Question bank — Rolling KE = ½mv² + ½Iω²

False. Same $v_{cm}$ hai lekin alag $\beta$ hai: ring ke liye $\frac{1}{2}M{v}^2(1+1)$ hai, sphere ke liye $\frac{1}{2}M{v}^2(1+\frac{2}{5})$ hai, to equal speed par ring zyada KE carry karta hai.

True. $KE=\frac{1}{2}M{v}_{cm}^2(1+\beta )$, $M$ mein linear hai, aur $\beta$ sirf shape par depend karta hai, mass par nahi — isliye mass seedha KE ko scale karta hai.

True. $v=\sqrt{2gh/(1+\beta )}$ aur $\beta =I_{cm}/M{R}^2$, kisi bhi solid sphere ke liye $\frac{2}{5}$ hota hai; radius puri tarah cancel ho jaata hai, to dono same $v_{cm}$ par pohonchte hain.

True — lekin sirf $\beta >1$ ke liye, yaani mass $R$ se pare concentrated ho (rare). Plain ring ke liye $\beta =1$ exactly $1/2$ deta hai; standard solid shapes isse neeche rehti hain.

False. König's split kisi bhi rigid body ke liye valid hai jo plane motion mein ho. Rolling-without-slipping sirf $v_{cm}=\omega R$ link karne ke liye chahiye taaki KE ek variable mein collapse ho sake.

False. Translational KE zero hai lekin $\frac{1}{2}{I}_{cm}{\omega }^2$ bahut bada hai — saari energy spin mein lock hai kyunki no-slip link toot chuki hai.

False. Zero friction hone par center ke baare mein koi torque nahi hota, to $\omega$ kabhi nahi badalta; ball slide karti hai neeche aur $\sqrt{2gh}$ par pohonchti hai, roll karne se zyada fast.

True. Contact point instantaneously rest par hota hai, isliye friction force har instant mein zero displacement ke through kaam karta hai — yeh torque deta hai lekin energy transfer nahi karta.

Galat: yeh spinning bhi hai. $\frac{1}{2}{I}_{cm}{\omega }^2$ add karna zaroori hai; isko ignore karne se KE kam count hoti hai aur galat predict hota hai ki sab shapes ramp bottom par equal speed par pohonchenge.

Double-counting. Ya to $\frac{1}{2}M{v}^2+\frac{1}{2}{I}_{cm}{\omega }^2$ use karo ya akela $\frac{1}{2}{I}_{contact}{\omega }^2$ — dono kabhi nahi. Parallel-axis theorem se $I_{contact}=I_{cm}+M{R}^2$ mein pehle se $\frac{1}{2}M{v}^2$ wala piece hai.

Mass irrelevant hai. PE ($Mgh$) aur KE ($\frac{1}{2}M{v}^2(1+\beta )$) dono $M$ ke saath scale karte hain, isliye cancel ho jaata hai; order sirf $\beta$ decide karta hai.

Yeh sirf slipping mein sach hai. Rolling without slipping mein constraint $v_{cm}=\omega R$ unhe lock kar deta hai — woh independently choose karne ke liye free nahi hain.

Nahi. ${\beta }_{solid}=\frac{2}{5}<{\beta }_{hollow}=\frac{2}{3}$, to solid sphere ($v=\sqrt{2gh/1.4}$) hollow sphere ($v=\sqrt{2gh/1.667}$) ko beat karta hai.

Nahi — yeh exactly aur hamesha vanish hota hai, kyunki $\sum m_i{\vec{u}}_i=M{\vec{u}}_{cm,rel}=0$: center of mass ki apne relative velocity zero hoti hai, chahe ${\vec{v}}_{cm}$ kitni bhi fast ho.

Galat $R$. Rolling constraint geometric radius use karta hai (jo ground ko touch karta hai). Radius of gyration $k$ (jahan $I=M{k}^2$) ek alag length hai jo $\beta$ ke andar use hoti hai, $v=\omega R$ mein nahi.

Ring ka bada $\beta$ released PE ka ek bada fraction ($\frac{1}{2}$ vs $\frac{2}{7}$) spinning mein divert kar deta hai, forward motion ke liye kam bachta hai, isliye uska $v_{cm}$ chhota hota hai.

Kyunki König's cross term khatam ho jaata hai ($\sum m_i{\vec{u}}_i=0$), do energies cleanly decouple ho jaati hain — sliding aur spinning pictures ke beech kuch share ya double-count nahi hota.

Har ramp ke liye $v\propto 1/\sqrt{1+\beta }$ hai; $h$ ya angle badalne se saari speeds same factor se rescale hoti hain, isliye $\beta$ ke hisaab se ordering kabhi nahi badlti.

Contact point ki relative sliding velocity zero hoti hai, jo exactly static friction ki condition hai; kinetic friction sirf tab aata hai jab surfaces ek doosre ke against slide karein.

Parallel-axis: $I_{contact}=I_{cm}+M{R}^2$, isliye $\frac{1}{2}{I}_{contact}{\omega }^2=\frac{1}{2}{I}_{cm}{\omega }^2+\frac{1}{2}M(\omega R{)}^2=\frac{1}{2}{I}_{cm}{\omega }^2+\frac{1}{2}M{v}^2$ — algebraically identical hai.

Bahut kam friction se wheel slip karta hai, isliye contact point surface ke relative move karta hai; $v_{cm}\ne \omega R$ aur tumhe $v$ aur $\omega$ ko alag-alag track karna padta hai.

Pure translation: $KE=\frac{1}{2}M{v}^2$ jahan $v=\sqrt{2gh}$ hai. Yeh $\beta \to 0$ limit / no-spin case hai, sabse fast possible descent.

Exactly half: $\beta /(1+\beta )=1/2$. Sliding aur spinning mein equal energy — common shapes mein extreme case.

$v=\sqrt{2gh/(1+\beta )}\to 0$: almost saari PE spin mein jaati hai, isliye translational descent crawl ki taraf slow ho jaata hai.

Agar yeh bhi spin nahi kar raha ($\omega =0$), to nahi. Lekin agar rest se rolling release ho to dono $t=0$ par zero hain; KE tabhi aati hai jab speed aata hai aur, constraint se, spin bhi saath aata hai.

$\beta =0$, isliye $KE=\frac{1}{2}M{v}^2$ aur $v=\sqrt{2gh}$ — yeh frictionless slider ki tarah behave karta hai, kyunki spin energy store karne ke liye koi moment of inertia nahi hai.