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Dot product in uniform circular motion question -- Finding angle?

  1. Feb 15, 2015 #1
    • Member warned: Template use is required.
    I've attached an image of part a of the question to this thread.

    My question is this (the solution to these former homework problems are posted to help us study for exam, which is why know this already):

    The angle between the two velocity vectors is determined to be pi/2. How? I know that dot product is abcos(Θ) where theta is the angle between the two velocity vectors. However, I don't know how to solve for theta when abcosΘ is not equal to anything?

    I never had a trig class (which is really hurting me) so is there some unit circle technique that I'm supposed to know about to determine what the dot product is equal to so I can solve for theta? Something involving the x and y components of the velocity vectors?

    Also, why can't I just use the magnitude of the velocity vector as the radius then solve using the regular T=2pi(r)/v equation? Velocity is always tangent to the trajectory, so shouldn't its magnitude always be the length of the radius?

    Edit: Oops, sorry. Velocity vector 1 is (4.00)i + (3.00)j and velocity vector 2 is (-3.00)i + (4.00)j

    Thanks,

    Colton
    Dot Product Uniform Circ Motion Screen.PNG
     
    Last edited: Feb 15, 2015
  2. jcsd
  3. Feb 15, 2015 #2

    gneill

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    Staff: Mentor

    The dot product is an operation on vectors that has a definition based on the vector components. The "abcosΘ" version is a handy way to find its value when you happen to know the magnitudes of the vectors and the angle between them. If you're given the raw vectors themselves though, it's much easier to just do the dot product using the vectors themselves.

    The dot product of two vectors is the sum of the products of the like vector components. So if your vectors were A = (a1,a2,a3) and B = (b1,b2,b3), then the dot product would be: A dot B = a1b1 + a2b2 + a3b3. Simple.

    If you compute the dot product in this way and also extract the individual vector magnitudes then you can solve for the angle between the vectors using the other formula.

    Regarding the velocity vector, its magnitude is most certainly NOT the same as the radius. In fact, in circular motion the magnitude of the velocity is equal to the radius multiplied by the angular velocity of the radius vector. That is, in scalar terms v = ωR, or in vector terms, v = ω x r, where v, ω, and r are vectors.
     
  4. Feb 15, 2015 #3
    Thank you! Very helpful reply, but upon working through it with your advice, I encounter another problem.

    So I did the dot product with the individual components, which yields (-12.00)i + (12.00)j. So A dot B = (-12.00)i + (12.00)j

    I then use Pythagorean Theorem to determine the addition result, which is 16.97, rounded to 17.00 for simplicity. So A dot B = 17.00.

    That should be equivalent to saying ABcosΘ=17.00 right? Then I can solve for the angle theta?

    I did so, and I got an angle of 47° (rounded). However, the answer key indicates that the angle between the two vectors in question is pi/2. When converting from degrees to radians, I get a value of 0.26pi which is most definitely not what the answer key says. In order for it to be pi/2 wouldn't the calculated angle have to be 90°?

    Is there a clear error in my calculation that I missed?
     
  5. Feb 15, 2015 #4

    gneill

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    Staff: Mentor

    Nope. The dot product has a scalar result. The sum of the terms is a scalar number, not a vector. You should end with a single number, not vector components. Refer to the symbolic example I gave, not an i or a j in sight in the result.
     
  6. Feb 15, 2015 #5
    Doh. It appears I forgot some basic physics concepts lol. Yeah I understand now. arccos(0) (which is 12-12 from the addition of like components) is 90° which gives a radian value of pi/2.

    Thanks!
     
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