程序代写代做代考 physics_msgs_inertia

physics_msgs_inertia

InertiaMsgsTest::SetPendulumInertia¶
This documents the effect of moment of inertia on the expected natural frequency in the pendulum test.

Pendulum dimensions¶

A pendulum is illustrated with distance $L$ between the pin joint and center of mass.
The pendulum is modeled as a box of mass $m$ with overall length $2L$ and width $L/5$.

Moment of inertia¶
Computing the moment of inertia requires specifying a location on the body and an axis direction.
In the following equation, the moment of inertia $I$ is computed
at the center of mass along an axis parallel to the axis of rotation:

$I = frac{m}{12} ((2L)^2 + (frac{L}{5})^2)$

$I = mL^2 (frac{1}{3} + frac{1}{300})$

$I = frac{101}{300} mL^2$

Natural frequency¶
With gravity $g$ and pendulum angle $ heta$, the equations of motion are given as:

$(I + mL^2) ddot{ heta} + mgL * sin( heta) = 0$

Factoring out $mL^2$ and dividing by $mgL$,

$frac{mL^2}{mgL} (frac{I}{mL^2} + 1) ddot{ heta} + sin( heta) = 0$

$frac{L}{g} (frac{I}{mL^2} + 1) ddot{ heta} + sin( heta) = 0$

With the value of $I$ computed above,

$ frac{401}{300} frac{L}{g} ddot{ heta} + sin( heta) = 0$

Then when $ heta$ is small, $sin( heta) approx heta$
and the pendulum will have an approximately sinusoidal trajectory.
The frequency $omega$ of the sinusoidal trajectory satisfies the following:

$ frac{401}{300} frac{L}{g} omega^2 = 1$

$ omega^2 = frac{300}{401} frac{g}{L} $

The frequency $f$ in Hz is then:

$ f = frac{1}{2pi} sqrt{frac{300}{401} frac{g}{L}}$

Modified natural frequency with larger inertia¶
Suppose the inertia $I$ is artificially increased by a factor of $2$:

$I = frac{101}{150} mL^2$

Then the natural frequency changes as follows:

$frac{L}{g} (frac{I}{mL^2} + 1) ddot{ heta} + sin( heta) = 0$

$ frac{251}{150} frac{L}{g} ddot{ heta} + sin( heta) = 0$

$ f = frac{1}{2pi} sqrt{frac{150}{251} frac{g}{L}}$

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