Question

What is the linear velocity if angular velocity $\vec{\omega} = 3\hat{i} - 4\hat{j} + \hat{k}$ and radius $\vec{r} = (5\hat{i} - 6\hat{j} + 6\hat{k})$ ?

• A. $(-30\hat{i} - 13\hat{j} - 38\hat{k})$
• B. $(8\hat{i} - 10\hat{j} + 7\hat{k})$
• C. $(-18\hat{i} - 13\hat{j} + 2\hat{k})$
• D. $(-2\hat{i} - 2\hat{j} - 5\hat{k})$

Hint:

Always use the order $\vec{\omega} \times \vec{r}$. Reversing the order changes the sign of the result.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The linear velocity $\vec{v}$ of a particle rotating with angular velocity $\vec{\omega}$ at a position vector $\vec{r}$ is given by the cross product of the two vectors.
Step 2: Key Formula or Approach:
The relation is given by:
\[ \vec{v} = \vec{\omega} \times \vec{r} \]
Step 3: Detailed Explanation:
Given:
$\vec{\omega} = 3\hat{i} - 4\hat{j} + \hat{k}$
$\vec{r} = 5\hat{i} - 6\hat{j} + 6\hat{k}$
Using the determinant method for the cross product:
\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -4 & 1
5 & -6 & 6 \end{vmatrix} \]
Expanding along the first row:
\[ \vec{v} = \hat{i} [(-4)(6) - (1)(-6)] - \hat{j} [(3)(6) - (1)(5)] + \hat{k} [(3)(-6) - (-4)(5)] \]
\[ \vec{v} = \hat{i} [-24 + 6] - \hat{j} [18 - 5] + \hat{k} [-18 + 20] \]
\[ \vec{v} = -18\hat{i} - 13\hat{j} + 2\hat{k} \]
Step 4: Final Answer:
The linear velocity is $(-18\hat{i} - 13\hat{j} + 2\hat{k})$.