Moment of inertia I = Σmᵢrᵢ² — concept: Worked examples

1.5.5 · D3 · Physics › Rotational Mechanics › Moment of inertia I = Σmᵢrᵢ² — concept

Ye worked-example clinic hai parent concept note ke liye. Wahan humne formula $I = \sum_{i} m_{i} r_{i}^{2}$ banaya tha. Yahan hum use har case class mein drive karte hain — taaki jab koi problem saamne aaye, tumne uska shape pehle se dekha hua ho.

Scenario matrix

Koi bhi arithmetic se pehle, yahan poora landscape hai. Neeche har worked example us cell ke saath tagged hai jo wo cover karta hai.

Cell	Case class	Kya tricky hai	Example
A	Ordinary discrete sum	bas formula seedha apply karo	Ex 1
B	Mass axis par ( $r = 0$ )	zero contribution — degenerate distance	Ex 2
C	Masses axis ke dono taraf (negative coordinates)	position ka sign matter karta hai kya?	Ex 3
D	Axis-dependence — same object, nayi axis	$I$ fixed nahi hota	Ex 4
E	Limiting behaviour — mass ko door push karo	$I$ $r^{2}$ ki tarah badhta hai, unbounded	Ex 5
F	Real-world word problem — figure skater	$I$ ko $ω$ se conservation ke zariye jodo	Ex 6
G	Exam twist — axis corner se, 2-D layout	plane mein perpendicular distance	Ex 7
H	Sanity/units + parallel-axis cross-check	submit karne se pehle errors pakdo	Ex 8

Notation reminder (koi bhi cheez define hone se pehle use nahi ki gayi):

$m_{i}$ = particle number $i$ ka mass, kilograms mein ( $kg$ ).
$r_{i}$ = particle $i$ se axis line tak ka perpendicular distance, metres mein ( $m$ ). "Perpendicular" matlab particle se axis tak ki sabse chhoti seedhi line, jo axis se right angle par milti hai.
$I$ = moment of inertia, $kg m^{2}$ mein. Ise "spin-stubbornness" samjho.
$ω$ (Greek letter omega) = angular velocity, kitni tezi se koi cheez ghoomti hai, radians per second mein.

Cell A — plain vanilla sum

Figure — Moment of inertia I = Σmᵢrᵢ² — concept

Worked example Example 1 (Cell A) — teen beads ek wire par

Teen beads ek seedhi wire par, axis wire ke left end par perpendicular hai: $m_{1} = 1 kg$ at $r_{1} = 1 m$ , $m_{2} = 2 kg$ at $r_{2} = 2 m$ , $m_{3} = 3 kg$ at $r_{3} = 3 m$ . $I$ find karo.

Forecast: Kya door wala $3 kg$ bead dominate karega? Padhne se pehle guess karo ki $I$ $10$ ke kareeb hoga ya $30 kg m^{2}$ ke kareeb.

Step 1 — sum ko term by term likho. $I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + m_{3} r_{3}^{2} .$ Ye step kyun? Discrete point masses ke liye koi integral nahi hota — definition literally ek sum hi hai. Figure s01 dekho: har bead axis tak $r_{i}$ length ka ek coloured spoke phenkta hai.

Step 2 — har reach ko square karo, phir apne mass se multiply karo. $I = 1 (1)^{2} + 2 (2)^{2} + 3 (3)^{2} = 1 + 8 + 27.$ Ye step kyun? Square hi poori baat hai (kinetic energy mein $v^{2}$ se liya gaya). Notice karo kaise $27$ pehle se hi $1$ ko bhaari padta hai: door wala bead jeet jaata hai.

Step 3 — jodo. $I = 36 kg m^{2} .$

Verify: Units: $kg \times m^{2} = kg m^{2}$ ✓. Tumhara forecast: ye $36$ hai, $30$ se kaafi aage — door wala mass bilkul waisi tarah dominate kiya jaise squared reach predict karta hai.

Cell B — axis PAR baitha hua mass

Worked example Example 2 (Cell B) — ek bead pivot par hai

Same wire, lekin ab axis $m_{1}$ se hoke gujarti hai. Toh $r_{1} = 0$ , aur (nayi axis se measure karte hue) $m_{2}$ $1 m$ par hai, $m_{3}$ $2 m$ par. Masses unchanged.

Forecast: Axis par baitha $1 kg$ bead kitna add karta hai — $1 kg m^{2}$ , ya kuch nahi?

Step 1 — $r_{1} = 0$ plug karo. $I = 1 (0)^{2} + 2 (1)^{2} + 3 (2)^{2} .$ Ye step kyun? Axis par baitha particle zero radius ke circle mein travel karta hai — jab body spin karti hai toh ye bilkul bhi swing nahi karta, isliye ye spinning resist nahi kar sakta. Uska term $m \cdot 0^{2} = 0$ hai.

Step 2 — evaluate karo. $I = 0 + 2 + 12 = 14 kg m^{2} .$

Verify: On-axis bead ne exactly zero contribute kiya — mass akela kabhi rotation fight nahi karta; sirf reach karta hai. Units $kg m^{2}$ ✓. Ye degenerate distance case hai: $r = 0$ allowed hai aur term ko zero kar deta hai.

Cell C — dono taraf masses (kya sign matter karta hai?)

Worked example Example 3 (Cell C) — dumbbell axis ke aas-paas

Ek rod par do $2 kg$ masses. Axis centre par hai. Ek mass position $x = + 0.4 m$ par baitha hai, doosra $x = - 0.4 m$ par (opposite side).

Forecast: Positions ke opposite signs hain. Kya ye partly cancel honge (jaise centre of mass mein), ya dono add honge?

Step 1 — perpendicular distance $r$ nikalte hain, position $x$ nahi. $r_{1} = ∣ + 0.4 ∣ = 0.4 m, r_{2} = ∣ - 0.4 ∣ = 0.4 m .$ Ye step kyun? $r$ ek distance hai — hamesha $\geq 0$ . Figure s02 mein dono masses axis line se $0.4 m$ door hain; wo kis taraf hain ye is baat se irrelevant hai ki wo kitne bade circle mein swing karte hain.

Step 2 — square sign koaise bhi khatam kar deta hai. $I = 2 (0.4)^{2} + 2 (- 0.4)^{2} = 2 (0.16) + 2 (0.16) .$ Ye step kyun? Agar tumne carelessly $x = - 0.4$ use kiya bhi, toh square $(- 0.4)^{2} = (+ 0.4)^{2} = 0.16$ bana deta hai. Toh $I$ dono taraf masses hone se reduce nahi ho sakta. Dono add hote hain.

Step 3 — evaluate karo. $I = 0.32 + 0.32 = 0.64 kg m^{2} .$

Verify: Centre of mass se compare karo, jo $\sum m_{i} x_{i}$ use karta hai ( $x$ mein linear): wahan $+ 0.4$ aur $- 0.4$ do cancel karte hain, centre of mass ko $x = 0$ par rakhte hain. Moment of inertia $x^{2}$ use karta hai, toh kuch cancel nahi hota — dono sides pile on karte hain. Units ✓.

Recall Inertia kabhi negative kyun nahi ho sakti?

Har term $m_{i} r_{i}^{2}$ hai: mass positive hai, aur kisi bhi real number ka square $\geq 0$ hota hai. Toh sum hamesha $\geq 0$ hota hai. Negative moment of inertia ka matlab hoga ::: impossible — matlab ek object khud ko spin karne mein help karta hai, yaani resistance zero se neeche. Physically forbidden hai.

Cell D — same object, axis move ho gayi

Worked example Example 4 (Cell D) — axis dependence

Ex 3 wala dumbbell lo (do $2 kg$ at $\pm 0.4 m$ ). Ab axis ko left mass se hoke le jaao.

Forecast: Centred $0.64 kg m^{2}$ se bada hoga ya chhota?

Step 1 — nayi axis se reaches recompute karo. Left mass: $r_{1} = 0$ . Right mass: ye $+ 0.4$ par tha, axis ab $- 0.4$ par hai, toh $r_{2} = 0.4 - (- 0.4) = 0.8 m$ . Ye step kyun? $I$ ek axis ke baare mein define hota hai. Axis badlo $\Rightarrow$ har $r_{i}$ dobara measure hota hai.

Step 2 — sum karo. $I = 2 (0)^{2} + 2 (0.8)^{2} = 0 + 2 (0.64) = 1.28 kg m^{2} .$

Verify: Same do masses, lekin $I$ $0.64$ se $1.28 kg m^{2}$ ho gaya — ye double ho gaya. Ye prove karta hai ki $I$ koi fixed property nahi hai. (Parallel axis theorem is jump ko exactly predict karta hai: axis ko centre of mass se $d = 0.4 m$ shift karne par central $0.64$ mein $M d^{2} = 4 (0.4)^{2} = 0.64$ add hota hai, jo $1.28$ deta hai ✓.)

Cell E — limiting behaviour

Worked example Example 5 (Cell E) — mass ko infinity tak push karo

Ek single $1 kg$ mass distance $r$ par. Track karo $I$ ko jaise $r$ $0.5 \to 1 \to 2 \to 4 m$ jaata hai.

Forecast: Har baar $r$ double hone par kya $I$ double, triple, ya quadruple hota hai?

Step 1 — $I = 1 \cdot r^{2}$ tabulate karo. $r = 0.5 : I = 0.25; r = 1 : I = 1; r = 2 : I = 4; r = 4 : I = 16.$ Ye step kyun? Ek particle ke saath, $I = m r^{2}$ ek clean parabola in $r$ hai. Figure s03 ise plot karta hai: curve jitna door jaate ho utna steep hota jaata hai.

Step 2 — growth rate padho. $r$ double karne par $I$ $2^{2} = 4$ se multiply hota hai. Toh $I$ without bound badhta hai jaise $r \to \infty$ . Ye step kyun? Ye limiting case hai: koi ceiling nahi hai. Real machines ise exploit karte hain — ek flywheel mass ko bade $r$ par rakhta hai bahut saari rotational stubbornness store karne ke liye.

Step 3 — doosra limit. Jaise $r \to 0$ , $I \to 0$ . Axis par baitha point ka zero moment of inertia hota hai (Cell B se match karta hai).

Verify: $r = 1$ ( $I = 1$ ) se $r = 2$ ( $I = 4$ ) tak: factor of $4$ , $2$ nahi ✓. Growth quadratic hai — plot s03 curve dikhata hai, straight line nahi.

Cell F — real-world word problem

Worked example Example 6 (Cell F) — figure skater arms kheench leti hai

Ek skater arms bahar karke $I_{1} = 6 kg m^{2}$ aur $ω_{1} = 2 rad/s$ se spin karti hai. Wo apne arms andar kheench leti hai, $I_{2} = 2 kg m^{2}$ tak aa jaata hai. Koi external torque nahi. Uska naya spin rate $ω_{2}$ nikalte hain.

Forecast: Faster ya slower? Kitne factor se?

Step 1 — angular momentum conservation invoke karo. Koi torque nahi toh, angular momentum $L = I ω$ constant hai: $I_{1} ω_{1} = I_{2} ω_{2} .$ Ye step kyun? Arms andar kheenchne mein sirf internal forces involve hain; ye $L$ change nahi kar sakte. Ye "no external push, momentum stays put" ka rotational analogue hai.

Step 2 — $ω_{2}$ ke liye solve karo. $ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{6 \times 2}{2} .$ Ye step kyun? Hum unknown isolate karne ke liye rearrange karte hain; denominator mein chhota $I_{2}$ $ω_{2}$ ko upar force karta hai.

Step 3 — evaluate karo. $ω_{2} = 6 rad/s .$

Verify: $L$ pehle $= 6 \times 2 = 12$ ; baad mein $= 2 \times 6 = 12$ ✓ (units $kg m^{2} / s$ dono sides). Wo 3× faster spin karti hai kyunki usne har $r$ shrink karke $I$ one-third kar liya. Physically sensible — isliye skaters tucking in karke accelerate karti hain.

Cell G — exam twist: 2-D layout mein ek corner se axis

Worked example Example 7 (Cell G) — square par char masses

Ek square ke corners par char $1 kg$ masses, side $a = 2 m$ . Axis plane ke perpendicular hai aur ek corner se gujarti hai (ise origin kaho). $I$ nikalo.

Forecast: Teen masses axis se door hain aur ek uske par hai. Guess karo ki $I$ $8$ ke kareeb hai ya $16 kg m^{2}$ ke kareeb.

Step 1 — coordinates rakho aur perpendicular distances padho. Corners $(0, 0), (2, 0), (0, 2), (2, 2)$ par hain. Axis woh line hai jo $(0, 0)$ se page ke perpendicular gujarti hai, toh $(x, y)$ par baitha particle ka us axis se perpendicular distance $r = x^{2} + y^{2}$ hai. Ye step kyun? Axis plane ko pierce karti hai; iske sabse chhota raasta plane mein hi rehta hai, toh uski length ordinary 2-D distance $x^{2} + y^{2}$ hai. Figure s04 dekho: har corner se origin tak dotted spokes.

Step 2 — har corner ke liye $r^{2}$ compute karo (square root skip karo — humein $r^{2}$ chahiye). $(0, 0) : r^{2} = 0; (2, 0) : r^{2} = 4; (0, 2) : r^{2} = 4; (2, 2) : r^{2} = 8.$ Ye step kyun? Formula ko $r^{2}$ chahiye, aur $r^{2} = x^{2} + y^{2}$ directly milta hai — koi root zaroori nahi. Axis par wala corner $0$ deta hai (phir se Cell B).

Step 3 — $m r^{2}$ sum karo. $I = 1 (0) + 1 (4) + 1 (4) + 1 (8) = 16 kg m^{2} .$

Verify: Units ✓. Forecast: ye $16$ hai, top end par, kyunki door wala diagonal corner ( $r^{2} = 8$ ) akele half total contribute karta hai. Axis par baitha corner ne kuch contribute nahi kiya.

Cell H — parallel-axis theorem ke saath sanity check

Worked example Example 8 (Cell H) — Example 4 cross-check karo

Hum claim karte hain ki dumbbell (do $2 kg$ at $\pm 0.4 m$ , total $M = 4 kg$ ) ka apne centre ke baare mein $I_{cm} = 0.64 kg m^{2}$ hai aur left mass ke baare mein $I = 1.28 kg m^{2}$ . Ek independent method se confirm karo.

Forecast: Kya dono exactly agree karne chahiye, ya ek doosre ka rounding hai?

Step 1 — parallel-axis theorem state karo. $I = I_{cm} + M d^{2},$ jahan $d$ centre of mass se nayi axis tak ka distance hai. Ye step kyun? Ye ek independent relation hai (Parallel axis theorem mein kahin aur prove kiya gaya), toh agar hamare direct sums ise satisfy karte hain, toh dono trustworthy hain.

Step 2 — $d = 0.4 m$ plug karo. $I = 0.64 + 4 (0.4)^{2} = 0.64 + 4 (0.16) = 0.64 + 0.64.$ Ye step kyun? Left mass centre se $0.4 m$ door hai, toh axis wahan shift karne par $d = 0.4$ se move hoti hai.

Step 3 — evaluate karo. $I = 1.28 kg m^{2} .$

Verify: Example 4 ke direct computation $1.28 kg m^{2}$ se exactly match karta hai ✓. Do raaste, ek jawab — results self-consistent hain. Units $kg m^{2}$ throughout ✓.

Matrix ka Recap

Recall Kaun sa example kaun sa trap cover karta tha?

Cell A plain sum ::: Example 1 ( $36 kg m^{2}$ ) Cell B mass on axis, $r = 0$ ::: Example 2 ( $14 kg m^{2}$ ) Cell C both sides / sign trap ::: Example 3 ( $0.64 kg m^{2}$ ) Cell D axis moved ::: Example 4 ( $1.28 kg m^{2}$ ) Cell E limiting $r \to \infty$ ::: Example 5 (quadratic growth) Cell F real skater ::: Example 6 ( $ω_{2} = 6 rad/s$ ) Cell G exam twist, 2-D corner ::: Example 7 ( $16 kg m^{2}$ ) Cell H parallel-axis cross-check ::: Example 8 ( $1.28 kg m^{2}$ )

Connections

Parent concept note — jahan formula derive kiya gaya hai
Kinetic energy ½mv² — woh $v^{2}$ jo $r^{2}$ banta hai
Rotational kinetic energy ½Iω² — jahan ye $I$ values use hoti hain
Angular velocity ω — har example mein shared $ω$
Centre of mass (first moment Σmr) — $r$ mein linear contrast (Ex 3)
Parallel axis theorem — cross-check tool (Ex 4, Ex 8)
Angular momentum L = Iω — skater ke peeche conservation law (Ex 6)
Torque τ = Iα — rotation ke liye Newton's second law