wizard85
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Angular velocity and vectors...
Given the following vectors:
\vec A=\hat i +\hat j - 2\hat k and \vec C=\hat j - 5\hat k
Let \vec A and \vec C be drown from a common origin and let \vec C rotate about \vec A with angular velocity \vec w of 2 \frac{rad}{s}. Find the velocity \vec v of the head of \vec C.
My step-by-step way for resolving it, is:
1)I know that \vec v= w \times \vec C
2) By multiplying: \vec A \times \vec C I'll find a vector parallel to \vec w namely D
3) Now, \vec D= \vec A\times \vec C=(\hat i +\hat j - 2\hat k) \times (\hat j - 5\hat k) = 7*\hat i -5*\hat j +\hat k
4) I also know that \vec w is obtained by a linear combination of \vec D's parameter. Then:
\vec w= a * \vec D=a * (7*\hat i -5*\hat j +\hat k)
but |\vec w|= 2 so a= \frac{2}{|\vec D|} --> a=\sqrt{75}. Finally \vect w= \frac{2}{\sqrt{75}} (7*\hat i -5*\hat j +\hat k)
Thus:
\vect v= \vect w \times \vect C = \frac{2}{\sqrt{75}} (7*\hat i -5*\hat j +\hat k) \times (\hat j - 5\hat k)
is that correct?
Thanks to all...
Homework Statement
Given the following vectors:
\vec A=\hat i +\hat j - 2\hat k and \vec C=\hat j - 5\hat k
Let \vec A and \vec C be drown from a common origin and let \vec C rotate about \vec A with angular velocity \vec w of 2 \frac{rad}{s}. Find the velocity \vec v of the head of \vec C.
Homework Equations
The Attempt at a Solution
My step-by-step way for resolving it, is:
1)I know that \vec v= w \times \vec C
2) By multiplying: \vec A \times \vec C I'll find a vector parallel to \vec w namely D
3) Now, \vec D= \vec A\times \vec C=(\hat i +\hat j - 2\hat k) \times (\hat j - 5\hat k) = 7*\hat i -5*\hat j +\hat k
4) I also know that \vec w is obtained by a linear combination of \vec D's parameter. Then:
\vec w= a * \vec D=a * (7*\hat i -5*\hat j +\hat k)
but |\vec w|= 2 so a= \frac{2}{|\vec D|} --> a=\sqrt{75}. Finally \vect w= \frac{2}{\sqrt{75}} (7*\hat i -5*\hat j +\hat k)
Thus:
\vect v= \vect w \times \vect C = \frac{2}{\sqrt{75}} (7*\hat i -5*\hat j +\hat k) \times (\hat j - 5\hat k)
is that correct?
Thanks to all...
