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How to find the acceleration with polar coordinates?

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data

    vCO12t3.jpg

    The quality of the image is bad so heres the statement:

    For an interval of motion the drum of radius b turns clockwise at a constant rate ω in radians per second and causes the carriage P to move to the right as the unwound lenght of the connecting cable is shortened. Use polar coordinates r and θ and serive expressions for the velocity v and acceleration a of P in the horizontal guide in terms of the angle θ. Check your solution with time of the relation x2+h2=r2

    2. Relevant equations

    I first found that the velocity of the carriage is v=bω/sin(θ)

    3. The attempt at a solution

    I attempted to directly derivate the equation which give me -bωcos(θ)/sin2(θ)

    However, in the book answers, the answer is supposer to be b2ω2/h *cot3(θ).

    I think I have to do something with the vectors, like derivate v=vrer + vθeθ but I don't understand much.
     
  2. jcsd
  3. Jan 24, 2013 #2

    fgb

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    It seems to me that you just forgot a term in the chain rule to differentiate your expressior for the speed. Realize that as r is changing, so is [itex]\theta[/itex], so [itex]\theta[/itex] is also a function of time.

    You have [itex] v = -\dfrac{b\omega}{\sin{\theta}} [/itex]

    [itex]a=\dfrac{dv}{dt} = \dfrac{b\omega\cos{\theta}}{\sin{\theta}^2}\dfrac{d\theta}{dt}[/itex]

    There is probably a simpler way to do this, but to obtain [itex]\frac{d\theta}{dt}[/itex] you can realize that [itex] r*\cos{\theta} = h[/itex], where h is a constant. If you differentiate both sides, you get

    [itex] \dfrac{dr}{dt}*cos{\theta} + r(-\sin{\theta})\dfrac{d\theta}{dt} = 0 [/itex]

    You can solve for [itex]\dfrac{d\theta}{dt}[/itex] and plug it into the expression for a, it gives you the right answer :) Realize that [itex]\frac{dr}{dt} = -b\omega[/itex], since it is simply the rate at which the rope is being pulled.

    Did I make any sense? :P
     
    Last edited: Jan 24, 2013
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