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Speed of a point

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment)


    2. Relevant equations



    3. The attempt at a solution
    I was able to calculate the velocity of point P in x direction (to the left). For the speed, I need to find the velocity in y direction (vertically up). Here is my attempt:
    At t=0, the distance of point P from O is l, let at time t, the angle POR be ##\theta##. The displacement is ##y=l(\sin \theta -1)##. Differentiating this equation, ##\frac{dy}{dt}=l \cos \theta \cdot \frac{d \theta}{dt}##. The term dθ/dt is equal to the angular velocity but I don't know the radius of curvature of its path which I have to use in the formula ω=vr, where ω is the angular velocity, v is the velocity in x direction and r is the radius of curvature.

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Feb 4, 2013 #2

    tms

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    How far does P get from O after t =0?
     
  4. Feb 4, 2013 #3
    ##l \cos \theta##.
     
  5. Feb 4, 2013 #4

    Curious3141

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    Homework Helper

    You're overcomplicating things. Let ∠SUT = 2θ (so you're looking for what happens when θ = 45 deg).

    Take O as the origin.

    Let ##u_x, u_y, p_x, p_y## be (respectively), the co-ordinates of U and P at time t.

    Can you find simple expressions for those in terms of l and θ? (except that ##u_y## is simply constant at zero).

    Can you find ##\frac{du_x}{dt}## in terms of ##\frac{d\theta}{dt}## using Chain Rule? Hence rearrange to determine a numerical value for ##\frac{d\theta}{dt}## at the instant of interest.

    Can you differentiate ##p_x, p_y## wrt t to find the instantaneous horizontal and vertical velocities of P?

    Now use Pythagoras theorem to find the resultant speed.
     
  6. Feb 4, 2013 #5

    ehild

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    Find the coordinates of P in terms of the angle. The velocity components are the time derivatives of coordinates.

    ehild

    Edit:Curious beat me ...
     
  7. Feb 5, 2013 #6
    Thanks a lot Curious, that solved the problem. :smile:

    The velocity of P in both the directions come out to be same!
     
  8. Feb 5, 2013 #7

    Curious3141

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    That's because the sine and cosine of 45 deg are the same! :wink:
     
  9. Feb 5, 2013 #8

    tms

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    Not the x coordinate, but the distance from O. I was just trying to point out that P moves in a circle.
     
  10. Feb 5, 2013 #9
    Yep, didn't notice that at all. Thanks!
     
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