Rolling KE = ½mv² + ½Iω²: Worked examples — Hinglish deep dive

1.5.14 · D3 · Physics › Rotational Mechanics › Rolling KE = ½mv² + ½Iω²

Yeh page Rolling KE ka exhaustive workbook hai. Parent note ne tumhe formula diya tha $K E = \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} I_{c m} ω^{2} = \frac{1}{2} M v_{c m}^{2} (1 + β), β = \frac{I _{c m}}{M R ^{2}} .$ Yahan hum har tarah ka problem cover karenge jisme yeh formula aata hai — har shape, woh slipping case jahan $v \neq = ω R$ , degenerate "point mass" limit, ek energy-loss twist, aur ek real-world word problem. Kuch bhi "khud samajh lena" ke liye nahi chhoda gaya.

Scenario matrix

Har rolling-KE problem in case classes mein se ek (ya blend) hota hai. Har cell ke liye neeche kam se kam ek worked example hai.

Cell	Case class	Kya special hai	Covered by
A	Pure energy split, given $v$	Ek known speed se dono KEs	Ex 1 (ring)
B	Charon standard shapes	Shapes mein $β$ compare karo	Ex 2
C	Incline from rest	PE → rolling KE, $v$ solve karo	Ex 3
D	Race / ordering	Mass aur radius cancel ho jaate hain	Ex 4
E	Slipping (constraint broken)	$v \neq = ω R$ , dono alag chahiye	Ex 5
F	Degenerate limits	$β \to 0$ (point mass) aur $β \to \infty$	Ex 6
G	Energy-loss twist	Rolling phir rough patch — friction energy hataata hai	Ex 7
H	Real-world word problem	Units, given power/mass, speed nikalo	Ex 8
I	Reverse problem	Given KE split, shape ( $β$ ) nikalo	Ex 9

Examples

Worked example Ex 1 — (Cell A) Rolling ring ka energy split

Ek ring jiska mass $M = 4 kg$ hai, $v_{c m} = 2 m/s$ par rolling without slipping kar rahi hai. Total KE, translational KE, aur rotational KE nikalo.

Forecast: Ring "worst spinner" hai — iska aadha energy spin mein locked hona chahiye. Guess: rotational KE $=$ translational KE.

Step 1 — $β$ identify karo. Ring $\Rightarrow β = 1$ . Yeh step kyun? Compact formula ko sirf $β$ hi shape input chahiye. Step 2 — Total KE. $K E = \frac{1}{2} M v^{2} (1 + β) = \frac{1}{2} (4) (2^{2}) (1 + 1) = \frac{1}{2} (4) (4) (2) = 16 J$ . Kyun? $K E = \frac{1}{2} M v^{2} (1 + β)$ ka direct plug-in. Step 3 — Split karo. Translational $= \frac{1}{2} M v^{2} = \frac{1}{2} (4) (4) = 8 J$ . Rotational $= 16 - 8 = 8 J$ . Kyun? $\frac{1}{2} M v^{2}$ term "1" wala part hai; $β$ wala part spin hai.

Verify: Rotational fraction $= β / (1 + β) = 1/2$ , matlab $16$ ka $50% = 8 J$ ✓. Forecast confirm hua: spin equals slide.

Worked example Ex 2 — (Cell B) Same speed, chaar shapes

Ring, disc, solid sphere, hollow sphere — sab ka $M = 1 kg$ , sab $v = 1 m/s$ par rolling kar rahe hain. Inki total KE rank karo.

Forecast: Zyada spin cost = same forward speed par zyada total KE. Toh ring ki KE sabse zyada honi chahiye, sphere ki sabse kam.

Step 1 — $β$ list karo. Ring $1$ , disc $\frac{1}{2}$ , solid sphere $\frac{2}{5}$ , hollow sphere $\frac{2}{3}$ . Kyun? Moment of Inertia ke standard results se. Step 2 — $K E = \frac{1}{2} (1) (1) (1 + β) = \frac{1}{2} (1 + β)$ apply karo. Kyun? $M = v = 1$ hone se KE sirf $β$ par depend karta hai.

Ring: $\frac{1}{2} (2) = 1.0 J$
Hollow sphere: $\frac{1}{2} (1.6667) = 0.8333 J$
Disc: $\frac{1}{2} (1.5) = 0.75 J$
Solid sphere: $\frac{1}{2} (1.4) = 0.70 J$ Step 3 — Rank karo. Ring $>$ hollow sphere $>$ disc $>$ solid sphere.

Verify: $β$ ka order = KE ka order, bilkul forecast jaisa — bada $β$ same $v$ par zyada energy carry karta hai ✓.

Worked example Ex 3 — (Cell C) Disc ramp se rest se neeche

Ek solid disc rest se $h = 1.4 m$ ki height se rolling without slipping neeche aata hai. $g = 10 m/s^{2}$ lo. Neeche $v_{c m}$ nikalo.

Forecast: Frictionless slide se slower, $2 g h = 28 \approx 5.29 m/s$ , kyunki kuch PE spin ban jaati hai.

Step 1 — Conservation of energy. $M g h = \frac{1}{2} M v^{2} (1 + β)$ (dekho Conservation of Energy on Inclines). Kyun? Rolling without slipping matlab contact point slide nahi karta, isliye friction koi work nahi karta — saari PE rolling KE mein convert hoti hai. Step 2 — $M$ cancel karo, $v$ solve karo. $v = \frac{2 g h}{1 + β}$ . Kyun? Mass dono sides par aata hai, isliye cancel ho jaata hai — answer mass-independent hai. Step 3 — Disc, $β = \frac{1}{2}$ . $v = \frac{2 ( 10 ) ( 1.4 )}{1.5} = \frac{28}{1.5} = 18.6 \overline{6} \approx 4.32 m/s$ .

Verify: $4.32 < 5.29$ ✓ (slide se slower). Units: $m^{2} / s^{2} = m/s$ ✓.

Worked example Ex 5 — (Cell E) SLIPPING: constraint broken

Ek disc ( $M = 2 kg$ , $R = 0.1 m$ ) ko ice par $ω = 20 rad/s$ tak spin kiya gaya jabki center sirf $v_{c m} = 1 m/s$ par move kar raha hai. Ice par yeh slips karta hai: $v \neq = ω R$ . Total KE nikalo.

Forecast: Kyunki $ω R = 2 m/s \neq = v = 1 m/s$ , hum compact $\frac{1}{2} M v^{2} (1 + β)$ formula nahi use kar sakte. Dono energies alag-alag use karni hongi.

Step 1 — Constraint check karo. Rolling Without Slipping ke liye $v = ω R$ chahiye. Yahan $ω R = (20) (0.1) = 2 \neq = 1$ . Yeh step kyun? Compact formula secretly $v = ω R$ assume karta hai; isse violate karne par woh formula galat ho jaata hai. Step 2 — Translational KE. $\frac{1}{2} M v^{2} = \frac{1}{2} (2) (1^{2}) = 1 J$ . Kyun? Yeh term sirf $v_{c m}$ use karta hai, slipping ho ya na ho. Step 3 — Rotational KE. $I_{c m} = \frac{1}{2} M R^{2} = \frac{1}{2} (2) (0. 1^{2}) = 0.01 kg\cdotm^{2}$ ; $\frac{1}{2} I ω^{2} = \frac{1}{2} (0.01) (2 0^{2}) = 2 J$ . Kyun? Rotational KE hamesha true $ω$ use karta hai, jo yahan $v$ se independent hai. Step 4 — Add karo (König's theorem abhi bhi valid hai). $K E = 1 + 2 = 3 J$ .

Verify: Agar hum galti se compact formula use karte, $\frac{1}{2} (2) (1) (1.5) = 1.5 J$ — aadha galat, kyunki usne $ω R = v$ force kiya ✓ (dikhata hai kyun slipping par split karna zaroori hai).

Worked example Ex 6 — (Cell F)

$β$ ke degenerate limits Rolling speed $v = 2 g h / (1 + β)$ do extremes mein kya ban jaati hai: (a) $β \to 0$ (ek "point mass" bead), aur (b) $β \to \infty$ (saari mass ek massless rim par bahut door)?

Forecast: $β \to 0$ frictionless block jaisa behave karna chahiye ( $2 g h$ ); $β \to \infty$ zero speed ki taraf crawl karna chahiye.

Step 1 — Case $β \to 0$ . $v \to \frac{2 g h}{1 + 0} = 2 g h$ . Kyun? Koi moment of inertia nahi matlab spin par koi energy divert nahi — pure sliding block hai. Step 2 — Case $β \to \infty$ . $v = \frac{2 g h}{1 + β} \to 0$ . Kyun? Saari PE rotation mein chali jaati hai; center ko aage move karne ke liye almost kuch nahi bachta. Step 3 — Sanity ordering. Finite shapes ke liye, $0 < β < \infty$ , isliye $0 < v < 2 g h$ hamesha — rolling hamesha in bounds ke beech rahegi.

Verify: $h = 1.4, g = 10$ se: $β = 0 \Rightarrow 28 \approx 5.29$ ; $β = 1000 \Rightarrow 28/1001 \approx 0.167 \to 0$ ✓. Har real shape (ring $4.32$ ? nahi— $3.74$ ; disc $4.32$ ; sphere $4.47$ ) $(0, 5.29)$ ke andar hai ✓.

Worked example Ex 7 — (Cell G) Energy-loss twist

Ek solid sphere without slipping $v = 3 m/s$ par roll kar raha hai ( $M = 2 kg$ ), ek aise patch se takraata hai jahan briefly skids karta hai aur apni saari rotational KE friction mein de deta hai, phir re-grip karta hai. Maanlo skid ke dauran forward speed $v$ unchanged rahi, toh original total KE ka kitna fraction bachta hai?

Forecast: Sphere apni zyaadatar energy translation mein rakhta hai ( $β = \frac{2}{5}$ chhota hai), isliye usse sirf ek modest slice khoona chahiye.

Step 1 — Original total KE. $\frac{1}{2} M v^{2} (1 + β) = \frac{1}{2} (2) (9) (1.4) = 12.6 J$ . Kyun? Skid se pehle standard rolling KE. Step 2 — Khoya hua rotational part. Fraction $\frac{β}{1 + β} = \frac{0.4}{1.4} = \frac{2}{7}$ , toh $K E_{r o t} = 12.6 \times \frac{2}{7} = 3.6 J$ remove hoti hai. Kyun? Problem kehta hai rotational KE skid se dissipate hoti hai. Step 3 — Surviving KE. $12.6 - 3.6 = 9.0 J$ , matlab fraction $9.0/12.6 = 5/7 \approx 0.714$ . Kyun? Jo bachta hai woh exactly translational share $1/ (1 + β)$ hai.

Verify: Translational KE $= \frac{1}{2} (2) (9) = 9 J$ survivor se match karta hai ✓; $\frac{5}{7} + \frac{2}{7} = 1$ ✓; sphere $71.4%$ rakhta hai, ring ke $50%$ se zyada — forecast holds ✓.

Worked example Ex 8 — (Cell H) Real-world word problem

Ek go-kart wheel (solid disc maano, $M = 3 kg$ , $R = 0.25 m$ ) rolling without slipping kar raha hai. Ek motor is ek wheel ko rest se steadily $63 W$ power deliver karta hai. $2 s$ baad wheel ka center kitni tezi se move kar raha hai, agar saari delivered energy rolling KE ban jaaye?

Forecast: Power $\times$ time energy hai; rolling-KE formula se speed mein convert karne par kuch m/s milne chahiye.

Step 1 — Delivered energy. $E = P t = 63 \times 2 = 126 J$ . Kyun? Power energy per second hai; time $t$ tak steady power se $E = P t$ milta hai. Step 2 — Rolling KE ke barabar set karo. $126 = \frac{1}{2} M v^{2} (1 + β)$ disc $β = \frac{1}{2}$ ke saath: $126 = \frac{1}{2} (3) v^{2} (1.5) = 2.25 v^{2}$ . Kyun? Saari input energy KE ban jaati hai (given), isliye equate karo. Step 3 — Solve karo. $v^{2} = 126/2.25 = 56$ , $v = 56 \approx 7.48 m/s$ . Kyun? $v$ isolate karo.

Verify: Units: $J / kg = m^{2} / s^{2} = m/s$ ✓. Check back: $2.25 \times 56 = 126 J$ ✓. Radius $R$ kabhi matter nahi kiya — Cell D ke saath consistent.

Recall Quick self-test

Ek body slipping ke saath roll kar rahi hai. Kya tum $\frac{1}{2} M v^{2} (1 + β)$ use kar sakte ho? ::: Nahi — woh formula $v = ω R$ assume karta hai. Slipping mein, true alag $ω$ ke saath $\frac{1}{2} M v^{2} + \frac{1}{2} I_{c m} ω^{2}$ use karo. $β \to 0$ hone par height $h$ se rolling speed kya ho jaati hai? ::: $2 g h$ — frictionless-slide value (spin par koi energy nahi jaati). Given total KE aur $M, v$ , shape kaise nikaalte ho? ::: $\frac{1}{2} M v^{2}$ compute karo; phir $β = \frac{K E}{\frac{1}{2} M v ^{2}} - 1$ ; $β$ ko table se match karo. Kya heavier ya larger wheel downhill race mein same shape wale se jeet jaata hai? ::: Nahi — tie hota hai; $v$ sirf $β$ par depend karta hai, $M$ ya $R$ par nahi.