# Reference frames in a rotating body and reference frame of a rotating body

It’s obvious that a point on a rotating body executes circular motion with angular velocity omega.If I consider an object rotating and choose to be in a reference frame of a particle present in it. How would I react to a path of a stationary object(say, block) kept next to me. How would I consider the motion of the block if I(the observer) was rotating instead?

My Approach —If we can consider rotation of an object about an axis as a complicated form of circular motion itself we can consider the block to be executing a circular motion around the observer.

Note-the object or observer is not executing CTRM.

Your Case 1 is what is usually called a rotating reference frame. Your Case 2 is unusual, but I will call it an orbiting reference frame. In both cases I will denote the inertial frame by coordinates $$(x,y,z)$$

## Case 1: rotating reference frame

I will denote the rotating frame by coordinates $$(X,Y,Z)$$. And let's suppose that the axes are initially aligned at $$t=0$$ and that the axis of rotation is the $$z$$ axis. Then, we have the standard rotating reference frame transformation that can be looked up in any number of references $$X= x \cos(\omega t) - y \sin(\omega t)$$$$Y = x \sin(\omega t) + y \cos(\omega t)$$$$Z=z$$ Now, once you have this transformation it is just a matter of a little algebra to obtain the reverse transform $$x=X \cos(\omega t) + Y \sin(\omega t)$$$$y=-X \sin(\omega t)+Y \cos(\omega t)$$$$z=Z$$

So if a block is at rest at $$(x,y,z)=(x_b,y_b,z_b)$$ then we can easily obtain$$X_b(t)= x_b \cos(\omega t) - y_b \sin(\omega t)$$$$Y_b(t) = x_b \sin(\omega t) + y_b \cos(\omega t)$$$$Z_b(t)=z_b$$which describes the motion in the observer's rotating frame. In this frame the block rotates around the $$Z$$ axis in a circle with a radius equal to the block's distance from the $$Z$$ axis and it spins on its own axis as it rotates so as to always keep the same face pointed towards the $$Z$$ axis.

## Case 2: Orbiting reference frame

I will now denote the orbiting reference frame by coordinates $$(X,Y,Z)$$. In this case, the axes will always be aligned, but they will be shifted by a time-varying amount. The amount that it is shifted is equal to the position of the observer, and since the position of the observer is given by $$(x,y)=(R \cos(\theta),R \sin(\theta))$$ we have: $$X=x+ R \cos(\omega t)$$$$Y=y+ R \sin(\omega t)$$$$Z=z$$ where $$R$$ is the distance of the observer from the fixed axis of rotation (the $$z$$ axis). As before, a little algebra gets us the inverse transform $$x=X- R \cos(\omega t)$$$$y=Z- R \sin(\omega t)$$$$z=Z$$

So if a block is at rest at $$(x,y,z)=(x_b,y_b,z_b)$$ then we can easily obtain$$X_b(t)= x_b + R \cos(\omega t)$$$$Y_b(t) = y_b + R \sin(\omega t)$$$$Z_b(t)=z_b$$which describes the motion in the observer's orbiting frame. In this frame the block orbits around an axis that is offset from the $$Z$$ axis. It orbits with a radius equal to the observer's distance from the $$z$$ axis and it does not spin on its own axis as it rotates so different faces point towards the orbital axis at different parts of the cycle.

• Could you give an intuitive approach and shed some light on the differences of the two frames? Dec 6, 2022 at 17:38
• @TheCuriosOne "shed some light" is way too vague. I will need far more guidance on what you want. Please edit the question to clarify what you want.
– Dale
Dec 6, 2022 at 17:52
• I have added a picture to further clarify the question Dec 6, 2022 at 18:35
• @TheCuriosOne oh, I see. You are interested in an orbiting frame instead of a spinning frame
– Dale
Dec 6, 2022 at 18:53
• @TheCuriosOne I have updated it to describe both
– Dale
Dec 6, 2022 at 19:15

you have two coordinate system body fixed (B-system) and inertial system (I-system)

assume the body is rotating about the arbitrary vector ($$~\vec n~$$), with the angular velocity $$~\omega~$$. first we put the body z axes toward the rotation axes $$~\vec n~$$ , the x and y axis are perpendicular to the z axes.

we put the inertial system parallel to the body system.

you are at the body fixed point U, thus your coordinates in inertial system are

$$\vec R_{IU}=\begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}=\vec R_{IB}+ \mathbf S_z\,\vec R_{BU}\tag 1$$

where $$~\mathbf S_z~$$ is the rotation matrix between the B-system and I-system

$$\mathbf S_z= \left[ \begin {array}{ccc} \cos \left( \omega\,t \right) &-\sin \left( \omega\,t \right) &0\\ \sin \left( \omega\,t \right) &\cos \left( \omega\,t \right) &0\\ 0&0&1 \end {array} \right]$$

and $$\vec R_{BU}= \begin{bmatrix} u_x \\ u_y \\ u_z \\ \end{bmatrix}_B\quad, \vec R_{IB}=\begin{bmatrix} 0 \\ 0 \\ Z \\ \end{bmatrix}_I$$

both vectors are constant .

so what you see is the the path of $$~X(t),Y(t),Z(t)$$

• Sorry, but I do not comprehend on what a rotation matrix is. Dec 7, 2022 at 1:47
• the rotation matrix rotate the body coordinate system, or transformed the vector components that given in body system to inertial system
– Eli
Dec 7, 2022 at 8:14