Question bank — Relation to linear quantities - v = rω, a_t = rα, a_c = rω²

True — $v=r\omega$ $\omega$ mein linear hai, toh $r$ fixed rakh ke $\omega$ double karne se exactly $v$ double hoti hai.

False — $a_c=r{\omega }^2$ $\omega$ ke squared par depend karta hai, toh $\omega$ double karne se $a_c$ chaar se multiply hoti hai, do se nahi.

True — angular velocity poori body ke turning ko describe karti hai; har point same angle per second sweep karta hai regardless of $r$.

False — $v=r\omega$ $r$ ke saath badhti hai, toh rim point ek inner point se aage nikal jaata hai chahe woh ek $\omega$ share karte hon.

False — velocity ek vector hai; uski direction ghoomti rehti hai, toh $a_c=r{\omega }^2\ne 0$ tab bhi jab $v$ kabhi nahi badalti.

True — $a_t=r\alpha$ speed ka change measure karta hai; constant speed matlab $\alpha =0$, toh $a_t=0$ jabki $a_c$ abhi bhi exist karta hai.

False — ${\vec{a}}_t$ track ke saath point karta hai aur ${\vec{a}}_c$ inward point karta hai, toh woh $90^{\circ }$ par milte hain. $90^{\circ }$ par do arrows right triangle ki legs banate hain jiska diagonal saccha total hai, toh woh Pythagoras se combine hote hain: $a=\sqrt{a_t^2+a_c^2}$ (master figure dekho). Unhe plain numbers ke roop mein add karna galat taur par assume karega ki woh same direction mein point karte hain.

True — slowing matlab speed decrease ho rahi hai, toh ${\vec{a}}_t$ (jo speed change karta hai) motion arrow $\vec{v}$ ke against point karta hai; $\alpha$ tab $\omega$ ke relative negative hota hai.

True — $a_c=r{\omega }^2$ shared $\omega$ par $r$ ke saath scale karta hai, toh ratio $0.6/0.2=3$ hai, exactly unki speeds $v$ ke ratio se match karta hua.

Wrong: $\omega$ ek rigid body par har point ke liye identical hota hai. Tezi rim speed poori tarah se $v=r\omega$ mein bade $r$ se aati hai, bade $\omega$ se nahi.

Wrong: yeh $a_c=r{\omega }^2$ hai. Units check karo: $r^2\omega$ deta hai ${\text{m}}^2/\text{s}$, lekin acceleration ${\text{m/s}}^2$ honi chahiye, jo $r{\omega }^2=\text{m}\cdot {\text{s}}^{-2}$ deliver karta hai.

Wrong at fixed $\omega$: $a_c=r{\omega }^2$ ke roop mein rewrite karo, jo $r$ ke saath badhta hai. $v^2/r$ form hide karta hai ki $v$ khud $r$ ke saath badhti hai, toh rigid body ke liye $a_c\propto r$ hai.

Wrong: formula radians ke liye bana hai. Degrees mein convert karna ise tod deta hai; agar degrees mein diya ho, pehle radians mein convert karo.

Wrong: $a_t$ mein $\omega$ hai hi nahi. Yeh sirf $r$ aur $\alpha$ par depend karta hai; ek tez lekin steadily-spinning wheel ($\alpha =0$) ka $a_t=0$ hoga.

Wrong: sirf $a_c$ inward point karta hai. Jab body speed up ya slow hoti hai ($a_t\ne 0$), total tangential part ki wajah se radius se tilt ho jaata hai.

Wrong: constant $\omega$ phir bhi matlab hai $a_c=r{\omega }^2\ne 0$, jiske liye point ko apne circle par rakhne ke liye ek inward (centripetal) net force chahiye.

Kyunki hum point ki distance-along-the-path differentiate karte hain aur $r$ ko derivative se bahar nikalte hain; ek rigid body $r$ ko fixed rakhti hai, toh koi extra $dr/dt$ term nahi aata.

$a_t$ speed changes handle karta hai aur $a_c$ direction changes handle karta hai; circle par ek point hamesha turning kar raha hai, toh uski velocity direction hamesha change ho rahi hai — $a_c$ kabhi vanish nahi ho sakta jab tak woh move kar raha hai.

Kyunki centripetal acceleration velocity ki direction ke turning se paida hota hai; sirf vector differentiation us change ko capture kar sakta hai, jo ek scalar speed akela miss kar deta.

Radian is tarah define kiya gaya hai ki arc length equals radius times angle (dekho Radian measure and arc length), jo $v=r\omega$ aur $a_t=r\alpha$ ko bina kisi conversion factor ke sahi banata hai — degrees clumsy $\pi /180$ terms inject kar dete.

$a_t=r\alpha$: angular acceleration poori body share karti hai, lekin bade $r$ se multiply karne par yeh ek bade tangential acceleration mein stretch ho jaata hai.

Master figure imagine karo lekin $\omega$ climbing ke saath: inward arrow ${\vec{a}}_c=r{\omega }^2$ spin ke square ke saath stretch hota hai, jabki tangential arrow ${\vec{a}}_t=r\alpha$ $\omega$ feel hi nahi karta. Toh inward arrow fixed-length tangential ke paas quickly enormous ho jaata hai, aur total acceleration almost fully center ki taraf swing kar jaata hai — geometrically diagonal inward leg par collapse ho jaata hai. Isliye fast-spinning rims center se door khichti hain.

Newton's laws aur kinetic energy linear language ($v$, $a$) mein rehte hain; radius $r$ woh dictionary hai jo body-wide angular quantities ko point-by-point linear ones mein convert karta hai.

Teeno zero hain, kyunki har ek $r$ times ek angular quantity hai — axis par point bilkul nahi hilta, chahe body kitni bhi tezi se ghoomti ho.

Sirf $a_c=r{\omega }^2$ survive karta hai; $a_t=r\alpha =0$. Motion uniform hai — steady speed, constant inward pull.

$a_c=r{\omega }^2=0$ us instant par, lekin $a_t=r\alpha \ne 0$; point momentarily direction turn nahi kar raha lekin speed gain kar raha hai.

Haan — $v=r\omega$ ek product hai, toh bada radius ek chote angular velocity ke liye compensate kar sakta hai, ek substantial tangential speed deta hua.

Nahi — $a_c=r{\omega }^2$ ${\omega }^2$ par depend karta hai, jo kisi bhi taraf positive hai; centripetal acceleration hamesha inward point karta hai regardless of spin direction.

$a_c=v^2/r$ use karke, jaise $r\to \infty$ centripetal acceleration $\to 0$ — path seedha ho jaata hai aur uniform straight-line motion ki tarah behave karta hai.

Haan, jab $r\alpha =r{\omega }^2$, yaani $\alpha ={\omega }^2$; wahan total acceleration exactly $45^{\circ }$ par tangent aur inward radius ke beech point karta hai (master figure ka right-hand panel).