Question

At a given instant of time the position vector of a particle moving in a circle with a velocity $3\hat i - 4\hat j + 5\hat k$ is $\hat i + 9\hat j - 8\hat k$ . What is its angular velocity at that time ?
A.$\dfrac{{13\hat i - 29\hat j - 31\hat k}}{{\sqrt {146} }}$
B. $\dfrac{{13\hat i - 29\hat j - 31\hat k}}{{146}}$
C. $\dfrac{{13\hat i + 29\hat j - 31\hat k}}{{\sqrt {146} }}$
D. $\dfrac{{13\hat i + 29\hat j + 31\hat k}}{{146}}$

Answer 1

Hint: The relation connecting linear velocity $\overrightarrow v $and angular velocity $\overrightarrow \omega $ is given as \[\overrightarrow \omega = \dfrac{{\overrightarrow R \times \overrightarrow v }}{{\left| R \right|\left| R \right|}}\] where $\overrightarrow R $ is the position vector If $\overrightarrow R = \;\;x\hat i + y\hat j + z\hat k$, then magnitude of vector, \[\left| R \right|\] is given by the equation,
\[\left| R \right| = \sqrt {{x^2} + {y^2} + {z^2}} \]
Let $\overrightarrow R = \;\;x\hat i + y\hat j + z\hat k$ and $\overrightarrow v = {v_x}\hat i + {v_y}\hat j + {v_z}\hat k$. Then cross product of these two vectors is given as
\[
  \overrightarrow R \times \overrightarrow v = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  x&y&z \\
  {{v_x}}&{{v_y}}&{{v_z}}
\end{array}} \right| \\
   = \hat i\left( {\left( {y \times {v_z}} \right) - \left( {{v_y} \times z} \right)} \right) - \hat j\left( {\left( {x \times {v_z}} \right) - \left( {z \times {v_x}} \right)} \right) + \hat k\left( {\left( {x \times {v_y}} \right) - \left( {{v_x} \times y} \right)} \right) \\
 \]

Complete step by step answer:
The relation connecting linear velocity $\overrightarrow v $and angular velocity $\overrightarrow \omega $ is given as
$\overrightarrow v = \overrightarrow \omega \times \overrightarrow R $ (1)., where $\overrightarrow R $ is the position vector
Given
$\overrightarrow R = \;\;\hat i + 9\hat j - 8\hat k$
$\overrightarrow v = 3\hat i - 4\hat j + 5\hat k$
From equation (1) we can find $\overrightarrow \omega $ as,
\[\overrightarrow \omega = \dfrac{{\overrightarrow R \times \overrightarrow v }}{{\left| R \right|\left| R \right|}}\] (2)
Let $\overrightarrow R = \;\;x\hat i + y\hat j + z\hat k$ Then magnitude of vector, \[\left| R \right|\] is given by the equation,
\[\left| R \right| = \sqrt {{x^2} + {y^2} + {z^2}} \]
Therefore, for the given position vector
\[
  \left| R \right| = \sqrt {{x^2} + {y^2} + {z^2}} \\
   = \sqrt {{1^2} + {9^2} + {{\left( { - 8} \right)}^2}} \\
   = \sqrt {146} \\
 \]
Now let us find the cross product.
Let $\overrightarrow R = \;\;x\hat i + y\hat j + z\hat k$ and $\overrightarrow v = {v_x}\hat i + {v_y}\hat j + {v_z}\hat k$. Then cross product of these two vectors is given as
\[
  \overrightarrow R \times \overrightarrow v = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  x&y&z \\
  {{v_x}}&{{v_y}}&{{v_z}}
\end{array}} \right| \\
   = \hat i\left( {\left( {y \times {v_z}} \right) - \left( {{v_y} \times z} \right)} \right) - \hat j\left( {\left( {x \times {v_z}} \right) - \left( {z \times {v_x}} \right)} \right) + \hat k\left( {\left( {x \times {v_y}} \right) - \left( {{v_x} \times y} \right)} \right) \\
 \]
Therefore substituting the given values of position and velocity vector we get,
\[
  \overrightarrow R \times \overrightarrow v = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  1&9&{ - 8} \\
  3&{ - 4}&5
\end{array}} \right| \\
   = \hat i\left( {\left( {9 \times 5} \right) - \left( { - 4 \times - 8} \right)} \right) - \hat j\left( {\left( {1 \times 5} \right) - \left( {3 \times - 8} \right)} \right) + \hat k\left( {\left( {1 \times - 4} \right) - \left( {3 \times 9} \right)} \right) \\
   = 13\hat i - 29\hat j - 31\hat k \\
 \]
On substituting these values in equation (2), we get
\[
  \overrightarrow \omega = \dfrac{{\overrightarrow v \times \overrightarrow R }}{{\left| R \right|\left| R \right|}} \\
   = \dfrac{{13\hat i - 29\hat j - 31\hat k}}{{146}} \\
 \]

Thus the answer is option B.

Note:
The cross product is anticommutative. It means that $a \times b = - \left( {b \times a} \right)$. Therefore, instead of taking \[\overrightarrow R \times \overrightarrow v \] if we take \[\overrightarrow v \times \overrightarrow R \] the answer will be different .hence take care of the order of vectors while taking cross product.