Question

If \[ \overline{V}=x^{2}y\,\overline{i}-yz^{2}\,\overline{j}+xyz\,\overline{k} \] is the linear velocity of a particle in circular motion at the point \((1,1,1)\), then its angular velocity is

• A. \(-\frac{1}{2}(\overline{i}+\overline{j}+\overline{k})\)
• B. \(0\)
• C. \(\frac{1}{2}(3\overline{i}-\overline{j}-\overline{k})\)
• D. \(\frac{1}{2}(\overline{i}+\overline{j}+\overline{k})\)

Hint:

Whenever a velocity field is given and angular velocity is required, remember the standard relation \[ \vec\omega=\frac12(\nabla\times\vec V). \] First compute the curl and then divide by \(2\).

Answer 1

The Correct Option is D

Concept: For a velocity field \[ \vec V=P\hat i+Q\hat j+R\hat k, \] the angular velocity vector is related to the curl of the velocity field by \[ \vec\omega=\frac{1}{2}\left(\nabla\times\vec V\right). \] This result comes from fluid kinematics and rotational motion. Therefore, to determine the angular velocity, we first compute the curl of the given velocity field and then divide it by \(2\).

Step 1: Identify the components of the velocity field.
Given, \[ \vec V=x^2y\,\hat i-yz^2\,\hat j+xyz\,\hat k. \] Hence, \[ P=x^2y,\qquad Q=-yz^2,\qquad R=xyz. \]

Step 2: Compute the curl of the velocity field.
\[ \nabla\times\vec V= \begin{vmatrix} \hat i & \hat j & \hat k \\ \dfrac{\partial}{\partial x} & \\ \dfrac{\partial}{\partial y} & \\ \dfrac{\partial}{\partial z} \\ x^2y & -yz^2 & xyz \end{vmatrix}. \] Expanding, \[ \nabla\times\vec V = \left(\frac{\partial R}{\partial y} -\frac{\partial Q}{\partial z}\right)\hat i - \left(\frac{\partial R}{\partial x} -\frac{\partial P}{\partial z}\right)\hat j + \left(\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y}\right)\hat k. \] Now, \[ \frac{\partial R}{\partial y}=xz, \qquad \frac{\partial Q}{\partial z}=-2yz, \] therefore \[ \hat i\text{-component}=xz+2yz. \] Also, \[ \frac{\partial R}{\partial x}=yz, \qquad \frac{\partial P}{\partial z}=0, \] thus \[ \hat j\text{-component}=-yz. \] Further, \[ \frac{\partial Q}{\partial x}=0, \qquad \frac{\partial P}{\partial y}=x^2, \] hence \[ \hat k\text{-component}=-x^2. \] Thus, \[ \nabla\times\vec V = (xz+2yz)\hat i-yz\hat j-x^2\hat k. \]

Step 3: Evaluate at the point \((1,1,1)\).
Substituting \(x=y=z=1\), \[ \nabla\times\vec V = 3\hat i-\hat j-\hat k. \]

Step 4: Determine the angular velocity.
\[ \vec\omega = \frac12(\nabla\times\vec V). \] Therefore, \[ \boxed{ \vec\omega= \frac12(3\hat i-\hat j-\hat k) } \]