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.


1.2: Reducing Fractions
Fractions can be reduced (expressed) in their lowest terms. This simply means that the fractions can no longer be divided by any other number.
In proper fractions, if the numerator and denominator can be divided by the same number, the fraction can be reduced.
Example: \(\dfrac{2}{4}=\dfrac{1}{2}\)
In the example above, this is accomplished by dividing the numerator and denominator by the number 2. Two fourths and one half represent the same number. They are just expressed differently. One half is a reduced form of two fourths.
Example: \(\dfrac{2}{4} \div \dfrac{2}{2}=\dfrac{1}{2}\)
At times a fraction can be divided multiple times in order to reduce it.
Example: \(\dfrac{8}{12} \div \dfrac{2}{2}=\dfrac{4}{6} \div \dfrac{2}{2}=\dfrac{2}{3}\)
This can also be performed in one step as follows. Both solutions provide the same answer.
Example: \(\dfrac{8}{12} \div \dfrac{4}{4}=\dfrac{2}{3}\)
In reducing improper fractions the denominator is divided into the numerator to create a mixed number. The remainder is written as a fraction.
Example:
Remember: The resulting mixed number may also need to be reduced further.
Example:
Exercise 1.2
Reduce the following fractions to their lowest terms.
- \(\dfrac{7}{2}\)
- \(\dfrac{10}{4}\)
- \(\dfrac{18}{4}\)
- \(\dfrac{20}{4}\)
- \(\dfrac{6}{5}\)
- \(\dfrac{100}{3}\)
- \(\dfrac{39}{6}\)
- \(1\dfrac{8}{12}\)
- \(2\dfrac{9}{4}\)
- \(\dfrac{4}{3}\)
- \(\dfrac{14}{3}\)
- \(\dfrac{140}{200}\)
- \(\dfrac{10}{20}\)
- \(7\dfrac{30}{40}\)


21.4: Assessment- Safety of Exercise Participation
To complete the assignment,
1. Go to the Canadian Society of Exercise Physiology webpage below and read the Physical Activity Readiness Questionnaire (PAR-Q),
http://www.csep.ca/cmfiles/publications/parq/par-q.pdf
2. Open a Word document and summarize the information from the PAR-Q (for example, I answered “yes” to __ questions and “no” to __ questions). If you answer “Yes” to one or more questions, you may need to consult with your healthcare provider prior to beginning the exercise phase. Therefore, explain any “Yes” answers and if you have already been cleared by your healthcare provider to exercise . To submit the assignment,
3. Save the document and upload it to the course. This assignment is worth 5 points maximum.
|
Criteria |
Answer absent |
Answer present |
| Question #1 |
0 Points |
0.8 Points If your answer is “yes” you must further explain the answer and if you have been cleared by a health care provider to exercise. Otherwise you will not receive credit for the answer. |
| Question #2 |
0 Points |
0.7 Points If your answer is “yes” you must further explain the answer and if you have been cleared by a health care provider to exercise. Otherwise you will not receive credit for the answer. |
| Question #3 |
0 Points |
0.7 Points If your answer is “yes” you must further explain the answer and if you have been cleared by a health care provider to exercise. Otherwise you will not receive credit for the answer. |
| Question #4 |
0 Points |
0.7 Points If your answer is “yes” you must further explain the answer and if you have been cleared by a health care provider to exercise. Otherwise you will not receive credit for the answer. |
| Question #5 | 0 Points |
0.7 Points If your answer is “yes” you must further explain the answer and if you have been cleared by a health care provider to exercise. Otherwise you will not receive credit for the answer. |
| Question #6 | 0 Points |
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| Question #7 | 0 Points |
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- Lifetime Fitness and Wellness. Provided by : Extended Learning Institute of Northern Virginia Community College. Located at : https://eli.nvcc.edu/ . License : CC BY: Attribution


18.8: XVIII Glossary
Chaplains: Trained professionals in hospitals, nursing homes, assisted living facilities, and hospices that assist with the spiritual, religious, and emotional needs of patients, families, and staff. Chaplains support and encourage people of all religious faiths and cultures and customize their approach to each individual’s background, age, and medical condition.
Holism: The concept that a human is composed of a mind, body, and soul integrated into an inseparable whole.
Religion: A unified system of beliefs, values, and practices that a person holds sacred or considers to be spiritually significant.  Some people associate religion with a place of worship (e.g., a synagogue or church), a practice (e.g., attending religious services, receiving communion, or going to confession), or a concept that guides one’s daily life (e.g., sin or kharma).
Spiritual distress: A state of suffering related to the impaired ability to experience meaning in life through connections with self, others, the world, or a superior being.
Spirituality: A dynamic and intrinsic aspect of humanity through which persons seek ultimate meaning, purpose, and transcendence and experience relationships to self, family, others, community, society, nature, and the significant or sacred. Spirituality is expressed through beliefs, values, traditions, and practice.
Spiritual well-being: A pattern of experiencing and integrating meaning and purpose in life through connectedness with self, others, art, music, literature, nature, and/or a power greater than oneself.
Transcendence: An understanding of being part of a greater picture or of something greater than oneself, such as the awe one can experience when walking in nature. It can also be expressed as a search for the sacred through subjective feelings, thoughts, and behaviors.


9.3: Measuring Volume
- Identify reasonable values for volume applications.
- Convert between like units of measures of volume.
- Convert between different unit values.
- Solve application problems involving volume.
Volume is a measure of the space contained within or occupied by three-dimensional objects. It could be a box, a pool, a storage unit, or any other three-dimensional object with attributes that can be measured in the metric unit for distance–meters. For example, when purchasing an SUV, you may want to compare how many cubic units of cargo the SUV can hold.
Cubic units indicate that three measures in the same units have been multiplied together. For example, to find the volume of a rectangular prism, you would multiply the length units by the width units and the height units to determine the volume in square units:
\[1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 1 \mathrm{~cm}=1 \mathrm{~cm}^3 \nonumber \]
Note that to accurately calculate volume, each of the measures being multiplied must be of the same units. For example, to find a volume in cubic centimeters, each of the measures must be in centimeters.
The formula used to determine volume depends on the shape of the three-dimensional object. Here we will limit our discussions to the area to rectangular prisms like the one in Figure
Given this limitation, the basic formula for volume is:
Reasonable Values for Volume
Because volume is determined by multiplying three lengths, the magnitude of difference between different cubic units is exponential. In other words, while a meter is 100 times greater in length than a centimeter, a cubic meter, m 3 , is times greater in area than a cubic centimeter, cm 3 . This relationship between benchmark metric volume units is shown in the following table.
| Units | Relationship | Conversion Rate |
| km 3 to m 3 | \(\begin{array}{l}\mathrm{km} \times \mathrm{km} \times \mathrm{km}=\mathrm{km}^3 \\ 1 \mathrm{~km}=1,000 \mathrm{~m} \\ 1,000 \mathrm{~m} \times 1,000 \mathrm{~m} \times 1,000 \mathrm{~m}=1,000,000,000 \mathrm{~m}^3\end{array}\) | 1 km 3 = 1,000,000,000 m 3 |
| m 3 to dm 3 | \(\begin{array}{l}\mathrm{m} \times \mathrm{m} \times \mathrm{m}=\mathrm{m}^3 \\ 1 \mathrm{~m}=10 \mathrm{dm} \\ 10 \mathrm{dm} \times 10 \mathrm{dm} \times 10 \mathrm{dm}=1,000 \mathrm{dm}^3\end{array}\) | 1 m 3 = 1,000 dm 3 |
| dm 3 to cm 3 | \(\begin{array}{l}\mathrm{dm} \times \mathrm{dm} \times \mathrm{dm}=\mathrm{dm}^3 \\ 1 \mathrm{dm}=10 \mathrm{~cm} \\ 10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 10 \mathrm{~cm}=1,000,000 \mathrm{~cm}^3\end{array}\) | 1 cm 3 = 1,000 dm 3 |
| cm 3 to mm 3 | \(\begin{array}{l}\mathrm{cm} \times \mathrm{cm} \times \mathrm{cm}=\mathrm{cm}^3 \\ 1 \mathrm{~cm}=10 \mathrm{~mm} \\ 10 \mathrm{~mm} \times 10 \mathrm{~mm} \times 10 \mathrm{~mm}=1,000 \mathrm{~mm}^3\end{array}\) | 1 cm 3 = 1,000 mm 3 |
To have an essential understanding of metric volume, you must be able to identify reasonable values for volume. When testing for reasonableness you should assess both the unit and the unit value. Only by examining both can you determine whether the given volume is reasonable for the situation.
A grandparent wants to send cookies to their grandchild away at college. Which represents a reasonable value for the volume of a box to ship the cookies:
- 3,375 km 3 ,
- 3,375 m 3 , or
- 3,375 cm 3 ?
- Answer
-
A volume of 3,375 km 3 is equivalent to a rectangular prism with dimensions of which is far too large for a shipping box. An area of 3,375 m 3 is equivalent to a surface of which is also too large. A reasonable value for the volume of the box is 3,375 cm 3 .
Which represents a reasonable value for the volume of a storage area:
8 m
3
, 8 cm
3
, or 8 mm
3
?
A food manufacturer is prototyping new packaging for one of its most popular products. Which represents a reasonable value for the volume of the box:
- 2 dm 3 ,
- 2 cm 3 , or
- 2 mm 3 ?
- Answer
-
A decimeter is equal to 10 centimeters. A box with a volume of 2 dm 3 might have the dimensions
dm 1 , or , which is reasonable. A box with a volume of 2 cm 3 or 2 mm 3 would be too small. dm × 1 dm × 2 dm
Which represents a reasonable value for the volume of a fish tank:
40,000 mm
3
, 40,000 cm
3
, or 40,000 m
3
?
A farmer has a hay loft. They calculate the volume of the hayloft as 64 cm 3 . Does the calculation make sense? Explain your answer.
- Answer
-
No. Centimeters are used to determine smaller distances, such as the length of a pencil. A hayloft is more than 64 centimeters long, so a more reasonable unit of value would be m 2 . A volume of 64 m 3 can be calculated using the dimensions 4 meters by 4 meters by 4 meters, which are reasonable dimensions for a hayloft. So, a more reasonable value for the volume of the hayloft is 64 m 3 .
An artist creates a glass paperweight. They decide they want to box and ship the paperweight, so they measure and determine that the volume of the cubic box is 125,000 mm 3 . Does their calculation make sense? Explain your answer.
Converting Like Units of Measures for Volume
Just like converting units of measure for distance, you can convert units of measure for volume. However, the conversion factor, the number used to multiply or divide to convert from one volume unit to another, is different from the conversion factor for metric distance units. Recall that the conversion factor for volume is exponentially relative to the conversion factor for distance. The most frequently used conversion factors are illustrated in Figure 9.10.
A pencil case has a volume of 1,700 cm 3 . What is the volume in cubic millimeters?
- Answer
-
Use multiplication to convert from a larger metric volume unit to a smaller metric volume unit. To convert from cm 3 to mm 3 , multiply the value of the volume by 1,000.
The pencil case has a volume of 1,700,000 mm 3 .
A jewelry box has a volume of 8 cm 3 . What is the volume of the jewelry box in cubic millimeters?
How to Convert Cubic Centimeters to Cubic Meters
A shipping container has a volume of 33.2 m 3 . What is the volume in cubic centimeters?
- Answer
-
Use multiplication to convert a larger metric volume unit to a smaller metric volume unit. To convert from m 3 to cm 3 , first multiply the value of the volume by 1,000 to convert from m 3 to dm 3 , and then multiply again by 1,000 to convert from dm 3 to cm 3 .
The shipping container has a volume of 32,200,000 cm 3 .
A gasoline storage tank has a volume of 37.854 m 3 . What is the volume of the storage tank in cubic centimeters?
A holding tank at the local aquarium has a volume of 22,712,000,000 cm 3 . What is the volume in cubic meters?
- Answer
-
Figure 9.10 indicates that when converting from a smaller metric volume unit to a larger metric volume unit you divide using the given conversion factor. To convert from cm 3 to m 3 , divide the value of the volume by 1,000 to first convert from cm 3 to dm 3 , then divide again to convert from dm 3 to m 3 .
The holding tank has a volume of 22,712 m 3 .
A warehouse has a volume of 465,000,000 cm 3 . What is the volume of the warehouse in cubic meters?
Understanding Other Metric Units of Volume
When was the last time you purchased a bottle of soda? Was the volume of the bottle expressed in cubic centimeters or liters? The liter (L) is a metric unit of capacity often used to express the volume of liquids. A liter is equal in volume to 1 cubic decimeter. A milliliter is equal in volume to 1 cubic centimeter. So, when a doctor orders 10 cc (cubic centimeters) of saline to be administered to a patient, they are referring to 10 mL of saline.
The most frequently used factors for converting from cubi
eters to liters are listed in Table 9.2.
| m 3 to L | m 3 to mL |
|---|---|
|
|
|
|
|
|
|
|
|
Converting Metric Units of Volume
A holding tank at the local aquarium has a volume of 22,712,000,000 cm 3 ? What is the capacity of the holding tank in liters?
- Answer
-
Use division to convert from cubic centimeters to liters. To determine the equivalent volume in liters, convert from cm 3 to L by dividing the value of the volume in cm 3 by 1,000.
The holding tank holds 22,712,000 L of water.
A gas can has a volume of 19,000 cm 3 . How much gas, in liters, does the gas can hold?
An airplane used 150 m 3 of fuel to fly from New York to Hawaii. How many liters of fuel did the airplane use?
- Answer
-
Because 1 liter is equivalent to 1 cubic decimeter, use multiplication to convert from m 3 to dm 3 . Multiply the value of the volume by 1,000 to convert from m 3 to dm 3 . Because
dm 3 = 1 L 1 dm 3 = 1 L , the resulting value is equivalent to the number of liters used.The airplane used 150,000 liters of fuel.
A gasoline storage tank has a volume of 37.854 m 3 . What is the volume of the storage tank in liters?
How many liters can a pitcher with a volume of 8,000,000 mm 3 hold?
- Answer
-
Use division to convert from a smaller metric volume unit to a larger metric volume unit. To convert from mm 3 to dm 3 ,
Step 1: Divide by 1,000 to convert from mm 3 to cm 3 .
Step 2: Divide again by 1,000 to convert from cm 3 to dm 3 .
Step 3: Use the unit value to express the volume in terms of liters.
000 mm 3 1 ,000 = 8, 000 cm 3 8 , 000 cm 3 1 ,000 = 8 dm 3 = 8 L 8,000, 000 mm 3 1 ,000 = 8, 000 cm 3 8 , 000 cm 3 1 ,000 = 8 dm 3 = 8 L The pitcher can hold 8 liters of liquid.
A glass jar has a volume of 800,000 mm 3 . How many mL of liquid can the glass jar hold?
Solving Application Problems Involving Volume
Knowing the volume of an object lets you know just how much that object can hold. When making a bowl of punch you might want to know the total amount of liquid a punch bowl can hold. Knowing how many liters of gasoline a car’s tank can hold helps determine how many miles a car can drive on a full tank. Regardless of the application, understanding volume is essential to many every day and professional tasks.
A cubic shipping carton’s dimensions measure
- Answer
-
Step 1: Determine the volume of the shipping carton.
Step 2: Use the appropriate conversion factor to convert the volume of the shipping carton from m 3 to cm 3 .
Step 3: Determine the volume of the smaller boxes.
Step 4: Divide the volume of the shipping carton, in cm 3 , by the volume of the smaller box, in cm 3 .
The shipping carton will hold 8,000 smaller boxes.
A factory can mill 300 cubic meters of flour each day. They package the flour in boxes that measure \(20\,{\text{cm}} \times 5\,{\text{cm}} \times 30\,{\text{cm}}\). How many boxes of flour does the factory produce each day?
A carton of juice measures 6 cm long, 6 cm wide and 20 cm tall. A factory produces 28,800 liters of orange juice each day. How many cartons of orange juice are produced each day?
- Answer
-
Step 1: Find the volume of the carton in cubic centimeters.
Step 2: Convert the volume in cm 3 to liters.
1 cm 3 = 1 mL 720 cm 3 = 720 mL 720 mL 1 ,000 = 0.72 L 1 cm 3 = 1 mL 720 cm 3 = 720 mL 720 mL 1 ,000 = 0.72 L Step 3: Divide the number of liters of orange juice produced each day by the volume of each carton.
The factory produces 40,000 cartons of orange juice each day.
An ice cream maker boxes frozen yogurt mix in boxes that measure 25 cm long, 8 cm wide and 35 cm tall. They produce 42,000 liters of frozen yogurt mix each day. How many boxes of frozen yogurt mix are produced each day?
A fish tank measures 60 cm long, 15 cm wide and 34 cm tall (Figure 9.11). The tank is 25 percent full. How many liters of water are needed to completely fill the tank?
- Answer
-
Step 1: Determine the volume of the fish tank in cubic centimeters.
Step 2: Convert the volume in cm 3 to volume in liters.
1 cm 3 = 1 mL 30 , 600 cm 3 = 30, 600 mL 30 , 600 mL 1 ,000 = 30.6 L 1 cm 3 = 1 mL 30 , 600 cm 3 = 30, 600 mL 30 , 600 mL 1 ,000 = 30.6 L Step 3: Since the tank is 25 percent full, the tank is 75 percent empty. Convert 75 percent to its decimal equivalent. Multiply the total volume by 75 percent expressed in decimal form to determine how many liters of water are required to fill the tank.
So, 22.95 liters of water are needed to fill the tank.
A fish tank measures 75 cm long, 20 cm wide and 25 cm tall. The tank is 50 percent full. How many liters of water are needed to completely fill the tank?
Take two large sheets of card stock. Roll one piece to tape the longer edges together to make a cylinder. Tape the cylinder to the other piece of card stock which serves as the base of the cylinder. Fill the cylinder to the top with cereal. Pour the cereal from the cylinder into a plastic storage or shopping bag. Remove the cylinder from the base and the tape from the cylinder. Re-roll the cylinder along the shorter edges a tape together. Attach the new cylinder to the base. Pour the cereal from the plastic bag into the cylinder. What do you observe? How does the shape of a container affect its volume?
Check Your Understanding
For the following exercises, determine the most reasonable value for each volume.
- Terrarium: 50,000 km 3 , 50,000 m 3 , 50,000 cm 3 , or 50,000 mm 3
- Milk carton: 236,000 L, 236 L, 236,000 mL, or 236 mL
- Box of crackers: 1,500 km 3 , 1,500 m 3 , 1,500 cm 3 , or 1,500 mm 3
For the following exercises, Convert the given volume to the units shown.
- 42,500 mm 3 = __________ cm 3
- 1.5 dm 3 = __________ mL
- 6.75 cm 3 = __________ mm 3
For the following exercises, determine the volume of objects with the dimensions shown.
- \(20\,{\text{cm}} \times 20\,{\text{m}} \times 20\,\text{cm}\)
- \(V =\) ________ L
- \(12\,{\text{mm}} \times 5\,{\text{cm}} \times 1.7\,{\text{cm}}\)
- \(V =\) ________ mL
- \(7.3\,{\text{m}} \times 3.2\,{\text{m}} \times 7\,{\text{m}}\)
- \(V =\) ________ m 3


2.12: Best Place and Time for Reading
Stop for a moment and think about yourself and the act of reading. Whether you read headlines or social media posts on a smartphone, or love to settle in with the Sunday paper, or devour stacks of hard-copy books, or read only when forced to, what does reading look like for you?
Check out some of the following tips as you figure out what works best for you:
- Consider your ideal spot for reading. Is it a favourite chair in your living room? Your bed? A coffee shop? The cool green grass in a local park? The library?
- What’s your favorite time of day for reading? Try and find when you study best – and then make use of that time.
- Do you read best in silence? Or do you like to have music playing, be around other people, or have some other sort of active distraction?
- Do you prefer hard-copy (printed) books, e-books, or audiobooks?
Figure 2.1: Best place and time to read: What works best for you?
Student Tip
Develop a Routine
Once you’ve found the perfect setting and time, use it. Develop a routine of reading and studying at about the same time and in the same place as much as you can. Doing this helps the activity become a habit, and once that happens, it will be even easier – and more effective.
Attribution statement
This chapter is an adaptation of: (with editorial changes)
The Word on College Reading and Writing by Carol Burnell, Jaime Wood, Monique Babin, Susan Pesznecker, and Nicole Rosevear, licensed under a Creative Commons Attribution-NonCommercial 4.0 International License , except where otherwise noted. Download for free at: openoregon.pressbooks.pub/wrd/


14.2: Structures and Names of Alkanes
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Learning Objectives
- To identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names.
We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). Saturated , in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules.
The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).
We previously introduced the three simplest alkanes—methane (CH 4 ), ethane (C 2 H 6 ), and propane (C 3 H 8 ) and they are shown again in Figure \(\PageIndex{1}\).
The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure \(\PageIndex{2}\)).
Methane (CH 4 ), ethane (C 2 H 6 ), and propane (C 3 H 8 ) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH 2 unit. The first 10 members of this series are given in Table \(\PageIndex{1}\).
Consider the series in Figure \(\PageIndex{3}\). The sequence starts with C 3 H 8 , and a CH 2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH 2 group) is called a homologous series. The members of such a series, called homologs , have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.
The principle of homology allows us to write a general formula for alkanes: C n H 2 n + 2 . Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C 8 H (2 × 8) + 2 = C 8 H 18 .
Key Takeaway
- Simple alkanes exist as a homologous series, in which adjacent members differ by a CH 2 unit.


<IP_ADDRESS>: Propositions
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Prepositions
Perhaps you had to memorize every preposition (there are well over 100) when you were in junior high school, but have you ever thought about the importance of these “little” words?
We might call them “glue” words because they bring other words together in ways that create meaning. Look at something you’ve read recently and take out the prepositions. Here is an example of a sentence written with and without prepositions:
With prepositions: As soon as I closed the door, the turkeys that had been waiting at the wood line ran to eat the corn I had just put out for the squirrels.
Without prepositions: As soon I closed the door, the turkeys that had been waiting the wood line ran eat the corn I had just put out the squirrels.
They may be generally small in letters, but prepositions are important words that give great meaning for time and place relationships among actions, objects, and ideas. It’s important to distinguish if you are throwing a ball to someone or at someone. Did you want your sandwich with or without onions? Do you need that zombie fort built at the end of summer or before the end of summer?
Even though they are small words, prepositions can be difficult—particularly for someone learning English—because their use isn’t always logical. As an example, most mid-Westerners speak of standing in line, but many on the East coast speak of standing on line. It would be a lot of fun and quite enlightening to write some sentences using some of the prepositions on the following page and then comparing those sentences with a friend from a different part of the country.
Common Prepositions
Although you no longer have to memorize the preposition list, it can be helpful to review the list of some of the most common prepositions and think about the meaning for each one.
Of course, it’s important to remember that many of these words can also be different parts of speech. English is fun, right?
Common Prepositions
- aboard
- about
- above
- across
- after
- against
- along
- amid
- among
- around
- as
- at
- before
- behind
- below
- beneath
- beside
- between
- beyond
- but
- by
- considering
- despite
- down
- during
- except
- excluding
- following
- for
- from
- in
- inside
- into
- like
- near
- of
- off
- on
- onto
- outside
- over
- past
- regarding
- since
- than
- though
- to
- toward
- under
- underneath
- until
- up
- upon
- verses
- with
- within
- without
Preposition Tips
As you work to make good decisions about your preposition choices, you should consider the following preposition tips:
-
It’s actually acceptable to end a sentence with a preposition. Contemporary scholars and writing style guides acknowledge the acceptability of ending a sentence with a preposition (Casagrande, 2006). It’s natural and conversational to write short sentences that end with prepositions.
- Did you find what you were looking for?
However, you should be aware that some professors prefer that you don’t end a sentence with a preposition. See how correctness is relative?
-
You should avoid unnecessary prepositions because using prepositions unnecessarily can make your writing wordy and even confusing.
- I am not [for] sure I have the answer.
- Frank apparently fell off [of] his horse while doing a stunt for a YouTube video.
- The difference between "beside" and "besides" can be confusing. Beside means next to. Besides means in addition to.
- The difference between "between" and "among" can also be confusing. You should use between when referring to two people or things, and you should use among when referring to more than two people or things.