# Rigid circular disc equation of motion

I am working on an assignment and I have a doubt in a step in my problem. Could someone please help me understand it?

Question and its solution is:

. My doubt is in the part which is underlined. How did the 2nd term become zero? Also. how did they calculate external moment vector. If someone can explain these steps it would be very helpful.

P.S: If this is not the correct forum to ask such question, I apologize.

• Just work out the vector and tensor products, using the fact that only one component of $\overrightarrow\omega$ is non-zero. Actually you should be able to see this without working anything out in full, because the vectors $\overrightarrow\omega$ and $\mathbf{J}\overrightarrow\omega$ are parallel. – alephzero Aug 4 '18 at 13:29
• Thanks. I was able to figure that one out. Stupid of me to not see it directly. But I still don't get the external moment vector. – Preeti Srivastava Aug 4 '18 at 14:37

How did the second term become zero?

Recall that the cross product of vectors $\vec{a}$ and $\vec{b}$ is a vector representing the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$ in the direction perpendicular on the parallelogram.

\begin{align} \vec{\omega}(t) \times \mathbf{J}\vec\omega(t) &= \begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix} \times \begin{pmatrix} J_x & 0 & 0 \\ 0 & J_y & 0 \\ 0 & 0 & J_z \end{pmatrix} \begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix} \\ &= \underbrace{\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix}}_{\vec{a}} \times \underbrace{\begin{pmatrix} 0 \\ J_y\omega_y \\ 0 \end{pmatrix}}_{\vec{b}} \\ &= \vec{0} \end{align}

Note that within this case the vectors $\vec{a}$ and $\vec{b}$ are parallel, hence the area of the parallelogram they span is $0$ and therefore the cross product is also zero.

How did they calculate the external moment vector?

There is only one external force present in this system, the bearing force $$\vec{F} = \begin{pmatrix} A_x \\ 0 \\ A_z \end{pmatrix}.$$ The moment this forces causes on the center of gravity is \begin{align} \vec{M}_C &= \vec{r}_C \times \vec{F}\\ &= \begin{pmatrix} r_x \\ r_y \\ r_z \end{pmatrix} \times \begin{pmatrix} A_x \\ 0 \\ A_z \end{pmatrix}\\ &= \begin{pmatrix} A_z r_y - 0 r_z \\ A_x r_z - A_z r_x \\ 0 r_x - A_x r_y \end{pmatrix} \end{align}

Assuming that $r_y = 0$ result in $$\vec{M}_C = \begin{pmatrix} 0 \\ A_x r_z - A_z r_x \\ 0 \end{pmatrix}$$

Note that the rotation $\varphi_y$ is clockwise and not counter clockwise as it would have been by using a right handed coordinate system. Therefore the external moment should be defined in the opposite direction $$\vec{\mathbf{I}} = -\vec{M}_C = \begin{pmatrix} 0 \\ -A_x r_z + A_z r_x \\ 0 \end{pmatrix}.$$