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Linear velocity of a rotating body

  1. Jul 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A flat rigid body is rotating with angular velocity 3 rads-1 about an axis in the
    direction of the vector (i + 2 j + 3 k) and passing through the point (1, 1, 0) on
    the body. Find the linear velocity of the point P = (1, 0, 1) on the body.
    (You may use the result v = w x r .)


    2. Relevant equations

    v= w x r

    3. The attempt at a solution

    i have no idea where to go with this - i need to find r, but not sure how to go about using the coordinates given.
     
  2. jcsd
  3. Jul 30, 2009 #2

    HallsofIvy

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    Do you mean 3 rads/sec (often written just "3 s-1") ?

    First you need to know the radius of the circle the point is moving in. Draw a line from (1, 0, 1) to the line x= 1+ t, y= 1+ 2t, z= 3t. The plane containing (1, 0, 1) and perpendicular to i+ 2j+ 3k is (x-1)+ 2y+ 3(z-1)= 0. The line passes through that plane at (1+ t- 1)+ 2(1+ 2t)+ 3(3t-1)= 14t- 1= 0 or t= 1/14. x= 1+ 1/14, y= 1+ 2/14, z= 3/14 or (15/14, 16/14, 3/14). The distance from that point to (1, 0, 1) is
    [tex]\sqrt{(1- 15/14)^2+ (-16/14)^2+ (1- 3/14)^2}[/tex]
    [tex]= \sqrt{1/196+ 256/196+ 121/196}[/tex]
    [tex]= 3\sqrt{42}/14[/tex]
    and that is the radius of the circle the point is moving in. (Better check my arithmetic- that looks peculiar.) From the radius you can calculate the distance corresponding to 3 radians and so the distance the particle moves in one second.
     
  4. Aug 1, 2009 #3

    how did you get the x, y and z to equal those three? and where did the t's come from?

    thanks for the help!
     
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