Ans.
Let us consider the position vector of the particle is given by,
\( \vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \)
and the angular velocity of that particle is given by,
\( \vec{\omega}={\omega}_x\hat{i}+{\omega}_y\hat{j}+{\omega}_z\hat{k} \)
Therefore, \( curl\vec{V}=\vec{\nabla}\times\vec{V} \)
\( =\vec{\nabla}\times(\vec{\omega}\times\vec{r}) \)
\( =\vec{\omega}(\vec{\nabla}\cdot\vec{r})-\vec{r}(\vec{\nabla}\cdot\vec{\omega}) \)
now
\( \vec{\nabla}\cdot\vec{r}=(\frac{\partial}{\partial{x}}\hat{i}+\frac{\partial}{\partial{y}}\hat{j}+\frac{\partial}{\partial{z}}\hat{k})\cdot(x\hat{i}+y\hat{j}+z\hat{k})\\=\frac{\partial{x}}{\partial{x}}+\frac{\partial{y}}{\partial{y}}+\frac{\partial{z}}{\partial{z}}\\=3 \)
and,
\( \vec{\nabla}\cdot\vec{\omega}=(\frac{\partial}{\partial{x}}\hat{i}+\frac{\partial}{\partial{y}}\hat{j}+\frac{\partial}{\partial{z}}\hat{k})\cdot({\omega}_x\hat{i}+{\omega}_y\hat{j}+{\omega}_z\hat{k})\\=\frac{\partial}{\partial{x}}{\omega}_x+\frac{\partial}{\partial{y}}{\omega}_y+\frac{\partial}{\partial{z}}{\omega}_z \)
Therefore,
\( curl\vec{V}=\vec{\omega}\cdot{3}-\vec{r}(\frac{\partial}{\partial{x}}{\omega}_x+\frac{\partial}{\partial{y}}{\omega}_y+\frac{\partial}{\partial{z}}{\omega}_z) \)
\( =\vec{\omega}\cdot{3}-(\frac{\partial}{\partial{x}}{\omega}_x+\frac{\partial}{\partial{y}}{\omega}_y+\frac{\partial}{\partial{z}}{\omega}_z)(x\hat{i}+y\hat{j}+z\hat{k}) \)
\( =3\vec{\omega}-(\hat{i}{\omega}_x\frac{\partial}{\partial{x}}x+\hat{j}{\omega}_y\frac{\partial}{\partial{y}}y+\hat{k}{\omega}_z\frac{\partial}{\partial{z}}z) \)
\( =3\vec{\omega}-{\omega}_x\hat{i}+{\omega}_y\hat{j}+{\omega}_z\hat{k} \)
\( =3\vec{\omega}-\vec{\omega} \)
\( =2\vec{\omega} \)
Again we know that the angular momentum \( \vec{L}=\vec{r}\times\vec{p} \)
or, \( \vec{L}=\vec{r}\times(m\vec{v}) \)
or, \( \vec{L}=m(\vec{r}\times\vec{v}) \)
or, \( \vec{L}=m(\vec{r}\times\vec{\omega}\times\vec{r}) \)
or, \( \vec{L}=m[\vec{\omega}(\vec{r}\cdot\vec{r})-\vec{r}(\vec{r}\cdot\vec{\omega})] \)
or, \( \vec{L}=m[\vec{\omega}r^2-\vec{r}\cdot{0}] \)
or, \( \vec{L}=mr^2\vec{\omega} \)
or, \( \vec{L}=constant \), since \( \vec{\omega} \) is constant.