May 27, 2015 #11

    D H

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    What's that two doing there? (That's a devil's advocate kind of question.)

    I find that it's best to work symbolically until the very, very end. If needed (sometimes it's not needed), I'll plug in the numerical values, but only at the very, very end.
     
  13. May 27, 2015 #12
    I have worked symbolically only . In fact what I did in post#7 is also correct , since 'a' is a constant .
     
    Last edited: May 27, 2015
  14. May 27, 2015 #13
    Is the tangential acceleration equal to ##4m/s^2## ?

    Is net acceleration equal to ##2\sqrt{6}m/s^2## ?
     
  15. May 27, 2015 #14

    SammyS

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    I don't mind that 2 so much, since early in the problem we're told the tangent line rotates with constant angular speed of 2 rad/s . Otherwise, we need to carry around some ω0, which I suppose isn't so bad. Plugging in a = 1/2 so early in the game could be a bad idea.
     
  16. May 27, 2015 #15

    SammyS

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    I did not get that.

    Can you show some intermediate steps?
     
  17. May 27, 2015 #16

    ehild

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    The curve is given in Cartesian coordinates, so the acceleration can be expressed by ##\ddot x(t), \ddot y(t)##. We also know that dx/dt =2a, constant.
     
  18. May 27, 2015 #17
    o:)

    Is ##4m/s^2## the net acceleration ?
     
  19. May 27, 2015 #18

    ehild

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    I think yes. Why aren't you sleep yet:wink:?
     
  20. May 27, 2015 #19
    Hooray!!! Now I can sleep peacefully :oldsmile: .

    Thanks ehild , Sammy , D H .
     
  21. May 27, 2015 #20

    D H

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    That's the answer.

    Good night!
     
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