9.7: Motion in rotating reference frames
We will now discuss methods for analysing motion of point masses in rotating and accelerating reference frames. These methods can for instance be used to analyse mechanisms with rotating parts, and can help to predict the effect of earth’s rotation on the motion of aircraft, vehicles and weather systems.
9.7.1 Rotating reference frame
The challenge we will deal with is illustrated in Figure 9.9. A ball \(B\), a point mass, rolls over the deck of a ship \(S\) and the captain of the ship measures its velocity \(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}\) and acceleration \(\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}\) using the \(x^{\prime} y^{\prime} z^{\prime}\) system that moves along with the ship. An observer \(O\) on the shore also measures the velocity and acceleration of the ball and finds values \(\overrightarrow{\boldsymbol{v}}_{B}\) and \(\overrightarrow{\boldsymbol{a}}_{B}\), which are different from the values of the captain because the ship moves and rotates. In this section we will discuss how to relate the velocity and acceleration of the ball \(B\) as measured by the captain \(\left(\overrightarrow{\boldsymbol{v}}_{B}^{\prime}, \overrightarrow{\boldsymbol{a}}_{B}^{\prime}\right)\) to those measured by the observer \(O\) on the shore \(\left(\overrightarrow{\boldsymbol{v}}_{B}, \overrightarrow{\boldsymbol{a}}_{B}\right)\), if we know the motion of the ship. So we know both its translation \(\left(\overrightarrow{\boldsymbol{v}}_{O^{\prime}}\right.\) and \(\left.\overrightarrow{\boldsymbol{a}}_{O^{\prime}}\right)\), and its angular velocity and acceleration \(\overrightarrow{\boldsymbol{\omega}}_{S}\) and \(\overrightarrow{\boldsymbol{\alpha}}_{S}\).
Since the mathematics to obtain the kinematic equations is rather involved, we directly present the equations here, and then derive them at the end of this chapter for interested readers.
9.7.2 Equations relating kinematics in rotating and fixed CS
With the following equations one can determine the velocity and acceleration vector of a point mass \(B\) as measured by the observer \(O\) if its velocity and acceleration as measured by the captain in the rotating and translating ship \(S\) are known:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{B}= & \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{v}}_{O^{\prime}}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}} \tag{9.77} \label{9.77}\\[4pt] \overrightarrow{\boldsymbol{a}}_{B}= & \overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{a}}_{O^{\prime}}+\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}} \tag{9.78} \label{9.78}\\[4pt] & +2 \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{\omega}} \times\left(\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right) \\[4pt] \overrightarrow{\boldsymbol{v}}_{B, 2 D}= & \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{v}}_{O^{\prime}}+\omega\left|\overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right| \hat{\boldsymbol{\phi}}_{O^{\prime}} \tag{9.79} \label{9.79}\\[4pt] \overrightarrow{\boldsymbol{a}}_{B, 2 D}= & \overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{a}}_{O^{\prime}}+\alpha\left|\overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right| \hat{\boldsymbol{\phi}}_{O^{\prime}} \tag{9.80} \label{9.80}\\[4pt] & +2 \omega\left(v_{B / O^{\prime}, \rho}^{\prime} \hat{\boldsymbol{\phi}}_{O^{\prime}}-v_{B / O^{\prime}, \phi}^{\prime} \hat{\boldsymbol{\rho}}_{O^{\prime}}\right)-\omega^{2} \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\end{align}\]
We provide both the \(3 \mathrm{D}\) version of the equations and the planar \(2 \mathrm{D}\) version. Note that \(\hat{\boldsymbol{\rho}}_{O^{\prime}}\) and \(\hat{\boldsymbol{\phi}}_{O^{\prime}}\) are the unit vectors in a cylindrical coordinate system with \(O^{\prime}\) as its origin. Let us now discuss the different components of these equations.
9.7.3 Object at rest in rotating reference frame
We first consider the situation where the ball is at rest on the deck of the ship at fixed position coordinate \(\overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\), with zero velocity \(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}=\overrightarrow{\mathbf{0}}\) and acceleration
\(\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}=\overrightarrow{\mathbf{0}}\) in the reference frame of the ship. Substituting this in Equation 9.77 and Equation 9.78 we find:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{B, \text { rest }} & =\overrightarrow{\boldsymbol{v}}_{O^{\prime}}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}} \tag{9.81} \label{9.81}\\[4pt] \overrightarrow{\boldsymbol{a}}_{B, \text { rest }} & =\overrightarrow{\boldsymbol{a}}_{O^{\prime}}+\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}+\overrightarrow{\boldsymbol{\omega}} \times\left(\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right) \tag{9.82} \label{9.82}\end{align}\]
Despite the fact that the ball is at rest on the ship, these equations show us that according to the observer on the shore it can have a substantial velocity and acceleration \(\overrightarrow{\boldsymbol{v}}_{B, \text { rest }}\) and \(\overrightarrow{\boldsymbol{a}}_{B, \text { rest }}\). Note also that the obtained equations are identical to those for a fixed point on a rigid body, Equation 9.26 and Equation 9.27 as might be expected.
9.7.4 Coriolis acceleration
Now let’s consider what happens when the ball \(B\) is given a velocity \(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}\) and acceleration \(\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}\) on the ship deck. If we compare Equation 9.81 and Equation 9.77 we see that these vectors are added to vectors in rest:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{B} & =\overrightarrow{\boldsymbol{v}}_{B, \text { rest }}+\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime} \tag{9.83} \label{9.83}\\[4pt] \overrightarrow{\boldsymbol{a}}_{B} & =\overrightarrow{\boldsymbol{a}}_{B, \text { rest }}+\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{a}}_{B, \mathrm{Cor}} \tag{9.84} \label{9.84}\end{align}\]
We see that a new acceleration term is added to the acceleration equation, which is called the Coriolis acceleration.
If a point mass has a non-zero velocity \(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}\) in a reference frame that rotates with angular velocity \(\overrightarrow{\boldsymbol{\omega}}\), this results in a Coriolis acceleration term that appears when the point mass is observed in a fixed reference frame:
\[\overrightarrow{\boldsymbol{a}}_{B, \text { Cor }}=2 \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime} \tag{9.85} \label{9.85}\]
It follows from the cross product, that the direction of the Coriolis acceleration is perpendicular to the velocity of the point mass and will therefore cause it to follow a curved circular path in the \(x y\)-plane. This rotational motion (Coriolis effect) can for instance be observed in the winds near a low pressure area as a consequence of the rotation of the earth.
9.7.5 Acceleration components in a rotating reference frame
Equation 9.78 can be subdivided in five components, let us discuss each of them:
\[\overrightarrow{\boldsymbol{a}}_{B}=\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{a}}_{B, \text { trans }}+\overrightarrow{\boldsymbol{a}}_{B, \mathrm{ang}}+\overrightarrow{\boldsymbol{a}}_{B, \mathrm{cptl}}+\overrightarrow{\boldsymbol{a}}_{B, \mathrm{Cor}} \tag{9.86} \label{9.86}\]
- The acceleration \(\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}\) of the ball inside the rotating ship.
- The translational acceleration \(\overrightarrow{\boldsymbol{a}}_{B, \text { trans }}=\overrightarrow{\boldsymbol{a}}_{O^{\prime}}\) due to the acceleration of the ship.
- The angular acceleration \(\overrightarrow{\boldsymbol{a}}_{B, \text { ang }}=\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\) due to angular acceleration of the ship.
- The centripetal acceleration \(\overrightarrow{\boldsymbol{a}}_{B, \mathrm{cptl}}=\overrightarrow{\boldsymbol{\omega}} \times\left(\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right)\).
- The Coriolis acceleration that is a consequence of the velocity \(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}\) the ball has inside the rotating ship: \(\overrightarrow{\boldsymbol{a}}_{B, \text { Cor }}=2 \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}\).
The equations up to now determine the acceleration \(\overrightarrow{\boldsymbol{a}}_{B}\) the observer on the shore measures. If one would like to determine the acceleration that the captain in the ship measures, this equation can be rewritten to:
\[\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}=\overrightarrow{\boldsymbol{a}}_{B}-\overrightarrow{\boldsymbol{a}}_{B, \text { trans }}-\overrightarrow{\boldsymbol{a}}_{B, \mathrm{ang}}-\overrightarrow{\boldsymbol{a}}_{B, \mathrm{cptl}}-\overrightarrow{\boldsymbol{a}}_{B, \mathrm{Cor}} \tag{9.87} \label{9.87}\]
Note that even if there are no forces acting on the ball, such that \(\overrightarrow{\boldsymbol{a}}_{B}=\overrightarrow{\mathbf{0}}\), the captain will measure an acceleration at a rate \(\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}\) given by Equation 9.87. If the captain would not know the ship is accelerating and rotating, the acceleration might be attributed to forces acting on the ball. Such forces due to measurements in a non-inertial reference frame are called pseudo-forces. Examples of such pseudo-forces are the centrifugal ’force’ \(-m \overrightarrow{\boldsymbol{a}}_{B, \mathrm{cptl}}\) and the Coriolis ’force’ \(-m \overrightarrow{\boldsymbol{a}}_{B, \text { Cor }}\).
A Example 9.5 In Figure 9.9 a ship \(S\) is making a pure rotation in the harbour with a constant angular velocity \(\overrightarrow{\boldsymbol{\omega}}_{S}=\omega_{S} \hat{\boldsymbol{k}}\) and \(\overrightarrow{\boldsymbol{\alpha}}_{S}=\overrightarrow{\mathbf{0}}\) around fixed position \(\overrightarrow{\boldsymbol{r}}_{O^{\prime}}\). A ball \(B\) rolls over the deck of the ship with constant velocity \(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}=v_{B}^{\prime} \hat{\boldsymbol{\imath}}^{\prime}\) as measured by the captain in the \(x^{\prime} y^{\prime} z^{\prime}\) coordinate system that moves along with the ship. A second ball \(C\) is thrown over the ship at constant velocity \(\overrightarrow{\boldsymbol{v}}_{C}=v_{C} \hat{\boldsymbol{\jmath}}\) as measured by an observer \(O\) on the shore. Both the captain and observer \(O\) determine the velocities and accelerations of the balls. Determine the expressions for the velocity and and acceleration vectors that the observer and captain measure for balls \(B\) and \(C\).
To determine \(\overrightarrow{\boldsymbol{v}}_{B}\) we use Equation 9.77 with \(\overrightarrow{\boldsymbol{v}}_{O^{\prime}}=\overrightarrow{\mathbf{0}}\) :
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{B} & =\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{\omega}}_{S} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}} \tag{9.88} \label{9.88}\\[4pt] & =v_{B}^{\prime} \hat{\boldsymbol{\imath}}^{\prime}+\omega_{S} \hat{\boldsymbol{k}}^{\prime} \times\left(x_{B}^{\prime} \hat{\boldsymbol{\imath}}^{\prime}\right) \tag{9.89} \label{9.89}\\[4pt] & =v_{B}^{\prime} \hat{\boldsymbol{\imath}}^{\prime}+\omega_{S} x_{B}^{\prime} \hat{\boldsymbol{\jmath}}^{\prime} \tag{9.90} \label{9.90}\end{align}\]
Note that you can either use the unit vectors from the \(x^{\prime} y^{\prime} z^{\prime}\) or the \(x y z\)-system, as long as you make sure not to take cross or dot products between them, since the relations like Equation 3.32 do not hold if the unit vectors are not perpendicular to each other.
To determine \(\overrightarrow{\boldsymbol{a}}_{B}\) we use Equation 9.78 with \(\overrightarrow{\boldsymbol{a}}_{O^{\prime}}=\overrightarrow{\mathbf{0}}, \overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}=\overrightarrow{\mathbf{0}}\) and \(\overrightarrow{\boldsymbol{\alpha}}_{S}=\overrightarrow{\mathbf{0}}\) :
\[\begin{align} \overrightarrow{\boldsymbol{a}}_{B} & =2 \overrightarrow{\boldsymbol{\omega}}_{S} \times \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}-\omega_{S}^{2} \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}} \tag{9.92} \label{9.92}\\[4pt] & =2 \omega_{S} \hat{\boldsymbol{k}}^{\prime} \times\left(v_{B}^{\prime} \hat{\boldsymbol{\imath}}^{\prime}\right)-\omega_{S}^{2}\left(x_{B}^{\prime} \hat{\boldsymbol{\imath}}^{\prime}\right) \tag{9.93} \label{9.93}\\[4pt] & =2 \omega_{S} v_{B}^{\prime} \hat{\boldsymbol{j}}^{\prime}-\omega_{S}^{2} x_{B}^{\prime} \hat{\boldsymbol{\imath}}^{\prime} \tag{9.94} \label{9.94}\end{align}\]
We see that the acceleration consists of a Coriolis contribution perpendicular to the velocity in the moving frame and a centripetal contribution pointing towards the moving origin \(O^{\prime}\). For the observer on the shore it is easier if we transform these vectors to the unit vectors in the \(x y z\)-system that she is using. We use the angle \(\phi_{S}=\arctan (3 / 4)\) or the properties of a 3:4:5 triangle for this projection:
\[\begin{align} \hat{\boldsymbol{\imath}}^{\prime} & =\cos \left(\phi_{S}\right) \hat{\boldsymbol{\imath}}+\sin \left(\phi_{S}\right) \hat{\boldsymbol{\jmath}} \tag{9.95} \label{9.95}\\[4pt] & =4 / 5 \hat{\boldsymbol{\imath}}+3 / 5 \hat{\boldsymbol{\jmath}} \tag{9.96} \label{9.96}\\[4pt] \hat{\boldsymbol{\jmath}}^{\prime} & =-\sin \left(\phi_{S}\right) \hat{\boldsymbol{\imath}}+\cos \left(\phi_{S}\right) \hat{\boldsymbol{\jmath}} \tag{9.97} \label{9.97}\\[4pt] & =-3 / 5 \hat{\boldsymbol{\imath}}+4 / 5 \hat{\boldsymbol{\jmath}} \tag{9.98} \label{9.98}\\[4pt] \hat{\boldsymbol{k}}^{\prime} & =\hat{\boldsymbol{k}} \tag{9.99} \label{9.99}\end{align}\]
However, an easier solution is to rotate the \(x y z\)-axes that the observer on the wall is using, since then we can directly set \(\hat{\boldsymbol{i}}^{\prime}=\hat{\boldsymbol{\imath}}, \hat{\boldsymbol{j}}^{\prime}=\hat{\boldsymbol{\jmath}}\) and \(\hat{\boldsymbol{k}}^{\prime}=\hat{\boldsymbol{k}}\).
Now we are going to determine the kinematics of ball \(C\) and vectors \(\overrightarrow{\boldsymbol{v}}_{C / O^{\prime}}^{\prime}\) and \(\overrightarrow{\boldsymbol{a}}_{C / O^{\prime}}^{\prime}\) as observed by the captain of the ship:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{C} & =\overrightarrow{\boldsymbol{v}}_{C / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{\omega}}_{S} \times \overrightarrow{\boldsymbol{r}}_{C / O^{\prime}} \tag{9.100} \label{9.100}\\[4pt] v_{C} \hat{\boldsymbol{\jmath}} & =\overrightarrow{\boldsymbol{v}}_{C / O^{\prime}}^{\prime}+\omega_{S} \hat{\boldsymbol{k}} \times\left(x_{C / O^{\prime}} \hat{\boldsymbol{\imath}}+y_{C / O^{\prime}} \hat{\boldsymbol{\jmath}}\right) \tag{9.101} \label{9.101}\\[4pt] \overrightarrow{\boldsymbol{v}}_{C / O^{\prime}}^{\prime} & =v_{C} \hat{\boldsymbol{\jmath}}-\omega_{S}\left(x_{C / O^{\prime}} \hat{\boldsymbol{\jmath}}-y_{C / O^{\prime}} \hat{\boldsymbol{\imath}}\right) \tag{9.102} \label{9.102}\\[4pt] & =\omega_{S} y_{C / O^{\prime}} \hat{\boldsymbol{\imath}}+\left(v_{C}-\omega_{S} x_{C / O^{\prime}}\right) \hat{\boldsymbol{j}} \tag{9.103} \label{9.103}\end{align}\]
The acceleration \(\overrightarrow{\boldsymbol{a}}_{B}\) of ball \(C\) is zero since it has a constant velocity, so we have:
\[\begin{align} 0 & =\overrightarrow{\boldsymbol{a}}_{C / O^{\prime}}^{\prime}+2 \overrightarrow{\boldsymbol{\omega}}_{S} \times \overrightarrow{\boldsymbol{v}}_{C / O^{\prime}}^{\prime}-\omega_{S}^{2} \overrightarrow{\boldsymbol{r}}_{C / O^{\prime}} \tag{9.104} \label{9.104}\\[4pt] -\overrightarrow{\boldsymbol{a}}_{C / O^{\prime}}^{\prime} & =2 \omega_{S} \hat{\boldsymbol{k}} \times\left(\omega_{S} y_{C / O^{\prime}} \hat{\boldsymbol{\imath}}+\left(v_{C}-\omega_{S} x_{C / O^{\prime}}\right) \hat{\boldsymbol{\jmath}}\right)-\omega_{S}^{2} \overrightarrow{\boldsymbol{r}}_{C / O^{\prime}} \\[4pt] \overrightarrow{\boldsymbol{a}}_{C / O^{\prime}}^{\prime} & =-2 \omega_{S}\left(\omega_{S} y_{C / O^{\prime}} \hat{\boldsymbol{\jmath}}-\left(v_{C}-\omega_{S} x_{C / O^{\prime}}\right) \hat{\boldsymbol{\imath}}\right)+\omega_{S}^{2}\left(x_{C / O^{\prime}} \hat{\boldsymbol{\imath}}+y_{C / O^{\prime}} \hat{\boldsymbol{\jmath}}\right) \\[4pt] & =2 \omega_{C} v_{C} \hat{\boldsymbol{\imath}}-\omega_{S}^{2}\left(x_{C / O^{\prime}} \hat{\boldsymbol{\imath}}+y_{C / O^{\prime}} \hat{\boldsymbol{j}}\right) \tag{9.105} \label{9.105}\\[4pt] & =2 \omega_{C} v_{C} \hat{\boldsymbol{\imath}}-\omega_{S}^{2} \overrightarrow{\boldsymbol{r}}_{C / O^{\prime}} \tag{9.106} \label{9.106}\end{align}\]
In the second step we used Equation 9.103. Again the acceleration consists of a Coriolis contribution perpendicular to the velocity vector in the IRF and a centripetal component that points towards the origin of the moving frame.
9.7.6 Transport theorem*
To derive Equation 9.77 and Equation 9.78, we will first introduce the transport theorem. The transport theorem relates the time derivative of a vector \(\frac{\mathrm{d}}{\mathrm{d} t} \overrightarrow{\boldsymbol{f}}\) in a fixed system \(x y z\) to its time derivative \(\frac{\mathrm{d}}{\mathrm{d} t} \overrightarrow{\boldsymbol{f}}^{\prime}\) as measured in a reference frame \(x^{\prime} y^{\prime} z^{\prime}\) that rotates with angular velocity vector \(\overrightarrow{\boldsymbol{\omega}}\) with respect to the fixed frame.
The time derivative of a vector function \(\overrightarrow{\boldsymbol{f}}(t)\) described in an IRF is related to the time derivative of the same vector function \(\overrightarrow{\boldsymbol{f}}(t)\) as measured in a reference frame \(S\) that is rotating with angular velocity vector \(\overrightarrow{\boldsymbol{\omega}}\), by the transport equation:
\[\frac{\mathrm{d} \overrightarrow{\boldsymbol{f}}}{\mathrm{d} t}=\left(\frac{\mathrm{d} \overrightarrow{\boldsymbol{f}}}{\mathrm{d} t}\right)^{\prime}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{f}} \tag{9.107} \label{9.107}\]
Let us explain why the transport theorem tells us that time derivatives of vectors are different in a rotating reference frame than in a fixed reference frame. A vector represents a certain magnitude and direction in space that is independent of the coordinate system in which it is measured. However, when taking a time derivative of such a vector one actually takes the difference between two vectors measured at two different times \(t\) and \(t+\mathrm{d} t\), e.g. for \(\overrightarrow{\boldsymbol{v}}(t)=\lim _{\mathrm{d} t \rightarrow 0}(\overrightarrow{\boldsymbol{r}}(t+\mathrm{d} t)-\overrightarrow{\boldsymbol{r}}(t)) / \mathrm{d} t\). If the coordinate system is rotating, the vector \(\overrightarrow{\boldsymbol{r}}(t+\mathrm{d} t)\) is measured with respect to coordinate axes that have a different orientation than those for which \(\overrightarrow{\boldsymbol{r}}(t)\) was measured. As a consequence, vectors that are time derivatives of other vectors, in particular \(\overrightarrow{\boldsymbol{v}}(t)\) and \(\overrightarrow{\boldsymbol{a}}(t)\), can be different \(\left(\overrightarrow{\boldsymbol{v}}^{\prime} \neq \overrightarrow{\boldsymbol{v}}\right)\) if they are measured in different, relatively rotating or translating, coordinate systems and the transport theorem provides a way to relate them. In this context it is important to note that Newton’s second
law defines forces in terms of an acceleration vector in a non-rotating inertial reference frame (IRF). By this definition it is ensured that force vectors (and moment vectors, see Ch. 10) are independent of the coordinate system and do not need to be transformed using the transport theorem.
Let us derive the transport theorem for the specific case of the vector function \(\overrightarrow{\boldsymbol{f}}(t)\) that is measured both in a fixed reference frame \(x y z\) and in a rotating coordinate system \(x^{\prime} y^{\prime} z^{\prime}\) that has constant angular velocity \(\overrightarrow{\boldsymbol{\omega}}\).
At every time, the vector can be described as \(\overrightarrow{\boldsymbol{f}}=f_{x}(t) \hat{\boldsymbol{\imath}}^{\prime}(t)+f_{j}(t) \hat{\boldsymbol{j}}^{\prime}(t)+\) \(f_{z}(t) \hat{\boldsymbol{k}}^{\prime}(t)\). The main point in the derivation is that for an observer that moves (and rotates) along with the rotating system \(x^{\prime} y^{\prime} z^{\prime}\) the unit vectors \(\hat{\boldsymbol{i}}^{\prime}, \hat{\boldsymbol{\jmath}}^{\prime}\) and \(\hat{\boldsymbol{k}}^{\prime}\) do not depend on time, whereas for an observer in a fixed reference frame their direction is time-dependent. The observer that moves along with the rotating coordinate system measures the time derivative of \(\overrightarrow{\boldsymbol{f}}\) to be:
\[\left(\frac{\mathrm{d} \overrightarrow{\boldsymbol{f}}}{\mathrm{d} t}\right)^{\prime}=\dot{f}_{x} \hat{\boldsymbol{\imath}}^{\prime}+\dot{f}_{y} \hat{\boldsymbol{\jmath}}^{\prime}+\dot{f}_{z} \hat{\boldsymbol{k}}^{\prime} \tag{9.108} \label{9.108}\]
The observer that is at rest in the fixed system \(x y z\) also measures the time derivative and finds, using the product rule:
\[\frac{\mathrm{d} \overrightarrow{\boldsymbol{f}}}{\mathrm{d} t}=\left(\dot{f}_{x} \hat{\boldsymbol{\imath}}^{\prime}+\dot{f}_{y} \hat{\boldsymbol{\jmath}}^{\prime}+\dot{f}_{z} \hat{\boldsymbol{k}}^{\prime}\right)+\left(f_{x} \frac{\mathrm{d} \hat{\boldsymbol{\imath}}^{\prime}}{\mathrm{d} t}+f_{y} \frac{\mathrm{d} \hat{\boldsymbol{\jmath}}^{\prime}}{\mathrm{d} t}+f_{z} \frac{\mathrm{d} \hat{\boldsymbol{k}}^{\prime}}{\mathrm{d} t}\right) \tag{9.109} \label{9.109}\]
When analysing the time derivatives of unit vectors in cylindrical coordinates we found in Equation 5.72 relations like \(\frac{\mathrm{d} \hat{\rho}}{\mathrm{d} t}=\overrightarrow{\boldsymbol{\omega}} \times \hat{\boldsymbol{\rho}}\), and it can be shown that these relations hold for any unit vector in a rotating frame that is observed from a fixed reference frame like \(\frac{\mathrm{d} \hat{\imath}^{\prime}}{\mathrm{d} t}=\overrightarrow{\boldsymbol{\omega}} \times \hat{\boldsymbol{\imath}}^{\prime}\), such that Equation 9.108 and (9.109) result in the transport theorem:
\[\begin{align} \frac{\mathrm{d} \overrightarrow{\boldsymbol{f}}}{\mathrm{d} t} & =\left(\frac{\mathrm{d} \overrightarrow{\boldsymbol{f}}}{\mathrm{d} t}\right)^{\prime}+\left(f_{x} \overrightarrow{\boldsymbol{\omega}} \times \hat{\boldsymbol{\imath}}^{\prime}+f_{y} \overrightarrow{\boldsymbol{\omega}} \times \hat{\boldsymbol{\jmath}}^{\prime}+f_{z} \overrightarrow{\boldsymbol{\omega}} \times \hat{\boldsymbol{\imath}}^{\prime}\right) \tag{9.110} \label{9.110}\\[4pt] & =\left(\frac{\mathrm{d} \overrightarrow{\boldsymbol{f}}}{\mathrm{d} t}\right)^{\prime}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{f}} \tag{9.111} \label{9.111}\end{align}\]
9.7.7 Derivation kinematics in a rotating reference frame*
Let us apply the transport theorem to the relative position vector setting \(\overrightarrow{\boldsymbol{f}}=\overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\) in Equation 9.107, this yields:
\[\begin{align} \frac{\mathrm{d} \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}}{\mathrm{d} t} & =\left(\frac{\mathrm{d} \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}}{\mathrm{d} t}\right)^{\prime}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}} \tag{9.112} \label{9.112}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}} & =\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}} \tag{9.113} \label{9.113}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B} & =\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{v}}_{O^{\prime}}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}} \tag{9.114} \label{9.114}\end{align}\]
Where we used \(\overrightarrow{\boldsymbol{v}}_{B / O}=\overrightarrow{\boldsymbol{v}}_{B}-\overrightarrow{\boldsymbol{v}}_{O^{\prime}}\) in the last step, thus deriving Equation 9.77.
\[\begin{align} \frac{\mathrm{d} \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}}{\mathrm{d} t}= & \left(\frac{\mathrm{d} \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}}{\mathrm{d} t}\right)^{\prime}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}} \tag{9.115} \label{9.115}\\[4pt] \overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}= & \frac{\mathrm{d}}{\mathrm{d} t}\left(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right)^{\prime} \tag{9.116} \label{9.116}\\[4pt] & +\overrightarrow{\boldsymbol{\omega}} \times\left(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right) \\[4pt] = & \overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}+\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}+2 \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime} \tag{9.117} \label{9.117}\\[4pt] & +\overrightarrow{\boldsymbol{\omega}} \times\left(\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right)\end{align}\]
Here we have substituted \(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}\) twice using Equation 9.113, used the product rule for differentiation, used \(\overrightarrow{\boldsymbol{a}}_{B / O^{\prime}}^{\prime}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}\right)^{\prime}\) and \(\overrightarrow{\boldsymbol{v}}_{B / O^{\prime}}^{\prime}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\overrightarrow{\boldsymbol{r}}_{B / O^{\prime}}\right)^{\prime}\), and used that \(\overrightarrow{\boldsymbol{\alpha}}=\overrightarrow{\boldsymbol{\alpha}}^{\prime}\) which can be shown by applying the transport theorem with \(\overrightarrow{\boldsymbol{f}}=\overrightarrow{\boldsymbol{\omega}}\) and \(\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{\omega}}=\overrightarrow{\mathbf{0}}\). With Equation 9.117 we have derived Equation 9.78.