1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Using v= omega cross r to find instantaneous velocity

  1. Oct 14, 2014 #1
    1. The problem statement, all variables and given/known data
    An object is rotating at 4 rad/s about an axis in direction of [itex](2 \hat{i} - 4 \hat{j} + 3\hat{k} ) [/itex] which then passes through a point (1,2,0)m .

    Calculate the instantaneous velocity at the point (2,0,3)m (Hint: use [itex] \vec{v} = \vec{\omega} \times \vec{r} [/itex])


    2. Relevant equations
    [itex]
    \hat{A} = \frac{\vec{A}}{|\vec{A}|}
    [/itex]
    3. The attempt at a solution
    First off, i would like to say I am extremely new (1-2 weeks new) to cross products, and this is the first time we have been given physical (rather than purely mathematical) problems to implement it.

    What I immediately thought of was the equation I put in the relevent equations section, how a unit vector is calculated and since all I need/want from the vector [itex](2 \hat{i} - 4 \hat{j} + 3\hat{k} ) [/itex] is the direction I turned that into a unit vector, and then to get the angular velocity vector times that by 4. So...
    [itex]
    \sqrt{2^2+4^2+3^2} = \sqrt{29} \\
    \therefore \vec{\omega} = (\frac{4 \cdot 2}{\sqrt{29}} \hat{i} - \frac{4 \cdot 4}{\sqrt{29}} \hat{j} \frac{4 \cdot 3}{\sqrt{29}} \hat{k}) rad/s = (\frac{8}{\sqrt{29}} \hat{i} - \frac{16}{\sqrt{29}} \hat{j} + \frac{12}{\sqrt{29}} \hat{k}) rad/s[/itex]
    And then converted to decimal to makes things easier
    [itex]
    \vec{\omega} = (1.486 \hat{i} - 2.971 \hat{j} + 0.557 \hat{k}) rad/s
    [/itex]

    And then to get [itex] \vec{r} [/itex] I did (1,2,0)-(2,0,3)=(-1,2,-3)

    And then did the cross product, I am not sure how to do matrices in latex but I put i,j,k on top row (of 3 by 3 matrx) and then on second row put the omega vector and then on third put (-1,2,-3).

    Then I got the determent and thus the velocity vector to be...
    [itex]
    \vec{v} = [(-2.971 \cdot -3)-(0.557 \cdot 2)] \hat{i} + [(1.486 \cdot -3)-(0.557 \cdot -1)] \hat{k} + [(1.486 \cdot 2)-(-2.971 \cdot -1)] \hat{k} \\
    \vec{v} = 7.8 \hat{i} -3.9\hat{k}+0.001\hat{k}
    [/itex]

    I don't know if my method is correct at all, I am suspicious of the low value for the k component for a start off. Oh and as the question asks for "the instantaneous velocity" I do not know whether they mean just the velocity vector or its magnitude; I assume if they meant the magnitude then they would have just said "speed" instead, though.
     
  2. jcsd
  3. Oct 14, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    This makes things a lot more complicated. It is easier to keep the prefactor ##\frac{4}{\sqrt{29}}## as it is (don't even put it to the components). It stays a common prefactor of everything, so you can continue to work with the nice integers (like 2,-4,3 for your rotation axis) and care about the ugly prefactor later. This also does not lead to the rounding error you see in your result.

    They probably mean velocity, but you can give both. You have the velocity, the speed is easy to calculate.
     
  4. Oct 14, 2014 #3
    ok, fair enough, thanks. But is my method correct and what I did after converting to decimal? Just so if I know its correct I can just go back an keep with the surds but follow the same steps.
     
  5. Oct 14, 2014 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Yes the approach is correct.
     
  6. Oct 14, 2014 #5
    Ok, thanks for your help :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Using v= omega cross r to find instantaneous velocity
  1. V = r times omega (Replies: 4)

Loading...