Exercises — Rotational kinetic energy = ½Iω²

KYA chahiye: Hume $K_{rot}$ chahiye, aur hume $I$ aur $\omega$ directly diye gaye hain. YEH formula kyun: $K_{rot}=\frac{1}{2}I{\omega }^2$ exactly woh quantity hai jo poochi gayi hai; kuch build karne ki zaroorat nahi. $K_{rot}=\frac{1}{2}(4)(10{)}^2=\frac{1}{2}(4)(100)=200\text{J}.$ Answer: $200\text{J}$. Dhyan do ki humne pehle $\omega$ ko square kiya ($10^2=100$), phir multiply kiya.

KYA: Same $I$, double $\omega$. KYU: $K$ depend karta hai ${\omega }^2$ par, toh $\omega$ double karne se $K$ multiply hoti hai $2^2=4$ se. $K_{rot}=\frac{1}{2}(4)(20{)}^2=\frac{1}{2}(4)(400)=800\text{J}.$ Answer: $800\text{J}$, jo L1·1 ki value ka $4\times$ hai. Square hi poori story hai.

Step 1 — $I$ nikalo. Ek solid disc apne centre ke baare mein $I=\frac{1}{2}M{R}^2$ hota hai. $I=\frac{1}{2}(3)(0.2{)}^2=\frac{1}{2}(3)(0.04)=0.06{\text{kg⋅m}}^2.$ Yeh shape formula kyun: disc ki mass $r=0$ se $r=R$ tak spread hai; factor $\frac{1}{2}$ uss spread par $r^2$ ka average hai (Moment of inertia se). Step 2 — KE formula apply karo. $K_{rot}=\frac{1}{2}I{\omega }^2=\frac{1}{2}(0.06)(50{)}^2=\frac{1}{2}(0.06)(2500)=75\text{J}.$ Answer: $I=0.06{\text{kg⋅m}}^2$, $K_{rot}=75\text{J}$.

Hoop: $I_{\text{hoop}}=M{R}^2=(2)(0.3{)}^2=0.18{\text{kg⋅m}}^2$. $K_{\text{hoop}}=\frac{1}{2}(0.18)(8{)}^2=\frac{1}{2}(0.18)(64)=5.76\text{J}.$ Disc: $I_{\text{disc}}=\frac{1}{2}M{R}^2=0.09{\text{kg⋅m}}^2$. $K_{\text{disc}}=\frac{1}{2}(0.09)(64)=2.88\text{J}.$ Answer: hoop $5.76\text{J}$, disc $2.88\text{J}$; hoop exactly double store karta hai, kyunki iski saari mass radius $R$ par baithi hai (sabse bada possible $\sum m{r}^2$). Figure 1 dekho — hoop ki mass outer rim par hai (red), disc ki andar spread hai (blue).

Naya tool — rolling link $v_{cm}=R\omega$. Kyun zaroorat hai: rolling spin ko forward speed se tie karta hai, toh ek akela $v_{cm}$ fix karta hai $\omega =v_{cm}/R$. Isse hum $R$ ko bilkul eliminate kar sakte hain. (a) Translational: poori mass ko $v_{cm}$ par move karte maano (yeh ordinary Kinetic energy of a particle $\frac{1}{2}m{v}^2$ hai). $K_{trans}=\frac{1}{2}M{v}_{cm}^2=\frac{1}{2}(5)(4{)}^2=40\text{J}.$ (b) Rotational: $\omega =v_{cm}/R$ ko $\frac{1}{2}I{\omega }^2$ mein substitute karo:

Naya tool — Work-energy theorem (rotational). Kyun: net work KE ke change ke barabar hota hai, $W=\Delta K_{rot}=K_{final}-K_{initial}$. Rest se start karte hain, $K_{initial}=0$, toh $W=\frac{1}{2}I{\omega }^2$. $\omega$ ke liye rearrange karo — hum formula invert karte hain kyunki $\omega$ unknown hai: $\omega =\sqrt{\frac{2W}{I}}=\sqrt{\frac{2(100)}{0.5}}=\sqrt{400}=20\text{rad/s}.$ Answer: $\omega =20\text{rad/s}$. Square root ${\omega }^2$ ke square ko undo karta hai; hum positive root lete hain kyunki spin ki speed ek magnitude hai.

Step 1 — energy bookkeeping. Slipping nahi hai matlab friction net work nahi karta, toh saari gravitational potential energy kinetic ban jaati hai: $Mgh=\underset{\text{translation}}{\underbrace{\frac{1}{2}M{v}^2}}+\underset{\text{rotation}}{\underbrace{\frac{1}{2}{I}_{cm}{\omega }^2}}.$ Step 2 — rolling link se $\omega$ ko khatam karo. $\omega =v/R$ ke saath: $\frac{1}{2}{I}_{cm}{\omega }^2=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right)(\frac{v}{R}{)}^2=\frac{1}{5}M{v}^2.$ Step 3 — combine aur solve karo. $Mgh=\frac{1}{2}M{v}^2+\frac{1}{5}M{v}^2=\frac{7}{10}M{v}^2\Rightarrow v=\sqrt{\frac{10gh}{7}}.$ Plug in: $v=\sqrt{\frac{10(9.8)(1.4)}{7}}=\sqrt{19.6}=4.427\text{m/s}$ (3 d.p. tak). Sliding block (koi spin nahi): saari PE translation mein jaati hai, $v_{slide}=\sqrt{2gh}=\sqrt{27.44}=5.238\text{m/s}$. Answer: rolling sphere $\approx 4.43\text{m/s}$ tak pahunchta hai, sliding block ke $\approx 5.24\text{m/s}$ se slower, kyunki iski kuch energy spin mein lock hai (Figure 2).

Step 1 — energy balance. Block ka lost PE ban jaata hai block ka translational KE plus flywheel ka rotational KE: $mgH=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2.$ Step 2 — $\omega$ ko $v$ se link karo. Cord bina slipping ke unwind hoti hai, toh rim speed block speed ke equal hai: $v=R\omega \Rightarrow \omega =v/R$. $\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)\frac{v^2}{R^2}=\frac{1}{4}M{v}^2.$ Step 3 — solve karo. $mgH=\frac{1}{2}m{v}^2+\frac{1}{4}M{v}^2=(\frac{1}{2}m+\frac{1}{4}M)v^2.$ $v=\sqrt{\frac{mgH}{\frac{1}{2}m+\frac{1}{4}M}}=\sqrt{\frac{(2)(9.8)(1.0)}{1+1}}=\sqrt{9.8}=3.130\text{m/s}.$ Answer: $v\approx 3.13\text{m/s}$. Agar flywheel massless hota toh block $\sqrt{2gH}=4.43\text{m/s}$ tak pahunchta; flywheel spin mein energy "chura" leta hai.

Step 1 — pivot (end) ke baare mein $I$ nikalo. Do tarike hain: end result $I=\frac{1}{3}M{L}^2$ quote karo, ya centre value se Parallel axis theorem ke saath build karo. Centre: $I_{cm}=\frac{1}{12}M{L}^2$. Axis ko $d=L/2$ se shift karo $I=I_{cm}+M{d}^2$ use karke: $I=\frac{1}{12}M{L}^2+M(\frac{L}{2}{)}^2=\frac{1}{12}M{L}^2+\frac{1}{4}M{L}^2=\frac{1}{3}M{L}^2.$ Yeh tool kyun: energy formula ko $I$ chahiye actual rotation axis (pivot) ke baare mein, centre ke baare mein nahi; parallel axis theorem bridge hai. Numerically: $I=\frac{1}{3}(1.2)(0.6{)}^2=\frac{1}{3}(1.2)(0.36)=0.144{\text{kg⋅m}}^2$. Step 2 — KE formula apply karo. $K_{rot}=\frac{1}{2}(0.144)(3{)}^2=\frac{1}{2}(0.144)(9)=0.648\text{J}.$ Answer: $I=0.144{\text{kg⋅m}}^2$, $K_{rot}=0.648\text{J}$.

General rolling result. Kisi bhi rolling body ke liye $I_{cm}=cM{R}^2$ likho (toh $c=\frac{2}{5}$ solid, $c=\frac{2}{3}$ hollow). Energy conservation with $\omega =v/R$:

(a) $c\to 0$: saari mass axis par $\Rightarrow$ negligible moment of inertia $\Rightarrow$ koi energy spin mein nahi jaati. Formula deta hai $v=\sqrt{2gh}$ — exactly frictionless sliding block limit. Acha: rolling formula smoothly sliding wale mein reduce hota hai jab spin karne ke liye kuch nahi hota. (b) Hoop: $c=1$ (saari mass radius $R$ par). Tab $v=\sqrt{\frac{2gh}{2}}=\sqrt{gh}$, aur rotational fraction hai: $\frac{K_{rot}}{K_{total}}=\frac{c}{1+c}=\frac{1}{2}=50\%.$ Ek hoop apni energy exactly aadhi translation, aadhi rotation mein split karta hai — common rollers mein sabse "spin-heavy." Answer: (a) sliding-block limit $v=\sqrt{2gh}$; (b) $c=1$, aadhi energy rotational hai.

$\omega$ double karne se $K$ kitne factor se multiply hoti hai? ::: Chaar (energy mein ${\omega }^2$ hota hai). Same $M,R,\omega$: hoop ya disc zyaada KE store karta hai? ::: Hoop — disc ka double, mass sab radius $R$ par. Rolling body ki total KE? ::: $\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2$ with $v_{cm}=R\omega$. Sphere height $h$ se neeche final speed? ::: $v=\sqrt{10gh/7}$, sliding $\sqrt{2gh}$ se slower. Rolling body with $I=cM{R}^2$ ki height $h$ se final speed? ::: $v=\sqrt{2gh/(1+c)}$. Kya ek fixed-axle flywheel spin karte waqt PE gain/lose karta hai? ::: Nahi — iski centre move nahi karta, sirf $K_{rot}$ change hoti hai. Rod apne end ke baare mein swing kare: kaun sa $I$? ::: $\frac{1}{3}M{L}^2$ (parallel axis use karo $\frac{1}{12}M{L}^2$ se).