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Special Relativity

  1. Nov 14, 2009 #1

    bon

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    Special Relativity URGENT!!

    1. The problem statement, all variables and given/known data

    A laser is placed on a turntable that rotates at 1200 rev/s. The laser, whose beam makes an angle of 30 degrees with the horizontal, shines on clouds 50km away. Calculate the speed with which the light spot on the clouds moves.

    2. Relevant equations



    3. The attempt at a solution

    I guessed that you have to use trig to find the radius of the circle that the laser ends up "sweeping out", then find the circumference, then find the speed..since speed = dist/time (and t = 2pi/w)..but it doesn't come to the right answer (3.3 x 10^8 m/s), so where am i going wrong?

    Thanks!
     
  2. jcsd
  3. Nov 14, 2009 #2

    Doc Al

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    Staff: Mentor

    Re: Special Relativity URGENT!!

    Your method sounds good to me. Show the details of your calculation.
     
  4. Nov 14, 2009 #3

    bon

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    Re: Special Relativity URGENT!!

    Well, using simple trig, i get the radius of the circle to be 50000tan60 meters. therefore the total dist. travelled is 2 pi (50000tan60)..

    and the time taken is 2pi/1200

    but when you divide the distance by the time, it doesn't give you the right speed (which should be around 3.3 x 10^8...

    any ideas?
     
  5. Nov 14, 2009 #4

    Doc Al

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    Staff: Mentor

    Re: Special Relativity URGENT!!

    The distance given is the distance to the cloud. You want the horizontal component of that distance.
     
  6. Nov 14, 2009 #5

    bon

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    Re: Special Relativity URGENT!!

    ok so 50000sin60..it still doesn't give me the right answer..
     
  7. Nov 14, 2009 #6

    Doc Al

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    Re: Special Relativity URGENT!!

    It does for me. Show the steps in your calculation.
     
  8. Nov 14, 2009 #7

    bon

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    Re: Special Relativity URGENT!!

    dist = 50000sin60 x 2pi

    time = 2pi/1200 s

    dist / time =/= 3.3 x 10^8
     
  9. Nov 14, 2009 #8

    Doc Al

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    Re: Special Relativity URGENT!!

    Right. The distance for one revolution.
    Not right. Note that the angular speed is given in rev/sec, not rad/sec.
     
  10. Nov 14, 2009 #9

    bon

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    Re: Special Relativity URGENT!!

    Ah ok great. Thanks for your help.

    so angular speed is in rev/sec...what the w then in: T = 2pi/w...

    please could you explain the difference..
    thanks
     
  11. Nov 14, 2009 #10

    Doc Al

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    Staff: Mentor

    Re: Special Relativity URGENT!!

    Well, if the angular speed is 1200 revs/sec, then 1 rev takes 1/1200 sec.

    The relationship T = 2pi/ω is only valid if ω is in rad/sec. 1200 rev/sec = 1200*2pi rad/sec.
    So: T = 2pi/ω = 2pi/(1200*2pi) = 1/1200 sec.
     
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