A sphere of radius \(a\), centered at the origin, rotates about the \(z\) axis at angular velocity \(\Omega \mathrm{rad} / \mathrm{s}\). The rotation direction is clockwise when one is looking in the positive \(z\) direction. ( \(a\) ) Using spherical components, write an expression for the velocity field, \(\mathbf{v}\), that gives the tangential velocity at any point within the sphere; \((b)\) convert to rectangular components.

To find the velocity field in spherical coordinates, we have to consider the tangential velocity, which is given by the cross product of the position vector and the angular velocity vector: $$ \mathbf{v} = \mathbf{r} \times \boldsymbol{\Omega} $$ For our problem, the position vector in spherical coordinates is given by: $$ \mathbf{r} = a\sin\theta\ \hat{\boldsymbol{\rho}} + a\cos\theta\ \hat{\boldsymbol{z}} $$ The angular velocity vector is given by: $$ \boldsymbol{\Omega} = \Omega \hat{\boldsymbol{z}} $$ Now we can find the cross product: $$ \mathbf{v} = a\sin\theta\ \hat{\boldsymbol{\rho}} \times \Omega \hat{\boldsymbol{z}} $$ From the properties of cross products: $$ \mathbf{v} = a\Omega\sin\theta\ (\hat{\boldsymbol{\rho}} \times \hat{\boldsymbol{z}}) $$ Since \(\hat{\boldsymbol{\rho}} \times \hat{\boldsymbol{z}} = -\hat{\boldsymbol{\phi}}\), we have: $$ \mathbf{v} = -a\Omega\sin\theta\ \hat{\boldsymbol{\phi}} $$

Now we need to convert the velocity field from spherical to rectangular coordinates. From the transformation formulas, we have: $$ \hat{\boldsymbol{\phi}} = -\sin\phi\ \hat{\boldsymbol{x}} + \cos\phi\ \hat{\boldsymbol{y}} $$ Now, substitute this into the expression for \(\mathbf{v}\): $$ \mathbf{v} = -a\Omega\sin\theta\ (-\sin\phi\ \hat{\boldsymbol{x}} + \cos\phi\ \hat{\boldsymbol{y}}) $$ Which simplifies to: $$ \mathbf{v} = a\Omega\sin\theta\sin\phi\ \hat{\boldsymbol{x}} - a\Omega\sin\theta\cos\phi\ \hat{\boldsymbol{y}} $$ So the rectangular components of the velocity field are: $$ \mathbf{v_x} = a\Omega\sin\theta\sin\phi $$ $$ \mathbf{v_y} = -a\Omega\sin\theta\cos\phi $$ The \(\mathbf{v_z}\) component does not exist because the particle moves in the \(xy\) plane, so we can say that: $$ \mathbf{v_z} = 0 $$