Finding minimum angular speed

  • Thread starter phys62
  • Start date
  • #1
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Homework Statement


The figure illustrates Michelson's setup for measuring the speed of light with the mirrors placed on Mt. San Antonio and Mt. Wilson in California, which are 35 km apart. Using a value of 3.00 x 108 m/s for the speed of light, find the minimum angular speed (in rev/s) for the rotating mirror.

Link to figure: http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c24/ch24f_12.gif


Homework Equations


delta t = 2d/c


The Attempt at a Solution


delta t = 2(3500)/3x10^8 = 2.33x10^-4

Now if I'm on the right track, then I'm not sure what my next step should be.. I feel like I should be dividing or multiplying by 8 since I was told the object is an octagon. Am I right??
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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35 km or 3.5 km?

Doesn't the wheel have to advance 1/8 of a revolution by the time it bounces to see it again in the next successive reflecting surface from when it was originated?
 
  • #3
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Oh, sorry I meant 2(35000)/3x10^8 = 2.333x10^-4

do I then just divide that by 8 to get 2.91x10^-5 ?
 
  • #4
LowlyPion
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Oh, sorry I meant 2(35000)/3x10^8 = 2.333x10^-4

do I then just divide that by 8 to get 2.91x10^-5 ?

You found Δt and that is 1/8 of a rev isn't it?

So how long to make a rev?
 
  • #5
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I took 2.333x10^-4 and multiplied that by 8 to get .0018664 rev/s but that answer is incorrect
 
  • #6
LowlyPion
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I took 2.333x10^-4 and multiplied that by 8 to get .0018664 rev/s but that answer is incorrect

The time for a revolution is not rev/s.
 
  • #7
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Awesome, got it! 535.79. Thanks for the help! :]
 

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