## Section11.2Rotating Frame

Sun appears to go around Earth making a round trip in 24 hours even though Sun is more or less fixed in position. This is a consequence of observing Sun from a rotating frame.

To relate kinematic quantities of motion of a particle in a rotating frame to those of a fixed frame, we will work on describing the motion of a point particle in a rotating frame and in a fixed frame. To distinguish variables in rotating frame, let us denote kinematic variables in rotating frame with a prime, e.g., $\vec r^{\prime}\text{,}$ $\vec v^{\prime}\text{,}$ and $\vec a^{\prime}$ be the position, velocity, and acceleration of the particle as observed in the rotating frame.

For simplicity we now first look at a particle at rest in the rotating frame. For more complex situation, we will need to work in vector notation, as explained below. Finally, we will work out a relation between acceleration observed in an inertial frame and a uniformly rotating frame.

### Subsection11.2.1A particle fixed at the $x$-axis of a uniformly rotating frame

Consider a particle fixed at $(x',0,0)$ as shown in Figure 11.2.1. Although this particle's position in the rotating frame is fixed, its position in the fixed frame is changing. As a matter of fact, as far as the fixed frame is concerned, the particle moving in $xy$-plane in a circle of radius equal to $x'$ with angular speed $\Omega\text{.}$

From the right side of Figure 11.2.1, we can easily find the position of the particle in the two frames.

\begin{align*} \amp x =x'\cos\Omega t,\ y=x'\sin\Omega t,\ z=0\\ \amp x' =\text{fixed},\ y'=0,\ z'=0 \end{align*}

Therefore, components of the velocity of the particle in the two frames, obtained by taking derivatives of the positions, will be

\begin{align*} \amp v_x =\frac{dx}{dt}=-\Omega x'\sin\Omega t,\ v_y=\frac{dy}{dt}=\Omega x'\cos\Omega t,\ v_z=\frac{dz}{dt}=0 \\ \amp v'_x = \frac{dx'}{dt}=0,\ v'_y=\frac{dy'}{dt}=0,\ v'_z=\frac{dz'}{dt}=0 \end{align*}

The acceleration of the particle in the two frames can be obtained by taking derivatives of the corresponding velocities.

\begin{align*} \amp a_x =\frac{dv_x}{dt}=-\Omega^2 x'\cos\Omega t,\ \ a_y=\frac{dv_y}{dt}=-\Omega^2 x'\sin\Omega t,\ \ a_z=\frac{dv_z}{dt}=0\\ \amp a'_x =\frac{dv'_x}{dt}=0,\ \ a'_y=\frac{dv'_y}{dt}=0,\ \ a'_z=\frac{dv'_z}{dt}=0 \end{align*}

The $x$ and $y$-components of acceleration in the inertial frame are equivalent to the radially inward acceleration, i.e. the centripetal acceleration. This makes sense, because from the perspective of the fixed frame, the particle moves in a uniform circular motion of radius equal to $x'$ at constant angular speed $\Omega\text{.}$ Therefore, the acceleration in the fixed frame of a body moving uniformly in a circle about the center is pointed radially inwards towards the center, as expected.

### Subsection11.2.2Vector Notation

It is instructive to write the kinematic quantities in the fixed and rotating frames given above in a vector notation. The example particle does not move with respect to the rotating frame. This particle will appear to rotate in a circle about the $z$-axis when observed from the fixed frame. Therefore, in the fixed frame the angular velocity vector will be pointed along the $z$-axis.

\begin{equation*} \vec \Omega = \Omega\hat u_z. \end{equation*}

The velocity of the particle in the fixed frame is tangential to the circle in the $xy$-plane and is given by the cross product of the angular velocity vector and the position vector.

\begin{equation*} \vec v = \frac{d\vec r}{dt}=\vec \Omega \times \vec r. \end{equation*}

Since acceleration is equal to the time-derivative of velocity, we obtain the following for the acceleration in the fixed frame.

\begin{align*} \amp \vec a =\frac{d\vec v}{dt} = \vec \Omega \times \frac{d\vec r}{dt}\\ \amp \Longrightarrow\ \ \vec a = \vec \Omega \times \left( \vec \Omega \times \vec r\right)\ \ \text{(Const )} \end{align*}

We find that, although the particle is not moving in the rotating frame, the particle has an acceleration in the fixed frame.

### Subsection11.2.3A particle moving in the $xy$-plane of a uniformly rotating frame

Let us consider a particle that is not fixed but moves in the $xy$-plane of a uniformly rotating frame $Ox'y'z'$ which is rotating with respect to a fixed frame $Oxyz$ about their common $z$-axis. Since the frame is rotating about the $z$-axis, the particles moves in the $xy$-plane of the fixed frame as well. The situation between time $t$ and $t+\Delta t$ is shown in Figure 11.2.2. For simplicity, let the axes of the two frames be coincident at time $t\text{.}$ Figure 11.2.2. The physical situations at times $t$ and $t+\Delta t\text{.}$ In duration $t$ to $t+\Delta t\text{,}$ the particle moves from P to Q, and the rotating frame rotates by an angle $\Omega t\text{.}$ The common $z$-axis pointed out-of-page.

Let $(x,y)$ and $(x',y')$ be the coordinates of the particle at some time $t$ in the $Oxyz$ and $Ox'y'z'$ frames respectively. The change of the coordinates over the interval from $t$ to $t+\Delta t$ are $(\Delta x,\Delta y)$ and $(\Delta x',\Delta y')$ respectively. What are their relations? The relation between the displacements in the two frames is more conveniently written in the vector notation. Therefore, we will work with vector notation here.

As shown in Figure 11.2.2, particle is at point $P$ along momentarily coincident $x$ and $x'$-axes. The non-rotating frame marks the location with a marker at $P$ at its $x$-axis and the rotating frame marks the same location in its frame as $P'$ at its $x$-axis. As time passes, the particle moves to another location shown as $Q$ at time $t+\Delta t\text{.}$ Meanwhile the rotating frame has also moved.

Therefore, from the perspective of the rotating frame, the displacement of the particle between $t$ and $t+\Delta t$ is the vector from $P'$ on its $x$-axis where the particle was at $t$ to the space point $Q\text{.}$ The same displacement of the particle from the perspective of the non-rotating frame is $PQ$ vector in space. From $\triangle PP'Q$ we obtain the following relation in space.

\begin{equation} \overrightarrow{PQ} = \overrightarrow{PP'} + \overrightarrow{P'Q}. \label{}\label{eq-rot-inert-fr-1}\tag{11.2.1} \end{equation}

Now, $\overrightarrow{PP'}$ is on the arc of a circle of radius $|\vec r|$ with the arc angle $\Omega\Delta t\text{,}$ therefore

\begin{equation*} \overrightarrow{PP'} = \left( \vec \Omega \Delta t\right) \times \vec r. \end{equation*}

The vector $\overrightarrow{PQ}$ is the displacement $\Delta \vec r$ of the particle in the fixed frame and $\overrightarrow{P'Q}$ is the displacement $\Delta \vec r^{\prime}$ of the particle in the rotating frame. Putting these quantities in Eq. (11.2.1) we have the following relation among displacements in the time interval $\Delta t\text{.}$

\begin{equation*} \Delta \vec r = \left( \vec\Omega\Delta t \right)\times \vec r + \Delta\vec r^{\prime}. \end{equation*}

Dividing by $\Delta t$ and taking the $\Delta t \rightarrow 0$ limit, we find that the velocity in the inertial frame is equal to the sum of the velocity of the particle in the rotating frame and an additional term resulting from the rotation of the frame.

\begin{equation} \vec v = \vec v^{\prime} + \vec\Omega\times \vec r. \label{eq-vel-relation-frames-1}\tag{11.2.2} \end{equation}

Rather than use prime for the quantities in the rotation frame, sometimes a different notation is use in which we attach a subscript “in” and “rot” for quantities in the fixed frame and the rotating frame respectively.

\begin{equation} \vec v_\text{in} = \vec v_\text{rot} + \vec\Omega\times \vec r. \label{eq-vel-relation-frames}\tag{11.2.3} \end{equation}

There is no need to put a subscript to $\vec r$ since $\vec r$ is a vector from the origin to the position of the particle and the origins of the two frames are at the same point. The result of relation between the velocity vectors is also often written as

\begin{equation*} \left(\frac{d\vec r}{dt}\right)_\text{in} = \left(\frac{d\vec r}{dt}\right)_\text{rot} + \vec\Omega\times \vec r. \end{equation*}

where the subscript “in” denotes the quantity in the inertial or non-rotating frame and the subscript “rot” for a quantity with respect to the rotating frame.

Note that to derive this relation, we only used the geometric property of the position vector. A similar argument can be made about the rate of change of any arbitrary vector $\vec w$ in the two frames.

\begin{equation*} \left(\frac{d\vec w}{dt}\right)_\text{in} = \left(\frac{d\vec w}{dt}\right)_\text{rot} + \vec\Omega\times \vec w. \end{equation*}

For instance, we can obtain the time-derivative of the velocity vector $\vec v_\text{in}$ in the inertial frame by setting $\vec w = \vec v_\text{in}$ in this equation.

\begin{equation} \left(\frac{d\vec v_\text{in}}{dt}\right)_\text{in} = \left(\frac{d\vec v_\text{in}}{dt}\right)_\text{rot} + \vec\Omega\times \vec v_\text{in}.\label{eq-rot-inert-eq-2}\tag{11.2.4} \end{equation}

The left hand side of this equation is the acceleration of the particle in the inertial frame since it is the rate of change of the corresponding velocity in the same frame. The first term on the right side of this equation is not acceleration of anything since this term mixes information from the two frames - this term is the rate of change of the velocity in the inertial frame as observed from the rotating frame.

Finally, by substituting $\vec v_\text{in}$ in the right-side of Eq. (11.2.4) in terms of $\vec v_\text{rot}$ as given in Eq. (11.2.3) we can work out the relation between accelerations in the two frames.

\begin{align*} \left(\frac{d\vec v_\text{in}}{dt}\right)_\text{in} \amp = \left[ \frac{d}{dt} \left( \vec v_\text{rot} + \vec\Omega\times\vec r_\text{in} \right) \right]_\text{rot} + \vec\Omega\times\left( \vec v_\text{rot} \times\vec\Omega\times\vec r_\text{in}\right)\\ \amp = \vec a_\text{rot} + 2\vec \Omega\times \vec v_\text{rot} + \vec \Omega \times \left( \vec\Omega \times \vec r \right) \end{align*}

Therefore, the accelerations of the particle in the two frames are related as follows.

\begin{equation} \vec a_\text{in} = \vec a_\text{rot} + 2\vec \Omega\times \vec v_\text{rot} + \vec \Omega \times \left( \vec\Omega \times \vec r \right)\ \ (\text{Constant }\vec\Omega). \label{eq-rot-frame-acc-main}\tag{11.2.5} \end{equation}

The second term on the right is called Coriolis acceleration and the third is called centrifugal acceleration. In the next section we will use this expression to get Newton's equation of motion in a rotating frame, such as an Earth-based system.

### Subsection11.2.4Newton's Second Law in Uniformly Rotating Frame

Newton's second law for a particle of fixed mass $m$ in inertial frame is

\begin{equation*} m\vec a_{in} = \vec F \end{equation*}

where $\vec F$ is the net force. Substituting $\vec a_{in}$ from Eq. (11.2.5) into this equation, we find that mass times acceleration in a rotating frame has the following form, which we will refer to as Newton's second law in a frame that is rotating uniformly with a constant angular velocity $\vec \Omega\text{.}$

\begin{equation} m \vec a_{rot}=\vec F - m \left[2\vec \Omega\times \vec v_{rot} + \vec \Omega \times \left( \vec\Omega \times \vec r \right)\right]. \label{eq-rot-frame-acc}\tag{11.2.6} \end{equation}

We see that acceleration observed in a rotating frame depends on forces as well as fators that depend on the rotaton of the frame. One of these extra terms depends on the velocity of the particle. This term is called Coriolis force.

\begin{equation*} \text{Coriolis Force } = -2m\vec \Omega\times \vec v_{rot}. \end{equation*}

Other term depends on the position of the particle and is called centrifugal force.

\begin{equation*} \text{Centrifugal Force } = -m\vec \Omega \times \left( \vec\Omega \times \vec r \right). \end{equation*}

While the direction of the Coriolis force is perpendicular to the axis of rotation and the velocity of the particle, the centrifugal force is always pointed away from the axis of rotation and remains perpendicular to the axis.

That is, in a rotating frame, mass times acceleration is not equal to the net real force on the particle. In addition to the net real force $\vec F\text{,}$ there are terms that have units of force but are dependent on the rotation of the frame. These are the fictitious forces or inertial forces. In a rotating frame, the “real” and “fictitious” forces are indistinguishable in their effect on the acceleration except that there appears to be no agent(s) for the fictitious forces as far as the rotating frame is concerned.

A large platform is rotating uniformly with an angular speed $\Omega$ about the $z$-axis of a fixed frame $Oxyz$ with the origin at the center of the platform. A man is on the platform at a distance $R$ from the center.

(a) What is the motion of the man in the fixed frame?

(b) What is the motion of the man with respect to a frame $O'x'y'z'$ that has the same origin and $z$-axis as the fixed frame but whose $x$ and $y$-axes rotate with the platform?

(c) Let the man be along the $x'$ axis at $t=0$ when he starts to walk directly towards the center at a constant speed $v'\text{.}$ That is, he is walking along the $x'$ axis towards origin with speed $v'$ with respect to the origin $O'\text{.}$ Assuming no change in the rotation of the platform, what is the motion of the person as seen from the fixed frame?

[The conservation of angular momentum would mean that the rotation speed will go up as the person walks towards the center. Here we assume that some torque maintains the angular speed.]

Hint

Definition.

See solution.

Solution

(a) The man moves in a circle of radius $R$ with constant speed $v = \Omega R\text{.}$

(b) Think of this frame as the frame of someone standing along the $z$-axis observing the person on the platform. The person on the platform will be at rest with respect to this observer.

(c) In the fixed frame you will notice a change in the rotation speed to conserve the angular momentum of the platform plus the person system. Here, for the sake of simplicity we are assuming that the rotation speed remain at $\Omega\text{.}$ This will mean that in the fixed frame the radius of the circle in which the person is moving changes as $R(t) = R - v' t\text{.}$ Now, the circular motion will have speed $v = \Omega (R - v' t)\text{.}$

A car in Los Angeles, California, USA, is moving with a velocity of $30\text{ m/s}$ towards North with respect to an observer on the ground. Find magnitude and direction of Coriolis force on the car if its mass is $3000\text{ kg}\text{.}$ Los Angeles, USA has latitude $34.0522^{\circ}\text{ N}\text{,}$ and the longitude $118.2428^{\circ}\text{ W}\text{.}$

Hint

Definition.

7.3\text{ N} towards the local East.

Solution

The Coriolis force will be

\begin{equation*} \vec F_c = -2 m \vec \Omega \times \vec v_{rot}. \end{equation*}

Here,

\begin{align*} \amp \vec \Omega = \frac{2\pi}{T} \hat k,\\ \amp \vec v_{rot} = -(30\ \text{m/s})\ \hat u_{\theta}, \end{align*}

where $T = 24\times 3600$ s, $\hat k$ is the unit vector from the origin towards the North pole,

and $\hat u_{\theta}$ is the unit vector tangent to the great circle through the point at LA and passing through the North and South pole, the polar unit vector for the spherical coordinate system, the other unit vector we need is the tangent to the circle for constant Latitude, which will be denoted as $\hat u_{\phi}\text{.}$ The vector $\hat u_{\phi}$ is pointed towards the local East direction.

The cross-product of the unit vectors $\hat k$ and $\hat u_{\theta}$ can be shown to be $\hat u_z \times \hat u_{\theta} =\sin\lambda \hat u_{\phi},$ where $\lambda$ is the latitude, the angle towards the North from the equator. The cross product gives the Coriolis force.

\begin{equation} \vec F_c = 2 m \frac{2\pi}{T}(30)\sin\lambda\, \hat u_{\phi}. \label{eq-Coriolis-Force-on-a-Car-in-Northern-Hemisphere-1}\tag{11.2.7} \end{equation}

Now, putting in the numbers we find $F_c = 7.3\text{ N}$ towards the local East. From the given longitude and latitude you can write the unit vector in terms of the unit vectors of the Cartesian axes.

A car in Canberra, Australia, is moving with a velocity of $30\text{ m/s}$ towards North with respect to an observer on the ground. Find magnitude and direction of Coriolis force on the car if its mass is $3000\text{ kg}\text{.}$ Canberra, Australia has latitude $35.2828^{\circ}\text{ S}\text{,}$ and longitude $149.1314^{\circ}\text{ E}\text{.}$

Hint

Definition.

$7.56\ \text{N, West}\text{.}$

Solution

Using the Eq. (11.2.7) in the last problem with $\lambda$ negative now gives the direction of the Coriolis force towards the local West with the magnitude

\begin{equation*} F_c = 2 m \frac{2\pi}{T}(30)|\sin\lambda| = 7.56\ \text{N}. \end{equation*}

A bead of mass $m$ can slide frictionlessly on a rotating ring of mass $M$ and radius $R$ which rotates at an angular speed of $\omega$ about a vertical axis through its center. When the bead is at a particular location, it does not slide. (a) Find this special position of the bead using calculations in a rotating frame. (b) Repeat the calculation in an inertial frame.

Hint

Write cartesian components of equation of motion.

$\theta = \cos^{-1} \left( \frac{g}{\omega^2 R} \right)\text{.}$

Solution 1 (a)

(a) The figure shows the choice of the fixed frame $Oxyz$ and $Ox'y'z\text{.}$ In the rotating frame, the bead is fixed in the $x'z$-plane as the bead rotates about the $z$-axis. The calculations in the rotating frame with $\vec v_{rot}=0$ and $\vec a_{rot} = 0$ gives

\begin{align*} \amp z\textrm{-component: }\ \ F_N\cos\theta-mg = 0\\ \amp x\textrm{-component: }\ \ F_N\sin\theta - m\omega^2 R\sin\theta = 0. \end{align*}

Therefore,

\begin{equation*} \theta = \cos^{-1} \left( \frac{g}{\omega^2 R} \right). \end{equation*}

Note that $\omega^2R>g$ for the cosine here. Therefore, the bead will settle at a non-zero value if $\omega^2R>g\text{.}$

Solution 2 (b)

(b) The calculations in the fixed frame with $v = |\vec v| =\omega R\sin\theta$ and $|\vec a| = v^2/R\sin\theta$ gives

\begin{align*} \amp z\textrm{-component: }\ \ F_N\cos\theta-mg = 0 \\ \amp \textrm{centripetal component: }\ \ F_N\sin\theta = m\frac{v^2}{R\sin\theta}. \end{align*}

Therefore, we get the same angle as in the rotating frame.

\begin{equation*} \theta = \cos^{-1} \left( \frac{g}{\omega^2 R} \right). \end{equation*}

A block of mass $M$ is attached to a string of length $l\text{.}$ The other end of the string is attached to a post at the center of a rotating platform. When the platform is rotating at a steady angular speed $\Omega$ the block moves in a circular motion.

Find the radius of the circular motion of the block performing the calculations in a rotating frame.

Hint

$\frac{l}{\omega^2 R}\sqrt{\omega^4 R^2 - g^2}\text{.}$
Let $T$ be the magnitude of the tension force in the string. This problem is analogous to Checkpoint 11.2.7 the following mapping.