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Linear velocity of rigid body with vectors

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    v= ω x r

    3. The attempt at a solution

    Magnitude of vector joining P and Q = |P-Q| = r

    .:. r = √6

    v= ω x r
    v= 2 x √6
    v = 4.8990m/s

    I'm assuming I'm completely wrong because the amount of work isn't enough for the amount of marks. I've seen similar questions online and I've noticed that x=1+t, y=1+2t and z=3t have been used in a number of the answers. Are these general rules or have they been derived from the question somehow? Any and all help would be greatly appreciated. Thanks.
     
  2. jcsd
  3. Jan 21, 2012 #2

    SammyS

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    You haven't taken into account the vector nature of r and ω.
     
  4. Jan 21, 2012 #3
    So P-Q = (1,-1,2) and w = (2,2,2). Therefore v = (6,-2,-4)m/s ?

    I understand your reply but it doesn't really tell me what to do.
     
  5. Jan 21, 2012 #4

    SammyS

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    Well, you did take the cross product of r and what you thought was ω. So, maybe I helped point you in the right direction -- which is all I wanted to do.

    By the way: ω ≠ (2,2,2). The magnitude of (2,2,2) is 2√(3) . The magnitude of ω is 2 .
     
  6. Jan 22, 2012 #5
    Erm... So what is the answer then? Am I even right in thinking that P-Q is equal to r? How do I get v? Sorry but you're being very cryptic and I just don't get it...
     
  7. Jan 22, 2012 #6
    Is it something to do with using P-Q as the normal to the plane and discovering the plane's equation with N.(x-Px) = 0? Or something like that?
     
  8. Jan 22, 2012 #7

    SammyS

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    Yes P-Q = r .

    Cryptic? I did mention that you had a problem with what you had for ω.

    What is wrong with ω specifically? It does have the correct direction, but the wrong magnitude.
    You multiplied i+j+k by 2 . However, the magnitude of the vector i+j+k is not 1 .​
     
  9. Jan 22, 2012 #8
    Ahhh so the magnitude is √3. Therefore ω= 2*√3. So v = 2√3 * √6 = 6√2 = 8.49m/s ?

    I'm sorry if I seemed annoyed or something. I was just confused and frustrated.

    Edit: I have r in vector form, I need ω in vector form. The magnitude of ω is 2. So I need to get the vector form from this. But how?
     
    Last edited: Jan 22, 2012
  10. Jan 22, 2012 #9

    SammyS

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    First of all: 2*√3 = √(12) ≠ √6 .

    The vector, (2, 2, 2) has a magnitude of 2*√3 , so multiply that by 1/√3 . What will that give you?
     
  11. Jan 22, 2012 #10
    The vector (2,2,2) doesn't have magnitude 2*√3, it has a magnitude of √(2^2 + 2^2 + 2^2) = √12, doesn't it? Is the intention to get the unit vector? I really don't understand at all. If you could please just tell me, I'm simply not getting it.

    v and r are vectors, ω is a scalar
    If P-Q = r, then r=(1,-1,2)
    ω = 2rad/s in the direction of (i,j,k)

    v=(6,-2,-4) from determinant method.
    |v|=√(6^2 + (-2)^2 + (-4)^2) = 2*√14=7.48m/s

    Am I even close?
     
  12. Jan 22, 2012 #11

    SammyS

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    Answer this quoted question.
     
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