Moment of inertia of - rod (about end, centre), disk, ring, sphere (solid, hollow), cylinder

1.5.8 · Physics › Rotational Mechanics

WHY / WHAT / HOW

WHAT hai $r$ ? Hamesha rotation axis tak perpendicular distance, centre tak distance NAHI.
WHY $r^{2}$ ? Ek particle ki kinetic energy hai $\frac{1}{2} m v^{2} = \frac{1}{2} m (ω r)^{2}$ . Sum karne par, $K E = \frac{1}{2} (\sum m r^{2}) ω^{2} = \frac{1}{2} I ω^{2}$ . $r^{2}$ naturally nikal aata hai — yeh defined hi aise hai taaki rotational KE, translational KE jaisi lagey.
HOW compute karein: ek mass element $d m$ lo, $d m$ ko linear/area/volume density ke terms mein likho, $r$ express karo, integrate karo.

Derivation 1 — Thin Rod about its Centre

Rod length $L$ , mass $M$ , linear density $λ = M / L$ . Axis ⟂ rod ke centre se guzarti hai.

Centre se $x$ distance par element $d x$ lo, toh $r = x$ , $- L /2$ se $L /2$ tak. $I = \int_{- L /2}^{L /2} x^{2} λ d x = λ [\frac{x ^{3}}{3}]_{- L /2}^{L /2} = λ \cdot \frac{2}{3} \frac{L ^{3}}{8} = \frac{λ L ^{3}}{12}$ Yeh step kyun? Humne $\int x^{2} d x = x^{3} /3$ use kiya aur symmetry ne half-integral ko double kar diya. Ab $λ L = M$ : $I_{centre} = \frac{1}{12} M L^{2}$

Derivation 2 — Rod about its End

Ab $x$ , $0 \to L$ tak jaata hai: $I = \int_{0}^{L} x^{2} λ d x = λ \frac{L ^{3}}{3} = \frac{1}{3} M L^{2}$ $I_{end} = \frac{1}{3} M L^{2}$ Zyada kyun? Mass, end axis se centre axis ke comparison mein average par zyada door hai. Parallel Axis Theorem se check karo: $I_{e n d} = I_{c m} + M d^{2} = \frac{1}{12} M L^{2} + M (L /2)^{2} = \frac{1}{12} M L^{2} + \frac{1}{4} M L^{2} = \frac{1}{3} M L^{2}$ . ✓

Derivation 3 — Ring about central axis (⟂ to plane)

Mass ka har hissa axis se same distance $R$ par hai. Toh $r = R$ constant hai: $I = \int R^{2} d m = R^{2} \int d m = M R^{2} \Rightarrow I_{ring} = M R^{2}$ Yeh step kyun? $R$ integral se bahar aa jaata hai — yeh diye gaye $M, R$ ke liye maximum hai kyunki saara mass sabse door possible point par baitha hai.

Derivation 4 — Disk about central axis

Disk radius $R$ , surface density $σ = M / (π R^{2})$ . Radius $r$ , thickness $d r$ ke thin rings mein baanto. Har ring ka mass $d m = σ (2 π r d r)$ hai aur saara distance $r$ par hai: $I = \int_{0}^{R} r^{2} (σ 2 π r) d r = 2 π σ \int_{0}^{R} r^{3} d r = 2 π σ \frac{R ^{4}}{4} = \frac{π σ R ^{4}}{2}$ $σ = M / (π R^{2})$ substitute karo: $I_{disk} = \frac{1}{2} M R^{2}$ Yeh step kyun? Humne ring result ko building block ki tarah reuse kiya — Feynman-style decomposition. Disk < ring kyunki disk ka mass andar ki taraf spread hota hai.

Solid cylinder apni long axis ke baare mein = disks ka stack ⇒ same answer $\frac{1}{2} M R^{2}$ (length irrelevant hai).

Derivation 5 — Hollow Sphere (spherical shell)

Mass $M$ , radius $R$ , surface density $σ = M / (4 π R^{2})$ . Axis se polar angle $θ$ use karke rings mein slice karo. Angle $θ$ par ring ka:

radius (axis se distance) $r = R sin θ$
width $R d θ$ , circumference $2 π R sin θ$
mass $d m = σ (2 π R sin θ) (R d θ)$

$I = \int_{0}^{π} (R sin θ)^{2} σ 2 π R^{2} sin θ d θ = 2 π σ R^{4} \int_{0}^{π} sin^{3} θ d θ$ Ab $\int_{0}^{π} sin^{3} θ d θ = \frac{4}{3}$ hai, toh $I = 2 π σ R^{4} \cdot \frac{4}{3} = \frac{8}{3} π σ R^{4}$ . $σ$ substitute karo: $I = \frac{8}{3} π R^{4} \cdot \frac{M}{4 π R ^{2}} = \frac{2}{3} M R^{2} \Rightarrow I_{hollow} = \frac{2}{3} M R^{2}$

Derivation 6 — Solid Sphere

Mass $M$ , radius $R$ , $ρ = M / (\frac{4}{3} π R^{3})$ . Thin spherical shells radius $r$ , thickness $d r$ , shell mass $d m = ρ 4 π r^{2} d r$ se banao. Har shell contribute karta hai (hollow result) $\frac{2}{3} r^{2} d m$ : $I = \int_{0}^{R} \frac{2}{3} r^{2} (ρ 4 π r^{2}) d r = \frac{8 π ρ}{3} \int_{0}^{R} r^{4} d r = \frac{8 π ρ}{3} \frac{R ^{5}}{5} = \frac{8 π ρ R ^{5}}{15}$ $ρ$ substitute karo: $I = \frac{8 π R ^{5}}{15} \cdot \frac{3 M}{4 π R ^{3}} = \frac{2}{5} M R^{2} \Rightarrow I_{solid} = \frac{2}{5} M R^{2}$ Solid < hollow kyun? Solid sphere mein bahut saara mass centre ke paas baitha hai ( $r < R$ ), jo average $r^{2}$ ko kam karta hai.

Figure — Moment of inertia of - rod (about end, centre), disk, ring, sphere (solid, hollow), cylinder

Summary Table (yeh 20% yaad karo)

Body	Axis	$I$
Rod	centre, ⟂	$\frac{1}{12} M L^{2}$
Rod	end, ⟂	$\frac{1}{3} M L^{2}$
Ring	central ⟂	$M R^{2}$
Disk / solid cylinder	central long	$\frac{1}{2} M R^{2}$
Hollow sphere	diameter	$\frac{2}{3} M R^{2}$
Solid sphere	diameter	$\frac{2}{5} M R^{2}$

Common Mistakes (Steel-manned)

Active Recall

Recall Quick self-test (answers cover karo)

Rod end vs centre ratio? → $4 : 1$ ( $\frac{1}{3}$ vs $\frac{1}{12}$ ).
Disk = half ring kyun? → disk ka mass andar $r < R$ tak spread hai.
Solid sphere derive karo? → hollow shells $\frac{2}{3} r^{2} d m$ ko integrate karke.

Recall Feynman: ek 12-saal ke bachche ko explain karo

Kisi cheez ko spin karna waisa hi hai jaise log circles mein daudaana. Merry-go-round ke centre se door khada ek bachcha bahut tez daudhna padta hai (bada circle) — woh "start karna mushkil" hai. Moment of inertia bas yahi add karta hai ki "spinning pole se har ek weight-piece kitna door hai, squared." Agar saara weight pole ke paas bunch karo (solid sphere), toh spin karna aasaan hai. Agar sab kuch rim par push kar do (ek ring), toh yeh sabse mushkil hai. Poori kahani yahi hai: door weight = spin karna mushkil.

Moment of inertia definition (integral)

I = \int r^{2} d m

, jahan

r

= rotation axis tak perpendicular distance; units

kg m^{2}

I

mein

r^{2}

kyun aata hai

K E = \frac{1}{2} m (ω r)^{2}

se; sum karne par

\frac{1}{2} (\sum m r^{2}) ω^{2} = \frac{1}{2} I ω^{2}

milta hai.

Rod about centre, axis ⟂

\frac{1}{12} M L^{2}

Rod about end, axis ⟂

\frac{1}{3} M L^{2}

(= centre value +

M (L /2)^{2}

Ring about central ⟂ axis

M R^{2}

(saara mass distance

R

par hai).

Disk / solid cylinder about long axis

\frac{1}{2} M R^{2}

(length-independent).

Hollow sphere about diameter

\frac{2}{3} M R^{2}

Solid sphere about diameter

\frac{2}{5} M R^{2}

Solid sphere < hollow sphere kyun

Solid mein mass centre ke paas hai (

r < R

) jo average

r^{2}

kam karta hai.

Disk integral ka building block

Thin rings

d m = σ 2 π r d r

, har ek poori tarah distance

r

par.

Solid sphere ka building block

Thin spherical shells, har ek

\frac{2}{3} r^{2} d m

contribute karta hai.

\int_{0}^{π} sin^{3} θ d θ

hollow sphere mein use hota hai

= \frac{4}{3}

Connections

Parallel Axis Theorem — $I$ ko centre of mass se shift karta hai.
Perpendicular Axis Theorem — planar body axes relate karta hai (disk: $I_{z} = 2 I_{d iam e t er}$ ).
Rotational Kinetic Energy — $r^{2}$ ka origin.
Torque and Angular Acceleration — $τ = I α$ yahi values use karta hai.
Rolling Motion — $I$ ratios rolling acceleration order set karte hain: sphere > cylinder > ring.
Radius of Gyration — $I = M k^{2}$ .