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Polar Coordinates [Finding the velocity]

  1. Sep 6, 2014 #1
    1. The problem statement, all variables and given/known data
    The projectile A is being tracked by the radar at O. At a given instant,
    the radar readings are θ = 30degrees, R = 2000m, dR/dt = 200 m/s, and d^2R/dt^2 = 20 m/s^2.
    Determine the speed of the projectile at that instant.

    2. Relevant equations
    I've used the Vr = dR/dt & Vθ= R(dθ/dt)

    3. The attempt at a solution
    I've worked out an equation which is
    cos30=x/2000 --> x = 1732.05
    dR/dt = 1732.05 (dθ/dt)(secθtanθ)
    200 = 1732.05 (dθ/dt)(sec30tan30)
    dθ/dt = 0.1732 deg/sec?
    And using V^2 = Vr + Vθ I didn't get 299.7m/s So I'am wrong.

    Attached Files:

  2. jcsd
  3. Sep 6, 2014 #2


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    Could you show please, your work in detail?
    And what do you mean on V^2 = Vr + Vθ???? The square of a vector is not the sum of its components.

  4. Sep 7, 2014 #3


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    The problem is very interesting, I want to keep it alive.

    R(t) is the time-dependent distance of the projectile from the origin. With the Cartesian coordinates of the projectile, ##R^2=x^2+y^2##. Differentiate the equation twice.

  5. Sep 7, 2014 #4


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    Differentiating the equation ##R^2=x^2+y^2## results ##R\dot R=x\dot x+y\dot y##. The second derivative is ##{\dot R }^2 +R\ddot R={\dot x}^2+{\dot y}^2+ x\ddot x +y\ddot y##.
    ##R##, ##\dot R ## and ##\ddot R ## are given. It is a projectile, the horizontal acceleration is zero, the vertical one is -g. And ##{\dot x}^2+{\dot y}^2=v^2##, the square of the speed.

  6. Sep 8, 2014 #5
    what are the values of x and its derivatives... same goes for y.. I got confused. But thanks sir.
  7. Sep 8, 2014 #6


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    The x,y coordinates of the projectile can be expressed with the polar coordinates R and θ, how? You gave already the x coordinate as x=Rcos(30°), what is y?
    The motion of a projectile is the resultant of a horizontal motion and with a vertical one. The only force is the gravity of Earth. What are the horizontal and vertical accelerations?

  8. Sep 8, 2014 #7


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    Follow the advice from ehild (somehow I had the impression ehild is a she).
    You get the solution handed to you on a silver platter: in the expression for the second derivative, you know everything except v:
    you don't need x, because it is multiplied with its second derivative, which is zero
    you don't need ##\dot x## or ##\dot y##: the sum of squares is v square and v is what you are after!
    all you need is to calculate y, which is almost trivial!

    Commenting on the equations in your solution:
    Looks like ##R \tan\theta \; \dot\theta## and I don't follow that. So from there on, you're lost as far as I can see.

    @ehild: sorry to have jumped in; thought MrM was awake and you were away; it's the other way around now.
    Nice exercise indeed.
  9. Nov 29, 2016 #8
    How to find the acceleration?
  10. Nov 30, 2016 #9


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    You should know how the Cartesian coordinates and the polar ones are related. x=Rcos(θ), y=Rsin(θ).
    You also know that the horizontal component of acceleration of the projectile is zero, ##\ddot x =0 ##, the vertical component is ##\ddot y = - g##. Use these values, together with the given data in the expression of the second derivative of R2, ##{\dot R }^2 +R\ddot R={\dot x}^2+{\dot y}^2+ x\ddot x +y\ddot y##.
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