Chapter 5.P2, Problem 1P

In the following problem, we ask the reader to some of the details left out of the above discussion, to analyse the closed-loop system for the stability properties, and to conduct a numerical simulation of the nonlinear system.

Work out the details leading toEq. (1).

$mL\theta \text{'}\text{'}=-\gamma \theta \text{'}-mg\sin \theta +m{\Omega }^{2}L\sin \theta \cos \theta$. (1)

To determine

The expression for $\theta$, where the model for the fly ball governor follows from taking into account all of the forces acting on the fly ball and applying Newton’s law, $ma=F$, and $\theta$ is the angle between the fly ball connecting arm and the vertical shaft about which the fly balls revolve. Assuming that the angular velocity of the vertical shaft and the rotational speed of the engine have the same value, $\Omega$, the centrifugal acceleration acting on the fly balls in the outward direction of magnitude ${\Omega }^{2}L\sin \theta$ is also the magnitude of the curvature of the motion, $1/L\sin \theta$, times the square of the tangential velocity, ${\Omega }^2{L}^2{\sin }^2\theta$. Damping force is of the magnitude $\gamma \theta \text{'}$.

Answer to Problem 1P

Solution:

The expression for the angle between the fly ball connecting arm and the vertical shaft is: $mL\theta "=-\gamma \theta \text{'}-mg\sin \theta +m{\Omega }^{2}L\sin \theta \cos \theta$.

Explanation of Solution

Given information:

The model for the fly ball governor follows from taking into account all of the forces acting on the fly ball and applying Newton’s law, $ma=F$ and $\theta$ is the angle between the fly ball connecting arm and the vertical shaft about which the fly balls revolve. Assuming that the angular velocity of the vertical shaft and the rotational speed of the engine have the same value, $\Omega$, the centrifugal acceleration acting on the fly balls in the outward direction of magnitude ${\Omega }^{2}L\sin \theta$ is also the magnitude of the curvature of the motion, $1/L\sin \theta$, times the square of the tangential velocity, ${\Omega }^2{L}^2{\sin }^2\theta$. Damping force is of the magnitude $\gamma \theta \text{'}$.

Explanation:

Let $m$ be the mass of the fly ball and $L$ be the length of the fly ball connecting arm.

By using the Newton’s second law of motion, $ma=F$.

There are two forces acting on the point mass, one is gravity, which is in downward direction, so it is $-mg$, and the other is tension in the connecting arm.

The magnitude of the sum is easily found from the figure as $F=-mg\sin \theta$.

Also, for upward direction, the gravitational force is $m{\Omega }^{2}L\sin \theta$, and the magnitude of the sum found from the figure is $m{\Omega }^{2}L\sin \theta \cos \theta$, and also the magnitude of the damping force is $\gamma \theta \text{'}$.

Thus, the total magnitude of the sum is $F=-\gamma \theta \text{'}-mg\sin \theta +m{\Omega }^{2}L\sin \theta \cos \theta$.

But Newton’s second law of motion tells that the net force is the mass times the acceleration.

Let $x$ be the distance travelled, which is the length of the arc traced out by the point mass, the arc length is related to the angle.

Thus, $x=L\cdot \theta$, and acceleration is $x"$

$x\text{'}=L\theta \text{'}$ and $x"=L\theta "$

$\Rightarrow mx"=-\gamma \theta \text{'}-mg\sin \theta +m{\Omega }^{2}L\sin \theta \cos \theta$

$\Rightarrow mL\theta "=-\gamma \theta \text{'}-mg\sin \theta +m{\Omega }^{2}L\sin \theta \cos \theta$

Therefore, the expression for the angle between the fly ball connecting arm and the vertical shaft is $mL\theta "=-\gamma \theta \text{'}-mg\sin \theta +m{\Omega }^{2}L\sin \theta \cos \theta$.