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.
If a point mass has a non-zero velocity \(\overrightarrow>_>\) 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: \[\overrightarrow_>=2 \overrightarrow<\boldsymbol<\omega>> \times \overrightarrow>_>^ <\prime>\tag \label\]
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.
A Example 9.5 In Figure 9.9 a ship \(S\) is making a pure rotation in the harbour with a constant angular velocity \(\overrightarrow<\boldsymbol<\omega>>_=\omega_ \hat>\) and \(\overrightarrow>_=\overrightarrow<\mathbf<0>>\) around fixed position \(\overrightarrow>_
To determine \(\overrightarrow>_\) we use Equation 9.77 with \(\overrightarrow>_
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.
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 \(\phi_=\arctan (3 / 4)\) or the properties of a 3:4:5 triangle for this projection:
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 \(\hat>^<\prime>=\hat<\boldsymbol<\imath>>, \hat>^<\prime>=\hat<\boldsymbol<\jmath>>\) and \(\hat>^<\prime>=\hat>\).
Now we are going to determine the kinematics of ball \(C\) and vectors \(\overrightarrow>_>^<\prime>\) and \(\overrightarrow_>^<\prime>\) as observed by the captain of the ship:
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.
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
The time derivative of a vector function \(\overrightarrow>(t)\) described in an IRF is related to the time derivative of the same vector function \(\overrightarrow>(t)\) as measured in a reference frame \(S\) that is rotating with angular velocity vector \(\overrightarrow<\boldsymbol<\omega>>\), by the transport equation:
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 \(t+\mathrm t\), e.g. for \(\overrightarrow>(t)=\lim _ <\mathrmt \rightarrow 0>(\overrightarrow>(t+\mathrm t)-\overrightarrow>(t)) / \mathrm t\). If the coordinate system is rotating, the vector \(\overrightarrow>(t+\mathrm t)\) is measured with respect to coordinate axes that have a different orientation than those for which \(\overrightarrow>(t)\) was measured. As a consequence, vectors that are time derivatives of other vectors, in particular \(\overrightarrow>(t)\) and \(\overrightarrow(t)\), can be different \(\left(\overrightarrow>^ <\prime>\neq \overrightarrow>\right)\) 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 \(\overrightarrow>(t)\) 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 \(\overrightarrow>=f_(t) \hat<\boldsymbol<\imath>>^<\prime>(t)+f_(t) \hat<\boldsymbol>^<\prime>(t)+\) \(f_(t) \hat>^<\prime>(t)\). 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 \(\hat>^<\prime>, \hat<\boldsymbol<\jmath>>^<\prime>\) and \(\hat>^<\prime>\) 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 \(\overrightarrow>\) to be:
The observer that is at rest in the fixed system \(x y z\) also measures the time derivative and finds, using the product rule:
When analysing the time derivatives of unit vectors in cylindrical coordinates we found in Equation 5.72 relations like \(\frac <\mathrm
Let us apply the transport theorem to the relative position vector setting \(\overrightarrow>=\overrightarrow>_>\) in Equation 9.107, this yields:
Where we used \(\overrightarrow>_=\overrightarrow>_-\overrightarrow>_
We now apply the transport theorem a second time to the velocity vector \(\overrightarrow>=\overrightarrow>_>\) to obtain the acceleration vector \(\overrightarrow_>\) :
This page titled 9.7: Motion in rotating reference frames is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Peter G. Steeneken via source content that was edited to the style and standards of the LibreTexts platform.
