enter image description hereIt’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.


2 Answers 2


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.

  • $\begingroup$ Could you give an intuitive approach and shed some light on the differences of the two frames? $\endgroup$ Dec 6, 2022 at 17:38
  • $\begingroup$ @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. $\endgroup$
    – Dale
    Dec 6, 2022 at 17:52
  • $\begingroup$ I have added a picture to further clarify the question $\endgroup$ Dec 6, 2022 at 18:35
  • $\begingroup$ @TheCuriosOne oh, I see. You are interested in an orbiting frame instead of a spinning frame $\endgroup$
    – Dale
    Dec 6, 2022 at 18:53
  • $\begingroup$ @TheCuriosOne I have updated it to describe both $\endgroup$
    – Dale
    Dec 6, 2022 at 19:15

enter image description here

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 $~X(t),Y(t),Z(t)$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.