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Angular velocity as car passes by man.

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Close call! A fast car barely misses a man by 3 meters, while traveling at 30 meters per second. The man's head is tracking the car. What's his head's angular speed when the car is 3 meters away?

    2. Relevant equations

    Angular velocity = change in angle / change in time

    3. The attempt at a solution

    I don't know which trig function to use on this one. When I draw a right triangle representing the car's position in relation to the man before or after the car's passing, one side of the triangle ends up having a speed (30 m/s), while the other has a distance (3m). I don't know how to reconcile the two. I want to use the tangent trig function, but it seems I should use the sine? Any help would be appreciated.
     
  2. jcsd
  3. Mar 11, 2008 #2

    tiny-tim

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    Hi DocZaius!

    Triangles don't have speeds, only distances.

    So take a typical instant, when the car is at an angle theta, and a distance x from the closest point.

    One side of the triangle is x, and the other is 3

    So x = 3.tan(theta).

    But dx/dt = 30.

    So d(theta)/dt = … ? :smile:
     
  4. Mar 11, 2008 #3
    As seen in the image, which I have hopefully attached, all you need to know is a distance, which you can decide and a, which is 3m. Using tan you can work out the angle between one of the up right and either diagonal (hyp.) if you know how long it took the vehicle to travel d you also know how long it took to travel the angle you calculated.

    Thus divide the angle by the time taken to travel d and you have your angular velocity. I hope that helps.
     

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  5. Mar 11, 2008 #4
    What I am having trouble with is knowing what the angular velocity is when the car is at the closest point to the man. There seems to be angular acceleration going on here.

    The man's angular velocity is not constant throughout the car's approach and departure. If you imagine watching a car come zooming by while standing next to a freeway, you will notice that your head's angular velocity is increasing to a maximum velocity when the car is closest and then slows down again.

    So I am having trouble understanding why finding out an angular velocity at a time when x is not 0 would help since that wouldn't be the angular velocity at the car's closest point which is when it is 3 m away from the man. All the more frustrating: it is only at that point that there is no more right triangle to help surmise a theta!

    Thanks for your help!
     

    Attached Files:

    Last edited: Mar 11, 2008
  6. Mar 12, 2008 #5

    tiny-tim

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    Hi DocZaius!

    There is angular acceleration going on here - but it doesn't matter!

    Once you have theta as a function of t, angular velocity = dthetat/dt, and angular acceleration = dtheta/dt^2.

    What worries you about that?

    (a) The triangle is artificial - you drew it in to calculate theta. But you could have started when it was at 45º, and drawn all your triangles from there, and used the sinA/sina = sinB/sinb formula. Then you would have a triangle at the closest point (though not at 45º)!

    (b) Alternatively, remember that a derivative is lim((theta(t+∂t) - theta(t))/∂t), so t and t+∂t do give you a triangle! :smile:
     
  7. Mar 12, 2008 #6
    Unfortunately that does not work in this problem. Your method only would give me the average angular velocity over that span of distance "which I can decide." Since this problem has an angular velocity which is not constant (there is angular acceleration), that won't help me find the angular velocity at a specific instant.

    If this problem had a constant angular velocity, your method would work, since over any span of time when angular velocity is constant, average angular velocity = angular velocity at an instant within that span of time.

    Since this isn't the case, arbitrarily choosing a distance d will give me different answers for the angular velocity. Try this yourself. Give yourself various different numbers for d, and you will see that angular velocity is never the same.

    I solved this problem by "cheating". I chose d to be 0.00000000000000000001 meters, and found the answer to approximate very closely 10 radians per second. I assumed that at the actual instant I am looking for, angular velocity is 10 radians per second. I consider this cheating because I had to resort to finding the angular velocity at a CLOSE moment to the point of tangential contact, but not AT that moment.

    Can someone give me the answer for solving AT the point of tangential contact? My current method would have d = 0 (a side of a triangle = 0), which of course is not possible. :wink:

    Thanks.
     
    Last edited: Mar 12, 2008
  8. Mar 13, 2008 #7

    tiny-tim

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    Hi DocZaius!

    You method isn't cheating, it's just a going-back-to-basics way of calculating the derivative, as a limit of a ratio of two things which tend to zero. :smile:

    Bu it's easier to calculate the derivative the usual way:

    dx/dt = 3.sec^2(theta)d(theta)/dt,

    So angular velocity = d(theta)/dt = 10.cos^2(theta).

    In particular, when theta = 0, angular velocity = 10. :smile:
     
  9. Mar 13, 2008 #8
    Thank you so much for your help tiny-tim! Now I get it. Some trig identities involved there...
     
  10. Mar 13, 2008 #9
    So to take this further, what steps should be taken to find the anti-derivative of the d(theta)/dt = 10.cos^2(theta) equation and learn the actual d(theta)/dt^2 acceleration equation?
     
  11. Mar 14, 2008 #10

    tiny-tim

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    angular acceleration

    Hi DocZaius! :smile:

    I'm not sure what you're asking here - the anti-derivative of d(theta)/dt is just theta - the derivative of d(theta)/dt is the angular acceleration, d^2(theta)/dt^2.

    (Though I'm not sure what you'd want the angular acceleration for.)

    Assuming it's the latter: you simply differentiate again with respect to t.

    You could do it from the original equation [tex] \theta\,=\,tan^{-1}10\,t[/tex], if you know what the derivative of [tex]tan^{-1}10\,t[/tex] is.

    But you may find it easier, and you're much less likely to make mistakes, if you do it direct from d(theta)/dt = 10.cos^2(theta), which will give you an extra d(theta)/dt on the right, which you can then substitute:
    [tex]d^2\theta/dt^2\,=\,d/dt(d\theta/dt)[/tex]
    [tex]= d/dt(10.cos^2\theta)\,=\,-20\,sin\theta\,cos\theta\,d\theta/dt[/tex]
    [tex]-20\,sin\theta\,cos\theta\,10.cos^2\theta[/tex]
    [tex]=\,-200\,tan\theta\,cos^4\theta\,,[/tex]
    which you can then convert back to t. :smile:
     
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