Question

This exercise demonstrates a connection between the curl vector and rotations. Let be a rigid body rotating about the z-axis. The rotation can be described by the vector\(w = \omega k,\)where is the angular speed of, 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 in the figure, show that the

velocity field of is given by \(v = w \times r\) .

(b) Show that \(v = - \omega yi + \omega xj\) .

(c) Show that\(v = 2w\).

Short answer

It is proved that,

(a) \(v = w \times r\)

(b)\(v = - \omega yi + \omega xj\)

(c)\(v = 2w\)

Step-by-step solution

Given information

The given equations to prove are:

\(v = w \times r\).

\(v = - \omega yi + \omega xj\) .

\(v = 2w\).

(a)Step 2: To prove the two vectors are equal

We have to prove

\(v = w \times r\)

To prove that two vectors are equal, it is sufficient and necessary to prove that both their magnitudes and directions are equal.

Proving magnitude of LHS and RHS are equal

Magnitude of v is speed of the point P

Therefore, magnitude of \(v = \omega d\)

Whereas the magnitude of RHS is

\(|w \times r| = |w||r|sin\theta \)

Note that

\(|w| = |\omega k| = \omega \)

And that

\(|r|sin\theta = d\;\)

Therefore,

\(\begin{array}{c}\left| {w \times r} \right| = \left| w \right|\left| r \right|\sin \theta \\ = \omega d\end{array}\)

Hence proved that the magnitudes of LHS and RHS are equal.

The velocity must be perpendicular to both r and w and should be given by right and rule, since it is the cross product of the two vectors.

Which is the case, as can be seen from the figure.

Alternate solution

The position vectors of the point P is

\({\bf{r}} = dcos\omega ti + dsin\omega tj + ck\)

Where c is a positive real number.

Whereas, it is given that \(w = \omega k\)

Therefore,

\(\begin{array}{l}w \times r = (\omega k) \times (dcos\omega ti + dsin\omega tj + ck)\\w \times r = \omega dcos\omega t(k \times i) + \omega dsin\omega t(k \times j) + \omega c(k \times k)\\w \times r = \omega dcos\omega t(j) + \omega dsin\omega t( - i) + \omega c(0)\\w \times r = - \omega dsin\omega ti + \omega dcos\omega tj\end{array}\)

Whereas we can also write

\(v = \frac{{dr}}{{dt}} = \frac{{d(dcos\omega ti + dsin\omega tj + ck)}}{{dt}} = - \omega dsin\omega ti + \omega dcos\omega tj\)

Hence proved.

(b)Step 5: recall from part (a) to prove part (b)

\(r = xi + yj + zk\)

We want to show that \(v = - \omega yi + \omega xj\)

\(v = w \times r\)

So,

\(\begin{array}{c}v = \left| {\begin{array}{*{20}{l}}i&j&k\\0&0&\omega \\x&y&z\end{array}} \right|\\ = i(0 - \omega y) - j(0 - \omega x) + k(0 - 0)\\ = - \omega yi + \omega xj\end{array}\)

Thus we showed that \(v = - \omega yi + \omega xj\).

(c)Step 6: The vector representing the point's position

From part (a), we know that

\(v = w \times r\)

Let the position vectors of the point P

\(r = xi + yj + zk\)

Whereas it is given that \(w = \omega k\)

Therefore,

\(\begin{array}{l}w \times r = (\omega k) \times (xi + yj + zk)\\w \times r = x\omega (k \times i) + y\omega (k \times j) + z\omega (k \times k)\\w \times r = x\omega (j) + y\omega ( - i) + \omega z(0)\end{array}\)

\(\begin{array}{c}v = w \times r\\ = - y\omega i + x\omega j\end{array}\)

find curl

Remember that

Curl \(F = \left( {\frac{{\partial R}}{{\partial y}} - \frac{{\partial Q}}{{\partial z}}} \right)i + \left( {\frac{{\partial P}}{{\partial z}} - \frac{{\partial R}}{{\partial x}}} \right)j + \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)k\)

Where \(F = Pi + Qj + Rk\)

Therefore,

\(v = \left( {\frac{{\partial (0)}}{{\partial y}} - \frac{{\partial (x\omega )}}{{\partial z}}} \right)i + \left( {\frac{{\partial ( - y\omega )}}{{\partial z}} - \frac{{\partial (0)}}{{\partial x}}} \right)j + \left( {\frac{{\partial (x\omega )}}{{\partial x}} - \frac{{\partial ( - y\omega )}}{{\partial y}}} \right)k\)

\(\begin{array}{c}curlv = (0 - 0)i + (0 - 0)j + (\omega - ( - \omega ))k\\ = 2\omega k\\ = 2w\end{array}\)

Hence proved.