Suppose you have one of these Centrifugal Governours
or a similar object. How to calculate the height that these balls have when the construct is spinning at a given speed?
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3 Answers
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Yeah, the way to calculate that is not obvious.
The circumnavigating weights are going around in a horizontal plane. The required centripetal force lies in that horizontal plane.
Other than the constraint force from the way the weights are suspended the only force that is available to be the ultimate source of centripetal force is the gravity that acts on the weights, and that gravity is acting perpendicular to the plane in which the weights are circumnavigating. The suspending arm is redirecting force, and thus centripetal force is provided.
So for the purpose of the calculation you will have to figure that out.
For instance:
About the suspending arm from the pivot point at the top to the weight. The force in that arm is larger than the force of gravity acting on the weight. The force in the direction parallel to that arm can be decomposed in a horizontal component and a vertical component.
At all angles of the arm the vertical component is equal to the actual weight of the weight. The magnitude of the horizontal component follows from the angle of the arm. The horizontal component is providing the required centripetal force.
Notation:
$a_c$ centripetal acceleration
$r$ radial distance
$\omega$ angular velocity
The following expression gives the required centripetal acceleration when an object is circumnavigating with an angular velocity $\omega$, at a distance $r$ to the center of rotation:
$$ a_c = \omega^2 r $$
The above elements are sufficient to set up the calculation.
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Below is a diagram of a centrifugal governor.
Each ball has three forces acting on it $m\vec g,\,\vec a$ and $\vec b$ as shown in the diagram.
Two free body diagrams for a ball are shown below.
The left-hand fbd is for the governor at rest so the net force on each ball is zero with the angle between an arm and the spindle $\theta_0$.
The right-hand fbd is for the governor rotating with $\theta>\theta_0$, ie the ball is higher up.
The net force, $\vec F$, is horizontal and it produces the centripetal acceleration of a ball.
Thus, as the governor rotates faster, the ball moves up, the angle $\theta$ increases as does the net force $F$.
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To obtain the Hight of the mass, you have to generate the equation of motion
you start with the position to the right mass
$$ \b P_R=\left[ \begin {array}{ccc} \cos \left( \omega\,t \right) &0&\sin \left( \omega\,t \right) \\ 0&1&0 \\ -\sin \left( \omega\,t \right) &0&\cos \left( \omega\,t \right) \end {array} \right] \,L\,\left[ \begin {array}{c} \sin \left( \psi \right) \\ -\cos \left( \psi \right) \\ 0 \end {array} \right] $$ the position to the left mass is $~\b P_L=\b P_R(-\psi)~$
from here you obtain
the velocities $$\b v_R=\frac{\partial\b P_R}{\partial \psi}\dot\psi\quad, \b v_L=\frac{\partial\b P_L}{\partial \psi}\dot\psi$$
the kinetic energy
$$T=\frac m2\left(\b v_R\cdot\b v_R+\b v_L\cdot\b v_L\right)$$
the potential energy
$$U=m\,g\,\left[\left(\b P_R\right)_y+\left(\b P_L\right)_y\right]$$
and with EL die equation of motion
$$\ddot\psi={\omega}^{2}\cos \left( \psi \right) \sin \left( \psi \right) -{\frac {\sin \left( \psi \right) g}{L}}\tag 1 $$
from equation (1) you obtain (numerically) $~\psi(t)~$ thus the Hight of the mass is $~\left[\b P_R(\psi(t)\right]_y~$
I assume that the rotation $~\omega~$ is constant
