# Rigid body motion and perpendicular directions

While studying Rotation of Rigid Bodies, I came across the following situation:

Consider a rigid body in pure rotational motion about a fixed axis (for example the z-axis). For any particle in the object, its linear velocity is given by $$v=r \omega=R \sin \theta \omega$$ where $$\mathbf{R}$$ is the position vector of the particle from the origin (see Fig. 7.9) and $$\theta$$ is the angle between the position vector and the $$z$$-axis. As shown in Fig. 7.9, the direction of $$y$$ is perpendicular to the plane formed by $$\boldsymbol{\omega}$$ and $$\mathbf{R}$$ where it can be verified using the right-hand rule.

Problem:

Why is the direction of $$y$$ perpendicular to the plane formed by $$\boldsymbol{\omega}$$ and $$\mathbf{R}$$? It is stated that this relation can be obtained by using the right hand rule. However, I am having trouble using the right hand rule in this scenario. Furthermore, I just cannot properly visualize the plane formed by $$\boldsymbol{\omega}$$ and $$\mathbf{R}$$. I would appreciate if someone could further elaborate an explanation regarding the above statements.

• I don’t think that the direction of 𝑦 is perpendicular to the plane formed by $\omega$ and R
– Eli
Commented Jul 5, 2021 at 12:28

Look at this figure. The plane formed by $$\boldsymbol{\omega}$$ and $$\mathbf{R}$$ is perpendicular to y’ axis, thus $$\vec e_{y’}\cdot \vec e_R=0$$.
• There is an easy way to identify the plane containing $\omega$ and $R$. In the OP's sketch, that plane contains a double headed arrow labelled $\Theta$ and the line segment labelled $r = Rsin\Theta$. Commented Jul 5, 2021 at 16:47
• I still cannot see it... Is $y^{\prime}$ equal to the $X$ axis? Commented Jul 5, 2021 at 18:49