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

1.5.7 · D4 · Physics › Rotational Mechanics › Perpendicular axis theorem — I_z = I_x + I_y — proof, restri

Shuru karne se pehle ek quick vocabulary reminder, taaki koi symbol unfamiliar na rahe:

Definition Woh words jo hum baar baar use karenge

Lamina / planar body — ek aisa body jo itna patla ho ki hum uski saari mass ek flat plane mein maante hain. Hum us plane ko $x y$ -plane lete hain, isliye ==har mass point ka $z = 0$ == hota hai.
Moment of inertia — yeh number $\int r_{⊥}^{2} d m$ batata hai ki ek chosen axis ke baare mein ek body ko angularly accelerate karna kitna mushkil hai (axis se door mass, pass waale mass se kahin zyada spinning resist karta hai). Dekho Moment of inertia — definition. Yahan $r_{⊥}$ = ek tiny mass bit $d m$ se axis tak ki sabse chhoti, seedhi distance.
$I_{z}$ — moment of inertia us axis ke baare mein jo plane se seedha bahar nikal raha hai (perpendicular axis).
$I_{x}$ , $I_{y}$ — moments of inertia un do lines ke baare mein jo plane ke upar drawn hain, ek doosre ke right angles par, dono $z$ -axis ko same point par cross karti hain (do in-plane axes).

Neeche ki picture har upar waale symbol ko anchor karti hai. Q1 se pehle ise dekho: teal aur plum arrows do in-plane axes hain, dark arrow perpendicular ( $z$ ) axis hai jo sheet se bahar aa raha hai, aur dashed segment $r_{⊥} = x^{2} + y^{2}$ hai — wahi quantity jise hum $I_{z}$ ke andar square karte hain.

Figure — Perpendicular axis theorem — I_z = I_x + I_y — proof, restrictions

Figure 1 dikhata hai ki kyun $z$ -axis distance $x^{2} + y^{2}$ hai: mass bit, origin, aur uske do coordinate projections ek right triangle banate hain, isliye Pythagoras hypotenuse deta hai. Is triangle ko yaad rakho — poora theorem ek drawing mein yahi hai.

Level 1 — Recognition

Kya tum spot kar sakte ho ki theorem kab apply hoti hai, aur pieces read kar sakte ho?

Recall Solution Q1

Theorem ki ek requirement hai ki har mass element ka $z = 0$ ho — ek flat lamina.

(a) CD disc → haan, essentially planar.
(b) solid ball → nahi, mass teeno dimensions mein hai, $z \neq = 0$ .
(c) sheet of paper → haan, ek ideal lamina.
(d) thick brick → nahi, $z$ mein real thickness hai. Answer: (a) aur (c).

Recall Solution Q2

Ring ki saari mass $z$ -axis se $R$ distance par hai, isliye $I_{z} = M R^{2}$ . $I_{z} = (2) (0.5)^{2} = 2 \times 0.25 = 0.5 kg m^{2} .$

Recall Solution Q3

Kisi axis tak ki distance baaki do coordinates use karti hai (Pythagoras us plane mein jo us axis ke perpendicular hai — exactly wahi right triangle jo Figure 1 mein draw hai):

$z$ -axis tak: $x^{2} + y^{2}$
$x$ -axis tak: $y^{2} + z^{2} \to y^{2}$ (kyunki $z = 0$ )
$y$ -axis tak: $x^{2} + z^{2} \to x^{2}$ (kyunki $z = 0$ ) Ye bilkul $I_{z}$ , $I_{x}$ , $I_{y}$ ke integrands hain.

Level 2 — Application

Theorem plug in karo aur number nikalo, usually symmetry shortcut ke saath.

Q4 se pehle — kyun symmetry hume $I_{x} = I_{y}$ likhne deti hai. Figure 2 ek uniform disc dikhata hai jisme do diameters drawn hain: ek horizontal aur ek vertical. Kyunki disc round aur uniformly dense hai, poori picture ko $9 0^{\circ}$ rotate karna horizontal diameter ko vertical par map karta hai aur mass pattern unchanged rehta hai. Do axes jo ek symmetry se related hain jo body ko identical rakhti hai, unka moment of inertia same hona chahiye — isliye $I_{x} = I_{y}$ . Yeh koi shortcut nahi hai jo hum faith par lete hain; yeh is fact se force hota hai ki $\int y^{2} d m$ aur $\int x^{2} d m$ ek baar picture symmetric hone par same number compute karte hain. (Dekho bhi Symmetry arguments in MOI.)

Recall Solution Q4

Koi bhi do perpendicular diameters identical dikhte hain (Figure 2: ek $9 0^{\circ}$ turn unhe swap karta hai aur uniform disc unchanged rehti hai), isliye $I_{x} = I_{y} = I_{dia}$ . $I_{z} = I_{x} + I_{y} = 2 I_{dia} \Rightarrow I_{dia} = \frac{1}{2} I_{z} = \frac{1}{4} M R^{2} .$ Numerically: $I_{dia} = \frac{1}{4} (4) (0.3)^{2} = \frac{1}{4} (4) (0.09) = 0.09 kg m^{2} .$

Recall Solution Q5

Ring mein disc jaisi hi rotational symmetry hai (Figure 2), aur uniform linear density hai, isliye $I_{x} = I_{y}$ . $I_{z} = M R^{2}$ ke saath, $I_{dia} = \frac{1}{2} I_{z} = \frac{1}{2} M R^{2}$ . $I_{dia} = \frac{1}{2} (3) (0.4)^{2} = \frac{1}{2} (3) (0.16) = 0.24 kg m^{2} .$

Recall Solution Q6

Yahan $a \neq = b$ hai, isliye $9 0^{\circ}$ turn plate ko khud par map nahi karta — isliye $I_{x} \neq = I_{y}$ . Koi symmetry shortcut nahi, lekin theorem phir bhi directly add karta hai: $I_{z} = I_{x} + I_{y} = \frac{1}{12} M (a^{2} + b^{2}) .$ $I_{z} = \frac{1}{12} (5) (0. 6^{2} + 0. 4^{2}) = \frac{1}{12} (5) (0.36 + 0.16) = \frac{1}{12} (5) (0.52) = 0.2167 kg m^{2} .$

Level 3 — Analysis

Sochna ki kaunsa axis, theorem reverse karna, ya doosre result ke saath combine karna.

Recall Solution Q7

Theorem reverse karo: $I_{z} = 2 I_{dia} = 2 (0.05) = 0.1 kg m^{2}$ . Kyunki $I_{z} = \frac{1}{2} M R^{2}$ : $R = \frac{2 I _{z}}{M} = \frac{2 ( 0.1 )}{2} = 0.1 \approx 0.316 m .$

Recall Solution Q8

Ek square $9 0^{\circ}$ turn se khud par map hota hai (Figure 2 ki logic ek square par apply karke), isliye dono in-plane central axes equal hain, $I_{x} = I_{y}$ : $I_{x} = \frac{1}{2} I_{z} = \frac{1}{2} \cdot \frac{1}{6} M a^{2} = \frac{1}{12} M a^{2} .$ Yeh direct plate result $\frac{1}{12} M a^{2}$ se match karta hai. ✓ Numerically: $I_{x} = \frac{1}{12} (1) (0.2)^{2} = \frac{0.04}{12} = 0.00333 kg m^{2} .$

Recall Solution Q9

Ring about a diameter: $I_{dia} = \frac{1}{2} M R^{2}$ . $M k^{2} = \frac{1}{2} M R^{2}$ set karke: $k = \frac{R}{2} .$ $k = \frac{0.5}{2} \approx 0.354 m .$

Level 4 — Synthesis

Perpendicular theorem ko Parallel axis theorem ke saath ya integration ke saath combine karo.

Recall Solution Q10

Do steps, do theorems: 1. Perpendicular axis theorem → central diameter MOI: $I_{dia} = \frac{1}{4} M R^{2}$ . 2. Parallel axis theorem → us in-plane axis ko $d = R$ se tangent tak shift karo: $I_{tan} = I_{dia} + M R^{2} = \frac{1}{4} M R^{2} + M R^{2} = \frac{5}{4} M R^{2} .$ Numerically: $I_{tan} = \frac{5}{4} (2) (0.3)^{2} = \frac{5}{4} (2) (0.09) = 0.225 kg m^{2} .$

Recall Solution Q11

$I_{z} = \frac{1}{2} M R^{2}$ (central perpendicular) se shuru karo, phir $d = R$ se parallel-axis shift karo: $I_{z, tan} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2} .$ $= \frac{3}{2} (2) (0.09) = 0.27 kg m^{2} .$ Q10 vs Q11 compare karo: in-plane axis shift karne se $\frac{5}{4} M R^{2}$ mila, perpendicular axis shift karne se $\frac{3}{2} M R^{2}$ mila — alag hain kyunki central starting values ( $\frac{1}{4}$ vs $\frac{1}{2}$ ) alag hain.

Recall Solution Q12

1. Central perpendicular: $I_{z, cm} = \frac{1}{12} M (a^{2} + b^{2})$ . 2. Parallel-axis shift ek corner tak, centre se distance $d = \frac{1}{2} a^{2} + b^{2}$ , isliye $d^{2} = \frac{1}{4} (a^{2} + b^{2})$ : $I_{z, corner} = \frac{1}{12} M (a^{2} + b^{2}) + M \cdot \frac{1}{4} (a^{2} + b^{2}) = M (a^{2} + b^{2}) (\frac{1}{12} + \frac{1}{4}) = \frac{1}{3} M (a^{2} + b^{2}) .$ Numerically: $a^{2} + b^{2} = 0.64 + 0.36 = 1.0$ , isliye $I_{z, corner} = \frac{1}{3} (6) (1.0) = 2.0 kg m^{2} .$

Level 5 — Mastery

Prove karo, generalise karo, ya ek subtle failure pakdo.

Recall Solution Q13

$I_{x} = \int y^{2} d m$ aur $I_{y} = \int x^{2} d m$ se, dono integrands squares hain → non-negative → $I_{x} \geq 0$ , $I_{y} \geq 0$ . Phir $I_{z} = I_{x} + I_{y} \geq I_{x}$ (kyunki hum ek non-negative $I_{y}$ add kar rahe hain), aur isi tarah $I_{z} \geq I_{y}$ . ∎ Equality $I_{z} = I_{x}$ ke liye $I_{y} = \int x^{2} d m = 0$ chahiye, yaani saari mass ka $x = 0$ ho — lamina $y$ -axis par collapse ho jaati hai (ek thin rod along $y$ ). Symmetrically $I_{z} = I_{y}$ matlab ek rod along $x$ . Isliye ek genuine 2-D lamina mein hamesha $I_{z}$ strictly har in-plane MOI se bada hota hai.

Recall Solution Q14

Sphere ke liye true $I_{z} = \frac{2}{5} M R^{2} = 0.4$ hai ( $M = R = 1$ ke saath). Student ka sum deta hai $\frac{4}{5} M R^{2} = 0.8$ . Ye agree nahi karte: $0.8 \neq = 0.4$ . Kahan break karta hai: proof step $I_{x} = \int (y^{2} + z^{2}) d m \to \int y^{2} d m$ ne $z = 0$ use kiya. Ek sphere teeno dimensions mein hai, isliye $z \neq = 0$ aur $\int z^{2} d m$ terms vanish nahi karte. Unhe rakhte hue: $I_{x} + I_{y} = \int (x^{2} + 2 z^{2} + y^{2}) d m = I_{z} + 2 \int z^{2} d m > I_{z} .$ Isliye ek 3-D body ke liye $I_{x} + I_{y}$ exactly $2 \int z^{2} d m$ se $I_{z}$ ko overcount karta hai. Theorem sirf planar bodies ke liye hai.

Khud ko score karo

Recall Har level ne kya test kiya

L1 — recognize karo ki $z = 0$ actually kab hold karta hai.
L2 — direct plug-in, symmetry ke saath aur bina.
L3 — theorem reverse karo, radius of gyration laao.
L4 — ise Parallel axis theorem ke saath chain karo.
L5 — ek inequality prove karo aur failure mode diagnose karo. Agar koi L4/L5 slip hua, to parent proof Step 4 dobara dekho — woh single $z = 0$ line hi sab kuch ka hinge hai.