Parallel axis theorem — I = I_CM + Md² — proof: Worked examples

1.5.6 · D3 · Physics › Rotational Mechanics › Parallel axis theorem — I = I_CM + Md² — proof

Yeh page ek workout hai. Parent proof ne sirf ek line $I = I_{C M} + M d^{2}$ banai thi. Yahan hum us ek line ko har tarah ki situation mein chalate hain — taaki jab exam ya koi real machine tumhare saamne koi case leke aaye, tum uska joda pehle se dekh chuke ho.

Shuru karne se pehle, ek reminder plain words mein. $I$ (yaani moment of inertia, dekho Moment of inertia — definition) ek aisa number hai jo batata hai ki kisi body ko space mein kisi chosen line (yaani "axis") ke around ghoomna shuru ya band karna kitna mushkil hai. $I_{C M}$ woh number hai jab woh line balance point — yaani center of mass (CM) — se guzarti hai. $M$ total mass hai. Aur $d$ perpendicular distance hai — CM-axis aur jis parallel axis ki tumhe zaroorat hai, unke beech ka seedha, right angle pe measure kiya hua, sabse chhota gap.

The scenario matrix

Is theorem se jo bhi problem ban sakti hai woh in cells mein se kisi ek mein aati hai. Neeche ke examples mein cell(s) label kiye gaye hain.

Cell	Case class	Kya cheez mushkil banati hai	Covered by
C1	Direct: jaana hua $I_{C M}$ → parallel axis distance $d$ par	bas $M d^{2}$ add karo	Ex 1, Ex 2
C2	Degenerate: $d = 0$ (axis CM-axis HI hai)	correction khatam ho jaati hai	Ex 3
C3	Reverse: jaana hua $I$ kisi point par → $I_{C M}$ nikalo (subtract karo)	ulta jaana padega	Ex 4
C4	Do non-CM axes ke beech	inhe directly connect NAHI kar sakte	Ex 5
C5	$d$ do components $(a, b)$ se bana ho	$d^{2} = a^{2} + b^{2}$ , na ki $a + b$	Ex 6
C6	Limiting / minimisation: kaun si axis sabse kam $I$ deti hai?	$M d^{2} \geq 0$ ⇒ CM wins	Ex 7
C7	Real-world word problem (physical pendulum)	words ko $d$ mein translate karo	Ex 8
C8	Exam twist: Perpendicular axis theorem ya Radius of gyration ke saath combine	do theorems chain mein	Ex 9, Ex 10

Worked example Example 1 — Rod apne end ke baare mein (Cell C1)

Thin uniform rod, mass $M$ , length $L$ . Diya hai $I_{C M} = \frac{1}{12} M L^{2}$ apne middle se guzarne wali perpendicular axis ke baare mein. $I$ nikalo ek ek end se guzarne wali perpendicular axis ke baare mein.

Forecast: padhne se pehle andaza lagao — end wali value middle wali se badi hogi ya chhoti, aur lagbhag kitni?

CM locate karo. Uniform rod ke liye balance point midpoint hota hai. Yeh step kyun? $d$ ko CM se measure kiya jaata hai, isliye hume pata hona chahiye ki CM kahan hai.
$d$ nikalo. Midpoint se end tak $d = \frac{L}{2}$ hai. Yeh step kyun? $d$ woh ⟂ distance hai jo (parallel) middle-axis aur end-axis ke beech hai — in dono axes ke liye woh gap bas aadha rod hai.
$M d^{2}$ add karo. $I_{e n d} = \frac{1}{12} M L^{2} + M (\frac{L}{2})^{2} = \frac{1}{12} M L^{2} + \frac{1}{4} M L^{2} = \frac{4}{12} M L^{2} = \frac{1}{3} M L^{2}$ Yeh step kyun? Ek axis (beech wali) CM se guzarti hai, isliye theorem bina kisi cross terms ke apply hota hai.

Verify karo: $\frac{1}{3} M L^{2}$ standard textbook value hai rod ke end ke baare mein (direct integration se). End wali value badi hai — rod ko uski tip se ghoomna beech se ghoomne se zyada mushkil hai. ✓ Units: $[kg] [m^{2}]$ throughout. ✓

Figure — Parallel axis theorem — I = I_CM + Md² — proof

Worked example Example 2 — Solid disc apne rim ke baare mein (Cell C1, geometric)

Solid disc, mass $M$ , radius $R$ . Perpendicular central axis deta hai $I_{C M} = \frac{1}{2} M R^{2}$ . $I$ nikalo ek perpendicular axis ke baare mein jo rim par kisi point se guzarti hai.

Forecast: rim centre se $R$ distance par hai. $M R^{2}$ ka multiplier guess karo.

Confirm karo ki axes parallel hain. Dono disc ke face se seedhe bahar nikalte hain (disc ke perpendicular), isliye woh parallel hain. Yeh step kyun? Theorem sirf parallel axes ko relate karta hai; agar parallel nahi hote toh hum ise use nahi kar sakte the.
$d$ read karo. Centre se rim $= R$ , isliye $d = R$ . Yeh step kyun? $d$ do parallel lines ke beech ka ⟂ gap hai, jo yahan simply radius hi hai.
Correction add karo. $I_{r im} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}$

Verify karo: $\frac{3}{2} M R^{2} = 1.5 M R^{2}$ . Sanity check: yeh $I_{C M} = 0.5 M R^{2}$ se zyada hona chahiye, aur hai — axis ko bahar le jaana hamesha inertia badhata hai. ✓

Worked example Example 4 —

$I_{C M}$ nikalte hue ulte kaam karna (Cell C3) Ek flat irregular plate ka mass $M = 2 kg$ hai. Ek axis ke baare mein measure kiya gaya jo uske CM se $d = 0.30 m$ door hai (desired CM-axis ke parallel), uska moment of inertia $I = 0.50 kg\cdotm^{2}$ hai. $I_{C M}$ nikalo.

Forecast: kya $I_{C M}$ $0.50$ se chhoti aanI chahiye? Kyun?

Theorem likho, phir rearrange karo. $I = I_{C M} + M d^{2} \Rightarrow I_{C M} = I - M d^{2}$ . Yeh step kyun? Hume door wali axis pata hai aur CM axis chahiye, isliye hum correction subtract karte hain, add nahi.
$M d^{2}$ compute karo. $M d^{2} = 2 \times (0.30)^{2} = 2 \times 0.09 = 0.18 kg\cdotm^{2}$ . Yeh step kyun? Yeh point-mass contribution hai jo hume hatana hai.
Subtract karo. $I_{C M} = 0.50 - 0.18 = 0.32 kg\cdotm^{2}$

Verify karo: $I_{C M} = 0.32 < 0.50$ ✓ — CM-axis sabse sasti honi chahiye, isliye chhoti honi chahiye. Wapas plug karo: $0.32 + 0.18 = 0.50$ ✓. Units: kg·m² ✓.

Worked example Example 5 — Do non-CM axes (Cell C4, classic trap)

Ek solid sphere, mass $M$ , radius $R$ , $I_{C M} = \frac{2}{5} M R^{2}$ . Axis A CM se $d_{A} = R$ door hai; Axis B CM se $d_{B} = 2 R$ door hai; dono ek doosre ke aur CM-axis ke parallel hain. $I_{B} - I_{A}$ nikalo.

Forecast: tempting shortcut kehta hai $I_{B} = I_{A} + M (d_{B} - d_{A})^{2} = I_{A} + M R^{2}$ . Kya yeh sahi hai?

Direct link refuse karo. Tum $I_{B} = I_{A} + M (gap)^{2}$ nahi likh sakte, kyunki A aur B mein se koi bhi CM se nahi guzarti, isliye cross terms khatam NAHI honge. Yeh step kyun? Theorem ke clean form ke liye ek axis CM par honi chahiye.
Dono ko CM se route karo. $I_{A} = \frac{2}{5} M R^{2} + M R^{2}, I_{B} = \frac{2}{5} M R^{2} + M (2 R)^{2} = \frac{2}{5} M R^{2} + 4 M R^{2}$ Yeh step kyun? CM woh shared reference hai jisse dono axes legally connect kar sakti hain.
Subtract karo. $I_{B} - I_{A} = (4 M R^{2}) - (M R^{2}) = 3 M R^{2}$

Verify karo: naive shortcut ne $M R^{2}$ diya tha; sahi jawab $3 M R^{2}$ hai. Woh alag hain, jo prove karta hai ki shortcut galat tha. $\frac{2}{5} M R^{2}$ terms difference mein cancel ho jaate hain, jaise hone chahiye (same body, same $I_{C M}$ ). ✓

Worked example Example 6 —

$d$ do components se (Cell C5) Ek thin plate ek perpendicular axis ke around ghoom rahi hai jo uske CM se guzarti hai, $I_{C M} = 0.040 kg\cdotm^{2}$ , mass $M = 0.50 kg$ . Nayi (parallel) axis plane ko us point $(a, b) = (0.30 m, 0.40 m)$ par pierce karti hai jo CM se measure kiya gaya hai. $I$ nikalo.

Forecast: dhyan do — kya $d = a + b = 0.70$ hai, ya kuch aur?

Geometry se $d$ banao. CM se point $(a, b)$ tak ⟂ distance hypotenuse hai: $d^{2} = a^{2} + b^{2}$ . Yeh step kyun? Plane mein distance Pythagoras se milti hai, do legs ka sum NAHI — yeh exactly woh $a^{2} + b^{2} = d^{2}$ line hai jo proof mein thi. $d^{2} = (0.30)^{2} + (0.40)^{2} = 0.09 + 0.16 = 0.25 m^{2} \Rightarrow d = 0.50 m$
$M d^{2}$ add karo. $I = 0.040 + 0.50 \times 0.25 = 0.040 + 0.125 = 0.165 kg\cdotm^{2}$ Yeh step kyun? Hume $M d^{2}$ chahiye, aur $d^{2} = 0.25$ seedha mil gaya hai, isliye square root lene ki bhi zaroorat nahi.

Verify karo: $d = 0.70$ use karne wala trap answer $M d^{2} = 0.50 \times 0.49 = 0.245$ deta, jo bahut zyada hai. (3,4,5) triangle $d = 0.5$ cleanly deta hai. Final $0.165 kg\cdotm^{2} > I_{C M}$ ✓.

Worked example Example 7 — Kaun si axis sabse sasti hai? (Cell C6, limiting behaviour)

Ek fixed direction ke parallel sabhi axes mein se, ek minimum moment of inertia deti hai. Ek rod ( $M$ , $L$ , $I_{C M} = \frac{1}{12} M L^{2}$ ) ke liye, $I (d) = \frac{1}{12} M L^{2} + M d^{2}$ ko plot karo jab axis CM se $d$ distance slide karti hai, aur woh jagah nikalo jahan $I$ sabse chhoti ho.

Forecast: graph ki shape aur uske lowest point ki location guess karo.

Function padho. $I (d) = I_{C M} + M d^{2}$ ek parabola hai $d$ mein, upar ki taraf khulta hua, kyunki $M d^{2} \geq 0$ hai. Yeh step kyun? Shape pehchanna bina calculus ke minimum bata deta hai.
Minimum nikalo. $M d^{2}$ sabse chhota $d = 0$ par hota hai, jo $I_{min} = I_{C M} = \frac{1}{12} M L^{2}$ deta hai. Yeh step kyun? Ek squared term kabhi negative nahi ho sakti, isliye lowest point wahan hai jahan woh zero ho — CM par.
Symmetry check karo. $I (+ d) = I (- d)$ : axis ko CM ke left ya right dono taraf same $d$ move karne par same $M d^{2}$ ka cost aata hai. Yeh step kyun? Parabola CM ke baare mein symmetric hai, jo confirm karta hai ki exactly ek hi sabse sasti axis hai.

Verify karo: $d = 0$ par, $I = \frac{1}{12} M L^{2} \approx 0.0833 M L^{2}$ ; $d = L /2$ par, $I = \frac{1}{3} M L^{2} \approx 0.333 M L^{2}$ (Ex 1 se match karta hai). Min sach mein CM par hai. ✓

Worked example Example 8 — Real-world: ek jhulta hua sign (Cell C7, physical pendulum)

Ek uniform rectangular shop sign, mass $M = 8.0 kg$ , height $h = 1.2 m$ , apni top edge ke along ek horizontal pivot se latakta hai aur ek Physical pendulum ki tarah jhulta hai. Uska CM geometric centre par hai. Ek thin flat plate ke liye CM se guzarne wali horizontal axis ke baare mein moment of inertia sign ke plane mein $I_{C M} = \frac{1}{12} M h^{2}$ hai. Top-edge pivot ke baare mein $I$ nikalo.

Forecast: pivot top edge par hai; centre se woh kitna door hai?

Words ko $d$ mein translate karo. CM centre par hai, pivot top edge par, isliye $d = \frac{h}{2} = 0.60 m$ . Yeh step kyun? "Top edge" phrase ek geometry clue hai jise hume $d$ ke liye ek number mein convert karna hai.
$I_{C M}$ compute karo. $I_{C M} = \frac{1}{12} (8.0) (1.2)^{2} = \frac{1}{12} (8.0) (1.44) = 0.96 kg\cdotm^{2}$ . Yeh step kyun? Correction add karne se pehle hume CM wali value chahiye.
$M d^{2}$ add karo. $M d^{2} = 8.0 \times (0.60)^{2} = 8.0 \times 0.36 = 2.88 kg\cdotm^{2}$ . $I_{p i v o t} = 0.96 + 2.88 = 3.84 kg\cdotm^{2}$ Yeh step kyun? Pivot axis CM-axis ke parallel hai (dono horizontal, top ke along), isliye theorem apply hota hai.

Verify karo: notice karo $I_{p i v o t} = \frac{1}{12} M h^{2} + M (h /2)^{2} = \frac{1}{3} M h^{2} = \frac{1}{3} (8.0) (1.44) = 3.84$ ✓ — rod-about-end jaisa hi $\frac{1}{3} M h^{2}$ form, kyunki apni height ke along ek thin plate rod ki tarah behave karti hai. Units kg·m² ✓.

Worked example Example 9 — Exam twist: perpendicular axis theorem ke saath chain (Cell C8)

Ek thin disc, mass $M$ , radius $R$ . Perpendicular axis theorem kehta hai ki flat plate ke liye $I_{z} = I_{x} + I_{y}$ , jahan $z$ plate ke perpendicular hai aur $x, y$ CM se hokar uske andar hain. Symmetry se $I_{x} = I_{y}$ , aur $I_{z} = \frac{1}{2} M R^{2}$ . $I$ nikalo ek diameter ke baare mein jo rim ki tangent ke along hai (disc ke plane mein ek line, edge ko tangent).

Forecast: do theorems ek baad ek. Pehle in-plane CM value nikalo, phir use tangent tak slide karo.

Perpendicular axis theorem → in-plane $I_{C M}$ . $I_{z} = I_{x} + I_{y} = 2 I_{x}$ , isliye $I_{x} = \frac{1}{2} I_{z} = \frac{1}{2} \cdot \frac{1}{2} M R^{2} = \frac{1}{4} M R^{2}$ . Yeh step kyun? Use slide karne se pehle hume ek diameter (in-plane, CM se) ke baare mein moment of inertia chahiye.
Tangent line tak $d$ identify karo. Rim par tangent line ek central diameter ke parallel hai aur $d = R$ door hai. Yeh step kyun? Parallel axis theorem ko parallel lines ke beech ⟂ gap chahiye; centre se tangent tak ek radius hai.
Parallel axis theorem apply karo. $I_{t an g e n t} = \frac{1}{4} M R^{2} + M R^{2} = \frac{5}{4} M R^{2}$

Verify karo: $\frac{5}{4} M R^{2} = 1.25 M R^{2}$ . Sanity check: yeh diameter wali value $\frac{1}{4} M R^{2}$ se zyada honi chahiye, aur hai. Dono theorems legally use kiye gaye (har stage par ek axis CM se). ✓

Worked example Example 10 — Exam twist: radius of gyration (Cell C8)

Radius of gyration $k$ define hota hai $I = M k^{2}$ se — woh distance jis par poori mass, ek point ki tarah, same $I$ deti. Ek solid sphere, $M = 3.0 kg$ , $R = 0.10 m$ , $I_{C M} = \frac{2}{5} M R^{2}$ , ek tangent axis ke around ghoom rahi hai (CM se $d = R$ door). Uski radius of gyration $k$ us tangent ke baare mein nikalo.

Forecast: kya $k$ , $R$ se bada hoga ya chhota?

Tangent $I$ ke liye parallel axis theorem. $I = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2}$ . Yeh step kyun? Tangent axis CM-axis ke parallel hai, $d = R$ door.
Definition se $k$ mein convert karo. $I = M k^{2} \Rightarrow k^{2} = \frac{I}{M} = \frac{7}{5} R^{2}$ . Yeh step kyun? Radius of gyration bas $I$ ko ek equivalent distance ke roop mein repackage karta hai. $k = R \frac{7}{5} = 0.10 1.4 \approx 0.1183 m$

Verify karo: $k \approx 0.118 m > R = 0.10 m$ ✓ — sensible hai, kyunki tangent axis sphere ki apni radius se bahar hai, isliye "equivalent point-mass distance" $R$ se zyada hogi. Numerically $1.4 \approx 1.1832$ , times $0.10 = 0.11832$ . ✓

Recall Sabhi cells par quick self-test

Direct add ( $C 1$ ) ka matlab hai... ::: $I = I_{C M} + M d^{2}$ , bas correction add karo. Agar $d = 0$ ( $C 2$ ) ho toh $I$ barabar hai... ::: exactly $I_{C M}$ ; koi correction nahi. Jaane hue far-axis $I$ se $I_{C M}$ nikalte hain ( $C 3$ ) toh... ::: subtract karo, $I_{C M} = I - M d^{2}$ . Do non-CM axes ( $C 4$ ) — safe route hai... ::: CM se alag-alag jaao, phir subtract karo. Diya hua $(a, b)$ , distance hai ( $C 5$ )... ::: $d = a^{2} + b^{2}$ , kabhi $a + b$ nahi. Sabse sasti axis ( $C 6$ ) hai... ::: CM se guzarne wali, kyunki $M d^{2} \geq 0$ .

Connections

Parent proof
Moment of inertia — definition
Center of mass — definition and computation
Perpendicular axis theorem
Radius of gyration
Rotational kinetic energy
Physical pendulum

Worked examples — Parallel axis theorem — I = I_CM + Md² — proof

The scenario matrix

Connections

Scenario Map