Rotational kinetic energy = ½Iω²: Worked examples

1.5.9 · D3 · Physics › Rotational Mechanics › Rotational kinetic energy = ½Iω²

Kisi bhi example se pehle, teen tools jinhe hum baar baar use karenge. Har ek plain words mein bataya gaya hai taaki koi symbol bina wajah na aaye:

Scenario matrix

Is topic ke har problem ko inhi cells mein se ek (ya mix) mein daala ja sakta hai:

#	Case class	Kya special hai	Example
A	Pure spin (axis fixed, centre still)	sirf $\frac{1}{2} I ω^{2}$	Ex 1
B	Spin + roll (translation add hota hai)	$\frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}$	Ex 2
C	Shape race (same $M, R$ , alag $I$ )	kaun pehle bottom pe pohunche	Ex 3
D	Unit trap ( $ω$ rev ya deg mein)	rad/s mein convert karna zaroori	Ex 4
E	Axis trap (off-centre axis)	parallel-axis theorem chahiye	Ex 5
F	Degenerate / limiting ( $R \to 0$ , saara mass centre pe, $ω \to 0$ )	formula ka edge behaviour	Ex 6
G	Real-world word problem (flywheel kisi device ko power karta hai)	energy → time / work	Ex 7
H	Exam twist (do bodies coupled, shared $ω$ find karo)	conservation + geometry	Ex 8

Neeche ke aath examples har cell ko cover karte hain.

Example 1 — Cell A: pure spin

Worked example Grinding wheel, centre fixed

Ek solid disc grinding wheel: $M = 4 kg$ , $R = 0.10 m$ , $ω = 150 rad/s$ pe ek fixed spindle par ghoom raha hai. Iska centre nahi hilt raha. $K_{r o t}$ nikalo.

Forecast: chhota radius hai, to $I$ tiny hoga — lekin $ω$ squared hota hai. Order of magnitude guess karo.

Step 1 — Solid disc ka $I$ centre ke baare mein nikalo. $I = \frac{1}{2} M R^{2} = \frac{1}{2} (4) (0.10)^{2} = 0.02 kg\cdotm^{2}$ Yeh step kyun? Centre still hai, isliye yeh pure rotation hai — sirf energy $\frac{1}{2} I ω^{2}$ hai, aur pehle $I$ chahiye. Disc formula $\frac{1}{2} M R^{2}$ Moment of inertia se aata hai.

Step 2 — Energy formula mein plug karo. $K_{r o t} = \frac{1}{2} I ω^{2} = \frac{1}{2} (0.02) (150)^{2} = 225 J$ Yeh step kyun? Direct substitution; kuch translate nahi karta isliye koi $\frac{1}{2} M v^{2}$ term nahi hai.

Verify: Units: $kg\cdotm^{2} \cdot (rad/s)^{2} = kg\cdotm^{2} / s^{2} = J$ ✓ (radian dimensionless hota hai). Sanity: $ω^{2} = 22500$ bada hai, to $0.02$ moment of inertia bhi ek respectble $225$ J deta hai.

Example 2 — Cell B: spin + roll

Figure — Rotational kinetic energy = ½Iω²

Worked example Cylinder ramp se neeche roll karta hua

Ek uniform solid cylinder ( $I_{c m} = \frac{1}{2} M R^{2}$ ) rest se height $h = 1.4 m$ se bina slipping ke roll karta hua neeche aata hai. Bottom par centre speed $v$ nikalo. $g = 9.8 m/s^{2}$ lo.

Forecast: ek sliding block $2 g h$ tak pohunchta. Cylinder faster hoga ya slower? Kyun?

Step 1 — Energy conservation DONO kinetic terms ke saath likho. $M g h = translation \frac{1}{2} M v^{2} + rotation \frac{1}{2} I_{c m} ω^{2}$ Yeh step kyun? Figure dekho: centre aage move karta hai (translation) aur body spin karti hai (rotation). Gravity ki stored energy dono mein split hoti hai. Conservation of energy se, no slipping ka matlab friction koi net work nahi karta.

Step 2 — Rolling link $ω = v / R$ se $ω$ hatao. $\frac{1}{2} I_{c m} ω^{2} = \frac{1}{2} (\frac{1}{2} M R^{2}) \frac{v ^{2}}{R ^{2}} = \frac{1}{4} M v^{2}$ Yeh step kyun? Ek unknown chahiye. Rolling link $v_{c m} = R ω$ se $ω$ ko $v$ se replace karte hain; $R^{2}$ cancel ho jaata hai, isliye answer kabhi radius par depend nahi karta.

Step 3 — Combine karo aur solve karo. $M g h = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2} \Rightarrow v = \frac{4 g h}{3}$ Numerically $v = \frac{4 ( 9.8 ) ( 1.4 )}{3} = 18.29 \dots = 4.28 m/s$ . Yeh step kyun? $M$ cancel ho jaata hai — mass matter nahi karta, sirf mass distribution (through $\frac{1}{2}$ in $I$ ) matter karta hai.

Verify: Ek slider $2 g h = 27.44 = 5.24 m/s$ leta. Cylinder slower hai (4.28 < 5.24) kyunki energy ka ek quarter spin mein gaya. ✓

Example 3 — Cell C: shape race

Worked example Hoop vs disc vs sphere, same ramp

Same mass aur radius ke teen bodies ek saath same height se release hote hain aur bina slipping ke roll karte hain. $I_{h oo p} = M R^{2}$ , $I_{d i sc} = \frac{1}{2} M R^{2}$ , $I_{s p h er e} = \frac{2}{5} M R^{2}$ . Unki bottom speeds rank karo.

Forecast: zyada mass bahar ki taraf matlab zyada $I$ . Kya bada $I$ matlab faster ya slower?

Step 1 — $I = c M R^{2}$ ke saath general result likho. Ex 2 ko $I = c M R^{2}$ se repeat karo (to $c$ shape factor hai): $M g h = \frac{1}{2} M v^{2} + \frac{1}{2} (c M R^{2}) \frac{v ^{2}}{R ^{2}} = \frac{1}{2} (1 + c) M v^{2} \Rightarrow v = \frac{2 g h}{1 + c}$ Yeh step kyun? Ek formula sabhi shapes cover karta hai; sirf $c$ badalta hai. Bada $c$ (mass bahar) matlab bada denominator, isliye chhota $v$ .

Step 2 — Har $c$ insert karo.

Sphere $c = \frac{2}{5}$ : $v = \frac{2 g h}{1.4} = \frac{10 g h}{7}$ — sabse fast.
Disc $c = \frac{1}{2}$ : $v = \frac{2 g h}{1.5} = \frac{4 g h}{3}$ — beech mein.
Hoop $c = 1$ : $v = \frac{2 g h}{2} = g h$ — sabse slow. Yeh step kyun? Hoop ka saara mass radius $R$ par hai, isliye sabse bada $I$ hai, to spin mein sabse zyada store hota hai aur forward motion ke liye sabse kam bachta hai.

Verify ( $g h$ ke ratios mein numbers): sphere $10/7 = 1.195$ , disc $4/3 = 1.155$ , hoop $1.000$ . Order sphere > disc > hoop ✓. Dhyan do radius aur mass kabhi appear nahi karte — ek marble same shape ke truck tyre ko bhi beat karta hai. ✓

Example 4 — Cell D: unit trap

Worked example Motor spec rev/min mein given

Ek flywheel disc, $M = 12 kg$ , $R = 0.30 m$ , $600 rev/min$ par ghoomti hai. $K_{r o t}$ nikalo.

Forecast: agar tum seedha $600$ plug in karo, to tumhara answer too big hoga ya too small?

Step 1 — rev/min ko rad/s mein convert karo. Ek revolution $2 π$ radians hai; ek minute $60$ s hai: $ω = 600 \cdot \frac{2 π}{60} = 20 π = 62.83 rad/s$ Yeh step kyun? Formula $v = r ω$ se derive hua hai, jo sirf radians mein sahi hai (dekho Angular velocity note aur parent ka [!mistake]). Yeh skip karne se number meaningless ho jaata hai.

Step 2 — Moment of inertia aur energy. $I = \frac{1}{2} (12) (0.30)^{2} = 0.54 kg\cdotm^{2}, K = \frac{1}{2} (0.54) (62.83)^{2} = 1066 J$ Yeh step kyun? Standard disc $I$ , phir $\frac{1}{2} I ω^{2}$ converted $ω$ ke saath.

Verify: Agar galti se $ω = 600$ use karte, to $\frac{1}{2} (0.54) (600)^{2} = 97, 200 J$ milta — lagbhag $91 \times$ zyada bada. Conversion optional nahi hai. $K \approx 1.07 kJ$ ✓.

Example 5 — Cell E: axis trap (parallel axis)

Worked example Rod ek end ke baare mein swing karta hua

Ek uniform rod, $M = 2 kg$ , length $L = 1.2 m$ , ek end par pivot hai aur swing karta hai to bottom par $ω = 3 rad/s$ hai. $K_{r o t}$ nikalo.

Forecast: $I$ end ke baare mein centre ke baare mein se bada hoga ya chhota? Same $ω$ par kaun zyada energy store karega?

Step 1 — Centre ke baare mein $I$ yaad karo, phir axis shift karo. Centre ke baare mein ek rod ka $I_{c m} = \frac{1}{12} M L^{2}$ hota hai. Pivot $d = L /2$ door hai, isliye Parallel axis theorem se: $I_{e n d} = I_{c m} + M d^{2} = \frac{1}{12} M L^{2} + M (\frac{L}{2})^{2} = \frac{1}{3} M L^{2}$ Yeh step kyun? $I$ ek fixed property nahi hai — yeh axis par depend karta hai (parent [!mistake]). Theorem centre axis se end axis par shift karne mein $M d^{2}$ add karke help karta hai.

Step 2 — Numbers. $I_{e n d} = \frac{1}{3} (2) (1.2)^{2} = 0.96 kg\cdotm^{2}, K = \frac{1}{2} (0.96) (3)^{2} = 4.32 J$ Yeh step kyun? Actual axis ke liye sahi $I$ ke saath, energy formula as usual apply hota hai.

Verify: Centre ke baare mein $I = \frac{1}{12} (2) (1.2)^{2} = 0.24$ milta, sirf $\frac{1}{2} (0.24) (9) = 1.08 J$ deta. End axis 4× zyada hold karta hai ( $0.96/0.24 = 4$ ) — mass average par pivot se door hai, isliye faster move karta hai aur zyada KE carry karta hai. ✓

Example 6 — Cell F: degenerate & limiting cases

Worked example Formula edges par kya karta hai

Ek spinning body par teen edge inputs test karo: (a) saara mass axis par collapse ho jaaye ( $r \to 0$ ), (b) $ω \to 0$ , (c) length $R$ ki string par ek point mass $m$ ko $ω$ par swing karo — kya $\frac{1}{2} I ω^{2}$ , $\frac{1}{2} m v^{2}$ se agree karta hai?

Forecast: check karne se pehle har answer guess karo. Kya (a) aur (b) dono zero denge?

Step 1 — Case (a): saara mass axis par. To har $r_{i} = 0$ , isliye $I = \sum m_{i} r_{i}^{2} = 0$ aur $K = \frac{1}{2} (0) ω^{2} = 0$ . Kyun? Spin axis par ek pencil-thin line of mass actually move hi nahi karta — axis par points ka $v = r ω = 0$ hota hai. Zero speed → zero KE. Formula agree karta hai.

Step 2 — Case (b): $ω \to 0$ . $K = \frac{1}{2} I (0)^{2} = 0$ chahe $I$ kuch bhi ho. Kyun? Koi spin nahi, koi motion nahi, koi kinetic energy nahi. ✓

Step 3 — Case (c): ek single point mass, ultimate sanity check. Radius $R$ par ek point mass ka $I = m R^{2}$ hota hai. Iska speed $v = R ω$ hai, isliye $K = \frac{1}{2} I ω^{2} = \frac{1}{2} (m R^{2}) ω^{2} = \frac{1}{2} m (R ω)^{2} = \frac{1}{2} m v^{2} .$ Yeh step kyun? Rotational formula ko ek particle ke liye plain Kinetic energy of a particle formula mein reduce karna chahiye — kyunki yahi to tha jise sum karke banaya gaya tha. Aur karta hai.

Verify (numbers): $m = 0.5 kg$ , $R = 2 m$ , $ω = 4 rad/s$ ke saath: $\frac{1}{2} I ω^{2} = \frac{1}{2} (0.5 \cdot 4) (16) = 16 J$ aur $\frac{1}{2} m v^{2} = \frac{1}{2} (0.5) (8)^{2} = 16 J$ . Identical ✓. Formula teeno edge tests pass karta hai.

Example 7 — Cell G: real-world word problem

Worked example Flywheel ek machine ko power karta hua

Ek flywheel $K = 1.125 \times 1 0^{5} J$ store karta hai (parent ka Example 1 flywheel). Ek machine steady power $P = 2500 W$ draw karti hai aur flywheel hi akela source hai. Yeh kitni der tak chal sakti hai jab tak energy khatam na ho jaaye?

Forecast: power energy per second hoti hai — guess karo answer seconds mein hoga ya minutes mein.

Step 1 — Stored energy, power, aur time ko relate karo. Power energy per unit time delivered hoti hai, isliye $P = \frac{K}{t}$ , jo deta hai $t = \frac{K}{P} = \frac{1.125 \times 1 0 ^{5}}{2500} = 45 s .$ Yeh step kyun? Work-energy theorem (rotational) kehta hai machine par kiya gaya work flywheel ke $K_{r o t}$ loss ke barabar hai; constant power par deliver karne se duration set hoti hai.

Step 2 — Interpret karo. Flywheel machine ko 45 s tak chalata rehta hai (jab tak $ω \to 0$ , yaani $K \to 0$ ). Kyun? Jab spin energy khatam hoti hai to flywheel ruk jaata hai — energy in equals energy out.

Verify: Units $J / W = J / (J/s) = s$ ✓. Cross-check: $P \cdot t = 2500 \times 45 = 1.125 \times 1 0^{5} J = K$ ✓.

Example 8 — Cell H: exam twist (coupled bodies)

Worked example Do discs ek saath clutch hote hue

Disc 1 ( $I_{1} = 0.6 kg\cdotm^{2}$ ) $ω_{1} = 40 rad/s$ par ghoom rahi hai. Ise ek stationary coaxial disc 2 ( $I_{2} = 0.2 kg\cdotm^{2}$ , $ω_{2} = 0$ ) se press kiya jaata hai jab tak dono lock hokar ek saath na ghoomne lagein. (a) common final $ω_{f}$ nikalo, aur (b) kitni kinetic energy lost hui.

Forecast: ek "sticking" collision mein, kya kinetic energy conserve hoti hai? Step 3 se pehle guess karo.

Step 1 — Angular momentum conserve karo, energy nahi. Clutching ke dauran koi external torque act nahi karta, isliye total angular momentum $L = I ω$ conserve hota hai: $I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{f} .$ Yeh step kyun? Clutch force internal hai; faces ke beech friction bhi internal hai, isliye energy waste kar sakta hai lekin total $L$ nahi badal sakta. Yeh perfectly inelastic collision ka rotational analogue hai.

Step 2 — $ω_{f}$ solve karo. $ω_{f} = \frac{I _{1} ω _{1}}{I _{1} + I _{2}} = \frac{0.6 \times 40}{0.8} = \frac{24}{0.8} = 30 rad/s .$ Kyun? Combined inertia $0.8$ fixed momentum $24$ share karta hai.

Step 3 — Kinetic energies compare karo. $K_{i} = \frac{1}{2} (0.6) (40)^{2} = 480 J, K_{f} = \frac{1}{2} (0.8) (30)^{2} = 360 J .$ Energy lost $= 480 - 360 = 120 J$ , slipping clutch faces se heat mein badal gayi. Yeh step kyun? Angular momentum conserve hota hai lekin kinetic energy nahi — sticking se $120 J$ waste hota hai (original ka 25%).

Verify: $L_{i} = 0.6 \times 40 = 24$ , $L_{f} = 0.8 \times 30 = 24$ — momentum match karta hai ✓. $K_{f} < K_{i}$ as expected ek inelastic lock ke liye ✓.

Recall Quick self-test

Kaunsa cell har phrase ka hai? "rev/min given" ::: Cell D (unit trap) — rad/s mein convert karo. "pivoted at the end" ::: Cell E (axis trap) — parallel axis theorem. "do gears lock together" ::: Cell H — angular momentum conserve karo, energy drop hogi. "same mass, kaun fastest roll karega" ::: Cell C — sabse chhota shape factor $c$ jeetta hai. "saara mass axis par" ::: Cell F — $I = 0$ , isliye $K = 0$ .

Connections

Parent: ½Iω² — woh formula jo har example use karta hai.
Moment of inertia — disc, hoop, sphere, rod ke liye $I$ supply karta hai.
Kinetic energy of a particle — woh $\frac{1}{2} m v^{2}$ jisme Example 6c reduce hota hai.
Rolling motion — Examples 2, 3 mein $v_{c m} = R ω$ link.
Parallel axis theorem — Example 5 mein axis shift.
Conservation of energy — Examples 2, 3, 7 ko power karta hai.
Work-energy theorem (rotational) — Example 7 mein stored $K$ ko run-time mein convert karta hai.
Angular velocity — Example 4 ke peeche radian requirement.