Velocity of a plane

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  • #1
bungie77
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Homework Statement


http://img90.imageshack.us/img90/2456/planeqc1.th.jpg [Broken]

This is the question.

Homework Equations


I am unsure how to relate the velocity of the plane to the angular velocity.


The Attempt at a Solution



I calculated r to be 11 540m. Therefore the distance in the i direction from the satellite must be 5770m and the distance in the j direction 10 000m. However when I try to calculate velocity of the plane using Vp = Vs + W*R (where W is angular velocity and R is distance from the plane to the satellite) I get an answer in terms of i and j even though I know there shouldn't be any velocity in the j direction. What am I doing wrong?
 
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Answers and Replies

  • #2
saket
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Ah! One will have to get inside the picture to read the question!
 
  • #4
Galileo
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Notice that h and theta give all the information you need to find the position of the plane.
Thereforce I suggest to work in these given variables as much as possible. The horizontal distance is also useful since its time derivative gives the velocity of the plane. r is really superfluous and the line is there only to define theta.

Let's call the horizontal distance from the satellite to the plane x. What are x and dx/dt in terms of h, theta and d(theta)/dt?
 
  • #5
saket
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However when I try to calculate velocity of the plane using Vp = Vs + W*R (where W is angular velocity and R is distance from the plane to the satellite) I get an answer in terms of i and j even though I know there shouldn't be any velocity in the j direction. What am I doing wrong?

Do you have any idea where this formula is used and why? As far as I know, it is used if a particle, P, is in pure rotation (angular velocity, W) about S, which itself may move with velocity Vs. Here, is the aeroplane in pure rotation about the radar station? Verify yourself that it is NOT, and thus the formula is, kind of, not applicable. (I mean, terms can be redefined to make it applicable!)


P.S.: Whenever you memorize a formula, please be careful about its usage and limitations!
 
  • #6
bungie77
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yeah i understand that. sorry don't crucify me i just wanted to show i had not just gone straight here for an answer. i really just wanted a point in the right direction as was done by the previous poster. thanks for your comment and ill have a better look at the problem and post back soon.
 
  • #7
saket
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Well, how about this approach: (It does not need knowledge of calculus, and shall I go as far as saying.. it is simpler!)
Let radar station be denoted by S, the plane be denoted by P.
I am using the following fact: Velocity along SP does not contribute to angular velocity of aeroplane about S. It is just the component of the velocity perpendicular to SP which contributes. Remember, [tex]\vec{v}[/tex] = [tex]\vec{w}[/tex] X [tex]\vec{r}[/tex]. (Let, angular velocity = w = 0.020 rad/s)

Component of aeroplane's velocity in a direction perpendicular to SP = vSin[tex]\theta[/tex]. Thus, w.r = vSin[tex]\theta[/tex].
Also, rSin[tex]\theta[/tex] = h.
Therefore, v = w.h / (Sin[tex]\theta[/tex])^2
 
  • #8
bungie77
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That is a lot simpler! Thanks for your reply that method makes sense to me and I get 960km/hr which I assume is correct. As for the other method though I am still a sketchy.
 
  • #9
saket
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Ah, that is simple as well, if you have knowledge of calculus.
From geometry, h = xtan(theta).
Differentiate once, and use the fact that d(theta)/dt = w and dx/dt = v and dh/dt = o, as h remains constant.
(x is horizontal distance traveled by aeroplane in time t)
 

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