Exercises — Rolling KE = ½mv² + ½Iω²

Total König's split hai: $KE=\underset{\text{translation (aage slide karna)}}{\underbrace{\frac{1}{2}M{v}_{cm}^2}}+\underset{\text{rotation (spinning)}}{\underbrace{\frac{1}{2}{I}_{cm}{\omega }^2}}$ Pehla term woh energy hai jo centre of mass aage slide karte waqt carry karta hai; doosra term woh energy hai jo us centre ke around spin mein locked hai. Aur kuch present nahi kyunki König's theorem guarantee karta hai ki cross term vanish ho jaata hai.

Definition se $\beta =\frac{I_{cm}}{M{R}^2}=\frac{\frac{1}{2}M{R}^2}{M{R}^2}=\frac{1}{2}$. Rotational fraction $=\frac{\beta }{1+\beta }=\frac{1/2}{3/2}=\frac{1}{3}$. Toh disc ki kinetic energy ka ek tehai spinning hai, do tehai forward motion hai.

$KE=\frac{1}{2}M{v}_{cm}^2(1+\beta )$ use karo: $KE=\frac{1}{2}(3)(2^2)\left(1+\frac{2}{5}\right)=\frac{1}{2}(3)(4)(1.4)=8.4\text{J}.$ Compact form kyun? Kyunki rolling $\omega$ ko $v$ se tie karti hai, toh humein $\omega$ ya $R$ alag se kabhi nahi chahiye.

Disc: $\beta =\frac{1}{2}$. (a) $KE=\frac{1}{2}(4)(25)(1.5)=75\text{J}$. (b) Translational $=\frac{1}{2}M{v}^2=\frac{1}{2}(4)(25)=50\text{J}$. (c) Rotational $=$ total $-$ translational $=75-50=25\text{J}$. Fraction check karo: $\frac{25}{75}=\frac{1}{3}=\frac{\beta }{1+\beta }$ ✓.

Energy conservation: saari PE rolling KE ban jaati hai, toh $Mgh=\frac{1}{2}M{v}^2(1+\beta )$. $v=\sqrt{\frac{2gh}{1+\beta }}=\sqrt{\frac{2(10)(1.4)}{1+2/5}}=\sqrt{\frac{28}{1.4}}=\sqrt{20}\approx 4.47\text{m/s}.$ Compare karo: ek frictionless slide $\sqrt{2gh}=\sqrt{28}\approx 5.29\text{m/s}$ deta — sphere slower hai kyunki kuch PE spin mein gayi. ✓

Neeche wali figure physical setup dikhati hai: sphere rest se $Mgh$ potential energy (lavender height arrow) ke saath start karta hai aur bottom par forward centre speed $v_{cm}$ (coral arrow) ke saath pahunchta hai. Dhyan do ki KE label $(1+\beta )$ factor carry karta hai — woh extra $\beta$ hi woh spin bucket hai jo speed chura leta hai.

Rolling constraint: $\omega =\frac{v_{cm}}{R}=\frac{6}{0.2}=30\text{rad/s}$. Ring: $\beta =1$, toh rotational/translational $=\beta =1$ — dono energies equal hain (50/50 split). Exactly $\beta$ kyun? Dono KEs likhte hain: rotational $=\frac{1}{2}{I}_{cm}{\omega }^2=\frac{1}{2}(\beta M{R}^2)(v/R{)}^2=\frac{1}{2}\beta M{v}^2$, translational $=\frac{1}{2}M{v}^2$. Unka ratio $\frac{\frac{1}{2}\beta M{v}^2}{\frac{1}{2}M{v}^2}=\beta$ hai — $\beta$ ke alawa sab cancel ho jaata hai kyunki rolling constraint ne $\omega R=v$ force kar diya.

Bottom-speed formula se shuru karo, jo khud $Mgh=\frac{1}{2}M{v}^2(1+\beta )$ se aayi thi — dono bodies ke liye same height $h$, toh $Mgh$ shared energy budget hai. $v=\sqrt{\frac{2gh}{1+\beta }}.$ Yeh proportionality kyun deta hai? Dono bodies ke liye numerator $2gh$ identical hai (same drop, same $g$); sirf $(1+\beta )$ alag hai. Toh $v$ shape par sirf $\frac{1}{\sqrt{1+\beta }}$ ke through depend karta hai, aur dono speeds divide karne se common $\sqrt{2gh}$ cancel ho jaata hai: $\frac{v_{\text{sphere}}}{v_{\text{ring}}}=\frac{\sqrt{2gh}/\sqrt{1+{\beta }_{\text{sphere}}}}{\sqrt{2gh}/\sqrt{1+{\beta }_{\text{ring}}}}=\sqrt{\frac{1+{\beta }_{\text{ring}}}{1+{\beta }_{\text{sphere}}}}=\sqrt{\frac{1+1}{1+2/5}}=\sqrt{\frac{2}{1.4}}=\sqrt{\frac{10}{7}}\approx \mathrm{1.195.}$ Sphere lagbhag 20% faster pahunchta hai. Mass $M$ cancel ho gaya kyunki woh PE ($Mgh$) aur KE ($\frac{1}{2}M{v}^2(1+\beta )$) dono ko multiply karta hai; radius $R$ kabhi bacha hi nahi kyunki rolling constraint ne use pehle hi remove kar diya tha. Isi liye result $M$ aur $R$ se independent hai.

Bar chart ordering visible karti hai: har shape ki final speed (frictionless $\sqrt{2gh}$ ke units mein, dashed lavender line) tab zyaada tall hai jab $\beta$ chota ho. Sphere (mint) beats disc (butter) beats ring (coral).

$v=\sqrt{\frac{2gh}{1+\beta }}$ se, dono sides square karo: $v^2=\frac{2gh}{1+\beta }$. $v^2=gh$ diya hai: $gh=\frac{2gh}{1+\beta }\Rightarrow 1+\beta =2\Rightarrow \beta =1$. $\beta =1$ ek ring / hollow cylinder hai. Yeh sabse slow common shape hai, exactly $v=\sqrt{gh}$ se match karta hai.

(a) $I_{cm}=\frac{1}{2}M{R}^2$, $\omega =v/R$: $KE=\frac{1}{2}M{v}^2+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)\frac{v^2}{R^2}=\frac{1}{2}M{v}^2+\frac{1}{4}M{v}^2=\frac{3}{4}M{v}^2.$ (b) Parallel axis: $I_{\text{contact}}=I_{cm}+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$. $\omega =v/R$ ke saath: $KE=\frac{1}{2}\left(\frac{3}{2}M{R}^2\right)\frac{v^2}{R^2}=\frac{3}{4}M{v}^2.$ Identical ✓. Contact-point view translational term ko ek bade moment of inertia mein absorb kar leta hai.

(a) Flat par, saari PE rolling KE ban gayi hai (no slipping ⇒ friction se koi energy loss nahi): $Mgh=\frac{1}{2}M{v}^2(1+\beta )\Rightarrow v=\sqrt{\frac{2gh}{1+\beta }}=\sqrt{\frac{2(10)(3)}{1.5}}=\sqrt{40}\approx 6.32\text{m/s}.$ (b) Rotational fraction $=\frac{\beta }{1+\beta }=\frac{1/2}{3/2}=\frac{1}{3}\approx 33.3\%$. Yeh shape fix hone ke baad change nahi hota.

Poore system ke liye energy conservation. Girta block PE lose karta hai; woh block ki translational KE plus sphere ki rolling KE ban jaati hai. $m_{b}gh=\frac{1}{2}{m}_b{v}^2+\underset{\text{rolling sphere}}{\underbrace{\frac{1}{2}m{v}^2(1+\beta )}}.$ $\beta =\frac{2}{5}$ ke saath: sphere factor $=1.4$. $m_{b}gh=\frac{1}{2}{v}^2[m_b+m(1+\beta )]=\frac{1}{2}{v}^2[1+2(1.4)]=\frac{1}{2}{v}^2(3.8).$ Left side: $1(10)(0.7)=7$. Toh $7=1.9v^2\Rightarrow v^2=\frac{7}{1.9}\approx 3.684\Rightarrow v\approx 1.92\text{m/s}.$

Pehle sign/direction convention (subtle part). Positive $x$-axis us direction mein set karo jis direction mein wheel ka top flung ho raha hai — yahi woh direction hai jisme contact point floor par slide karta hai. Kinetic friction hamesha contact point ki sliding ko oppose karta hai, toh friction wheel ke base par $+x$ direction mein point karta hai, matlab woh centre ko forward push karta hai (forward motion mein spin-up karta hai). Sath hi woh same forward-pointing friction force, bottom par act karke, centre ke baare mein ek torque produce karta hai jo spin ko oppose karta hai — woh $\omega$ ko slow down karta hai. Toh friction simultaneously: $v$ ko $0$ se badhata hai, $\omega$ ko ${\omega }_0$ se ghataata hai. Rolling us instant shuru hoti hai jab yeh $v=\omega R$ par milte hain. Is sign ko sahi samajhna hi banata hai ki $\omega$ $\frac{1}{2}{\omega }_0$ tak gire, badhne ki bajaye.

(a) Kyunki friction contact point par horizontal hai, iska us point ke baare mein zero moment arm hai, toh contact point ke baare mein angular momentum conserved hai (isi liye hum woh axis choose karte hain — yeh baahri horizontal force ke torque ko khatam kar deta hai). Initial (pure spin, cm rest par): $L_i=I_{cm}{\omega }_0=M{R}^2{\omega }_0$. Final (rolling): $L_f=I_{cm}{\omega }_f+MR{v}_f=M{R}^2{\omega }_f+MR({\omega }_{f}R)=2M{R}^2{\omega }_f$ ($v_f={\omega }_{f}R$ use karke). Equal set karo: $M{R}^2{\omega }_0=2M{R}^2{\omega }_f\Rightarrow {\omega }_f=\frac{1}{2}{\omega }_0$, toh $v_f={\omega }_{f}R=\frac{1}{2}{\omega }_{0}R=\frac{1}{2}(20)(0.5)=5\text{m/s}.$ (b) Initial KE (pure spin): $\frac{1}{2}{I}_{cm}{\omega }_0^2=\frac{1}{2}M{R}^2{\omega }_0^2$. Final KE (rolling ring): $KE=\frac{1}{2}M{v}_f^2(1+\beta )$ use karke $\beta =1$ ke saath $K{E}_f=\frac{1}{2}M{v}_f^2(2)=M{v}_f^2=M(\frac{1}{2}{\omega }_{0}R{)}^2=\frac{1}{4}M{R}^2{\omega }_0^2$ milta hai. Toh $\frac{K{E}_f}{K{E}_i}=\frac{\frac{1}{4}M{R}^2{\omega }_0^2}{\frac{1}{2}M{R}^2{\omega }_0^2}=\frac{1}{2}$. Aadhi kinetic energy friction mein lost ho gayi — kyunki yahan woh rolling shuru hone tak slip karta hai. Energy conservation fail ho jaati hai; angular-momentum conservation sahi tool hai.

Figure do states dikhati hai: left par ring spin-in-place kar rahi hai (${\omega }_0$, centre rest par, contact point slip karta hai — coral dot); friction (grey arrow) base par forward act karta hai; right par ring ${\omega }_f=\frac{1}{2}{\omega }_0$ aur $v_f=\frac{1}{2}{\omega }_{0}R$ ke saath rolling reach kar chuki hai.

Equal final speed ka matlab equal $v^2$ hai: $\frac{2g{h}_s}{1+{\beta }_s}=\frac{2g{h}_d}{1+{\beta }_d}$. $2g$ cancel karo: $\frac{h_s}{1+{\beta }_s}=\frac{h_d}{1+{\beta }_d}$. $h_d=h_s\cdot \frac{1+{\beta }_d}{1+{\beta }_s}=1.0\cdot \frac{1+1/2}{1+2/5}=\frac{1.5}{1.4}\approx 1.071\text{m}.$ Disc ko thoda zyaada ooncha ramp chahiye kyunki woh zyaada energy spin mein lock karti hai, toh sphere ki speed match karne ke liye use thodi zyaada PE chahiye.

L1 — Q1: $KE=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2$. Q2: $\beta =\frac{1}{2}$, rotational fraction $\frac{1}{3}$. Q3: $KE=8.4\text{J}$. L2 — Q1: total $75\text{J}$, translational $50\text{J}$, rotational $25\text{J}$. Q2: $v=\sqrt{20}\approx 4.47\text{m/s}$. Q3: $\omega =30\text{rad/s}$, KE ratio $=1$. L3 — Q1: $v_{\text{sphere}}/v_{\text{ring}}=\sqrt{10/7}\approx 1.195$. Q2: $\beta =1$ (ring). Q3: dono methods $\frac{3}{4}M{v}^2$ dete hain. L4 — Q1: $v=\sqrt{40}\approx 6.32\text{m/s}$, rotational fraction $\frac{1}{3}$. Q2: $v\approx 1.92\text{m/s}$. L5 — Q1: $v_f=5\text{m/s}$, aadha KE lost. Q2: $h_d\approx 1.071\text{m}$.