Classical-Mechanics · Unit 20 · Video 4 · Interactive Practice

Moment of Inertia of a Disc and the Parallel Axis Theorem

IKey Formulas

Formula	Name	What you need
$\sigma = \dfrac{M}{\pi R^2}, \qquad dm = \sigma\, r\, dr\, d\theta$	Polar mass element	A uniform disc; the patch of area $r\, dr\, d\theta$ sits at perpendicular distance $r$ from the axle
$I_{\text{cm}} = \dfrac{M}{\pi R^2}\displaystyle\int_0^R\!\!\int_0^{2\pi} r^3\, d\theta\, dr = \tfrac{1}{2}MR^2$	Uniform disc about its centre	Mass and radius; a thin hoop of the same $M$ and $R$ has the full $MR^2$
$I_S = I_{\text{cm}} + m\, d_{S,\text{cm}}^2$	Parallel axis theorem	The centre-of-mass moment and the perpendicular separation $d_{S,\text{cm}}$ of the two axes
$I_{\text{end}} = \tfrac{1}{3}mL^2 = 4\, I_{\text{cm}}, \qquad I_{\text{edge}} = \tfrac{3}{2}MR^2$	The two worked results	Rod about a perpendicular axis through one end; disc about a perpendicular axis through its rim

Key Insight: $m\, d_{S,\text{cm}}^2$ is never negative, so of all parallel axes the one through the centre of mass gives the smallest moment of inertia — and since the theorem connects an axis only to that one, moving from $S$ to $S'$ means subtracting $m\, d_{S,\text{cm}}^2$ and then adding $m\, d_{S',\text{cm}}^2$ .

IIVisualization 1 — Building the Disc Ring by Ring

The mass inside radius $a$ grows like $a^2$ , but the moment of inertia it carries grows like $a^4$ .

Radius reached (a/R) 0.45

IIIVisualization 2 — Sliding the Axis off the Centre

Every parallel axis costs $M d^2$ more than the axis through the centre of mass.

Axis offset (d/R) 1.00

💡 The theorem links an axis to the centre-of-mass axis and to no other: to go from $S$ straight to a second displaced axis $S'$ , subtract $m\, d_{S,\text{cm}}^2$ first and then add $m\, d_{S',\text{cm}}^2$ .

IVVisualization 3 — The Rod About Its End, Two Ways

The axis through the end of the rod: once by the theorem, once by a fresh integral.

Step centre axis

Step 1 — The centre-of-mass axis

I_{\text{cm}} = \lambda\int_{-L/2}^{L/2} x^2\, dx = \frac{m}{L}\cdot\frac{L^3}{12} = \tfrac{1}{12}mL^2

The shaded area is

I/\lambda

: the rod is only ever

L/2

from this axis.

Step 2 — Move the axis to the end

d_{S,\text{cm}} = \tfrac{L}{2}

I_{\text{end}} = \tfrac{1}{12}mL^2 + m\left(\tfrac{L}{2}\right)^2 = \tfrac{1}{12}mL^2 + \tfrac{3}{12}mL^2 = \tfrac{1}{3}mL^2

Step 3 — The same result by integration

x' = x + \tfrac{L}{2}, \qquad 0 \le x' \le L

I_{\text{end}} = \lambda\int_{0}^{L} (x')^2\, dx' = \frac{m}{L}\cdot\frac{L^3}{3} = \tfrac{1}{3}mL^2

Step 4 — Four times the centre-of-mass value

\frac{I_{\text{end}}}{I_{\text{cm}}} = \frac{\tfrac{1}{3}mL^2}{\tfrac{1}{12}mL^2} = 4

One addition, or a second integral with new limits — the same number either way.

VQuiz Questions

Problem 1 · Disc, Axis Halfway to the Rim

Given: a uniform disc of mass $M$ and radius $R$ has $I_{\text{cm}} = \tfrac{1}{2}MR^2$ about the perpendicular axis through its centre; it is remounted on a parallel axle through the point halfway from the centre to the rim — find $I_S$ .

✅ Correct! With

d_{S,\text{cm}} = R/2

the theorem adds

M(R/2)^2 = \tfrac{1}{4}MR^2

\tfrac{1}{2}MR^2

❌ That is only the added term.

M d^2

is what the shift adds to

I_{\text{cm}}

; it does not replace it.

❌ That is the value about the centre. The new axis is a distance

R/2

away, and every displaced axis costs an extra

M d^2

❌ That is the rim value. Halfway out means

d = R/2

, so

d^2 = R^2/4

, not

R^2

❌ Not quite. Use

I_S = I_{\text{cm}} + M d^2

with

d = R/2

Show solution

The axis separation is half the radius:

d_{S,\text{cm}} = \tfrac{R}{2} \quad\Longrightarrow\quad M d_{S,\text{cm}}^2 = M\left(\tfrac{R}{2}\right)^2 = \tfrac{1}{4}MR^2

The parallel axis theorem adds that to the centre-of-mass value:

I_S = \tfrac{1}{2}MR^2 + \tfrac{1}{4}MR^2 = \tfrac{3}{4}MR^2

Between the centre value $\tfrac{1}{2}MR^2$ and the rim value $\tfrac{3}{2}MR^2$ , as it must be — the added term grows with $d^2$ , so halfway out costs only a quarter of what the rim costs.

Problem 2 · Rod, a Quarter of the Way In

Given: a uniform rod of mass $m$ and length $L$ has $I_{\text{cm}} = \tfrac{1}{12}mL^2$ about the perpendicular axis through its centre and $I_{\text{end}} = \tfrac{1}{3}mL^2$ about the parallel axis through one end; a perpendicular axis is now placed through the point a distance $L/4$ from that end — find the moment of inertia about it.

✅ Correct! The new axis is

L/2 - L/4 = L/4

from the centre of mass, so

\tfrac{1}{12}mL^2 + \tfrac{1}{16}mL^2 = \tfrac{7}{48}mL^2

❌ The theorem was started from the wrong axis.

I_S = I_{\text{cm}} + m d^2

measures

d

from the centre-of-mass axis; adding

m(L/4)^2

I_{\text{end}}

has no meaning.

❌ Subtracting from $I_{\text{end}}$ does not work either. To leave the end axis you must subtract the separation it already carries,

m(L/2)^2

, which returns you to

I_{\text{cm}}

— then add

m(L/4)^2

❌ Check which distance matters. The separation is measured to the centre of mass, and the point sits

L/4

from the near end, hence

L/4

from the middle — not

3L/4

❌ Not quite. Locate the new axis relative to the centre of mass first, then add

m d^2

\tfrac{1}{12}mL^2

Show solution

Step 1: find the separation from the centre-of-mass axis. The centre of mass is at the midpoint, $L/2$ from either end, and the new axis is $L/4$ from one end:

d_{S,\text{cm}} = \tfrac{L}{2} - \tfrac{L}{4} = \tfrac{L}{4}

Step 2: apply the theorem from the centre-of-mass value.

I_S = \tfrac{1}{12}mL^2 + m\left(\tfrac{L}{4}\right)^2 = \tfrac{1}{12}mL^2 + \tfrac{1}{16}mL^2 = \tfrac{4}{48}mL^2 + \tfrac{3}{48}mL^2 = \tfrac{7}{48}mL^2

Why $I_{\text{end}}$ is the wrong starting point: the theorem holds only between an arbitrary axis and the parallel axis through the centre of mass. Starting at the end axis, the legal route is

I_S = I_{\text{end}} - m\left(\tfrac{L}{2}\right)^2 + m\left(\tfrac{L}{4}\right)^2 = \tfrac{16}{48}mL^2 - \tfrac{12}{48}mL^2 + \tfrac{3}{48}mL^2 = \tfrac{7}{48}mL^2

which passes through $I_{\text{cm}} = \tfrac{4}{48}mL^2$ on the way and gives the same answer.

Problem 3 · Disc Spinning About Its Rim

Given: the uniform disc of mass $M$ and radius $R$ ( $I_{\text{cm}} = \tfrac{1}{2}MR^2$ ) spins with angular speed $\omega$ about a perpendicular axis through a point on its rim — find its moment of inertia about that axis and its kinetic energy.

Moment of inertia about the rim axis?

Kinetic energy at angular speed $\omega$ ?

✅ Correct!

I_{\text{edge}} = \tfrac{3}{2}MR^2

and

K = \tfrac{1}{2}I_{\text{edge}}\omega^2 = \tfrac{3}{4}MR^2\omega^2

— three times the

\tfrac{1}{4}MR^2\omega^2

the same disc has about its centre at the same

\omega

❌ Check the added term. The rim is a distance

d = R

from the centre of mass, so the theorem adds a full

MR^2

\tfrac{1}{2}MR^2

❌ Check the energy.

K = \tfrac{1}{2}I\omega^2

must use the moment of inertia about the rotation axis, and that factor of

\tfrac{1}{2}

is easy to drop.

Show solution

Step 1: the moment of inertia about the rim axis. The separation between the rim axis and the parallel axis through the centre is the radius, $d_{S,\text{cm}} = R$ :

I_{\text{edge}} = I_{\text{cm}} + MR^2 = \tfrac{1}{2}MR^2 + MR^2 = \tfrac{3}{2}MR^2

Step 2: the kinetic energy uses the moment about the rotation axis.

K = \tfrac{1}{2}I_{\text{edge}}\,\omega^2 = \tfrac{1}{2}\cdot\tfrac{3}{2}MR^2\,\omega^2 = \tfrac{3}{4}MR^2\omega^2

Compare the centre axis at the same $\omega$ :

K_{\text{cm}} = \tfrac{1}{2}\cdot\tfrac{1}{2}MR^2\,\omega^2 = \tfrac{1}{4}MR^2\omega^2

The added term $MR^2$ is by itself twice $I_{\text{cm}}$ , so spinning about the rim stores three times the energy at the same angular speed.

Problem 4 · Hoop Against Disc, Both About the Rim

Given: a thin hoop and a uniform disc share the same mass $M$ and radius $R$ , with $I_{\text{cm}}^{\text{hoop}} = MR^2$ and $I_{\text{cm}}^{\text{disc}} = \tfrac{1}{2}MR^2$ ; each spins at the same angular speed $\omega$ about a perpendicular axis through a point on its own rim — find the ratio $K_{\text{hoop}}/K_{\text{disc}}$ .

✅ Correct! Both bodies gain the same

MR^2

, so the ratio

2MR^2 : \tfrac{3}{2}MR^2 = \tfrac{4}{3}

is closer to 1 than the

2:1

of their centre-of-mass values.

❌ That ratio is upside down. The hoop has the larger moment of inertia about either axis, so at equal

\omega

it must have the larger kinetic energy.

❌ That is the ratio about the centres. Both axes were moved to the rim, and the identical

MR^2

each one gains shrinks the ratio.

❌ The hoop needs the theorem too.

MR^2

is the hoop about its centre; about a rim axis it is

MR^2 + MR^2 = 2MR^2

❌ Not quite. Apply

I_S = I_{\text{cm}} + MR^2

to each body, then take the ratio — the equal

\omega

and the factor

\tfrac{1}{2}

cancel.

Show solution

Step 1: move each axis to the rim. The separation is $d_{S,\text{cm}} = R$ in both cases:

I_{\text{edge}}^{\text{hoop}} = MR^2 + MR^2 = 2MR^2, \qquad I_{\text{edge}}^{\text{disc}} = \tfrac{1}{2}MR^2 + MR^2 = \tfrac{3}{2}MR^2

Step 2: take the ratio of the energies. With the same $\omega$ , the factor $\tfrac{1}{2}\omega^2$ cancels:

\frac{K_{\text{hoop}}}{K_{\text{disc}}} = \frac{\tfrac{1}{2}(2MR^2)\omega^2}{\tfrac{1}{2}\left(\tfrac{3}{2}MR^2\right)\omega^2} = \frac{2}{\tfrac{3}{2}} = \frac{4}{3}

What the theorem did to the comparison: about their centres the hoop beats the disc $2:1$ , because the disc keeps mass in near the axle. Displacing both axes by the same $R$ adds the same $MR^2$ to each, and adding equal amounts to two unequal numbers pushes their ratio toward 1 — here from $2$ down to $\tfrac{4}{3}$ .

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