## Fictitious forces in rotating systems

### Uniform circular motion

Consider a particle of mass $m$ moving uniformly around a circle of radius $r$. The time for one rotation $T$ is called the period. The angle traversed per unit time is the angular velocity $\omega$: $$ \omega = \frac{2 \pi}{T} \tag {1} $$ The number of revulutions per time unit is the frequence $f$: $$ f = \frac{1}{T} \tag {2} $$ The angle $\theta$ swept out in time $t$ is : $$ \theta(t) = \omega t \tag {3} $$ The position of the particle at time $t$ is $$ \mathbf r(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} r \cos \theta \\ r \sin \theta \end{bmatrix} = \begin{bmatrix} r \cos \omega t \\ r \sin \omega t \end{bmatrix} \tag {4} $$ The velocity is: $$ \mathbf v(t) = \mathbf {\dot r}(t) = \begin{bmatrix} \dot x(t) \\ \dot y(t) \end{bmatrix} = \begin{bmatrix} - \omega r \sin \omega t \\ \omega r \cos \omega t \end{bmatrix} \tag {5} $$ The acceleration is: $$ \mathbf a(t) = \mathbf {\ddot r}(t) = \begin{bmatrix} \ddot x(t) \\ \ddot y(t) \end{bmatrix} = \begin{bmatrix} - \omega^2 r \cos \omega t \\ - \omega^2 r \sin \omega t \end{bmatrix} \tag {6} $$ The velocity is othogonal to the position vector because the scalar product of these two vectors is zero: $$ \mathbf r(t) \cdot \mathbf v(t) = 0 \tag {7} $$ So, the velocity is tangential to the trajectory circle. Its magnitude is called the "speed" $v$. It is constant: $$ v = | \mathbf v(t) | = \omega r. \tag {8} $$ The acceleration is obviously parallel to the position vector: $$ \mathbf a(t) = - \omega^2 \mathbf r(t) \tag {9} $$ and is directed towards the center of the circle. Its magnitude is constant: $$ a = | \mathbf a(t) | = \omega^2 r. \tag {10} $$

### Centripetal and centrifugal forces

The force causing the acceleration is the centripetal force : $$ \mathbf F(t) = m \mathbf a(t) = - m \omega^2 \mathbf r(t) \tag {11} $$ It is directed towards the center of the circle. Its magnitude is constant: $$ F = ma = m \omega^2 r = m \frac{v^2}{r} \tag {12} $$ Some typical causes of the centripetal force are:

- Gravitational force acting on a orbiting satellite.
- Pulling force of a string to which the rotating body is attached, e.g. on a carussel or merry-go-round.
- Pushing force of the wall of a centrifuge.

In our everyday life we observe and experience centrifugal forces: forces that act on the rotating body and are directed from the rotation center outwards. In the literature it is often claimed that there are no centrifugal forces, because only the centripetal force is responsible for the circular motion. We should not confuse the centrifugal forces with the reaction force of the body to the centripetal force. The body $m$ exerts a reactional pulling force on the string or a pushing force on the wall of a centrifuge. But these are forces acting not the body $m$ but on other bodies. In our understanding the centrifugal force acts on the body $m$ and tries to move it outwards from the center.

### Example: Chain carousel

The external observer, Alice A, sees Bob B rotating about the central pole of the chain carousel.

She sees two forces acting on B: the gravitation $G$ and the tension $T$. The resulting force is the centripetal force $F = G + T$ directing to the center and causing the rotation of Bob. The chain that is declined by the angle $α$ from the vertical: $$ \tan \alpha = \frac{F}{G} = \frac{m \omega^2 r}{m g} = \frac{\omega^2 r}{g} \tag {13} $$

Bob B sitting on the chair and concentrating on the carousel does not see any motion of his chair.

The chair is at rest with respect to the carousel. Bob is aware of his weight $G$, the tension $T$ and angle $θ$ with respect to the central pole. Because he is at rest there must be a third force $H$ that compensate $G + T$. He feels this centrifugal force $H = - (G + T) = - F$ that pushes him outwards. He sees Alice rotating around him.

The situation is similar to the situation of the accelerating car and can be explained by considering the respective frames of reference.

### Rotational frame of reference

The mathematical treatment is facilitated by embedding the rotational plane in the 3D space. We define the rotation vector as $$ \pmb \omega = (0, 0, \omega) \tag {14} $$ This vector is orthogonal to the rotational plane and its length is the angular speed. The velocity $v$ in eq. (5) can be written as the vector product $$ \mathbf{ \dot r} = \pmb \omega \times \mathbf r. \tag {15} $$

We will denote the stationary (invariant) system by $S$ and the rotating system by $S'$. The systems are given by their respective unit vectors $\mathbf e_i, \mathbf e'_i$.

We will also use the summation convention that we sum implicitly over identical indices, e.g. the position of the particle $m$ in the system S is $$ \mathbf r = \sum_{i=1}^3 r_i \mathbf e_i \equiv r_i \mathbf e_i \tag {16} $$ The same vector expressed in system $S'$ is $$ \mathbf r = r'_i \mathbf e'_i \tag {17} $$ The velocity in the system $S$ is: $$ \mathbf {\dot r} = \dot r_i \mathbf e_i \tag {18} $$ The velocity in the system $S'$ is obtained by differentiating eg. (17): $$\begin{align} \mathbf{\dot r} & = \dot r'_i \mathbf e'_i + r'_i \mathbf{\dot e}'_i \\ & = \dot r'_i \mathbf e'_i + r'_i \pmb \omega \times \mathbf e'_i \\ & = \dot r'_i \mathbf e'_i + \pmb \omega \times \mathbf r \tag {19} \end{align}$$ We will denote the velocity of the particle as seen by an observer in frame $S$ by $$ \left( \frac{d \mathbf r}{dt} \right) _{S} = \dot r_i \mathbf e_i \tag {20} $$ Similarly, the velocity as seen by an observer in frame $S'$ is just $$ \left( \frac{d \mathbf r}{dt} \right) _{S'} = \mathbf v' = \dot r'_i \mathbf e'_i \tag {21} $$ Hence, the two observers measure different velocities $$ \left( \frac{d \mathbf r}{dt} \right) _{S} = \left( \frac{d \mathbf r}{dt} \right) _{S'} + \pmb \omega \times \mathbf r \tag {22} $$ The accelerations also obviously differ in the two systems. In $S$ we have $$ \mathbf {\ddot r} = \ddot r_i \mathbf e_i \tag {23} $$ To get the acceleration in $S'$ we differentiate (19): $$\begin{align} \mathbf{\ddot r} & = \ddot r'_i \mathbf e'_i + 2 r'_i \pmb \omega \times \mathbf e'_i + r'_i \pmb \omega \times (\pmb \omega \times \mathbf e'_i) \tag {24} \end{align}$$ The acceleration seen by the observer in $S$ is $\ddot r'_i \mathbf e'_i$ while the acceleration seen by the observer in $S'$ is $\ddot r'_i \mathbf e'_i$. Equating the two equations (23) and (24) above gives: $$ \left( \frac{d^2 \mathbf r}{dt^2} \right) _{S} = \left( \frac{d^2 \mathbf r}{dt^2} \right) _{S'} + 2 \pmb \omega \times \left( \frac{d \mathbf r}{dt} \right) _{S'} + \pmb \omega \times (\pmb \omega \times \mathbf r) \tag {25} $$ or using primed symbols for accelaration and velocity in $S'$: $$ \mathbf a = \mathbf a' + 2 \pmb \omega \times \mathbf v' + \pmb \omega \times (\pmb \omega \times \mathbf r) \tag {26} $$ This equation relates the acceleration $\mathbf a$ in the invariant system $S$ to the acceleration $\mathbf a'$ in the rotating system $S'$.

### Fictitious forces

We multiply the eq. (25) by the mass $m$,
note that
$$
m \mathbf a = \mathbf F
$$
and rearrange to get
$$
m \mathbf a' =
\mathbf F
- 2 m \pmb \omega \times \mathbf v'
- m \pmb \omega \times (\pmb \omega \times \mathbf r) \tag {27}
$$
An observer in $S'$ explains the motion in his system by adding so called
*fictitios forces* to the *real force* $\mathbf F$,
namely the Coriolis force
$$
\mathbf F_{cor} = - 2 m \pmb \omega \times \mathbf v' \tag {28}
$$
and the centrifugal force
$$
\mathbf F_{cfg} = - m \pmb \omega \times (\pmb \omega \times \mathbf r) . \tag {29}
$$
In the above derivation we assumed that the rotation vector $\omega$
is constant. If this vector varied with time, a third fictitios force,
the *Euler force* would be added to the right hand of eq. (27):
$$
\mathbf F_{eul} = - m \dot {\pmb \omega} \times \mathbf r \tag {30}
$$
Let us note that the 'fictitious forces' are in fact 'real forces'
in the rotating frame and on equal terms as the force $\mathbf F$.

The Coriolis force acts on the moving particle orthogonal to its velocity and bends the particle's path to the right. The value of the force is: $$ F_{cor} = 2 m \omega v' \tag {31} $$

The centrifugal force acts on the particle in the direction of its position vector outwards of the rotation axis. Its value is: $$ F_{cfg} = m \omega^2 r \tag {32} $$

The equations of motion of a particle $\mathbf r' (t) = (x'(t), y'(t), 0)$ in a rotating 2D-system ($z = 0$) are $$ m \begin{bmatrix} \ddot x' \\\ddot y' \end{bmatrix} = \mathbf F_{inertial} - 2 m \omega \begin{bmatrix} - \dot y' \\ \dot x' \end{bmatrix} + m \omega^2 \begin{bmatrix} x' \\ y' \end{bmatrix} \tag {33} $$ The first correction force term is the Coriolis force, the second term is the centrifugal force.

### Links

- Circular motion (Wikipedia)
- Central forces (D. Tong, University of Cambridge)
- Non-Inertial Frames (David Tong, University of Cambridge)
- Non-inertial systems - Pseudo Force (Saurabh Basu, Indian Institute of Technology Guwahati)
- Motion in accelerated reference frames (Hanno Essén, Royal Institute of Technology, Sweden)
- Motion in a Noninertial Reference Frame (Martin Houde)
- Does centrifugal force exist? (StackExchange)
- Non Inertial Systems (Saurabh Basu)
- Rotating reference frame (Wikipedia)
- Fictitious force (Wikipedia)

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