Finding the Speed of an Object in Circular Motion

A rigid body rotates with a constant angular speed of \(\omega \) radians per second about a line through the origin in the direction of the vector \(2 \mathbf{i}+\mathbf{j}+2\mathbf{k}\). Find the speed of an object on this body at the instant the object passes through the point \((1,3,5)\). Assume the distance is in meters.

Solution The vector \( {\omega }\) is parallel to the axis of rotation of the rigid body, and so it is in the direction of the vector \(2 \mathbf{i}+\mathbf{j}+2\mathbf{k}\). Since \(\dfrac{2\mathbf{i}+\mathbf{j}+2 \mathbf{k}}{ \sqrt{2^{2}+1^{2}+2^{2}}}=\dfrac{2}{3}\mathbf{i}+\dfrac{1}{3} \mathbf{j}+\dfrac{2}{3}\mathbf{k}\) is a unit vector in the direction of \(2 \mathbf{i}+\mathbf{j}+2\mathbf{k}\), the angular velocity \( {\omega }\) is \[ \begin{equation*} {\omega }=\omega \left( \frac{2}{3}\,\mathbf{i}+\frac{1}{3}\,\mathbf{j }+\frac{2}{3}\,\mathbf{k}\right) \end{equation*} \]

The position of the object is \(\mathbf{r}=\mathbf{i}+3\mathbf{j}+5\mathbf{k}\). So, the velocity \(\mathbf{v}\) of the object is \[ \begin{eqnarray*} \mathbf{v}&=& {\omega }\times \mathbf{r}=\left[ \omega \left( \frac{2}{ 3}\,\mathbf{i}+\frac{1}{3}\,\mathbf{j}+\frac{2}{3}\,\mathbf{k}\right) \right] \times \left[ \mathbf{i}+3\mathbf{j}+5\mathbf{k}\right] =\,\omega \left\vert \begin{array}{l@{\quad}l@{\quad}l} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{2}{3} & \dfrac{1}{3} & \dfrac{2}{3} \\ 1 & 3 & 5 \end{array} \right\vert\\[5pt] &=&\omega \left( -\dfrac{1}{3}\mathbf{i}-\dfrac{8}{3}\mathbf{j}+ \dfrac{5}{3}\mathbf{k}\right) \end{eqnarray*} \]

The speed \(v\) of the object is \(v=\left\Vert \mathbf{v}\right\Vert =\omega \sqrt{\dfrac{1}{9}+\dfrac{64}{9}+\dfrac{25}{9}}= \sqrt{10}\,\omega\) meters per second.