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 $\theta$, the equations of motion are given as:
$(I + mL^2) \ddot{\theta} + mgL * sin(\theta) = 0$
Factoring out $mL^2$ and dividing by $mgL$,
$\frac{mL^2}{mgL} (\frac{I}{mL^2} + 1) \ddot{\theta} + sin(\theta) = 0$
$\frac{L}{g} (\frac{I}{mL^2} + 1) \ddot{\theta} + sin(\theta) = 0$
With the value of $I$ computed above,
$ \frac{401}{300} \frac{L}{g} \ddot{\theta} + sin(\theta) = 0$
Then when $\theta$ is small, $sin(\theta) \approx \theta$
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}{2\pi} \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:
$\frac{L}{g} (\frac{I}{mL^2} + 1) \ddot{\theta} + sin(\theta) = 0$
$ \frac{251}{150} \frac{L}{g} \ddot{\theta} + sin(\theta) = 0$
$ f = \frac{1}{2\pi} \sqrt{\frac{150}{251} \frac{g}{L}}$
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