MHT-CET Physics · Rotational Dynamics

Moment of Inertia and Radius of Gyration

Moment of inertia I = Σmr² measures how hard a body is to spin up — it depends on the mass AND on how far that mass sits from the axis; the radius of gyration k is the single distance at which all the mass would give the same I.

All Rotational Dynamics notes MHT-CET Physics strategy

Why this matters

23 PYQs, six HARD — all six rebuild a body (melt it, recast it, bend a rod into a ring) or scale one made of the same material, and ask for the new I. The rest compare standard bodies of the same mass, or ask for a radius of gyration. Know the six standard results cold and every question is a ratio.

Concept 1 of 3: Moment of Inertia of Standard Bodies

The further the mass is from the axis, the larger I. A ring keeps all its mass at the rim (MR²); a disc spreads it inward (MR²/2); a solid sphere spreads it through the volume (2MR²/5). For the same mass and radius, the hollow body always beats the solid one.

Definition

Ring about its axis: $MR^2$ ; about a diameter: $\dfrac{MR^2}{2}$ .
Disc (or solid cylinder) about its axis: $\dfrac{MR^2}{2}$ ; disc about a diameter: $\dfrac{MR^2}{4}$ . Annular disc: $\dfrac{M}{2}(R_1^2 + R_2^2)$ .
Solid sphere about a diameter: $\dfrac{2}{5}MR^2$ ; thin spherical shell: $\dfrac{2}{3}MR^2$ .
Rod about its centre: $\dfrac{ML^2}{12}$ ; about an end: $\dfrac{ML^2}{3}$ . Square plate about a perpendicular central axis: $\dfrac{Ma^2}{6}$ .
Semicircular wire about its diameter: same as the full ring's $\dfrac{MR^2}{2}$ , with $R = \dfrac{L}{\pi}$ .
Same torque on a disc and a ring of equal mass and radius: the disc (smaller I) gains angular speed faster.

Definition

I = \sum m_i r_i^2, \qquad I_{\text{ring}} = MR^2,\; I_{\text{disc}} = \tfrac{1}{2}MR^2,\; I_{\text{sphere}} = \tfrac{2}{5}MR^2

Worked example

A solid sphere and a thin spherical shell have the same mass and radius. Ratio of their moments of inertia about a diameter?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q29Moderate

Example 1 · Rotational Dynamics · Moment of Inertia and Radius of Gyration

A square lamina of side '

b

' has same mass as a disc of radius '

R

'. The moment of inertia of the two objects about an axis perpendicular to the plane and passing through the centre is equal. The ratio

\frac{b}{R}

'Same material' does not mean 'same radius'

A solid sphere and a hollow one of the same MASS and material cannot have the same radius: the hollow one is bigger, so

I_h > I_s

even more clearly than the standard formulas suggest.

Drill 5 more on moment of inertia of standard bodies

Concept 2 of 3: Radius of Gyration

Squash all of a body's mass into a thin ring of radius k and it would have the same moment of inertia: I = Mk². So k is just √(I/M), and comparing k's is comparing I's with the mass divided out.

Definition

$k = \sqrt{\dfrac{I}{M}}$ .
Ring about its axis: $k = R$ ; disc about its axis: $\dfrac{R}{\sqrt{2}}$ ; disc about a diameter: $\dfrac{R}{2}$ ; solid sphere: $R\sqrt{\dfrac{2}{5}}$ ; rod about an end: $\dfrac{L}{\sqrt{3}}$ .
Disc to ring, same M and R: $k_d : k_r = 1 : \sqrt{2}$ . Disc, axis versus diameter: $\sqrt{2} : 1$ .
Four equal particles at the corners of a square of side L, central perpendicular axis: $k = \dfrac{L}{\sqrt{2}}$ — the distance of each from the centre.

Radius of gyration

I = Mk^2 \;\Rightarrow\; k = \sqrt{\frac{I}{M}}

Worked example

Radius of gyration of a solid sphere of radius 10 cm about a diameter?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q25Easy

Example 2 · Rotational Dynamics · Moment of Inertia and Radius of Gyration

Ratio of radius of gyration of a circular disc to that of circular ring each of same mass and radius around their respective axes is

Comparing I and reporting it as k

The disc-to-ring ratio of I is 1 : 2, but of k it is

1 : \sqrt{2}

— k is a square root. Both ratios are offered.

Drill 4 more on radius of gyration

Concept 3 of 3: Rebuilt Bodies: Recasting, Bending and Scaling

When a body is melted and recast, or a rod is bent into a ring, the MASS stays and the shape changes. Find the new radius from volume (or length), then put it into the new body's formula. For bodies made of the same stuff, the mass itself scales with size, so I climbs steeply with radius.

Definition

Same wire, loops of radius R: $M \propto R$ , $I = MR^2 \propto R^3$ . $\dfrac{I_P}{I_Q} = 27 \Rightarrow \dfrac{R_P}{R_Q} = 3$ .
Same material and thickness, discs: $M \propto R^2$ , $I \propto R^4$ . Same mass and thickness, different densities: $I \propto \dfrac{1}{d}$ .
Equal-mass spheres of densities $\rho_A, \rho_B$ : $\dfrac{I_B}{I_A} = \left(\dfrac{\rho_A}{\rho_B}\right)^{2/3}$ .
Disc of thickness $\dfrac{R}{6}$ recast into a sphere: $r = \dfrac{R}{2}$ , $I_{\text{sphere}} = \dfrac{I}{5}$ . One sphere into $n$ equal ones: each $\dfrac{I}{n^{5/3}}$ — 27 spheres give $\dfrac{I}{243}$ .
Rod ( $\dfrac{ML^2}{12}$ about its centre) bent into a ring of radius $\dfrac{L}{2\pi}$ : about a diameter $\dfrac{ML^2}{8\pi^2}$ , a ratio $\dfrac{I_{\text{ring}}}{I_{\text{rod}}} = \dfrac{3}{2\pi^2}$ .

Scaling for the same material

\text{loop: } I \propto R^3, \qquad \text{disc (same thickness): } I \propto R^4

Worked example

Two loops are made from the same wire, radii in the ratio 1 : 2. Ratio of their moments of inertia about their axes?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q15Hard

Example 3 · Rotational Dynamics · Moment of Inertia and Radius of Gyration

A disc of radius

R

and thickness

\frac{R}{6}

has moment of inertia

I

about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter is

Keeping the radius when the material fixes the mass

For loops of the same wire,

I \propto R^3

, not

R^2

— the bigger loop also has more wire. A ratio of 27 in I is 3 in radius; answering

\sqrt{27}

forgets the mass.

Drill 11 more on rebuilt bodies: recasting, bending and scaling

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Moment of Inertia of Standard Bodies
Definition
$I = \sum m_i r_i^2, \qquad I_{\text{ring}} = MR^2,\; I_{\text{disc}} = \tfrac{1}{2}MR^2,\; I_{\text{sphere}} = \tfrac{2}{5}MR^2$
Radius of Gyration
Radius of gyration
$I = Mk^2 \;\Rightarrow\; k = \sqrt{\frac{I}{M}}$
Rebuilt Bodies: Recasting, Bending and Scaling
Scaling for the same material
$\text{loop: } I \propto R^3, \qquad \text{disc (same thickness): } I \propto R^4$

Watch out for (3)

'Same material' does not mean 'same radius'
→ Moment of Inertia of Standard Bodies
Comparing I and reporting it as k
→ Radius of Gyration
Keeping the radius when the material fixes the mass
→ Rebuilt Bodies: Recasting, Bending and Scaling

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