力学 Classical Mechanics · Laboratory experiment

Moment of Inertia of a Flywheel

Aim: to determine the moment of inertia $I$ of a heavy flywheel about its horizontal axis of rotation by converting gravitational potential energy $mgh$ into rotational kinetic energy $\frac12 I \omega^2$ and accounting for bearing friction.

Flywheel Rig & Revolution Counter

Cord Wound · Ready to Release

Click "Release Mass" to drop the weight. Watch the cord unwind, detach, and the flywheel coast under bearing friction.

Axle Vernier Caliper (Radius $r$)

Energy Partition at Detachment

Potential energy $mgh$ splits into Flywheel Rotation ($\sim 96\%$), Falling Mass Kinetic ($\sim 1\%$), and Descent Bearing Friction ($\sim 3\%$).

Observation Table

Theory of Flywheel Energy Conservation

A flywheel of mass $M$ and radius $R$ is mounted on a horizontal axle of radius $r$. A cord with a mass $m$ attached to its lower end is wrapped $N_1$ times around the axle and released from rest at height $h$.

As the mass falls, its initial potential energy $mgh$ transforms into linear kinetic energy of the falling mass, rotational kinetic energy of the flywheel, and work done against axle bearing friction:

$$ m g h = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 + W_1 $$

where $v = r \omega$ is the linear speed at detachment. Let $W_f$ be the frictional work dissipated in one complete revolution of the flywheel. During the fall ($N_1$ revolutions):

$$ W_1 = N_1 W_f $$

After the cord detaches, the flywheel continues to coast, making $N_2$ complete revolutions in time $t_2$ before bearing friction brings it to rest. The rotational kinetic energy at detachment is completely dissipated in these $N_2$ turns:

$$ \frac{1}{2} I \omega^2 = N_2 W_f \implies W_f = \frac{\frac{1}{2} I \omega^2}{N_2} $$

Substituting $W_1 = \left(\frac{N_1}{N_2}\right)\left(\frac{1}{2} I \omega^2\right)$ into the energy equation:

$$ m g h = \frac{1}{2} m r^2 \omega^2 + \frac{1}{2} I \omega^2 \left(1 + \frac{N_1}{N_2}\right) $$

Solving for the moment of inertia $I$:

$$ I = \frac{2 m g h - m r^2 \omega^2}{\omega^2 \left(1 + \frac{N_1}{N_2}\right)} = \frac{N_2}{N_1 + N_2} \left[\frac{2 m g h}{\omega^2} - m r^2\right] $$

Measurement of Detachment Speed & Precautions

Since the frictional deceleration $\alpha$ is approximately constant, the angular velocity decreases linearly from $\omega$ to $0$ in time $t_2$:

$$ \theta_2 = 2\pi N_2 = \omega_{\text{avg}} t_2 = \frac{\omega + 0}{2} t_2 \implies \omega = \frac{4\pi N_2}{t_2} $$

The frictional torque opposing rotation is:

$$ \tau_f = I \alpha = I \left(\frac{\omega}{t_2}\right) = \frac{4\pi N_2 I}{t_2^2} $$

Precautions

The string should be thin, light, and wrapped evenly in a single layer without overlapping coils.
The loop slipped over the axle peg must detach freely at the bottom of the fall without tugging or jerking the wheel.
The axle diameter $2r$ must be measured in multiple directions using a Vernier caliper to account for slight eccentricity.

Viva Questions

Why is the mass of the flywheel concentrated mostly in its rim?

Moment of inertia depends on $\int r^2 dm$. Concentrating mass at the outer perimeter ($r = R$) maximizes $I$ for a given total mass, storing maximum rotational kinetic energy for smoothing cyclic variations in engines.

Why must we subtract the term $m r^2 \omega^2$?

The falling mass retains kinetic energy $\frac{1}{2} m v^2 = \frac{1}{2} m r^2 \omega^2$ at the instant it detaches. This energy hits the floor and is not transferred into the flywheel, so it must be subtracted from $mgh$.