3. Flywheel — Moment of Inertia and Frictional Couple¶

Module

Rotational Dynamics · Module 1

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Aim¶

To determine the moment of inertia of a flywheel about its own axis, and the frictional couple opposing its rotation, by the falling-mass method.

Theoretical Background¶

A flywheel is a heavy wheel mounted on a horizontal axle so that it can rotate with very little friction. A cord is wound several times around the axle and a known mass m is attached to its free end. As the mass falls through a height h, it unwinds the cord and sets the flywheel spinning; when the cord finally leaves the axle (after n₁ complete turns), the flywheel continues to spin under its own inertia, gradually decelerating due to friction at the bearings and air resistance, and comes to rest after a further n₂ revolutions.

Energy conservation over the falling phase relates the mass's loss of potential energy to the kinetic energy gained by both the falling mass and the spinning flywheel, plus the work done against the (assumed constant) frictional torque over that same phase. The deceleration phase, in which the flywheel alone loses all its rotational kinetic energy to friction over n₂ revolutions, gives a second, independent equation for the frictional torque.

Combining the two phases eliminates the unknown frictional torque and yields the flywheel's moment of inertia I purely in terms of the falling mass, axle radius, height of fall, angular velocity at detachment, and the numbers of revolutions n₁ and n₂ — quantities that are all directly measurable, without needing to measure friction independently.

This experiment nicely illustrates that friction, far from being a mere nuisance to be minimised, can be measured and incorporated quantitatively into an energy-balance calculation — a technique used throughout rotating machinery design, from turbine rotors to hard-disk spindles.

Governing Formulas¶

Angular velocity when the cord leaves the axle (n₁ turns in time t):

\[ \omega = \dfrac{2\pi n_{1}}{t} \]

Moment of inertia of the flywheel:

\[ I = \dfrac{m r^{2}\left(2 g h - \omega^{2} r^{2}\right)}{\omega^{2}}\times\dfrac{n_{2}}{n_{1}+n_{2}} \]

Frictional couple:

\[ \tau_{f} = \dfrac{I\,\omega}{2\pi n_{2}\,t_{2}} \times 2\pi \;=\; \dfrac{I\,\omega^{2}}{4\pi n_{2}} \]

where \(m\) is the falling mass, \(r\) the axle radius, \(h\) the height of fall, \(t\) the fall time, \(n_1\) the number of cord turns on the axle, and \(n_2\) the number of extra revolutions made by the flywheel after the cord detaches, before it stops.

Labelled apparatus layout — original schematic prepared for this guide (not a photograph). A heavy wheel free to rotate on a nearly frictionless horizontal axle, with a cord wound around the axle carrying a falling mass over a pulley or directly off the axle, and a marker to count revolutions.

Simulator Controls¶

Control	Symbol	Range	Unit
Falling mass	m	50 – 300	g
Axle radius	r	0.5 – 2.5	cm
Height of fall	h	50 – 150	cm
Cord turns on axle	n₁	5 – 15	turns

Readouts:

Quantity	Symbol	Unit
Time of fall	t	s
Extra revolutions to stop	n₂	rev

Procedure¶

Launch the Flywheel panel; wind the cord (set n₁) and choose a falling mass m, axle radius r, and height of fall h.
Release the mass (conceptually) and record the fall time t shown on the readout, along with the number of extra revolutions n₂ the flywheel makes before stopping.
Repeat for at least five different values of the falling mass m, keeping r, h and n₁ fixed, tabulating t and n₂ each time.
For each trial compute ω = 2πn₁/t, then calculate I using the governing formula.
Verify that the calculated I remains reasonably constant across trials with different m — this is the internal consistency check for this method.

Observation Table¶

Table: Variation of fall time and stopping revolutions with falling mass

S. No.	m (g)	h (cm)	t (s)	n₂ (rev)	ω = 2πn₁/t (rad/s)	I (g·cm²)
1
2
3
4
5
6

Graph¶

Plot 1/t² (y-axis) against Falling mass, m (x-axis).

Expected trend: an increasing, roughly linear trend — larger falling masses accelerate the system faster, shortening the fall time.

1/t² vs m, used to cross-check the consistency of the calculated I

15×15 grid — choose your own scale per square; plot from your observation table

Calculation¶

For each trial, compute ω and then I from the formula above.
Take the mean of I over all trials as the best estimate of the flywheel's moment of inertia.
Using the mean I, compute the frictional couple τ_f for each trial and take its mean.

Result¶

Table: Moment of inertia and frictional couple of the flywheel

Quantity	Mean value	Unit
1
2

Maximum Permissible Error¶

The axle radius r and fall time t both enter squared in the governing formula, so their fractional errors are doubled before summing with the others.

Precautions¶

Wind the cord evenly and tightly, in a single layer, so the effective axle radius r does not change during unwinding.
Make sure the cord detaches cleanly from the axle at the end of n₁ turns rather than slipping or tangling.
Count revolutions during the deceleration phase carefully; this count (n₂) is just as important as the timed fall itself.

Self-Check Quiz¶

1. If the axle radius r is doubled (all else fixed), the angular velocity ω at cord detachment, for the same fall time t, is:

A) Doubled
B) Halved
C) Unchanged, since ω = 2πn₁/t does not involve r directly
D) Quadrupled

Answer: C — ω depends only on the number of turns n₁ and the time t, not directly on r — though a different r will generally change t itself for a given falling mass and height.

2. The frictional couple in this experiment is determined from:

A) A separate direct measurement with a spring balance
B) The deceleration phase, where the flywheel's rotational KE is dissipated entirely by friction over n₂ revolutions
C) Assuming friction is exactly zero
D) The colour of the flywheel

Answer: B — The whole point of the n₂ (post-detachment) measurement is to let friction alone be responsible for stopping the flywheel, giving an energy-balance equation for the frictional torque.

3. If the flywheel had no friction at all, what would n₂ be?

A) Zero
B) Infinite — it would never stop
C) Exactly equal to n₁
D) Undefined

Answer: B — With zero friction there is nothing to remove the flywheel's rotational kinetic energy, so it would continue spinning indefinitely.

Viva-Voce Questions¶

Why is the falling-mass method able to determine both I and the frictional torque without an independent friction measurement?
What assumption is made about the frictional torque during the two phases of the motion, and how reasonable is it?
How would you experimentally locate the moment when the cord just leaves the axle?
Why is it important to measure n₂ (extra revolutions) as well as the fall time t?
How would the result change if the cord were wound in more than one layer?
Why do larger flywheels (larger I) take longer to both spin up and spin down?