# [Physics] Measuring the moment of inertia of a flywheel using simple pendulum motion

classical-mechanicsmoment of inertia

I've seen a method for experimentally determining the moment of inertia of a flywheel and I'm not sure whats the reasoning behind it. You attach a small weight $m_1$ to the flywheel's edge, it's important that $m_1 \ll M$, $M$ being the mass of the flywheel, so we can approximate that the small mass does not change the MOI of the flywheel. You then move $m_1$ through a small angle and let it oscillate like a simple pendulum. We measure $T$, the period of oscillation.

Then $I$, the MOI of the flywheel along its perpendicular axis is:

$$I=\left( \frac{T^2 g}{4 \pi ^2R} – 1\right) m_1 R^2$$

Where $R$ is the radius of the flywheel and $g$ is the standard acceleration due to gravity.

I'm stumped as to why this works, the only thing I can identify is that $\frac{T^2 g}{4 \pi ^2R} – 1 \backsimeq 0$ because $m_1$ describes a simple pendulum, so $T^2=4\pi^2\frac{R}{g}$.

The force $m_1g\sin\theta$ provides the moment $Rm_1g\sin\theta$ ($R$ is the radius of the flywheel), so the equation of motion becomes:

$I\ddot\theta+Rm_1g\sin\theta=0$.

For small $\theta$, then $\sin\theta \approx \theta$, so we get:

$\large{\ddot\theta+\frac{Rm_1g}{I}\theta=0}$.

The solution of this classic DE is:

$\theta(t)=\theta_0 \cos(\sqrt{\frac{Rm_1g}{I}}t)$, with period $T$:

$T=2\pi\sqrt{\frac{I}{Rm_1g}}$, so that:

$\Large{I=\frac{Rm_1gT^2}{4\pi^2}}$.

I've seen a method for experimentally determining the moment of inertia of a flywheel and I'm not sure whats the reasoning behind it. You attach a small weight m1 to the flywheel's edge, it's important that m1≪M, M being the mass of the flywheel, so we can approximate that the small mass does not change the MOI of the flywheel.

The derivation above is for that case described in your question. But it doesn't correspond to the formula you gave.

For the formula you gave, the inertial moment of $m_1$ is actually taken into account as follows. The actual inertial moment, including that of $m_1$ is $I'$:

$I'=I+m_1R^2$.

All we have to do now is substitute $I$ with $I'$, so that:

$T=2\pi\sqrt{\frac{I+m_1R^2}{Rm_1g}}$, so that:

$\large{I+m_1R^2=\frac{Rm_1gT^2}{4\pi^2}}$

$\Large{I=\left( \frac{T^2 g}{4 \pi ^2R} - 1\right) m_1 R^2}$.

As in your formula, but only if you do allow for the effect of $m_1$ on the total moment of inertia.