Conservation of angular momentum — conditions: Worked examples

1.5.12 · D3 · Physics › Rotational Mechanics › Conservation of angular momentum — conditions

Kuch bhi shuru karne se pehle, teen plain-word reminders (koi symbol use nahi hoga jab tak woh earn na ho):

Scenario matrix

Is topic ke har problem mein se ek yeh cells hain. Neeche ke examples mein cell label kiya gaya hai, aur milke poora grid fill karte hain.

#	Cell (case class)	Usmein tricky kya hai	Example
A	$I$ shrinks, koi mass add nahi hua (internal reshape)	$ω$ badhna chahiye; KE badhti hai (kaam hota hai)	Ex 1
B	$I$ badhta hai, koi mass add nahi hua	$ω$ girna chahiye; KE girती hai	Ex 2
C	Mass add hoti hai chipakke (inelastic)	naya $m r^{2}$ aata hai; KE heat mein jaati hai	Ex 3
D	Central force, curved path (planet)	$τ = 0$ kyunki $r ∥ F$ ; $v r$ constant	Ex 4
E	Do spinning bodies couple hote hain (opposite spins)	$ω$ ke signs matter karte hain, zero tak cancel ho sakte hain	Ex 5
F	Ek axis ke baare mein conserved, doosre ke baare mein nahi	sahi origin/axis choose karna hoga	Ex 6
G	Degenerate / limiting: $r \to 0$ , $I \to \infty$ , $ω \to 0$	dekho ki formula toot toh nahi raha	Ex 7
H	Exam twist: energy vs momentum trap	yeh maanna galat hai ki KE conserved hai	Ex 8

Example 1 — Cell A: internal reshape, $I$ shrinks

Worked example Skater arms andar kheenchti hai (numbers ke saath revisit)

Ek skater $ω_{1} = 3 rad/s$ pe spin karti hai, $I_{1} = 8 kg\cdotm^{2}$ ke saath. Woh apne arms andar kheenchti hai toh $I_{2} = 2 kg\cdotm^{2}$ ho jaata hai. $ω_{2}$ nikalo, aur kinetic energies compare karo.

Forecast: padhne se pehle andaaza lagao — kya $ω$ badhega ya ghategaa? Roughly kitne guna?

Condition check. Ice frictionless hai; uske muscles internal forces hain. Toh vertical spin axis ke baare mein koi external torque nahi ⟹ us axis ke baare mein $L$ conserved hai. Yeh step kyun? Hume $I_{1} ω_{1} = I_{2} ω_{2}$ likhne ka haq earn karna hoga; woh equation tabhi sach hai jab $τ_{e x t} = 0$ .
Conservation likho. $I_{1} ω_{1} = I_{2} ω_{2}$ Yeh step kyun? Constant $L$ ka matlab hai ki $I ω$ ka number pehle = baad mein.
$ω_{2}$ solve karo. $ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{8 \times 3}{2} = 12 rad/s$ Yeh step kyun? $I$ quarter ho gaya, toh product fixed rakhne ke liye $ω$ chaar guna badhna chahiye.
Kinetic energy (Rotational kinetic energy): $K E = \frac{1}{2} I ω^{2}$ . $K E_{1} = \frac{1}{2} (8) (3)^{2} = 36 J, K E_{2} = \frac{1}{2} (2) (12)^{2} = 144 J$ Yeh step kyun? Dikhane ke liye ki $L$ conserved hone ka matlab $K E$ conserved hona nahi hai.

Verify: $L_{1} = 8 \times 3 = 24$ , $L_{2} = 2 \times 12 = 24$ ✓ (units $kg\cdotm^{2} / s$ ). $K E$ chaar guna ho gayi — muscles ne $108$ J kaam kiya mass ko andar kheenchke. Sanity: arms andar ⟹ tez ⟹ chair-and-books wali feeling se match karta hai.

Example 2 — Cell B: internal reshape, $I$ badhta hai

Worked example Merry-go-round, bacche baahir jaate hain

Ek merry-go-round (ek disc) jisme bacche hain, $I_{1} = 100 kg\cdotm^{2}$ hai aur $ω_{1} = 4 rad/s$ pe spin kar raha hai. Bacche baahir jaate hain, jisse yeh $I_{2} = 250 kg\cdotm^{2}$ ho jaata hai. $ω_{2}$ nikalo.

Forecast: tez ya dhima? Yeh Ex 1 ka mirror image hai.

Condition check. Bacchon ka chalna system ke liye internal hai; pivot smooth hai. Axis ke baare mein koi external torque nahi ⟹ $L$ conserved hai. Yeh step kyun? Pehle jaisi permission — internal motion kabhi bhi total $L$ nahi badalta.
Conservation: $I_{1} ω_{1} = I_{2} ω_{2}$ .
Solve karo: $ω_{2} = \frac{100 \times 4}{250} = 1.6 rad/s$ Yeh step kyun? Mass baahir gaya ⟹ $I$ badha ⟹ $I ω$ fixed rakhne ke liye $ω$ girna chahiye.

Verify: $L_{1} = 400 = L_{2} = 250 \times 1.6 = 400$ ✓. $K E_{1} = \frac{1}{2} (100) (16) = 800$ J, $K E_{2} = \frac{1}{2} (250) (2.56) = 320$ J — KE giri; bacchon ke legs ne difference absorb kiya. Phaila hua ⟹ dhima — intuition se match karta hai.

Example 3 — Cell C: mass chipakti hai (inelastic)

Worked example Putty turntable pe girta hai

Turntable $I_{t} = 0.5 kg\cdotm^{2}$ hai aur $ω_{0} = 10 rad/s$ pe spin kar raha hai. Putty jiska mass $m = 0.2 kg$ hai, seedha neeche girta hai aur radius $r = 0.5 m$ par chipak jaata hai. Final $ω_{f}$ aur KE loss nikalo.

Forecast: kya mass add karne se tez hoga ya dhima?

Condition check. Gravity neeche point karti hai, spin axis ke parallel, toh woh us vertical axis ke baare mein zero torque banati hai ( $F ∥$ axis ⟹ koi sideways twist nahi). Chipakne ki force internal hai. ⟹ Axis ke baare mein $L$ conserved hai. Yeh step kyun? Chahe collision violent ho, $τ_{e x t} = 0$ is axis ke baare mein, toh $L$ impact ke baad bhi bachta hai.
Putty ka moment of inertia add karo: ek point mass $m r^{2}$ contribute karta hai. $I_{new} = I_{t} + m r^{2} = 0.5 + 0.2 (0.5)^{2} = 0.55 kg\cdotm^{2}$
Conservation (putty pehle spin nahi kar raha tha, toh woh koi $L$ add nahi karta): $I_{t} ω_{0} = (I_{t} + m r^{2}) ω_{f}$
Solve karo: $ω_{f} = \frac{0.5 \times 10}{0.55} = \frac{5}{0.55} \approx 9.09 rad/s$ Yeh step kyun? Bada $I$ ⟹ chhota $ω$ , lekin thoda hi kyunki putty halka hai.

Verify: $L$ pehle $= 5$ , baad mein $= 0.55 \times 9.09 = 5.0$ ✓. $K E_{i} = \frac{1}{2} (0.5) (100) = 25$ J, $K E_{f} = \frac{1}{2} (0.55) (9.09)^{2} \approx 22.7$ J. Lagbhag $2.3$ J heat/sound mein gaya — expected, kyunki yeh inelastic stick hai.

Example 4 — Cell D: central force, curved path

Worked example Planet perihelion vs aphelion par (numbers ke saath)

Ek comet Sun se $r_{1} = 1$ AU door hai aur perihelion (sabse paas ka point) par $v_{1} = 60 km/s$ se chal raha hai. Aphelion par yeh $r_{2} = 5$ AU door hai. Wahan ki speed $v_{2}$ nikalo. Apsides par $v ⊥ r$ hai.

Forecast: door pe — tez ya dhima?

Condition check. Gravity hamesha seedha Sun ki taraf point karti hai: $F$ , $r$ ke along hai. Tab $τ = r \times F = 0$ (ek vector ko parallel vector se cross karo toh zero aata hai — Central forces). ⟹ $L$ conserved hai. Yeh step kyun? Yahi wajah hai ki Kepler's 2nd law kaam karta hai.
Apsides par $v ⊥ r$ hai, toh $L = m v r$ (koi complex angle nahi). Yeh step kyun? Figure dekho: sirf wahan jahan velocity radius ke perpendicular ho, $L$ simple product $m v r$ mein collapse hoti hai.
Conservation (mass $m$ cancel ho jaata hai): $m v_{1} r_{1} = m v_{2} r_{2} \Rightarrow v_{2} = v_{1} \frac{r _{1}}{r _{2}} = 60 \times \frac{1}{5} = 12 km/s$ Yeh step kyun? Paanch guna door ⟹ paanchwa hissa speed. Paas = tez, door = dhima.

Verify: $v_{1} r_{1} = 60 \times 1 = 60$ ; $v_{2} r_{2} = 12 \times 5 = 60$ ✓ (dono $km\cdotAU/s$ mein). Equal-area law se match karta hai: swept area rate $\frac{1}{2} r v_{⊥}$ dono ends par same hai.

Example 5 — Cell E: do spins couple hoti hain, signs matter karte hain

Worked example Do discs clutch ho jaate hain, opposite spins

Disc A: $I_{A} = 3 kg\cdotm^{2}$ , $+ 8 rad/s$ pe spin kar raha hai (anticlockwise ko positive maan lo). Disc B: $I_{B} = 1 kg\cdotm^{2}$ , $- 8 rad/s$ pe spin kar raha hai (clockwise). Unhe ek hi axle par press karke grip karaya jaata hai. Final common $ω_{f}$ nikalo.

Forecast: in numbers ke saath, kya woh tez, dhima, ya bilkul ruk jaenge?

Condition check. Discs ke beech clutch/friction ek internal force pair hai; axle externally frictionless hai. ⟹ Axle ke baare mein $L$ conserved hai. Signs yahan physics hain, toh inhe rakho. Yeh step kyun? Angular momentum ek vector hai — opposite spins opposite signs carry karte hain aur subtract ho sakte hain.
Pehle total $L$ (signed contributions add karo): $L = I_{A} ω_{A} + I_{B} ω_{B} = 3 (+ 8) + 1 (- 8) = 24 - 8 = 16$
Grip karne ke baad woh $I_{A} + I_{B}$ share karte hain ek $ω_{f}$ pe: $(I_{A} + I_{B}) ω_{f} = 16 \Rightarrow ω_{f} = \frac{16}{4} = + 4 rad/s$ Yeh step kyun? Wahi conservation, lekin ab "before" ek signed sum hai, single body nahi.

Verify: $L_{after} = 4 \times 4 = 16 = L_{before}$ ✓. Final sign $+$ hai: A ka spin jeeta kyunki usne zyada $L$ carry kiya. (Agar $I_{B}$ ki spin $- 24$ hoti, toh total $L$ $0$ hota ⟹ $ω_{f} = 0$ , woh ruk jaate — ek valid degenerate outcome.)

Example 6 — Cell F: ek axis ke baare mein conserved, doosre ke baare mein nahi

Worked example Cylinder ek ramp pe roll karta hua — axis choose karo

Ek cylinder gravity ke under ek incline pe roll karta hua neeche jaata hai. Kya uska angular momentum conserved hai?

Forecast: haan, nahi, ya "axis pe depend karta hai"?

Cylinder ke apne centre ke baare mein: gravity centre ke through act karti hai (centre ke baare mein zero torque), lekin ground ki friction rim pe act karti hai — friction ka lever arm hai, toh $τ \neq = 0$ . ⟹ Centre ke baare mein $L$ conserved nahi hai (woh spin up hota hai). Yeh step kyun? $τ = r F_{⊥}$ : radius $R$ par friction centre ke baare mein ek real twist banata hai.
Incline ke contact line par ek fixed point ke baare mein: ab gravity ka downhill component bhi ek lever arm rakha hai, toh phir $τ \neq = 0$ .
Conclusion — yahan koi single axis $τ = 0$ nahi deta, toh $L$ is system ke liye kisi bhi obvious axis ke baare mein conserved nahi hai. Yeh step kyun? Yeh Cell F ka honest lesson hai: conservation ek per-axis statement hai. Parent ki warning " $L$ necessarily har axis ke baare mein conserved nahi hoti" bilkul yahi hai — aur yahan woh sabke liye fail karti hai kyunki gravity+friction ek net twist chhodte hain.

Verify (numbers): $3 0^{\circ}$ angle ke liye gravity component, $m = 2$ kg, $g = 10$ : $F_{∥} = m g sin 3 0^{\circ} = 2 \times 10 \times 0.5 = 10 N$ — contact line ke baare mein nonzero lever arm ke saath ek nonzero force, toh $τ \neq = 0$ aur $L$ genuinely change hota hai. ✓ (Ex 3 se contrast: wahan sirf bahari force axis ke parallel thi, jisse $τ = 0$ mila.)

Example 7 — Cell G: degenerate aur limiting inputs

Worked example Formula edges par kya karta hai

Putty problem (Ex 3) lo aur $r$ aur $ω_{0}$ ke teen extreme choices test karo.

Forecast: kaunse limits formula tod dete hain, aur kaunse sirf boring lekin sahi answer dete hain?

$r \to 0$ (putty exact axis par girta hai). $ω_{f} = \frac{I _{t} ω _{0}}{I _{t} + m ( 0 ) ^{2}} = \frac{I _{t} ω _{0}}{I _{t}} = ω_{0} = 10 rad/s$ Yeh step kyun? Axis par mass ka zero lever arm hota hai ⟹ koi $I$ add nahi hota ⟹ koi slowdown nahi. Formula perfectly behave karta hai.
$r \to \infty$ (putty bahut door)। $I_{t} + m r^{2} \to \infty$ , toh $ω_{f} \to 0$ . Yeh step kyun? Infinitely phaila mass spin karna infinitely mushkil hai ⟹ spin khatam ho jaata hai. Sensible limit.
$ω_{0} = 0$ (turntable rest mein). Kisi bhi $r$ ke liye $ω_{f} = 0$ . Yeh step kyun? Koi spin andar nahi, koi spin baahir nahi — still table par putty girne se woh ghoomna shuru nahi ho sakta (gravity koi axis-torque nahi banati).

Verify: $r = 0$ par: $ω_{f} = 5/0.5 = 10$ ✓. Jab $r \to \infty$ , denominator diverge karta hai, $ω_{f} \to 0$ ✓. $ω_{0} = 0$ par, numerator $= 0$ ✓. Division-by-zero kabhi nahi hoti kyunki $I_{t} > 0$ hamesha rahta hai.

Example 8 — Cell H: exam trap (energy vs momentum)

Worked example "Dono conserved hain, hai na?" — classic galat mod

Ek student Ex 3 solve karta hai kinetic energy conserved maan ke, angular momentum ki jagah, likhta hai $\frac{1}{2} I_{t} ω_{0}^{2} = \frac{1}{2} (I_{t} + m r^{2}) ω_{f}^{2}$ . Isse kaunsa $ω_{f}$ milta hai, aur yeh galat kyun hai?

Forecast: kya bogus answer true $9.09 rad/s$ se bada hoga ya chhota?

Bogus KE equation: $ω_{f}^{bad} = ω_{0} \frac{I _{t}}{I _{t} + m r ^{2}} = 10 \frac{0.5}{0.55} \approx 9.53 rad/s$
Yeh galat kyun hai. Putty chipakna inelastic hai — heat banti hai, toh KE conserved nahi hai. Valid law hai $τ_{e x t} = 0 \Rightarrow L$ conserved, jisse $9.09 rad/s$ milta hai. Yeh step kyun? [!mistake]-style trap: conservation laws ki alag alag conditions hoti hain. $L$ ko zero external torque chahiye (yahan sach hai); KE ko koi energy loss nahi chahiye (yahan galat hai).
Dono ek saath nahi ho sakte — agar hote, toh koi energy loss nahi hoti, jo chipakne se contradict karta.

Verify: bogus $9.53 >$ true $9.09$ ✓ (KE-conservation speed over-predict karta hai kyunki woh energy loss ko forbid karta hai jo actually hoti hai). Gap $= 25 - 22.7 = 2.3$ J exactly wahi heat hai jo Ex 3 mein mili.

Active Recall

Recall Matrix par khud test karo

Table cover karo. Har ek ke liye cell name karo, phir trick: skater tucks; kids spread out; putty sticks; comet swings; discs clutch; cylinder rolls; putty on the axle; energy trap.

Skater arms andar kheenchti hai,

I : 8 \to 2

ω_{1} = 3

— final

ω

ω_{2} = 8 \times 3/2 = 12 rad/s

; KE badhti hai

36 \to 144

J (muscles kaam karte hain).

Merry-go-round

I : 100 \to 250

ω_{1} = 4

— final

ω

ω_{2} = 400/250 = 1.6 rad/s

Putty

m = 0.2

r = 0.5

I_{t} = 0.5

par

ω_{0} = 10

pe chipakta hai — final

ω

ω_{f} = 5/0.55 \approx 9.09 rad/s

Comet

v_{1} = 60

r_{1} = 1

AU —

r_{2} = 5

AU par speed?

v_{2} = 60/5 = 12 km/s

(

v_{1} r_{1} = v_{2} r_{2}

se).

Discs

I_{A} = 3

+ 8

I_{B} = 1

- 8

clutch hote hain — final

ω

ω_{f} = (24 - 8) /4 = + 4 rad/s

; signs matter karte hain.

Cylinder ramp pe roll karte hue

L

conserved kyun nahi hoti?

Rim par friction (aur gravity ka downhill hissa) har natural axis ke baare mein net external torque deta hai.

Putty

r = 0

par girta hai — slowdown kyun nahi?

Axis par woh zero

m r^{2}

add karta hai, toh

I

unchanged rehta hai aur

ω_{f} = ω_{0}

Putty problem mein KE conservation maan lene se galat (bahut zyada)

ω

kyun milta hai?

Chipakna inelastic hai — KE heat ke roop mein jaati hai; sirf

L

conserved hai. KE assumption us loss ko forbid karta hai aur speed over-predict karta hai.

Connections

Parent: Conservation of angular momentum — conditions — woh rule jise yeh examples apply karte hain.
Moment of inertia — woh $I$ jo Ex 1, 2, 3, 5, 7 mein change hota hai.
Torque — woh twist jiska absence har conservation step ko license deta hai.
Newton's third law — isliye internal muscle/clutch/stick forces kabhi bhi total $L$ nahi badlate.
Kepler's laws — Ex 4 equal-area law disguise mein hai.
Central forces — Ex 4 ka $r ∥ F$ zero-torque case.
Rotational kinetic energy — isliye $L$ conserved $\neq =$ KE conserved (Ex 1, 3, 8).
Conservation of linear momentum — same bookkeeping ka straight-line analogue.