Posts on Physics

Water surface on a rotating bucket


Posted by Diego Assencio on 2015.04.22 under Physics (Mechanics)

Consider a bucket holding some viscous, homogeneous and incompressible fluid (e.g. water). The bucket starts to rotate with constant angular frequency $\omega$ around a vertical axis passing through the center of its bottom surface as shown in figure 1. Due to friction with the walls, the fluid is influenced by the rotation of the bucket and at some point will also be rotating with angular frequency $\omega$. What is the shape of the fluid surface at this point?

The easiest way to solve this problem is by using a non-inertial frame of reference which rotates with the bucket. The vertical axis $z'$ of this rotating frame is chosen to be the axis of rotation of the bucket (see figure 1). At the point at which the fluid is rotating with the bucket with angular frequency $\omega$, the fluid is seen as being at rest in this rotating frame.

Fig. 1: A rotating bucket holding some viscous, homogeneous and incompressible fluid. The reference frame $S$ sees the bucket rotating with angular frequency $\omega$ around the $z$ axis, while in the frame $S'$, the bucket is at rest since $S'$ is also rotating with respect to $S$ with angular frequency $\omega$.

Let $S$ be a reference frame in which the bucket is rotating and $S'$ a frame which rotates with the bucket as shown in figure 1. As proven in a previous post, the following equation relates the force ${\bf F}_S$ experienced by a fluid element of mass $m$ in $S$ and the effective force ${\bf F}^{\textrm{eff}}_{S'}$ experienced by this same fluid element in $S'$ (terms which trivially vanish were omitted below): $$ {\bf F}^{\textrm{eff}}_{S'} = m{\bf a}_{S'} = {\bf F}_S - m{\pmb\omega} \times ({\pmb\omega}\times{\bf x}') = {\bf F}_S + m\omega^2 \rho\hat{\pmb\rho} \label{post_195438f696c2c11e15340392ea2bf1cc_eq_forces1} $$ where ${\bf a}_{S'}$ and ${\bf x}'$ are the acceleration and position of the fluid element as measured in $S'$ respectively, ${\pmb\omega} = \omega\hat{\bf z}$ is the angular velocity of the bucket as measured in $S$, $\rho$ is the distance between the fluid element and the $z$ (or $z'$) axis and $\hat{\pmb\rho}$ is the unit vector which points radially outwards from the $z'$ axis (see figure 2).

Fig. 2: Forces acting on a fluid element of mass $m$ on the surface of the fluid (assumed to be already rotating with the bucket). The figure shows a cross section of the bucket which passes through the $z$ axis. The fluid height $h(\rho)$ at a given point on the surface depends only on the distance $\rho$ between this point and the $z$ axis.

The term ${\bf F}_S$ is the sum of the only two forces experienced in the non-rotating frame $S$: the gravitational force and the force exerted on the fluid element by the surrounding fluid (up to this point, we have not explicitly assumed that the fluid element is on the surface, so the equation below applies to any fluid element): $$ {\bf F}_S = m{\bf g} + {\bf F}_{\textrm{fluid}} \label{post_195438f696c2c11e15340392ea2bf1cc_eq_forces2} $$ Since in $S'$ the fluid is at rest, ${\bf a}_{S'} = {\bf 0}$. This fact, together with equations \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_forces1} and \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_forces2}, yields the following: $$ {\bf 0} = {\bf F}_{\textrm{fluid}} + m\left({\bf g} - {\pmb\omega} \times ({\pmb\omega}\times{\bf x}')\right) = {\bf F}_{\textrm{fluid}} + m\left({\bf g} + \omega^2 \rho\hat{\pmb\rho}\right) \label{post_195438f696c2c11e15340392ea2bf1cc_eq_acc} $$ Figure 2 shows these three forces. In equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_acc}, the centrifugal acceleration term $\omega^2 \rho\hat{\pmb\rho}$ can be interpreted as an additional component which, together with the gravitational acceleration ${\bf g}$, yields an effective gravitational acceleration ${\bf g}^{\textrm{eff}}$ which is no longer homogeneous in space: $$ {\bf g}^{\textrm{eff}} = \bf g - {\pmb\omega} \times ({\pmb\omega}\times{\bf x}') = \bf g + \omega^2 \rho\hat{\pmb\rho} \label{post_195438f696c2c11e15340392ea2bf1cc_effec_gravity} $$ The term ${\bf F}_{\textrm{fluid}}$ is a very interesting one since it is always orthogonal to the surface of the fluid (see figure 2). To understand this point, imagine the case in which there is no rotation, i.e., ${\pmb\omega} = {\bf 0}$. In this case, the frames $S$ and $S'$ coincide and the fluid is at rest on both. From equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_acc}, we see that ${\bf F}_{\textrm{fluid}} = - m{\bf g}$, which means ${\bf F}_{\textrm{fluid}}$ is the force exerted by the surrounding fluid against the gravitational force: the fluid element is not moving, so the surrounding fluid must be providing the force ${\bf F}_{\textrm{fluid}} = -m{\bf g}$ to keep it in place. Since the fluid surface is horizontal in this case, then ${\bf F}_{\textrm{fluid}}$ is orthogonal to the fluid surface. If the fluid rotates, ${\bf F}_{\textrm{fluid}}$ plays exactly the same role, but now the effective gravitational force given in equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_effec_gravity} is no longer homogeneous in space but depends explicitly on the position ${\bf x}'$: ${\bf F}_{\textrm{fluid}}$ will still be orthogonal to the surface of the fluid for every fluid element on the surface; a tangential force component would set the fluid element in motion since it would not be able to resist this force (but a perpendicular force is resisted by a pressure gradient created inside the fluid).

As can be seen on figure 2, the height $h(\rho)$ of a point on the surface of the fluid depends only on the distance between this point and the $z$ axis due to the cylindrical symmetry of the bucket around the $z$ axis. Since the derivative $dh/d\rho$ is the tangent to the curve $h(\rho)$ at a point which is at a distance $\rho$ from $z$, then: $$ \displaystyle\tan\phi = \frac{dh}{d\rho} = \frac{m\omega^2\rho}{mg} = \frac{\omega^2\rho}{g} \Longrightarrow \frac{dh}{d\rho} = \frac{\omega^2\rho}{g} \label{post_195438f696c2c11e15340392ea2bf1cc_eq_deriv} $$ Integrating both sides of equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_deriv} gives us the explicit form of $h(\rho)$: $$ \boxed{ \displaystyle h(\rho) = h(0) + \frac{\omega^2\rho^2}{2g} } \label{post_195438f696c2c11e15340392ea2bf1cc_eq_parab} $$ where $h(0)$ is the height of the water at $\rho = 0$, i.e., at the surface fluid element which is right at the rotation axis. Equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_parab} gives us the shape of the fluid surface: it is parabolic!

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Convert between Fahrenheit and Celsius the easy way


Posted by Diego Assencio on 2014.11.26 under Physics (Units)

Suppose you are calling someone from another part of the world and the person asks you "how is the weather there?". You answer promptly: "it's really nice today: $75^\circ\textrm{F}$", only to hear the confused reply "wait, what's that in Celsius?". Oops, what should you do now?

If you ever learned to convert temperature values between the Fahrenheit and Celsius scales, you probably learned to do it using the equation below: $$ \displaystyle\frac{{}^\circ\textrm{F} - 32}{9} = \frac{{}^\circ\textrm{C}}{5} $$ where ${}^\circ\textrm{F}$ is the temperature in the Fahrenheit scale and ${}^\circ\textrm{C}$ is the temperature in the Celsius scale. More concise ways to express the equation above are: $$ {}^\circ\textrm{F} = {}^\circ\textrm{C} \times 1.8 + 32 \label{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_exact} $$ and its inverse: $$ {}^\circ\textrm{C} = ({}^\circ\textrm{F} - 32)/1.8 \label{post_8a4390f653cdd7dca0e05db14bd8f760_F_to_C_exact} $$ However, both equations are unnecessarily complicated to be used for temperatures which one experiences on a daily basis because dividing or multiplying by 1.8 and subtracting or adding 32 are not trivially easy to do. Consider, for instance, the following equation: $$ \boxed{ {}^\circ\textrm{F} = {}^\circ\textrm{C}\times 2 + 30 } \label{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_approx} $$ and its inverse (just memorize the form which you find easier): $$ \boxed{ {}^\circ\textrm{C} = ({}^\circ\textrm{F} - 30)/2 } \label{post_8a4390f653cdd7dca0e05db14bd8f760_F_to_C_approx} $$ Much less daunting, aren't they? Dividing or multiplying by $2$ is much easier than by $1.8$, and subtracting or adding $30$ is much easier than it is with $32$.

During the vast majority of the year, and on most regions where the Fahrenheit scale is used, temperatures are in the range $[-10^\circ\textrm{C}, 35^\circ\textrm{C}]$ = $[14^\circ\textrm{F}, 95^\circ\textrm{F}]$. Interestingly, equation \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_approx} (or, equivalently, equation \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_F_to_C_approx}) works very well for converting between ${}^\circ\textrm{C}$ and ${}^\circ\textrm{F}$ over this temperature range (see figure 1). In fact, when converting from ${}^\circ\textrm{C}$ to ${}^\circ\textrm{F}$, the largest difference (in magnitude) between the values computed using equations \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_exact} and \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_approx} is only $5^\circ\textrm{F}$ at $35^\circ\textrm{C}$: the exact value is $95^\circ\textrm{F}$ but equation \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_approx} yields $100^\circ\textrm{F}$. A difference of only $4^\circ\textrm{F}$ occurs at the other end of the temperature range: equation \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_approx} yields $10^\circ\textrm{F}$ at $-10^\circ\textrm{C}$ but the exact value is $14^\circ\textrm{F}$.

Before you start thinking "well, $5^\circ\textrm{F}$ is not negligible", consider that as you walk over your house, you might already experience a difference of a few ${}^\circ\textrm{F}$ (or ${}^\circ\textrm{C}$ if you prefer). Some rooms will be warmer than others and you might not even notice the difference. Additionally, temperatures you see reported on the Internet, TV, radio etc. are just the values measured at some location near you and often differ from what you would experience in your garden by a few ${}^\circ\textrm{F}$ (${}^\circ\textrm{C}$). Finally, notice that the errors discussed above happen in the extremes of the given temperature range: "very hot" ($35^\circ\textrm{C}$ or $95^\circ\textrm{F}$) and "very cold" ($-10^\circ\textrm{C}$ or $14^\circ\textrm{F}$). For temperatures in between, the errors are smaller. For instance, $20^\circ\textrm{C}$ is exactly $68^\circ\textrm{F}$ but equation \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_approx} yields $70^\circ\textrm{F}$. Not so bad, right?

Not surprisingly, equation \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_F_to_C_approx} converts from ${}^\circ\textrm{F}$ to ${}^\circ\textrm{C}$ within an error of $2-3{}^\circ\textrm{C}$ on the temperature range we chose, with the largest errors happening at the extremes (very hot and very cold). In other words, you can use equations \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_approx} and \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_F_to_C_approx} to convert between ${}^\circ\textrm{F}$ and ${}^\circ\textrm{C}$ and obtain good approximate answers with little effort.

Fig. 1: Exact and approximate conversions from ${}^\circ\textrm{C}$ to ${}^\circ\textrm{F}$. These curves are described by equations \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_exact} and \eqref{post_8a4390f653cdd7dca0e05db14bd8f760_C_to_F_approx} respectively.
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Non-inertial frames of reference


Posted by Diego Assencio on 2014.10.17 under Physics (Mechanics)

Some physical systems can be studied much more easily when we consider their equations of motion on non-inertial frames of reference. As an example, studying ocean tides is easier if we use a non-inertial frame with origin at the center of the Earth; this frame is non-inertial since the Earth is under the action of gravitational forces caused by other bodies in the solar system such as the Sun and the Moon.

Consider a fixed inertial frame of reference $S$ with origin at a point $O$ and a non-inertial frame of reference $S'$ whose origin $O'$ is, at time $t$, positioned at ${\bf R}(t)$ as measured in $S$ (see figure 1). Our goal is to study the motion of a particle of mass $m$ on both $S$ and $S'$ and then find how the positions, velocities and accelerations of the particle as measured on each of these frames are related. After that, we will study the so-called fictitious forces which arise on non-inertial frames of reference. The axes of the fixed frame $S$ are labeled as $x_i$ while the axes of $S'$ are labeled as $x_i'$.

Fig. 1: A fixed inertial frame of reference $S$ and a non-inertial frame of reference $S'$. At time $t$, the position of the origin $O'$ of $S'$ is positioned at ${\bf R}(t)$ as measured in the frame $S$.

From figure 1, we see that: $$ \boxed{ {\bf x}(t) = {\bf R}(t) + {\bf x}'(t) } \label{post_51416695075e0ba12a5bb95029450fdb_x_x_prime} $$ This equation relates the coordinates ${\bf x}(t)$ of the particle on $S$ to its coordinates ${\bf x}'(t)$ on $S'$.

With respect to the fixed frame $S$, the motion of the frame $S'$ over an infinitesimal amount of time $dt$ can be broken into two parts: the translation of its origin and an infinitesimal rotation of its axes around an axis which passes through the origin of $S'$. To make this clearer, consider the case in which the origins of $S$ and $S'$ always coincide, i.e., ${\bf R}(t) = {\bf 0}$ for all $t$. Over an infinitesimal amount of time $dt$, the axes of $S'$ can do nothing but to rotate an infinitesimal angle $d\theta$ around some axis which passes through its origin. This axis, which is called the instantaneous axis of rotation of $S'$, can vary over time. This means the infinitesimal rotations can be done around different axes at different times. By letting the origin $O'$ of $S'$ move as well, we see that the full motion of $S'$ can be fully described by the translation of its origin plus the rotation of its axes.

Consider now how both $S$ and $S'$ see changes on ${\bf x}'(t)$ over an infinitesimal amount of time $dt$ (from now on, the dependency on $t$ will be omitted unless confusion would otherwise arise). If $S'$ sees a change $(d{\bf x}')_{S'}$, and since the axes of $S'$ might undergo an infinitesimal rotation during the period of time $dt$, then $S$ sees a change of $(d{\bf x}')_{S'}$ plus this small rotation: $$ (d{\bf x}')_S = (d{\bf x}')_{S'} + d{\pmb\theta} \times {\bf x}' \label{post_51416695075e0ba12a5bb95029450fdb_dx_prime_S_S_prime} $$ where above $d{\pmb\theta} = d{\pmb\theta}(t)$ is a vector which is parallel to the rotation axis of $S'$ at time $t$ and has magnitude $d\theta$. Dividing both sides of the equation above by $dt$, we obtain: $$ \left(\frac{d{\bf x}'}{dt}\right)_S = \left(\frac{d{\bf x}'}{dt}\right)_{S'} + \frac{d{\pmb\theta}}{dt} \times {\bf x}' = \left(\frac{d{\bf x}'}{dt}\right)_{S'} + {\pmb\omega} \times {\bf x}' \label{post_51416695075e0ba12a5bb95029450fdb_v_prime_S_S_prime} $$ where ${\pmb\omega} = {\pmb\omega}(t)$ is the angular velocity of the axes of $S'$ as measured in $S$ at time $t$. Notice that the derivation of \eqref{post_51416695075e0ba12a5bb95029450fdb_dx_prime_S_S_prime} does not explicitly use the fact that ${\bf x}'$ is the position of some particle as measured in $S'$. This means a change $(d{\bf q})_{S'}$ on any vector ${\bf q}$ as measured in $S'$ during $dt$ is seen by $S$ as the same change $(d{\bf q})_{S'}$ plus an additional term which comes from the infinitesimal rotation of $S'$ during $dt$: $$ (d{\bf q})_S = (d{\bf q})_{S'} + d{\pmb\theta} \times {\bf q} \label{post_51416695075e0ba12a5bb95029450fdb_dq_prime_S_S_prime} $$ Dividing both sides by $dt$, we obtain: $$ \displaystyle\left(\frac{d{\bf q}}{dt}\right)_S = \left(\frac{d{\bf q}}{dt}\right)_{S'} + \left(\frac{d{\pmb\theta}}{dt}\right)_S \times {\bf q} = \left(\frac{d{\bf q}}{dt}\right)_{S'} + {\pmb\omega} \times {\bf q} \label{post_51416695075e0ba12a5bb95029450fdb_dq_dt_S_S_prime} $$ Using equation \eqref{post_51416695075e0ba12a5bb95029450fdb_v_prime_S_S_prime}, we can relate the velocities of the particle measured in the frames $S$ and $S'$. Consider the perspective from frame $S$. From equation \eqref{post_51416695075e0ba12a5bb95029450fdb_x_x_prime}, we have that: $$ \displaystyle \left(\frac{d{\bf x}}{dt}\right)_S = \left(\frac{d{\bf R}}{dt}\right)_S + \left(\frac{d{\bf x}'}{dt}\right)_S \label{post_51416695075e0ba12a5bb95029450fdb_v_step1} $$ But using equation \eqref{post_51416695075e0ba12a5bb95029450fdb_v_prime_S_S_prime} on \eqref{post_51416695075e0ba12a5bb95029450fdb_v_step1}, we obtain: $$ \displaystyle \left(\frac{d{\bf x}}{dt}\right)_S = \left(\frac{d{\bf R}}{dt}\right)_S + \left(\frac{d{\bf x}'}{dt}\right)_{S'} + {\pmb\omega} \times {\bf x}' $$ A better way to write the equation above is: $$ \boxed{ {\bf v}_S = \dot{\bf R} + {\bf v}_{S'} + {\pmb\omega}\times{\bf x}' } \label{post_51416695075e0ba12a5bb95029450fdb_v_final} $$ where:

${\bf v}_S$is the velocity of the particle as measured in the inertial frame $S$
$\dot{\bf R}$is the velocity of the origin of $S'$ with respect to $S$
${\bf v}_{S'}$is the velocity of the particle as measured in the non-inertial frame $S'$
${\pmb\omega}$is the (instantaneous) angular velocity of the axes of $S'$
${\bf x}'$is the position of the particle as measured in $S'$

We can now relate the accelerations of the particle as measured in the frames $S$ and $S'$. Consider again the perspective from frame $S$. From equation \eqref{post_51416695075e0ba12a5bb95029450fdb_v_final}, we have that: $$ \displaystyle\left(\frac{d{\bf v}_S}{dt}\right)_S = \left(\frac{d\dot{\bf R}}{dt}\right)_S + \left(\frac{d{\bf v}_{S'}}{dt}\right)_S + \left(\frac{d{\pmb\omega}}{dt}\right)_S \times{\bf x}' + {\pmb\omega} \times \left(\frac{d{\bf x}'}{dt}\right)_S \label{post_51416695075e0ba12a5bb95029450fdb_a_step} $$ Using equation \eqref{post_51416695075e0ba12a5bb95029450fdb_dq_dt_S_S_prime} with ${\bf q}$ replaced by ${\bf v}_{S'}$ and equation \eqref{post_51416695075e0ba12a5bb95029450fdb_v_prime_S_S_prime} on equation \eqref{post_51416695075e0ba12a5bb95029450fdb_a_step}, we get: $$ \begin{eqnarray} \displaystyle\left(\frac{d{\bf v}_S}{dt}\right)_S = \left(\frac{d\dot{\bf R}}{dt}\right)_S + \left(\frac{d{\bf v}_{S'}}{dt}\right)_{S'} &+& {\pmb\omega}\times{\bf v}_{S'} + \left(\frac{d{\pmb\omega}}{dt}\right)_S \times{\bf x}' \nonumber\\[5pt] &+& {\pmb\omega} \times \left[\left(\frac{d{\bf x}'}{dt}\right)_{S'} + {\pmb\omega}\times{\bf x}'\right] \nonumber\\[5pt] \label{post_51416695075e0ba12a5bb95029450fdb_a_step2} \end{eqnarray} $$ Using our definition above for ${\bf v}_{S'}$: $$ {\bf v}_{S'} = \left(\frac{d{\bf x}'}{dt}\right)_{S'} $$ equation \eqref{post_51416695075e0ba12a5bb95029450fdb_a_step2} can finally be written in a simpler form: $$ \boxed{ {\bf a}_S = \ddot{\bf R} + {\bf a}_{S'} + \dot{\pmb\omega} \times{\bf x}' + {\pmb\omega} \times ({\pmb\omega}\times{\bf x}') + 2{\pmb\omega}\times{\bf v}_{S'} } \label{post_51416695075e0ba12a5bb95029450fdb_a_final} $$ where:

${\bf a}_S$is the acceleration of the particle as measured in the inertial frame $S$
$\ddot{\bf R}$is the acceleration of the origin of $S'$ with respect to $S$
$\dot{\pmb\omega}$is the (instantaneous) angular acceleration of the axes of $S'$
${\bf a}_{S'}$is the acceleration of the particle as measured in the non-inertial frame $S'$

Equation \eqref{post_51416695075e0ba12a5bb95029450fdb_a_final} is very important since it allows us to compute effective forces on non-inertial frames of reference. Before we do that, notice that Newton's second law of motion states that, in an inertial frame of reference $S$, the net force ${\bf F}_S$ on an object with constant mass $m$ satisfies: $$ {\bf F}_S = m{\bf a}_S \label{post_51416695075e0ba12a5bb95029450fdb_newton_2nd_law} $$ Using equations \eqref{post_51416695075e0ba12a5bb95029450fdb_a_final} and \eqref{post_51416695075e0ba12a5bb95029450fdb_newton_2nd_law}, we can then define an effective force acting on the mass $m$ as measured in $S'$: $$ \boxed{ {\bf F}^{\textrm{eff}}_{S'} = m{\bf a}_{S'} = {\bf F}_S - m\ddot{\bf R} - m\dot{\pmb\omega} \times{\bf x}' - m{\pmb\omega} \times ({\pmb\omega}\times{\bf x}') - 2m{\pmb\omega}\times{\bf v}_{S'} } \label{post_51416695075e0ba12a5bb95029450fdb_Feff} $$ According to Newton's laws, the real force is the one experienced in the inertial frame $S$, so on equation \eqref{post_51416695075e0ba12a5bb95029450fdb_Feff} all terms on the right-hand side except for ${\bf F}_S$ are fictitious forces which arise from the fact that we wrote an equation in the form ${\bf F}_{S'} = m{\bf a}_{S'}$ on the non-inertial frame of reference $S'$ even though Newton's second law states that ${\bf F} = m{\bf a}$ only on inertial frames of reference!

To finalize our study, let's analyze each one of the fictitious forces on equation \eqref{post_51416695075e0ba12a5bb95029450fdb_Feff}:

$-m\ddot{\bf R}$arises from the fact that $S'$ can be accelerating with respect to $S$
$-m\dot{\pmb\omega} \times{\bf x}'$arises from the fact that the axes of $S'$ can be rotating non-uniformly when seen from $S$, i.e., this term can only be nonzero if the angular velocity ${\pmb\omega}$ changes over time
$-m{\pmb\omega} \times ({\pmb\omega}\times{\bf x}')$this term is often called the centrifugal force because it always points outwards from the axis of rotation; its magnitude is $m\omega^2\rho$, where $\rho$ is the distance between the particle and the axis of rotation (see figure 2)
$- 2m{\pmb\omega}\times{\bf v}_{S'}$this term is called the Coriolis force and arises from the motion of the particle as measured in $S'$; this force always points to the right of the direction along which the particle travels if ${\pmb\omega}$ is taken as the "upwards" direction (see figure 3)
Fig. 2: The centrifugal force. From the figure, one can see that this force always points outwards from the axis of rotation. The magnitude of the centrifugal force is $m\omega^2\|{\bf x'}\|\sin\phi = m\omega^2\rho$, where $\rho = \|{\bf x}'\|\sin\phi$ is the distance between the axis of rotation and the particle.
Fig. 3: The Coriolis force. The velocity of the particle on the frame $S'$ is ${\bf v}_{S'} = {\bf v}^{\parallel}_{S'} + {\bf v}_{S'}^{\perp}$, where ${\bf v}^{\parallel}_{S'}$ and ${\bf v}^{\perp}_{S'}$ are the components of ${\bf v}_{S'}$ which are parallel and orthogonal to ${\pmb\omega}$ respectively. Since ${\pmb\omega}\times{\bf v}^{\parallel}_{S'} = {\bf 0}$, we see that the Coriolis force is equal to $-2m{\pmb\omega}\times{\bf v}^{\perp}_{S'}$, which always points to the right of ${\bf v}_{S'}$ if the system is seen from the perspective in which ${\pmb\omega}$ points to the observer.

References

[1]S. Thornton, J. Marion, Classical Dynamics of Particles and Systems, Thomson Brooks/Cole; 5th edition (2003)
[2]H. M. Nussenzveig, Curso de Física Básica - Vol. 1 (Mecânica), Edgard Blücher; 4th edition (2002)
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