This exercise demonstrates a connection between the curl vector and rotations. Let B be a rigid body rotating about the z-axis. The rotation can be described by the vector w = ωk, where ω is the angular speed of B, that is, the tangential speed of any point P in B divided by the distance d from the axis of rotation. Let r = ⟨x, y, z⟩ be the position vector of P.(a) By considering the angle θ in the figure, show that the velocity field of B is given by v = w × r.(b) Show that v = −ω y i + ω x j.(c) Show that curl v = 2w.

A vector field is a vector field's curl. The tendency of particles at point P to spin about the axis that faces the direction of the curl at P is measured by the curl of a vector field at P. If and only if its curl is zero, a vector field with a simply linked domain is considered conservative.

Let v represent the particle's velocity as a vector.

vector v=ui+vj+wk

curl v =

curl v = [tex]\left[\begin{array}{ccc}i&j&k\\d/dx&d/dy&d/dz\\u&v&w\end{array}\right][/tex]

= (dw/dy - dv/dz )i - (dw/dx - du/dz)j + (dv/dx - du/dy )k

Now by comparing with rotational matrix ,

We know that w = rotational matrix

[tex]w_{x}[/tex] = 1/2 (dw/dy - dv/dz )

[tex]w_{y}[/tex] = 1/2 (dw/dx - du/dz)

[tex]w_{z}[/tex] = 1/2 (dv/dx - du/dy )

w = [tex]w_{x} i +w_{y} j + w_{z} k[/tex]

w = 1/2[ (dw/dy - dv/dz )i + (dw/dx - du/dz)j + (dv/dx - du/dy )k]

w = 1/2 curl v , rotation = 1/2 curl v

let B be a rigid rotating about z axis

w = w.k

tangential speed =wk *di

v = wdj

curl of v= [tex]\left[\begin{array}{ccc}i&j&k\\d/dx&d/dy&d/dz\\0&w_{x} &0\end{array}\right][/tex]

v = wdj ,

The relation between curl vector and rotations is v = wdj

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