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Homework Help: Conceptual doubt in angular velocity

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data

    I am having a few doubts while writing relationship between angular speed and linear speed ,in the reference frame of a moving observer .

    Suppose at an instant, car A is heading towards car B with velocity 'v'. Car B is moving towards right with velocity 'u' ,i.e along positive x-axis . The instantaneous distance between A and B is 's' .

    What is the relationship between the different parameters at this instant ?

    2. Relevant equations

    3. The attempt at a solution

    In the reference frame of car A ,the car B is rotating with angular velocity ω ,and translating with velocity ucosθ-v .

    So , uT = sω where uT=usinθ is the tangential speed of car B .

    Now,my confusion is angular speed is the time rate of change of angle .But which angle ? Is it θ or α or something else?

    If it is θ then $$ s\dotθ = -usinθ $$ . If it is α ,then $$ s\dotα = usinθ $$ .

    While dealing with angular speeds,I have difficulty in determining the angle ,whose time rate of change is ω.

    I am not sure if what I have written makes sense .

    I would be grateful if somebody could reflect his/her thoughts on this .

    Attached Files:

    • cars.PNG
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  2. jcsd
  3. Oct 19, 2013 #2
    "Which angle" is really up to you, in two ways. First, the location of the "zero" angle is completely arbitrary, it can be the x-axis as depicted, the y-axis, or any other position. Second, you need to choose the "positive" direction. The choice is usually made so that the direction of the angular velocity vector will complete a right-handed triad with the x and y-axes. See http://en.wikipedia.org/wiki/Right-hand_rule . In 2D motion, this simply means that the positive direction is counter-clockwise.
  4. Oct 19, 2013 #3


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    Since A is heading towards B, A's velocity has no affect on the bearing of B from A. It's not clear how you are defining alpha and theta. If they are defined by the relative positions of A and B, as in the diagram, at any time t, then both your equations are valid. ##\dot \alpha = -\dot \theta##.
  5. Oct 19, 2013 #4
    Hi Voko...

    Does that mean I can arbitrarily choose any line in the X-Y plane and consider angle between that line and AB ( say,β ) .Now $$ u_T =s\dot \beta $$ or $$ u_T = - s\dotβ $$ ,depending on whether I choose positive to be clockwise or anticlockwise .

    Is it correct ?
  6. Oct 19, 2013 #5
    Yes, that is correct.
  7. Oct 19, 2013 #6
    Great...So, whether it is α , β ,or θ $$ |\dot\alpha|=|\dot\beta|=|\dot\theta| = \omega $$ where ω is the angular speed of B with respect to A.

    Is it correct ?
  8. Oct 19, 2013 #7
    This is pretty much the definition of angular speed.
  9. Oct 19, 2013 #8
    Thanks Voko...
  10. Oct 19, 2013 #9
    I didn't understand this .

    Does this mean - If the velocity of A is towards +y axis (v) instead of towards B ,the relation will be

    $$s\dot\theta = -(usin\theta + vcos\theta)$$ Is it correct ?

    Alpha is the angle which line joining A to B makes with Y-axis .Theta is the angle with the X-axis .
  11. Oct 20, 2013 #10
    haruspex...please repond to the above post.
  12. Oct 20, 2013 #11
    I believe its the "bearing" term that is the cause of problem. I quote Wikipedia :

    what haruspex meant was that direction of velocity of A and displacement vector from A to B has angle zero between them. I hope this helps!!!
  13. Oct 20, 2013 #12


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    I meant that, instantaneously, the fact that A is moving has no affect on alpha, because the direction of A's movement happens to be towards B.
    Yes, but that could be just the instantaneous condition. Since we are concerned with changes in alpha it's important to be clear that that definition applies at all times, even as A and B move. For example, it could have been that A is on a constant heading and only happens to be heading towards B at that instant. In that case alpha could be defined as that constant heading, and by definition its rate of change would have been zero.
  14. Oct 20, 2013 #13
    In my judgement, the easiest way to do this problem is to use vectors. Here is an example:

    velocity vector for car A[itex]=vcosθ\vec{i}+vsinθ\vec{j}[/itex]
    velocity vector for car B [itex]=u\vec{i}[/itex]
    position vector of car A [itex]=-s\sinθ\vec{j}[/itex]
    position vector of car B [itex]=s\cosθ\vec{i}[/itex]

    Relative velocity vector of car B relative to car A [itex]=\vec{V_r}=u\vec{i}-(vcosθ\vec{i}+vsinθ\vec{j})=(u-vcosθ)\vec{i}-vsinθ\vec{j}[/itex]
    Relative position vector of car B relative to car A [itex]=s\cosθ\vec{i}+s\sinθ\vec{j}=s(\cosθ\vec{i}+\sinθ\vec{j})[/itex]
    Unit vector in direction of car B relative to car A [itex]=\vec{U_r}=(\cosθ\vec{i}+\sinθ\vec{j})[/itex]
    Component of relative velocity vector of car B relative to car A in the direction of car B relative to car A = [itex](\vec{V_r}\centerdot \vec{U_r})\vec{U_r}=(ucosθ-v)\vec{U_r}[/itex]
    Component of relative velocity vector of car B relative to car A in the direction perpendicular to the direction of car B relative to car A = [itex]\vec{V_r}-(\vec{V_r}\centerdot \vec{U_r})\vec{U_r}=(u-vcosθ)\vec{i}-vsinθ\vec{j}-(ucosθ-v)(\cosθ\vec{i}+\sinθ\vec{j})=u\sin^2θ\vec{i}-u\sinθcosθ\vec{j}=usinθ(sinθ\vec{i}-cosθ\vec{j})[/itex]
    The magnitude of this perpendicular relative velocity component is just u sinθ.

    Therefore, the angular velocity of car B relative to car A is given by:

  15. Oct 21, 2013 #14


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    Chest is right, the best to do it with vectors.

    The angular velocity of a point mass with respect to point O is

    [tex]\vec \omega = \frac {\vec r \times \vec v }{r^2}[/tex].

    ##\vec r## is the position vector, ##\vec v## is the velocity. θ is the angle of ##\vec v## with respect to the direction of ##\vec r##. See picture.

    If you determine the position vector and the relative velocity of B with respect to A you get the angular velocity of B with respect to A or vice versa from the cross product.


    Attached Files:

  16. Oct 21, 2013 #15
    Yes. Use of the cross product is another good way to get the same result. Also, once I saw what the answer was and considered it in comparison with the figure, I could have kicked myself. I realized that the answer could have been written down almost on inspection. All that really needed to be done was to resolve u into components parallel and perpendicular to the line joining A and B; v was already aligned along the parallel direction.

  17. Oct 21, 2013 #16
    Thanks Chet for your excellent input .This is a very good and systematic way of handling the problem .You have given a very nice insight in using vectors to deal with the problems .

    Thank you ehild .
  18. Oct 21, 2013 #17
    Here is an extended version of the problem.

    "Car A moves uniformly with velocity ‘v’ so that its velocity is continually towards car B which in turn moves uniformly with velocity ‘u’ in x-direction (u<v). At t=0 , car A is at origin and car B points towards origin at a distance L . After how much time car B would catch up with car A ?"

    This is one way of approaching the problem .

    At any instant , let the distance between the cars be ‘s ‘ and the angle between the two velocities be θ .

    Then ds/dt = -(v-ucos θ) .

    Integrating L = vT – ∫ucosθdt (1)

    uT = ∫vcosθdt (2)

    So , T = (vl)/(v2 – u2)

    Is there any other way of thinking about this problem ? Can this problem be done vectorially or some other approach ? A longer method is perfectly alright .
    Last edited: Oct 21, 2013
  19. Oct 21, 2013 #18
    You can't do it that way because θ is changing with time.
    I recommend using polar coordinates centered at car A, and starting with the results from the previous example.

    [tex]\frac{dr}{dt}=-(v-ucos θ)[/tex]

    with r = L and θ = θ0 at t = 0.

    Divide one equation by the other:

    [tex]\frac{dr}{dθ}=\frac{r(v-ucos θ)}{usinθ}[/tex]

    Then integrate, and substitute result into r equation. Of course, for θ0 approaching zero, t = L/(v-u).
  20. Oct 21, 2013 #19
    $$ \frac{dr}{r}=\frac{v}{u}cosecθ-cotθ $$

    $$ lnr = \frac{v}{u}lntan\frac{θ}{2} - lnsinθ +C $$

    Now,initially $$ r = L , θ = \frac{\pi}{2}$$

    So , C= ln(L)

    and , $$ r = \frac{L[tan\frac{θ}{2}]^\frac{v}{u}}{sinθ} $$

    How to proceed from here ?
  21. Oct 21, 2013 #20
    Just for the record, the curve traced by A is known as the "dog curve", which is a particular case of a radiodrome, which are a subset of the so called "pursuit curves".
  22. Oct 21, 2013 #21
    There are math errors in the above. Anyhow, I get


    You can substitute this into the equation for dθ/dt, and integrate to find θ as a function of time. I don't know how to do the integration analytically, but, from some approximate analysis I did at long times, I can say that both r and θ become equal to zero simultaneously after a finite amount of time. Of course, the calculation can be done numerically for various values of v/u.

  23. Oct 21, 2013 #22
    Thanks for the info :)

    But , [tex]L\frac{(1-cosθ)^\frac{v}{u}}{(sinθ)^{(1+\frac{v}{u})}} = \frac{L[tan\frac{θ}{2}]^\frac{v}{u}}{sinθ}[/tex]
    So , I think we agree on value of r .
    Last edited: Oct 21, 2013
  24. Oct 21, 2013 #23
    Sorry. You're right.

    Try making a graph of r vs θ at constant v/u to see what the functionality looks like, particularly at long times (small θ).

  25. Oct 21, 2013 #24
    Good news. I think for the case where v = 2u, the equations can be readily solved analytically. Give it a try, and see what you get. Substitute the solution for r as a function of θ into the differential equation for θ, and solve for dt.

  26. Oct 22, 2013 #25


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    The equation ##\frac{dθ}{dt}=-\frac{usinθ}{r}## is integrable for all v/u>1 after you substituted for r. Use ##\sinθ=2\frac{\tan(θ/2)}{1+\tan^2(θ/2)}## and integral with substitution tan(θ/2) =z. (You can cheat ... with Wolframalpha)

    And the result is the same you got in post #17, Tania! You are a genius!

    Last edited: Oct 22, 2013
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