Perpendicular axis theorem — I_z = I_x + I_y — proof, restrictions: Worked examples

1.5.7 · D3 · Physics › Rotational Mechanics › Perpendicular axis theorem — I_z = I_x + I_y — proof, restri

Yeh page parent theorem ka drill floor hai. Hum ek hi rule $I_{z} = I_{x} + I_{y}$ lete hain aur uske saamne har tarah ki problem laate hain — symmetric shapes, non-symmetric shapes, ek degenerate flat rod, ek real-world word problem, ek trap jo 3-D lagti hai, aur ek exam twist jo ise Parallel axis theorem ke saath combine karta hai. Har worked example batata hai ki woh scenario matrix ka kaun sa cell cover karta hai.

Scenario matrix

#	Case class	Kya khaas hai	Kaun sa Example hit karta hai
A	Planar, symmetric, common point	$I_{x} = I_{y}$ , aadha kar do	Ex 1 (disc), Ex 2 (ring)
B	Planar, unequal in-plane axes	$I_{x} \neq = I_{y}$ , koi shortcut nahi	Ex 3 (rectangular plate)
C	Planar, ulta solve karo (do flats se $I_{z}$ nikalo)	pehle forecast, phir add	Ex 4 (annulus)
D	Degenerate input: ek "plate" jo line ban gayi	ek in-plane $I = 0$	Ex 5 (thin rod as $b \to 0$ )
E	Trap: ek 3-D body — theorem ko refuse karo	$z \neq = 0$	Ex 6 (solid sphere)
F	Real-world word problem	words ko axes mein translate karo	Ex 7 (spinning coin sign)
G	Exam twist: perpendicular phir parallel axis	axis offset → do theorems	Ex 8 (disc, tangent perpendicular axis)
H	Limiting behaviour check	ratios aur sanity limits	Ex 9 (square, aur Ex 3 ka $a \to b$ collapse)

Neeche jo bhi hai woh sirf $\int r^2\,dm$ definition, symmetry argument, aur Moment of inertia of standard bodies ke results par based hai.

Worked Examples

Worked example 1. Disc about a diameter —

Cell A (planar, symmetric) Ek uniform disc ka mass $M$ aur radius $R$ hai. Uske face se seedha bahar nikalti central axis ke baare mein MOI ka jaana-pehchana value hai $I_{z} = \frac{1}{2} M R^{2}$ . Diameter (centre se guzarti in-plane axis) ke baare mein MOI nikalo.

Forecast: Padhne se pehle andaza lagao — kya answer $I_{z}$ se bada hoga, chhota hoga, ya aadha? Kyun?

Axes ka naam rakho. Disc ko $x y$ -plane mein rakho, centre origin $O$ par. $z$ -axis face se bahar nikaalo; $x$ - aur $y$ -axes do perpendicular diameters hain (figure mein red aur yellow). Yeh step kyun? Theorem ko teen mutually perpendicular axes chahiye jo ek hi point se guzarti hon — hum woh point centre par fix karte hain.
Symmetry use karo. Disc ko $z$ ke around kisi bhi angle se rotate karo, woh same dikhti hai, isliye koi bhi diameter khaas nahi hai: $I_{x} = I_{y}$ . Yeh step kyun? Symmetry ek integral ki jagah le leti hai. Agar shape $x$ aur $y$ dono ke along same lagti hai, toh unki MOIs barabar hongi.
Theorem apply karo. $I_{z} = I_{x} + I_{y} = 2 I_{x}$ . Yeh step kyun? Theorem ka yahi toh poora point hai — out-of-plane spin, dono in-plane spins ka sum hai.
Solve karo. $I_{x} = \frac{I _{z}}{2} = \frac{1}{2} \cdot \frac{1}{2} M R^{2} = \frac{1}{4} M R^{2}$ .

Verify: Diameter axis face axis se "zyada aasaan" spin hai, isliye $I_{x} < I_{z}$ — aur sach mein $\frac{1}{4} < \frac{1}{2}$ . ✓ Units: $[kg] [m^{2}]$ throughout. ✓

Worked example 2. Ring about a diameter —

Cell A (planar, symmetric, saara mass ek hi radius par) Ek thin ring (saara mass rim par) ka mass $M$ , radius $R$ hai, aur $I_{z} = M R^{2}$ hai. Diameter ke baare mein MOI nikalo.

Forecast: Ring ka mass disc se zyada door hai. Kya diameter MOI, $I_{z}$ ka bada ya chhota fraction hoga disc ke $\frac{1}{2}$ se compare karke?

Symmetry. Disc jaisi hi — har diameter same dikhti hai, isliye $I_{x} = I_{y}$ . Yeh step kyun? Circular symmetry phir se dono in-plane axes ko barabar karne par majboor karti hai.
Theorem apply karo. $I_{z} = 2 I_{x} \Rightarrow I_{x} = \frac{I _{z}}{2} = \frac{1}{2} M R^{2}$ . Yeh step kyun? Ex 1 jaisa hi algebra; sirf input $I_{z}$ badla.

Verify: Fraction $I_{x} / I_{z} = \frac{1}{2}$ disc jaisa hi same hai. Yeh expected hai: theorem ka aadha karna symmetry se aata hai, mass kahan hai usse nahi. ✓

Worked example 3. Rectangular plate —

Cell B (planar, $I_{x} \neq = I_{y}$ , koi shortcut nahi) Ek uniform rectangular plate ki side $a$ along $x$ hai, side $b$ along $y$ hai, mass $M$ hai. Standard in-plane results (centre se) hain $I_{x} = \frac{1}{12} M b^{2}$ aur $I_{y} = \frac{1}{12} M a^{2}$ . Central perpendicular axis ke baare mein $I_{z}$ nikalo.

Forecast: $a \neq = b$ ke saath dono axes alag hain. Kya hum phir bhi unhe sirf add kar sakte hain? Ya theorem ko $I_{x} = I_{y}$ chahiye?

Applicability check karo. Plate flat hai, saari axes centre se hain — theorem apply hoti hai. Symmetry zaroori nahi hai. Yeh step kyun? Proof ne sirf $z = 0$ use kiya, $I_{x} = I_{y}$ kabhi nahi. Symmetry Ex 1–2 mein ek suvidha thi, requirement nahi.
Seedha add karo. $I_{z} = I_{x} + I_{y} = \frac{1}{12} M b^{2} + \frac{1}{12} M a^{2} = \frac{1}{12} M (a^{2} + b^{2})$ . Yeh step kyun? $\int$ ki linearity hume do integrals combine karne deti hai; theorem baaki kaam kar leti hai.

Verify: $a = b$ set karo (ek square) → $I_{z} = \frac{1}{12} M (2 a^{2}) = \frac{1}{6} M a^{2}$ , jo jaana-pehchana square-plate result hai. ✓ Dimensions: mass × length². ✓

Worked example 4. Annulus about a diameter —

Cell C (ulta solve karo, symmetric) Ek flat washer (annulus) ka inner radius $R_{1}$ , outer radius $R_{2}$ , mass $M$ hai. Uski perpendicular-axis MOI $I_{z} = \frac{1}{2} M (R_{1}^{2} + R_{2}^{2})$ hai. Diameter ke baare mein MOI nikalo.

Forecast: Kaunsi jaani-pehchani quantity "known" hai aur kaun si target? Hum kisse 2 se divide kar rahe hain?

Symmetry. Circular symmetry phir se $I_{x} = I_{y}$ deti hai. Yeh step kyun? Annulus $z$ ke around rotationally symmetric hai, isliye saare diameters equivalent hain.
$I_{z}$ ko aadha karo. $I_{z} = 2 I_{dia} \Rightarrow I_{dia} = \frac{I _{z}}{2} = \frac{1}{4} M (R_{1}^{2} + R_{2}^{2})$ . Yeh step kyun? Bilkul disc wali logic; annulus sirf ek hole wali disc hai, phir bhi planar aur symmetric hai.

Verify: $R_{1} \to 0$ karo (hole sikar ke kuch nahi → solid disc): $I_{dia} \to \frac{1}{4} M R_{2}^{2}$ , jo Ex 1 se match karta hai. ✓ $R_{1} \to R_{2}$ karo (thin ring): $I_{dia} \to \frac{1}{4} M (2 R_{2}^{2}) = \frac{1}{2} M R_{2}^{2}$ , jo Ex 2 se match karta hai. ✓ Do independent limits dono pehle ke answers par land karti hain — strong evidence hai.

Worked example 5. Degenerate input — plate jo rod ban gayi —

Cell D Ex 3 ki rectangular plate lo aur side $b \to 0$ karo: plate ek thin rod ban jaati hai jiska length $a$ hai aur jo $x$ -axis ke along lie karti hai. Uski perpendicular-axis MOI $I_{z}$ aur uski apni length ( $I_{x}$ ) ke baare mein MOI predict karo.

Forecast: Rod ki koi width nahi hai. Uski apni axis ( $x$ ) ke baare mein MOI kya honi chahiye? Zero, ya kuch chhota?

$b$ wale terms hatao. Ex 3 se, $I_{x} = \frac{1}{12} M b^{2} \to 0$ jab $b \to 0$ . Yeh step kyun? $I_{x} = \int y^{2} d m$ ; jab har point $x$ -axis par ho, toh $y = 0$ , isliye integral zero ho jaata hai. Ek line ko us line ke around spin karna free hai.
$a$ wala term bachao. $I_{y} = \frac{1}{12} M a^{2}$ baka rehta hai (mass abhi bhi $x$ ke along spread hai). Yeh step kyun? $I_{y} = \int x^{2} d m$ mein nonzero $x$ hai; yahi "end-over-end flip" ki asli difficulty hai.
Add karo. $I_{z} = I_{x} + I_{y} = 0 + \frac{1}{12} M a^{2} = \frac{1}{12} M a^{2}$ . Yeh step kyun? Theorem abhi bhi hold karta hai; uski do ingredients mein se ek sirf zero ho gayi.

Verify: $\frac{1}{12} M a^{2}$ ek thin rod ki centre ke baare mein textbook MOI hai — theorem ise ek limit ke roop mein reproduce karta hai. ✓ Aur yahan $I_{z} = I_{y}$ isliye hai kyunki degenerate rod ka out-of-plane spin hi uska end-over-end spin hai. ✓

Worked example 6. 3-D trap — theorem refuse karo —

Cell E Ek student likhta hai: "Mass $M$ , radius $R$ wale solid sphere ke liye, centre se guzarne wali teeno axes barabar hain, $I_{x} = I_{y} = I_{z} = \frac{2}{5} M R^{2}$ . Toh $I_{z} = I_{x} + I_{y}$ deta hai $\frac{2}{5} M R^{2} = \frac{4}{5} M R^{2}$ ." Galti dhundho.

Forecast: Theorem ki hidden assumption kahan hai, aur kya sphere use satisfy karta hai?

Ek condition yaad karo. Proof ne $z = 0$ set kiya tha har mass element ke liye. Sphere mein mass $- R$ se $+ R$ tak saare $z$ par hai. Yeh step kyun? Yeh proof ki woh single line hai (parent mein Step 4) jahan "flat" use kiya gaya tha. Ise toro aur baad mein sab kuch invalid hai.
Dikhao ki numbers reconcile nahi ho sakti. $I_{x} + I_{y} = \frac{4}{5} M R^{2} \neq = \frac{2}{5} M R^{2} = I_{z}$ . Yeh step kyun? Yeh contradiction proof hai ki theorem galat apply ki gayi — yeh solids ke liye rule nahi hai.

Verify: General 3-D identity hai $I_{x} + I_{y} + I_{z} = 2 \int (x^{2} + y^{2} + z^{2}) d m$ ; sirf jab $z = 0$ har jagah ho tab yeh $I_{z} = I_{x} + I_{y}$ mein collapse hoti hai. Sphere ke liye $z \neq = 0$ , isliye sahi statement hai $I_{x} + I_{y} + I_{z} = 3 \cdot \frac{2}{5} M R^{2}$ , $I_{z} = I_{x} + I_{y}$ nahi. ✓ Jawaab hai: theorem apply nahi hoti — sphere planar nahi hai.

Worked example 7. Real-world word problem —

Cell F Ek coin (uniform disc, mass $M = 8.0 g$ , radius $R = 1.2 cm$ ) table par flat rakhi hai. Aap (a) ise centre se guzarti vertical line ke baare mein fidget top ki tarah spin karte ho, aur (b) centre se guzarti horizontal line ke baare mein face par se end-over-end flip karte ho. Kaun si spin ki MOI zyada hai, aur kitne factor se? Dono numeric values do.

Forecast: Vertical top-spin $z$ -axis hai; flip ek diameter hai. Ex 1 se, compute karne se pehle factor guess karo.

Words ko axes mein translate karo. Top-spin = perpendicular axis → $I_{z} = \frac{1}{2} M R^{2}$ . Flip = diameter → $I_{dia} = \frac{1}{4} M R^{2}$ (Ex 1). Yeh step kyun? Word problems ki poori difficulty axis ka naam rakhna hai; ek baar naam rakh diya toh formulas automatic hain.
Numbers. $M = 8.0 \times 1 0^{- 3} kg$ , $R = 1.2 \times 1 0^{- 2} m$ ke saath: $I_{z} = \frac{1}{2} (8.0 \times 1 0^{- 3}) (1.2 \times 1 0^{- 2})^{2} = 5.76 \times 1 0^{- 7} kg m^{2}$ $I_{dia} = \frac{1}{4} (8.0 \times 1 0^{- 3}) (1.2 \times 1 0^{- 2})^{2} = 2.88 \times 1 0^{- 7} kg m^{2}$ Yeh step kyun? Dono formulas mein plug in karo; consistent SI units SI answers denge.
Factor. $I_{z} / I_{dia} = 2$ . Yeh step kyun? Theorem kisi bhi symmetric disc ke liye exactly yahi ratio predict karta hai — ek clean sanity check.

Verify: $I_{z} = I_{dia} + I_{dia} = 2.88 \times 1 0^{- 7} \times 2 = 5.76 \times 1 0^{- 7}$ . ✓ Top-spin do guna mushkil hai — yeh rozmarra ki feeling se match karta hai ki coin ko edge-over-edge flip karna fast top-spin lane se zyada aasaan hai. ✓

Worked example 8. Exam twist: perpendicular

phir parallel — Cell G Ek disc (mass $M$ , radius $R$ ) ki MOI us axis ke baare mein nikalo jo uske plane ke perpendicular ho lekin uski rim par ke ek point se guzre (ek tangent-point perpendicular axis).

Forecast: Perpendicular axis theorem akela centre values deta hai. Yeh axis offset hai. Kaun si extra theorem chahiye?

Centre perpendicular MOI nikalo. $I_{z}^{centre} = \frac{1}{2} M R^{2}$ (given/standard). Yeh perpendicular-axis family se aaya. Yeh step kyun? Parallel axis ko target ke parallel centre-of-mass axis ke baare mein MOI chahiye — woh central perpendicular axis hai.
Parallel axis theorem se shift karo. Rim axis centre axis ke parallel hai, $d = R$ offset ke saath: $I = I_{z}^{centre} + M d^{2} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2} .$ Yeh step kyun? Perpendicular-axis theorem sirf ek common point se guzarti axes ko relate karta hai; kisi alag point par jaane ke liye hume Parallel axis theorem use karna hoga. Kabhi bhi perpendicular-axis ko axis sideways shift karne ke liye use mat karo.

Verify: $\frac{3}{2} M R^{2} > \frac{1}{2} M R^{2}$ — axis ko centre se door le jaana MOI hamesha badhata hai, jaisa $+ M d^{2} \geq 0$ guarantee karta hai. ✓ Units: mass × length². ✓ Yeh classic exam trap hai: perpendicular-axis ek point par axis ka direction badalne ke liye, parallel-axis axis ki position badalne ke liye.

Worked example 9. Limiting behaviour — square aur

$a \to b$ collapse — Cell H (i) Side $a$ , mass $M$ wali ek square plate ke liye, Ex 3 se $I_{z}$ aur diameter (in-plane centre) MOI nikalo. (ii) Confirm karo ki Ex 3 ka rectangular result square par correctly collapse hota hai jab $a \to b$ .

Forecast: Square ke liye, kya $I_{x} = I_{y}$ hai? Woh diameter MOI ko $I_{z}$ ka kitna fraction banaata hai?

Ex 3 se square. $b = a$ set karo: $I_{z} = \frac{1}{12} M (a^{2} + a^{2}) = \frac{1}{6} M a^{2}$ . Yeh step kyun? Square rectangle ka symmetric special case hai, isliye uska result substitution se nikalna chahiye.
Symmetry se in-plane axis. Square $x$ aur $y$ dono ke along same dikhta hai, isliye $I_{x} = I_{y}$ , aur $I_{x} = \frac{I _{z}}{2} = \frac{1}{12} M a^{2}$ . Yeh step kyun? Square ke liye symmetry wapas aati hai (general rectangle ke unlike), jo hume $I_{z}$ ko aadha karne deti hai.
Collapse check. Ex 3 mein, $I_{x} - I_{y} = \frac{1}{12} M (b^{2} - a^{2}) \to 0$ jab $a \to b$ — dono axes ek hi value mein merge ho jaate hain. Yeh step kyun? Ek continuous limit test: unequal-axis rectangle smoothly equal-axis square ban jaana chahiye.

Verify: $\frac{1}{12} M a^{2} \times 2 = \frac{1}{6} M a^{2} = I_{z}$ . ✓ Dono in-plane axes $\frac{1}{12} M a^{2}$ ke barabar hain; unka sum exactly perpendicular value hai. ✓

Active Recall

Recall Kaun si theorem, kaun sa cell?

Har ek ke liye cell (A–H) aur zaroori tool batao.

Disc about a diameter ::: Cell A — perpendicular axis + symmetry ( $I_{z}$ ko aadha karo).
Rectangular plate $I_{z}$ do unequal in-plane axes se ::: Cell B — perpendicular axis, koi symmetry nahi.
Disc about a rim perpendicular axis ::: Cell G — pehle perpendicular value phir parallel axis theorem.
"Yeh sphere ke liye bhi kaam karta hai" ::: Cell E — refuse karo; sphere 3-D hai, $z \neq = 0$ .
Thin rod MOI ek collapsing plate se ::: Cell D — degenerate limit, ek in-plane MOI $\to 0$ .

Connections

Parent (Hinglish) — woh proof jo yeh examples exercise karti hai.
Parallel axis theorem — Ex 8 mein axis ko centre se door le jaane ke liye zaroori.
Moment of inertia — definition — woh $\int r^{2} d m$ jis par har answer tikaa hai.
Moment of inertia of standard bodies — disc, ring, plate, annulus, rod ke base values.
Symmetry arguments in MOI — Cells A, C, H mein use kiya gaya $I_{x} = I_{y}$ trick.
Radius of gyration — in MOIs ko repackage karne ka ek related tarika.