The centrifugal force is thus nothing but minus the centripetal force, which we already encountered for uniform rotational motion in equation {eq}`Fcp`, and as the second term in {eq}`radialforce`. The centrifugal force is the force you 'feel' pushing you sidewards when your car makes a sharp turn, and is also responsible for creating the parabolic shape of the water surface in a spinning bucket, see problem {numref}`ch:generalrotationalmotion`.\ref{prob:parabolicwatersurface}.
The centrifugal force is thus nothing but minus the centripetal force, which we already encountered for uniform rotational motion in equation {eq}`Fcp`, and as the second term in {eq}`radialforce`. The centrifugal force is the force you 'feel' pushing you sidewards when your car makes a sharp turn, and is also responsible for creating the parabolic shape of the water surface in a spinning bucket, see {numref}`pb:parabolicwatersurface`.
The Coriolis force is present whenever a particle is moving with respect to the rotating coordinates, and tends to deflect particles from a straight line (which you'd get in an inertial reference frame). We have $\bm{\omega} \cdot \bm{F}_\mathrm{Cor} = \bm{v} \cdot \bm{F}_\mathrm{Cor} = 0$, so the Coriolis force is perpendicular to both the rotation and velocity vectors - note that this is the velocity in the rotating frame. In the two-dimensional case, we had $\bm{F}_\mathrm{Cor} = 2 m \dot{\rho} \omega \bm{\hat{\theta}}$ (equation {eq}`FCoriolis`), which for a velocity in the radial direction, $\dot{\rho}$, gives a force in the angular direction $\bm{\hat{\theta}}$.
The azimuthal force occurs when the rotation vector of our rotating system changes - i.e., when the rotation speeds up or decelerates, or the plane of rotation alters. In either case we have $\bm{r} \cdot \bm{F}_\mathrm{Cor} = 0$, so the force is perpendicular to the position vector. If it is the magnitude of the rotation vector that changes, and we again take the rotation vector to lie along the $z$-axis in the rotating frame, $\dot{\bm{\omega}}$ also lies along the $z$-axis, and we get
**Leonhard Euler** (1707-1783) was a Swiss mathematician who made major contributions to many different branches of mathematics, and, by application, physics. He also introduced much of the modern mathematical terminology and notation, including the concept of (mathematical) functions. Euler was possibly the most prolific mathematician who ever lived, and likely is the person with the most equations and formulas named after him. Although his father, who was a pastor, encouraged Euler to follow in his footsteps, Euler's tutor, famous mathematician (and family friend) Johann Bernoulli convinced both father and son that Euler's talent for mathematics would make him a giant in the field. Famous examples of Euler's work include his contributions to graph theory (the K\"oningsberg bridges problem), the relation $e^{i\pi}+1=0$ between five fundamental mathematical numbers named after him, his work on power series, a method for numerically solving differential equations, and his work on fluid mechanics (in which there is also an 'Euler's equation').
**Leonhard Euler** (1707-1783) was a Swiss mathematician who made major contributions to many different branches of mathematics, and, by application, physics. He also introduced much of the modern mathematical terminology and notation, including the concept of (mathematical) functions. Euler was possibly the most prolific mathematician who ever lived, and likely is the person with the most equations and formulas named after him. Although his father, who was a pastor, encouraged Euler to follow in his footsteps, Euler's tutor, famous mathematician (and family friend) Johann Bernoulli convinced both father and son that Euler's talent for mathematics would make him a giant in the field. Famous examples of Euler's work include his contributions to graph theory (the Köningsberg bridges problem), the relation $e^{i\pi}+1=0$ between five fundamental mathematical numbers named after him, his work on power series, a method for numerically solving differential equations, and his work on fluid mechanics (in which there is also an 'Euler's equation').
```{figure} images/portraits/Euler.jpg
---
width:300
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@@ -275,7 +275,7 @@ For rotations about a principal axis, the torque is zero (by construction), so E
\end{align*}
```
If we rotate about axis $1$, then $\omega_2$ and $\omega_3$ are (at least initially) very small, so equation {eq}`Eulernotorque`A gives $\dot{\omega}_1 = 0$. We can then derive an equation for $\omega_2$ by taking the time derivative of {eq}`Eulernotorque`B and using {eq}`Eulernotorque`C for $\dot{\omega}_3$, which gives:
If we rotate about axis $1$, then $\omega_2$ and $\omega_3$ are (at least initially) very small, so equation {eq}`Eulernotorque`a gives $\dot{\omega}_1 = 0$. We can then derive an equation for $\omega_2$ by taking the time derivative of {eq}`Eulernotorque`b and using {eq}`Eulernotorque`c for $\dot{\omega}_3$, which gives:
@@ -307,7 +307,7 @@ Two Delft students wish to re-create Galilei's experiment dropping objects with
```
````{exercise} Foucault's pendulum
:label: prob:Foucaultspendulum
:label: pb:Foucaultspendulum
:class: dropdown
```{index} Foucault's pendulum
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@@ -317,7 +317,7 @@ Two Delft students wish to re-create Galilei's experiment dropping objects with
:name: fig:Foucaultpendulum
Foucault's pendulum. (a) Coordinates in Paris. (b) For a pendulum with a long string and small amplitude, the velocity of the bob will be almost horizontal. (c) The replica of Foucault's original pendulum at the Panthéon <sup>[^4]</sup>.
```
Paris is not on the North pole, but it does lie on the Northern hemisphere, so the pendulum will still appear to rotate clockwise, just at a slower frequency. We'll calculate this precession frequency in this problem.
Paris is not on the North pole, but it does lie on the Northern hemisphere, so the pendulum will still appear to rotate clockwise, just at a slower frequency. We'll calculate this precession frequency in this problem.
1. First, we need the angular velocity in Paris, in a useful coordinate system. Define the $\bm{\hat{z}}$ axis as pointing upwards in Paris, and $\bm{\hat{x}}$ as the tangent to the planet due North (see {numref}`fig:Foucaultpendulum`a). Express $\bm{\omega}$ in these coordinates.
1. If the pendulum has a very long string (the original Foucault one is 67 m) compared to its amplitude, the velocity $v$ of the weight will be roughly in the horizontal direction, see {numref}`fig:Foucaultpendulum`b. Argue why, in this case, the component of $\bm{\omega}$ in the $\bm{\hat{x}}$ direction will not change the frequency at which the plane of the pendulum precesses.
1. The pendulum's plane rotates with a (precession) frequency $\bm{\omega}_\mathrm{P} = \omega_\mathrm{P} \bm{\hat{z}}$ with respect to the Earth's frame fixed in Paris. This precession frequency must exactly compensate for the Earth's rotation in the frame of the pendulum (as in that frame, there are no forces acting on the pendulum, and thus its plane of oscillation stays fixed). Show that these considerations imply that $\omega_\mathrm{P} = - \omega \cos\theta$.
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@@ -330,7 +330,7 @@ An alternative way to show the effect of the rotation of the Earth involves only
```
````{exercise}
:label: prob:parabolicwatersurface
:label: pb:parabolicwatersurface
:class: dropdown
The centrifugal force emerges in a rotating coordinate frame, and famously causes the parabolic shape of the surface of water in a rotating bucket. As the centrifugal force is always perpendicular to the rotation axis, we can pick coordinates such that the rotation axis coincides with the $z$-axis, $\bm{\omega} = \omega \bm{\hat{z}}$, and we can express the centrifugal force in cylindrical coordinates as $\bm{F}_\mathrm{cf} = m \omega^2 \rho \bm{\hat{\rho}}$ (equation {eq}`centrifugalcylindrical`).
