Worked examples — Moment of inertia of - rod (about end, centre), disk, ring, sphere (solid, hollow), cylinder

1.5.8 · D3 · Physics › Rotational Mechanics › Moment of inertia of - rod (about end, centre), disk, ring,

Shuru karne se pehle, ek word jo hum baar baar use karte hain: axis. Axis woh imaginary line hai jiske around body spin karti hai. Upar ki definition batati hai ki sirf ek hi distance matter karti hai — $r_{\perp }$, us line par perpendicular drop — na centre tak ki distance, na body ke saath saath ki distance. Agar tum axis galat draw karo, toh har $r_{\perp }$ galat hoga aur har number galat hoga. Isliye har example axis draw karne aur $r_{\perp }$ identify karne se shuru hota hai.

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The scenario matrix

Is topic ke har problem ka type in case classes mein se ek hoga. Neeche ke examples mein har cell ka tag diya hua hai.

Cell	Case class	Kya cheez tricky banati hai	Covered by
A	Single body, standard axis	bas formula pick karo	Ex 1
B	Axis centre se shift hua	Parallel Axis Theorem chahiye	Ex 2, Ex 6
C	Axis plane mein (diameter)	Perpendicular Axis Theorem chahiye	Ex 3
D	Composite body (add / subtract parts)	$I$ add hota hai; holes subtract karte hain	Ex 4, Ex 5
E	Degenerate / limiting input ($R\to 0$, $L\to 0$, point mass)	check karo ki formula break na ho	Ex 7
F	Zero-distance mass (mass axis par)	contributes $0$	Ex 4 (hub)
G	Real-world word problem (energy / rolling)	words → axis → $I$ mein translate karo	Ex 8
H	Exam twist (two bodies, ratio, "which spins easier")	compare karo, sirf compute mat karo	Ex 9

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Ex 1 — Cell A: pure formula, single body

Forecast: Aage padhne se pehle order of magnitude guess karo. Kya yeh $0.01$ ke kareeb hoga ya $10\mathrm{kg}{\mathrm{m}}^2$?

Kya dekhna hai: red circle disk hai; centre par black dot woh jagah hai jahan axis use pierce karti hai (axis seedha page se bahar, tumhari taraf point kar rahi hai). $R$ label wala black arrow ek radius hai — notice karo ki rim par bahar wale ek mass element ka perpendicular drop $r_{\perp }=R$ hai, jabki beech mein aadhe wale element ka $r_{\perp }=R/2$ hai. Formula ka $\frac{1}{2}$ un sab par $r_{\perp }^2$ ko average karne se aata hai.

1. Axis identify karo. Figure dekho: red disk ko centre par axis pierce karti hai jo page se bahar aa rahi hai. Har mass element ka $r_{\perp }$ us centre dot se seedhi line ki distance hai. Yeh step kyun? Cell A tab hi kaam karta hai jab axis ek standard axis ho. Confirm karne par hum directly boxed formula use kar sakte hain, haath se $\int r_{\perp }^{2}dm$ integrate karne ki zaroorat nahi.
2. Formula pick karo. Us axis ke baare mein disk ke liye, parent note deta hai $I=\frac{1}{2}M{R}^2$. Yeh step kyun? Mass ka har tukda $r_{\perp }$ par kahin $0$ aur $R$ ke beech baitha hai; $\frac{1}{2}$ area ke hisaab se weighted $r_{\perp }^2$ ka average hai — hum woh integral dobara nahi karte, use reuse karte hain.
3. Numbers plug karo. $I=\frac{1}{2}(2)(0.30{)}^2=\frac{1}{2}(2)(0.09)=0.09\mathrm{kg}{\mathrm{m}}^2$ Yeh step kyun? Diye gaye $M$ aur $R$ substitute karne se symbolic law us ek number mein badal jaata hai jo question actually maangta hai; $R$ ko square karna zaruri hai (double nahi karna) kyunki $I$ distance ke square par depend karta hai.

Verify: Units: $\mathrm{kg}\cdot {\mathrm{m}}^2$ ✓. Sanity: same radius ki $2\mathrm{kg}$ ring hogi $M{R}^2=0.18$; disk exactly half hai kyunki uska mass andar ki taraf spread hai — "Ring 1, Disk ½" se match karta hai. ✓

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Ex 2 — Cell B: axis shifted (parallel axis)

Forecast: Ex 1 ke $0.09$ se bada hoga ya chhota? Kitna?

Kya dekhna hai: "CM" mark wala chhota black dot purani central axis hai; red rim par black square NAYA axis hai. Un dono ke beech double-headed arrow gap $d$ hai — aur yahan yeh exactly ek radius $R$ hai. Ab har mass element rim axis ke perpendicular measure kiya jaata hai, isliye har $r_{\perp }$ badh jaata hai aur $I$ increase hota hai.

1. Pehchaano ki yeh standard axis NAHI hai. Axis ab centre of mass se nahi guzarti, isliye koi boxed formula directly $I$ nahi deta. Yeh step kyun? "Off-centre" pehchaanna Parallel Axis Theorem use karne ka trigger hai, na ki table lookup ka.
2. Theorem state karo. $I=I_{cm}+M{d}^2$, jahaan $I_{cm}$ centre of mass se guzarne wali parallel axis ke baare mein moment hai, aur $d$ dono parallel axes ke beech ka gap hai. Yeh step kyun? Yeh kahta hai: axis ko $d$ se door shift karna hamesha $M{d}^2$ add karta hai, kyunki ab har mass element ka $r_{\perp }$ same offset se bada hota hai — theorem bas $\int r_{\perp }^{2}dm$ hai shifted line se re-measure kiya hua.
3. $I_{cm}$ aur $d$ identify karo. Parallel central axis deta hai $I_{cm}=\frac{1}{2}M{R}^2=0.09$ (Ex 1 se). Centre se rim tak gap hai $d=R=0.30\mathrm{m}$. Yeh step kyun? $d$ axes ke beech ki distance hai, jo yahan exactly radius hai — figure mein red segment dekho.
4. Add karo. $I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2=1.5(2)(0.09)=0.27\mathrm{kg}{\mathrm{m}}^2$ Yeh step kyun? Theorem ne jo do contributions diye unhe combine karte hain; kyunki $d=R$ hai, extra term $M{d}^2$ ek poora $M{R}^2$ ke barabar hai, original half ke upar doubling karta hai — arithmetic mein $R$ poore time squared rehna chahiye.

Verify: $0.27=3\times 0.09$, yani central value ka teen guna. Yeh bada hua, exactly jaisa theorem promise karta hai (CM se door jaane par $I$ kabhi chhota nahi ho sakta). Units ✓.

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Ex 3 — Cell C: axis plane mein (perpendicular axis)

Forecast: Central perpendicular axis ke liye ring ka $I$ hai $M{R}^2=0.25$. Diameter ke baare mein — same, zyada, ya kam?

Kya dekhna hai: red circle ring hai, horizontal black arrow diameter axis hai jo ring ke plane mein flat hai. Dashed vertical line sabse upar wale point ke liye perpendicular drop $r_{\perp }=R$ dikhata hai, jabki axis par baitha black dot $r_{\perp }=0$ dikhata hai. Kyunki ring ke around $r_{\perp }$, $0$ se $R$ tak vary karta hai, hum "sab mass $R$ par hai" nahi keh sakte — yahi wajah hai ki hume perpendicular-axis theorem ki zaroorat hai.

1. Dekho ki hume ek naya tool kyun chahiye. Diameter axis ke liye, ring ke alag alag points ke alag alag perpendicular distances $r_{\perp }$ hain (axis par wale points $0$ par hain, top aur bottom $R$ par). Ab hum "sab mass $R$ par hai" nahi keh sakte. Yeh step kyun? Yeh Cell C ko ring ke easy central-axis case se alag karta hai, jahaan har $r_{\perp }=R$ hota hai.
2. Perpendicular Axis Theorem invoke karo. Kisi bhi flat (planar) body ke liye jo $x$–$y$ plane mein hai, $I_z=I_x+I_y$, jahaan $z$ plane ke perpendicular hai aur $x,y$ usi point se guzarne wali dono in-plane axes hain. Yeh step kyun? Ring planar hai, isliye yeh theorem integrate kiye bina easy $z$-axis ko diameter axes se jodta hai.
3. Symmetry use karo. Dono diameters $x$ aur $y$ symmetry se identical hain, isliye $I_x=I_y=I_d$. Tab $I_z=2I_d$. Yeh step kyun? Symmetry ek unknown ko factor of 2 mein badal deta hai — koi integration nahi chahiye.
4. Solve karo. $I_z=M{R}^2$ (central perpendicular). Toh $I_d=\frac{1}{2}{I}_z=\frac{1}{2}M{R}^2=\frac{1}{2}(1)(0.5{)}^2=0.125\mathrm{kg}{\mathrm{m}}^2$ Yeh step kyun? $I_z=2I_d$ rearrange karne par $I_d$ directly milta hai, aur known $I_z=M{R}^2$ mein diye gaye numbers substitute karne par answer milta hai; halving isliye valid hai kyunki dono diameters poora $I_z$ equally share karte hain.

Verify: Diameter value ($0.125$) central value ($0.25$) se kam hai — sahi hai, kyunki ring ke aadhe mass ab diameter line ke kareeb baitha hai. Units ✓.

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Ex 4 — Cells D & F: composite body jisme hub axis par hai

Forecast: Teeno objects mein se kaun sabse zyada contribute karta hai? Kaun kuch nahi contribute karta?

1. Moments of inertia add hote hain. Same axis ke baare mein kai masses ke liye, $I_{\text{total}}=I_1+I_2+\dots$ — yeh bas $\int r_{\perp }^{2}dm$ (equivalently $\sum m{r}_{\perp }^2$) hai jo object ke hisaab se regroup kiya gaya hai. Yeh step kyun? Kyunki defining integral ek plain sum hai mass par, tum har piece alag se compute karke add kar sakte ho.
2. Disk. $I_d=\frac{1}{2}{M}_d{R}^2=\frac{1}{2}(100)(2{)}^2=200\mathrm{kg}{\mathrm{m}}^2$. Yeh step kyun? Standard disk formula, axis uski central axis hai, isliye har element ka $r_{\perp }$ exactly $0$ se $R$ tak run karta hai jaisa $\frac{1}{2}M{R}^2$ average assume karta hai.
3. Baccha (point mass rim par). Point ki tarah treat karo: $I_c=m{r}_{\perp }^2=50\cdot (2{)}^2=200\mathrm{kg}{\mathrm{m}}^2$. Yeh step kyun? Ek insaan ride ke comparison mein chhota hota hai, isliye defining $I=\int r_{\perp }^{2}dm$ ek single $m{r}_{\perp }^2$ mein collapse ho jaati hai jisme $r_{\perp }=R$ hai.
4. Hub-cap axis par — Cell F. Uski axis se perpendicular distance $r_{\perp }=0$ hai, isliye $I_{\text{hub}}=10\cdot 0^2=0$. Yeh step kyun? Spin line par baitha mass bilkul koi circle travel nahi karta ($r_{\perp }=0$ integral mein), isliye woh exactly zero contribute karta hai. Yeh degenerate zero-distance case hai.
5. Add karo. $I=200+200+0=400\mathrm{kg}{\mathrm{m}}^2$ Yeh step kyun? Step 1 ne parts sum karne ki permission di, isliye total bas teeno contributions combine karna hai; hub ka zero explicitly rakha gaya hai yeh dikhane ke liye ki woh genuinely drop out ho jaata hai.

Verify: Hub $10\mathrm{kg}$ hone ke bawajood kuch contribute nahi karta — yahi rotation mein "position beats mass" ka poora point hai. Units sab $\mathrm{kg}{\mathrm{m}}^2$ ✓.

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Ex 5 — Cell D: ek hole subtract karo (disk jisme se bite li gayi hai)

Forecast: Kya hole same $R$ ki poori disk ke comparison mein $I$ per kg bada banata hai ya chhota?

Kya dekhna hai: red-shaded annulus real material hai; beech mein white circle (radius $a$) missing hole hai. Do black arrows $R$ (outer) aur $a$ (inner) mark karte hain. Ise $R$ radius ki badi red disk socho jisme se $a$ radius ka same-density plug nikaala gaya — yahi "big disk minus plug" picture neeche ke algebra mein exactly hai.

1. "Big disk minus small disk" ke roop mein model karo. Socho ki hole bhar do $R$ radius ki poori disk banane ke liye, phir $a$ radius ka plug subtract karo. Dono ka same axis hai. Yeh step kyun? $\int r_{\perp }^{2}dm$ add aur subtract karta hai, isliye ek hole bas ek negative disk hai jisme real material jaisi same surface density $\sigma$ hai.
2. Surface density mein kaam karo. Maano $\sigma =M/(\pi (R^2-a^2))$ real annulus ki density hai. $R$ radius ki poori disk ka $I=\frac{1}{2}(\sigma \pi R^2)R^2$ hai; plug ka $I=\frac{1}{2}(\sigma \pi a^2)a^2$ hai. Yeh step kyun? Filled disk aur plug dono same stuff se bane hain, isliye same $\sigma$ unhe describe karta hai; tab hi hum cleanly subtract kar sakte hain.
3. Subtract karo. $I=\frac{1}{2}\sigma \pi R^4-\frac{1}{2}\sigma \pi a^4=\frac{1}{2}\sigma \pi (R^4-a^4)$ Ab $\sigma \pi =M/(R^2-a^2)$ substitute karo: $I=\frac{1}{2}\frac{M}{R^2-a^2}(R^4-a^4)=\frac{1}{2}M\frac{(R^2-a^2)(R^2+a^2)}{R^2-a^2}=\frac{1}{2}M(R^2+a^2)$ Yeh step kyun? Factorisation $R^4-a^4=(R^2-a^2)(R^2+a^2)$ se annoying denominator cancel ho jaata hai, aur clean annulus formula milta hai $I=\frac{1}{2}M(R^2+a^2)$.
4. Numbers ($R=0.06$, $a=0.015$): $I=\frac{1}{2}(0.020)(0.06^2+0.015^2)=0.010(0.0036+0.000225)=3.825\times 10^{-5}\mathrm{kg}{\mathrm{m}}^2$ Yeh step kyun? Diye gaye radii (metres mein convert karke taaki units $\mathrm{kg}{\mathrm{m}}^2$ mein aayein) step 3 ke compact formula mein substitute karte hain; dono radii ko square karna distance-squared law ko honour karta hai aur dono squared terms simply add ho jaate hain.

Verify: Agar $a\to 0$ (koi hole nahi) toh formula ban jaata hai $\frac{1}{2}M{R}^2$, poori disk — ✓ consistent. Agar $a\to R$ (sab rim, ek thin ring) toh ban jaata hai $\frac{1}{2}M(2R^2)=M{R}^2$, ring — ✓. Dono limits sahi hain, isliye general formula trustworthy hai.

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Ex 6 — Cell B: rod apne end ke baare mein, do tarikon se check kiya

Forecast: Centre value hai $\frac{1}{12}M{L}^2$. End value — kitne guna bada?

1. Direct formula. $I_{\text{end}}=\frac{1}{3}M{L}^2=\frac{1}{3}(0.6)(1.0{)}^2=0.20\mathrm{kg}{\mathrm{m}}^2$. Yeh step kyun? Yeh parent derivation ka standard end-axis result hai; $M$ aur $L$ substitute karne se requested number milta hai.
2. Parallel Axis Theorem se cross-check karo. $I_{\text{end}}=I_{cm}+M{d}^2$ jisme $I_{cm}=\frac{1}{12}M{L}^2$ aur $d=L/2$ (centre to end): $I=\frac{1}{12}(0.6)(1{)}^2+0.6{\left(\frac{1}{2}\right)}^2=0.05+0.15=0.20\mathrm{kg}{\mathrm{m}}^2$ Yeh step kyun? Do independent tarikon se same number aana dono formula aur axis placement ko prove karta hai; yahan $d=L/2$ hai kyunki CM rod ke midpoint par hai.

Verify: $\frac{1}{3}:\frac{1}{12}=4:1$, toh end value centre value ka $4\times$ hai — parent ke active-recall fact se match karta hai. ✓

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Ex 7 — Cell E: degenerate & limiting inputs

Forecast: Inmen se kaunsa $I=0$ dega, aur kaunsa finite nonzero answer dega?

1. (a) Rod point mein shrink ho rahi hai. $I_{\text{end}}=\frac{1}{3}M{L}^2\to \frac{1}{3}M(0{)}^2=0$. Yeh step kyun? Apne pivot par baitha ek point mass ka har jagah $r_{\perp }=0$ hai — travel karne ke liye koi circle nahi, isliye $I=0$. Formula sahi se collapse ho jaata hai.
2. (b) Sphere point mein shrink ho rahi hai. $I=\frac{2}{5}M{R}^2\to \frac{2}{5}M(0{)}^2=0$. Same reasoning: sab mass axis par. Yeh step kyun? Confirm karta hai ki sphere formula sensibly ek point mass mein degenerate hota hai ($r_{\perp }\to 0$).
3. (c) Sab mass rim par. Ex 5 ka annulus formula $\frac{1}{2}M(R^2+a^2)$ at $a=R$ deta hai $\frac{1}{2}M(2R^2)=M{R}^2$ — yani ring. Yeh step kyun? Mass ke har bit ko $R$ radius par bahar push karna maximum-$I$ ring reproduce karna chahiye; karta hai.

Verify: (a) aur (b) → $0$ (mass axis par, Cell F logic), (c) → $M{R}^2$ (ring, maximum). Teeno limits physically exactly wahi hain jo hum expect karte hain, koi formula blow up ya contradict nahi ho raha.

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Ex 8 — Cell G: real-world word problem (energy)

Forecast: Roughly kitne joules — kuch handful, ya hundreds?

1. $I$ nikalo. Long axis ke baare mein solid cylinder: $I=\frac{1}{2}M{R}^2=\frac{1}{2}(8)(0.20{)}^2=\frac{1}{2}(8)(0.04)=0.16\mathrm{kg}{\mathrm{m}}^2$. Yeh step kyun? Energy ke liye pehle $I$ chahiye; axis long axis hai, isliye disk/cylinder formula use hoga (length-independent).
2. Rotational KE apply karo. $KE=\frac{1}{2}I{\omega }^2$ — yeh $\frac{1}{2}m{v}^2$ ka rotational twin hai. Yeh step kyun? Yahi formula wajah hai ki $r_{\perp }^2$ $I$ ke andar rehta hai (parent note ka "Why $r^2$" dekho).
3. Compute karo. $KE=\frac{1}{2}(0.16)(30{)}^2=\frac{1}{2}(0.16)(900)=72\mathrm{J}$ Yeh step kyun? Step 1 ka $I$ aur diya hua $\omega$, $\frac{1}{2}I{\omega }^2$ mein substitute karne par stored energy milti hai; $\omega$ square hota hai kyunki energy spin rate ke square ke saath badhti hai.

Verify: Units $\mathrm{kg}{\mathrm{m}}^2\cdot (rad/s{)}^2=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$ ✓. $72\mathrm{J}$ ek modest lekin real store hai — ek chhote flywheel ke liye reasonable.

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Ex 9 — Cell H: exam twist ("pehle neeche kaun pahunchega?")

Forecast: Derive karne se pehle gut trust karo: solid ya hollow?

Pehle axis identify karo. Jab har sphere roll karta hai, woh apne khud ke centre of mass se guzarne wali axis ke baare mein spin karta hai, ground ke parallel aur travel direction ke perpendicular. Yeh diameter axis hai, isliye relevant $I$ values parent note se diameter formulas hain: solid $\frac{2}{5}M{R}^2$, hollow $\frac{2}{3}M{R}^2$. (Har $r_{\perp }$ $\int r_{\perp }^{2}dm$ mein us central spin axis se measure kiya jaata hai.)

1. Har $I$ ko $I=\beta M{R}^2$ likho. Solid sphere: $\beta =\frac{2}{5}=0.4$. Hollow sphere: $\beta =\frac{2}{3}\approx 0.667$. Yeh step kyun? Har rolling body ka behaviour ek single pure number $\beta =I/(M{R}^2)$ se control hota hai — Radius of Gyration squared over $R^2$. Ise bahar nikalne par dono spheres ko ek formula se treat kar sakte hain.
2. Energy do tarikon mein split karo (rolling constraint). Bina slip kiye rolling ka matlab contact point momentarily still hai, jo force karta hai $v=\omega R$, yani $\omega =v/R$. Total kinetic energy translation plus rotation hai: $KE=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}M{v}^2+\frac{1}{2}(\beta M{R}^2){\left(\frac{v}{R}\right)}^2=\frac{1}{2}M{v}^2(1+\beta )$ Yeh step kyun? Yahaan moment of inertia actually bite karta hai: $\frac{1}{2}I{\omega }^2$ term (parent note ka Rotational Kinetic Energy) woh energy hai jo spinning mein lock ho jaati hai aur forward speed mein nahi ja sakti. Bada $\beta$ matlab zyada lock.
3. Dropped height se speed nikalo energy conservation se. Height $h$ utarne ke baad, gravity ne $Mgh$ kaam kiya hai, poora us kinetic energy mein badal gaya: $Mgh=\frac{1}{2}M{v}^2(1+\beta )\Rightarrow v^2=\frac{2gh}{1+\beta }$ Yeh step kyun? Friction koi net kaam nahi karta (contact point slide nahi hota), isliye mechanical energy conserved hai. Yeh $v$ ko $\beta$ ke terms mein isolate karta hai — aur $M$, $R$ cancel ho gaye, ek key exam insight.
4. Ramp ke neeche linear acceleration nikalo. Rest se constant acceleration $a$ ke saath $s$ distance mein motion ke liye, $v^2=2as$, aur drop ki height $h=s\sin \theta$ hai. $v^2$ substitute karke: $2as=\frac{2g(s\sin \theta )}{1+\beta }\Rightarrow a=\frac{g\sin \theta }{1+\beta }$ Yeh step kyun? "Final speed" ko "acceleration" mein convert karne par compare kar sakte hain kaun poore raste tezzi se speed up karta hai, jo race decide karta hai — sirf finish speed nahi.
5. Dono $\beta$ plug karo aur compare karo. $a_{\text{solid}}=\frac{g\sin \theta }{1+0.4}=\frac{g\sin \theta }{1.4},a_{\text{hollow}}=\frac{g\sin \theta }{1+0.667}=\frac{g\sin \theta }{1.667}$ Solid sphere ka chhota denominator hai, isliye badi acceleration hai: $\frac{a_{\text{solid}}}{a_{\text{hollow}}}=\frac{1+\frac{2}{3}}{1+\frac{2}{5}}=\frac{5/3}{7/5}=\frac{25}{21}\approx 1.19$ Yeh step kyun? Dono ke liye same $g$, $\theta$ hai, isliye sirf $\beta$ alag hai: chhota $\beta$ (mass centre ke kareeb hoarded) → bada $a$.
6. Acceleration ko descent time mein convert karo. Rest se $s=\frac{1}{2}a{t}^2$ ke saath, same ramp length $s$ cover karne ka time $t=\sqrt{2s/a}$ hai, isliye $t\propto 1/\sqrt{a}\propto \sqrt{1+\beta }$: $\frac{t_{\text{solid}}}{t_{\text{hollow}}}=\sqrt{\frac{1+\frac{2}{5}}{1+\frac{2}{3}}}=\sqrt{\frac{7/5}{5/3}}=\sqrt{\frac{21}{25}}\approx 0.917$ Yeh step kyun? Yahi actual quantity hai jo question maangta hai — descent time. Solid sphere ka time hollow sphere ke time ka approximately $92\%$ hai, matlab woh pehle pahunchta hai.

Answer & interpretation: Solid sphere jeet jaata hai — woh $\approx 19\%$ zyada tezzi se accelerate karta hai aur hollow sphere ke time ka $\approx 92\%$ mein finish karta hai. Physically, hollow sphere apna sab mass $R$ radius par bahar push karta hai, usse bada $\beta$ milta hai, isliye gravity ki zyada energy spin mein trap ho jaati hai ($\frac{1}{2}I{\omega }^2$) aage drive karne ki bajaye. Result $M$, $R$, aur $\theta$ se independent hai — classic exam takeaway.

Verify: Sanity check ke liye, ek frictionless sliding block ka $\beta =0$ aur $a=g\sin \theta$ hai, dono spheres se tez — sahi hai, kyunki woh spinning mein koi energy waste nahi karta. Aur $t_{\text{solid}}/t_{\text{hollow}}<1$ confirm karta hai ki solid sphere pehle hai. ✓

Recall Race kis quantity ne decide ki?

Not mass, not radius, but $\beta =I/(M{R}^2)$ ::: woh pure shape fraction. Chhota $\beta$ = mass axis ke kareeb = badi acceleration = chhota descent time.

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Connections

• Parallel Axis Theorem — Ex 2 aur Ex 6 ka engine.
• Perpendicular Axis Theorem — Ex 3 ka engine.
• Rotational Kinetic Energy — Ex 8 aur Ex 9 mein use hua.
• Rolling Motion — Ex 9 ki race ke peeche physics.
• Radius of Gyration — $\beta$ fraction repackaged.
• Torque and Angular Acceleration — yeh $I$ values kaise forces ko spin mein convert karti hain.