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Angular velocity.

  1. Oct 12, 2011 #1
    a particle A has a circular movement with radius R and angular velocity of 2w, a particle B has a radius of 2R and an angular velocity of w, both velocities remain constant. particle B rotates clockwise and particle A rotates anticlockwise for an interval of t = n/(2w)

    what angle is formed by the positional vectors of these two particles?

    the options are: 0, 3pi/2, pi/2, pi.

    note: the vector of these two particle's movement have the same origin, their movement make two circles, which have the same origin. particle B's circle has double the radius of particle A.

    They don't say absolutely anything about the value of n.

    from a sketch I've made, I say the angle is pi/2. but again, it all depends on the value of n.

    any ideas?
     
  2. jcsd
  3. Oct 12, 2011 #2
    Some information is missing since the angle between the positional vectors will always be changing. It seems that the radius of each circle doesn't matter in terms of the angle between the position vectors.

    EDIT: You're right, the answer is in terms of n, but none of the options have n in them.
    w * t for particle A = -n radians
    w * t for particle B = n/2 radians.
     
  4. Oct 12, 2011 #3
    I have written absolutely everything that they gave me, I'm also finding it impossible to solve this problem.
     
  5. Oct 12, 2011 #4
    Why would your teacher give you R?
     
  6. Oct 12, 2011 #5
    if for example n = 1, the angle would be 3pi/2. do you agree?
     
  7. Oct 12, 2011 #6
    I guess it depends on how you define negative and positive, but yes I agree.
    It would be 3pi/2 if you measured clockwise from A to B.
     
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