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Expression for the instantaneous angular velocity

  1. May 16, 2017 #1
    1. The problem statement, all variables and given/known data
    The directed beam from a small but powerful searchlight placed on the ground tracks a small plane flying horizontally at a fixed height h above the ground with a uniform velocity v. If the search light starts rotating with an instantaneous angular velocity ##\omega_0## at time ##t=0## when the plane was directly overhead, then at a later time t, its instantaneous angular velocity ##\omega(t)## is given by?
    2. Relevant equations
    ##\omega## = ##\omega_0##+##\alpha t##
    ##\omega^2## = ##\omega_0^2## + 2##\alpha \theta##
    ##\theta## = ##\omega_0 t##+1/2##\alpha t^2##
    3. The attempt at a solution
    I am not sure where to begin the solution other than the equations I have written above.
    The answer given is ##\frac{\omega_0}{1+\omega_0^2t^2}##. I tried differentiating but I don't seem to get anywhere with it.
     
    Last edited: May 16, 2017
  2. jcsd
  3. May 16, 2017 #2

    ehild

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    You tried differentiating - what?
    ω need not be either a linear or quadratic function of time.
    Make a picture, and everything will be clear.
    How does the angle theta depend on time? How is the angular velocity ω defined?
    upload_2017-5-16_9-44-10.png


    The blue line represent the light beam at t=0, the red line is the light beam at a later time t.
     
  4. May 16, 2017 #3
    That to an extent was helpful however embarrassing to me I still couldn't get the right answer.
    I took sine of the angle theta, so ##\sin\theta = vt/l## where l is the line connecting the searchlight and the plane
    ##\theta = \sin^{-1}(vt/l)## I used ##v = l\omega## to make the substitution in vt/l
    differentiating it I get##\omega = \frac{\omega}{\sqrt{1-\omega^2t^2}}##
     
  5. May 16, 2017 #4

    ehild

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    l is changing, use vt/h = tan (theta) instead.
     
  6. May 16, 2017 #5
    I did use Tan at first however I got an answer with the terms l and h which couldn't figure out a way to eliminate. So with tan I get
    ##\theta = \tan^{-1}{vt/h}##
    on differentiating I get ## \omega = \frac{\omega }{h/l+l\omega^2t^2/h}##
    and I do not know how to simplify it further hence I used sine in my previous post
     
    Last edited: May 16, 2017
  7. May 16, 2017 #6

    haruspex

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    That doesn't make much sense. Did you mean ω0 on the right?
    Please post your working.
     
  8. May 16, 2017 #7

    ehild

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    I don not follow you.
    You have ##\theta = \tan^{-1}({vt/h})##
    Differentiate it with respect to t. What do you get? Remember v and h are constants.
     
  9. May 16, 2017 #8
    Differential of ##Tan^{-1}x = \frac{1}{1+x^2}##
    so then ##\omega = (1/1+(vt/h)^2 )* v/h##
    substituting ##v=l\omega##
    ##\omega = 1/1+(l\omega t/h)^2 * v/h##
    and I end up with ## \omega = \frac{\omega }{h/l+l\omega^2t^2/h}##

    I found where I made mistake 'v' should be, ##v=h\omega_0##
    and with that substitution
    I get the answer ## \omega = \frac{\omega_0}{1+\omega_0^2t^2}##
     
    Last edited: May 16, 2017
  10. May 16, 2017 #9

    ehild

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    correct so far...
    That is wrong. You've just derived the expression for ω, it is not v/L.
    You have to find ω0, the angular velocity at t=0. Substitute t=0 into the equation for ω, what do you get?
     
  11. May 16, 2017 #10
    Seems like you did not see my update in post #8.
     
  12. May 16, 2017 #11

    ehild

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    Well, I did not see. You got it at last!
    Next time remove the wrong part from your post.
     
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