# Representing vectors in non-inertial rotating frame of reference

Suppose an object is moving in a circular path about a certain point $$O$$, with an angular velocity $$\omega\hat{k}$$ or linear velocity $$v\hat{\theta}$$. This is basically describing an anticlockwise orbit.

Now consider a frame, that is rotating anticlockwise at an angular velocity $$\omega_0\hat{k}$$. The origin $$O'$$ of this rotating frame coincides with the origin of our inertial frame. Let us try to find the velocity of the object in this new frame :

$$\vec{v_r}=\vec{v_i}-(\omega_0\hat{k} \times \vec{r})$$

This is where my confusion begins. Let $$(\hat{r},\hat{\theta},\hat{k})$$ be the unit vectors in our inertial frame $$S_i$$. Similarly $$(\hat{r'},\hat{\theta'},\hat{k'})$$ are the unit vectors in our rotating frame $$S_r$$. Since the new frame is rotating about the $$z$$ axis of the old frame, we can write $$\hat{k'}=\hat{k}$$

Let $$\vec{r}_i$$ be the position vector in the $$S_i$$ frame basis, and $$\vec{r}_r$$ be the position vector in the $$S_r$$ frame basis. We have :

$$\vec{r}=\vec{r}_i=r\hat{r}=r'\hat{r}'=\vec{r}_r$$

Plugging this into our expression for velocity, we get :

$$\vec{v}_r=\omega r\hat{\theta}-(\omega_0\hat{k}\times r\hat{r})=r\omega\hat{\theta}-r\omega_0\hat{\theta}=r(\omega-\omega_0)\hat{\theta}$$

Hence, we have managed to find an expression for velocity in the $$S_r$$ frame. However, this is still expressed using $$\hat{\theta}$$ which is a unit vector in the $$S_i$$ frame. How do I express this velocity in the rotating frame, in terms of the rotating unit vectors ?

I suppose, in this case, it is rather simple, as the body is moving in a circle in both the frames. However, suppose the body is moving in a random direction in the inertial frame. How do I find the velocity of the object in the rotating frame, and express it using the new unit vectors ? I suppose we'd then have to express both $$\vec{v}_i$$ and $$\vec{r}$$ using the new rotating vectors and use that in the expression. How do I do that? If I have been given the vectors in the inertial frame basis, how do I find their expressions in the rotating frame basis?

When we write $$\vec{F}_{centrifugal}=-m\vec{\omega}\times(\vec{\omega}\times \vec{r})$$, we insert $$\vec{\omega}=\omega\hat{k}$$ and $$\vec{r}=r\hat{r}$$. Then we get the expression :

$$\vec{F}_{centrifugal}=m\omega^2 r\hat{r}$$

In this expression, isn't $$\hat{r}$$ and $$\hat{k}$$, unit vectors of the inertial frame ? If so, how do we express the centrifugal force and other forces using the unit vectors of the rotating frame ? How do I find a relation between $$\hat{r}$$ and $$\hat{r}'$$ ?

Any help understanding this would be very highly appreciated. Thanks.

Consider two frames of reference, O and O*. O is the origin of a fixed point in an inertial frame. O* is the origin of a fixed point in a noninertial frame. In general O* can be accelerating with respect to O.

As a special case, let P be the position of a particle in O that moves in a circle with respect to O. Define O* such that the particle is at rest in O*; that is, O* is a non-inertial frame that rotates with respect to the O frame such that in O* the particle does not move. See the figure below.

In polar coordinates O has the unit vectors, $$\hat {r}$$, $$\hat {\theta}$$, and $$\hat {z}$$. In polar coordinates O* has the unit vectors $$\hat {r^*}$$, $$\hat {\theta^*}$$, and $$\hat {z^*}$$. $$\hat {z}$$ and $$\hat {z^*}$$ are the same (same direction). $$\hat {\theta^*}$$ is not the same as $$\hat {\theta}$$, and $$\hat {r^*}$$ is not the same as $$\hat {r}$$, since in O* we are rotating with the particle; specifically, $$\hat {\theta^*}$$ and $$\hat {r^*}$$ in O* are constant (do not change direction), but $$\hat {\theta}$$ and $$\hat {r}$$ and in O are not constant (change direction).

With respect to O, there is a force $$\vec F_{applied}$$ that causes circular motion, that being the centripetal force supplied, for example, by a string attached from O to the particle. $$\vec F = m\vec a$$ so $$(1)\vec F_{applied} = -mr\omega^2\hat {r}$$ using the expression for acceleration in polar coordinates for this special case of uniform circular motion.

With respect to O* the particle is at rest so $$\vec a^*$$, the acceleration of the particle in the O* frame, is zero, $$\vec F^* = m\vec a^*= 0$$. Therefore, in addition to $$\vec F_{applied}$$, in O* there is a fictious force $$\vec F_{cent}$$, called the centrifugal force, such that $$\vec F_{applied} + \vec F_{cent}= m\vec a^*= 0$$. The centrifugal force here is $$+mr\omega^2\hat {r}$$. Therefore $$(2)\vec F_{applied} + mr\omega^2\hat {r} = 0$$

Results (1) and (2) are the same, but are based on different frames of reference. (1) is using an inertial frame and there is a net external force causing acceleration in that frame. (2) is using a non-inertial frame where in that frame there is no acceleration; in this frame there is no net force so the applied force must be countered by the fictitious centrifugal force.

Both the inertial and non-inertial frames have a set of (typically) orthogonal unit vectors and in general all three unit vectors are different (have different directions) in the two frames. See the figure below. In general, O* can be accelerating with respect to O and the unit vectors in O* can be rotating with respect to O. In the general case, other fictitious forces appear in the non-inertial frame in addition to the centrifugal force, such as the Coriolis force.

The unit vectors in O* depend on how you define the motion of O*. For example, consider the point P in general motion in O. You can define O* as fixed at P and moving with P but with unit vectors in O* not rotating with respect to the unit vectors in O. If you have a rigid body whose center of mass is accelerating with respect to O and that is rotating with respect to O, you can define a set of body axes, fixed in O*, with O* at the center of mass, of the moving body, and evaluate the rotational motion about O* coupled with the translational motion of the CM at O*. 