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

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.