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Movement of light from a lighthouse

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data
    In a lighthouse that is 2km from the shore, the searchlight rotates clockwise once every 50s. How fast is the spot of light on the shore moving: a) When it's directly opposite the island? b) When it is 1km from the point opposite the island?



    2. Relevant equations
    implicit differentation



    3. The attempt at a solution
    I started with a right triangle diagram, where the right angle is the point opposite the lighthouse (I included a diagram to show what I drew). My problem seems to be incorporating the time of rotation into the question. I can easily get an equation by using Pythagorean's Theorem, but I don't know how time would be used in this equation. Any help would be appreciated, thanks in advance.
     

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  3. Apr 10, 2009 #2

    tiny-tim

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    Hi Emethyst! :smile:

    Hint: what is the distance along the shore (from the directly opposite point) when the angle is θ? :wink:
     
  4. Apr 10, 2009 #3
    All I could think of is it being 2tan(theta); otherwise I don't a clue :tongue:. I'm completely stumped on this question, on a) especially (I am assuming I need to complete a) before attempting b).) I really have no idea how to proceed.
     
  5. Apr 11, 2009 #4

    tiny-tim

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    HI Emethyst! :smile:

    (copy my θ :wink:)
    ok … distance = 2tanθ … now what is that in terms of t? … and so what is the speed? :smile:
     
  6. Apr 11, 2009 #5
    Ohh I think I might see it now. Velocity will be 2sec^2θ, so t=tanθ/sec^2θ. After arriving at that I am then assuming I need to find the rate at which the angle is changing, though if I try that I think I can see the dθ/dt cancelling out :tongue:. At least I think i'm now on the right track.
     
  7. Apr 11, 2009 #6

    tiny-tim

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    Whoa! :bugeye:

    θ isn't time …

    you must differentiate 2tanθ with respect to t

    so what is 2tanθ in terms of t? :smile:
     
  8. Apr 11, 2009 #7
    Ohh my mistake, I was using the formula v=d/t, and simply finding v then solving the equation for t :tongue:. Hmm so if I have to make it in terms of t, will I not have to convert the 50s into radians? That then gets me 2tanθ in terms of t. I'm thinking it will be 50s*2pi, or something along those lines. Is this way correct or am I still missing something? :smile:
     
  9. Apr 11, 2009 #8

    tiny-tim

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    You only need 2π if you're given revolutions … the 2π converts from revolutions to radians.

    The ordinary formula v = rω will do here. :smile:
     
  10. Apr 11, 2009 #9
    Sorry for asking for so much help here tiny-tim, but what is that formula? I'm in physics as well and I haven't even heard of that formula yet :confused:. I am assuming it will help in converting time into radians, I just can't see how to do it yet (I'm learning as i'm going here :tongue:)
     
  11. Apr 12, 2009 #10

    tiny-tim

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    This is basic geometry, rather than physics … if you draw a diagram, it becomes pretty obvious. :wink:

    v = rω …

    (tangential) speed = radius times angular velocity.

    Works for distance and acceleration also …

    (tangential) distance = radius times angular distance.

    (tangential) acceleration = radius times angular acceleration. :smile:
     
  12. Apr 12, 2009 #11
    Ok, so applying that to the question, the radius at the point directly across from the lighthouse would be 2km, while the angular distance would be the opposite side of the triangle, which is 2tanθ...I think I am missing something here still :confused:
     
  13. Apr 13, 2009 #12

    tiny-tim

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    hmm … looking back, I got confused as to what the question was …
    Since the question says "the searchlight rotates clockwise once every 50s", we are given the angular speed in revolutions per second (1/50 rev ps) …

    so yes you must convert that to radians per second by multiplying by 2π …

    that gives you your θ in terms of t, so you can then put 2tanθ in terms of t, and differentiate with respect to t :smile:
     
  14. Apr 13, 2009 #13
    Ohh I had it right from the start :tongue:, I just divided 2π by 50s to get dθ/dt instead of converting it into radians first. I have tried the question again and it still seems I am getting the wrong answer. I know the answer is supposed to be -80π, but all I end up with is 2π/25. What I did was differentiate 2tanθ with respect to the time to get 2sec^2θ(dθ/dt). Because the light is directly across from the lighthouse, I assumed θ would be 0. Plugging this into my formula I ended up with 2π/25. I think my mistake might be where you said put 2tanθ in terms of t first (I just assumed θ was a time variable for this question, so I might be missing something still).
     
  15. Apr 13, 2009 #14

    tiny-tim

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    erm :redface: … 2π/25 km/s is 80π m/s :wink:
     
  16. Apr 13, 2009 #15
    *smacks myself* can't believe I missed that :redface: Thanks for all of that assistance tiny-tim, it was a huge help :smile:
     
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