Question bank — Parallel axis theorem — I = I_CM + Md² — proof

False. Ek axis CM se guzarni chahiye. Cross terms $-2a\sum m_i{x}_i$ tabhi vanish hote hain jab origin CM ho; do off-CM axes ke beech woh survive karte hain aur clean formula toot jaata hai.

True. Kyunki $I=I_{CM}+M{d}^2$ aur $M{d}^2\ge 0$, koi bhi parallel axis ek non-negative amount add karta hai. Minimum $d=0$ par hota hai, yaani CM axis khud.

False. Correction $M{d}^2$ hai, isliye $d$ double karne par woh chaar guna ho jaata hai ($M(2d{)}^2=4M{d}^2$). Distance squared mein enter hoti hai — door jaana bahut mehenga padta hai.

False. $I_{CM}$ body ki ek fixed property hai apni CM axis ke baare mein. Sirf added term $M{d}^2$ badalta hai jab aap outer axis move karte ho; $I_{CM}$ kabhi nahi badlta.

True. $I=I_{CM}+M{d}^2$ sirf $I_{CM}$, $M$, aur $d$ par depend karta hai — detailed shape ke baare mein kuch nahi jab yeh teen match kar jaayein.

False. Derivation ne sirf $\sum m_i{x}_i=0$ at the CM use kiya, jo kisi bhi rigid body ke liye kisi bhi shape mein hold karta hai. Koi bhi symmetry assume nahi ki gayi.

True. Proof ne $z_i$ coordinate ko bilkul ignore kiya (distance to a $z$-axis sirf $x,y$ use karta hai). Theorem puri tarah 3D hai; axis ko bas CM axis ke parallel hona chahiye.

Error $d$ mein hai. Ek uniform rod ka CM uske midpoint par hota hai, isliye end sirf $d=L/2$ door hai. Sahi: $I_{end}=\frac{1}{12}M{L}^2+M(L/2{)}^2=\frac{1}{3}M{L}^2$.

Galat — aap do off-CM axes ko directly connect nahi kar sakte. CM ke through route karo: $I_A=I_{CM}+M{R}^2$, $I_B=I_{CM}+M(2R{)}^2$, isliye $I_B-I_A=3M{R}^2$, na ki $M{R}^2$.

$d$ kisi arbitrary point se measure nahi hoti. Yeh do parallel axes ke beech ki perpendicular distance hai, equivalently CM-to-outer-axis. Ek corner irrelevant hai jab tak axis us se na guzre.

Theorem sirf parallel axes ke liye valid hai. Agar axes ek doosre ke relative tilted hain, toh poori $r_i$ geometry badal jaati hai aur $I=I_{CM}+M{d}^2$ bilkul apply nahi hota.

Impossible. $I_{CM}$ minimum hai, isliye koi bhi parallel axis kam nahi de sakta. Agar aapko chhhota mila, toh ya toh aapne mislabel kiya ki kaun sa term $I_{CM}$ hai ya theorem do non-CM axes ke beech use kiya.

Woh $\frac{1}{2}M{R}^2$ $I_{CM}$ hai (central axis). Rim axis $d=R$ door hai, isliye $I_{rim}=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$. $M{d}^2$ mat bhuulo.

Kyunki origin CM par rakha gaya tha, aur CM ki definition hi yeh hai ki $x_{CM}=\frac{\sum m_i{x}_i}{M}=0$, jo $\sum m_i{x}_i=0$ force karta hai ($y$ ke liye bhi same). Yahi akela reason hai ki formula clean hai.

Aapko ek chahiye taaki cross terms vanish ho sakein. Do parallel axes generally dono CM se nahi guzar saktiyin (sirf ek line us fixed direction mein CM se guzarti hai), aur zero rakhne par ugly cross terms rah jaate hain.

Kyunki $M{d}^2$ literally mass $M$ ke ek single point ka moment of inertia hai perpendicular distance $d$ par axis se. Theorem inertia ko "CM ke baare mein spread" aur "poora mass CM par ek point ke roop mein" mein split karta hai.

Added term $M{d}^2$ hai, aur dono $M>0$ aur $d^2\ge 0$ hain, isliye yeh sirf add hi kar sakta hai. $I$ ko uske minimum ki taraf kam karne ka ek hi tarika hai — axis ko CM ki taraf move karo.

$z$-axis tak ek point ki distance sirf uske $x$ aur $y$ par depend karti hai (yeh $\sqrt{x^2+y^2}$ hai). Kisi point ko $z$ ke along slide karne se vertical axis tak uski distance nahi badlti, isliye $z_i$ kabhi $I$ mein enter nahi karta.

Kyunki $I_{CM}$ standard shapes (rod, disc, sphere) ke liye ek baar tabulate hota hai, aur theorem aapko ek simple term $M{d}^2$ add karke kisi bhi parallel axis tak pahunchne deta hai — koi re-integration nahi chahiye. Isliye yeh Physical pendulum aur Radius of gyration ko underpin karta hai.

$I=I_{CM}+0=I_{CM}$. "Outer" axis CM axis ke saath coincide karta hai — theorem gracefully trivial identity par reduce ho jaata hai, confirm karta hai ki CM value hi floor hai.

Ek point mass ka $I_{CM}=0$ apne se guzarne wale kisi bhi axis ke baare mein hota hai (uska saara mass zero distance par hai). Toh $I=0+M{d}^2=M{d}^2$, seedha point-mass formula recover karta hai — theorem ka building block.

Nahi. $I_{CM}=\sum m_i{r}_i^2$ masses times squares ka sum hai, isliye yeh hamesha $\ge 0$ hai. $M{d}^2\ge 0$ add karna har axis ke liye $I$ ko non-negative rakhta hai.

Correction $M{d}^2\to 0$, isliye $I\to I_{CM}$, jo khud zero mass ke liye $0$ hai. Sab kuch zero inertia par degenerate ho jaata hai — consistent, kyunki rotation resist karne ke liye koi mass hi nahi hai.

Haan. CM ek geometric point hai jo $\sum m_i{\vec{r}}_i=0$ se define hota hai; yeh material ke andar hona zaroori nahi. Proof ne sirf woh mathematical property use ki, isliye theorem unchanged apply hota hai.

Formally $I=I_{CM}+M{d}^2\to \infty$ jab $d\to \infty$. Physically sensible: poore mass ko enormous radius par swing karna enormous rotational inertia maangta hai — $M{d}^2$ term dominate karta hai aur $I_{CM}$ negligible ho jaata hai.