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

Hume circle ke saath speed chahiye, toh tool hai $v=r\omega$ (power one). $v=r\omega =0.15\times 40=6\text{m/s}$ $\omega$ pehle se rad/s mein hai, isliye directly multiply karo — koi conversion nahi chahiye.

Velocity ki direction change ho rahi hai, isliye hume $a_c=r{\omega }^2$ chahiye (power two). $a_c=r{\omega }^2=0.15\times 40^2=0.15\times 1600=240{\text{m/s}}^2$

Tangential acceleration speed ki magnitude ko change karta hai, aur yeh $\alpha$ use karta hai: tool hai $a_t=r\alpha$. $a_t=r\alpha =0.3\times 8=2.4{\text{m/s}}^2$ Yahan $\alpha >0$ hai, isliye $a_t>0$: point speed up ho raha hai, aur $a_t$ $\vec{v}$ ke saath aage point karta hai.

Step 1 — KYA & KYUN: rev/min ko rad/s mein convert karo. Formula $v=r\omega$ ko $\omega$ radians per second mein chahiye, kyunki yeh arc-length relation $s=r\theta$ se aata hai (jahan $s$ upar define ki gayi arc length hai), jo khud tabhi hold karti hai jab $\theta$ radians mein ho (dekho Radian measure and arc length). Ek revolution $=2\pi$ radians, aur ek minute $=60$ s. $\omega =33.333\times \frac{2\pi }{60}=3.491\text{rad/s}$ Step 2 — bridge apply karo. $v=r\omega =0.12\times 3.491=0.419\text{m/s}$

Step 1 — $\alpha$ nikalo. Angular acceleration $\omega$ ke change ki rate hai: $\alpha =\frac{\Delta \omega }{\Delta t}=\frac{60-0}{12}=5{\text{rad/s}}^2$ Step 2 — $a_t=r\alpha$ apply karo. $a_t=0.4\times 5=2{\text{m/s}}^2$

Kyunki body rigid hai, dono ants ek hi $\omega$ share karte hain. Toh $v=r\omega$ mein sirf $r$ alag hai, aur speed radius ke proportional hai: $\frac{v_B}{v_A}=\frac{r_B\omega }{r_A\omega }=\frac{r_B}{r_A}=\frac{0.20}{0.05}=4$ Ant B 4 times zyada fast hai.

(a) Centripetal (centre ki taraf): $a_c=r{\omega }^2=0.5\times 20^2=200{\text{m/s}}^2$ (b) Tangential (circle ke saath): $a_t=r\alpha =0.5\times 5=2.5{\text{m/s}}^2$ (c) Yeh dono arrows perpendicular hain ($a_c$ radial, $a_t$ tangential), isliye Pythagoras se combine karo: $a=\sqrt{a_c^2+a_t^2}=\sqrt{200^2+2.5^2}\approx 200.016{\text{m/s}}^2$ Notice karo $a_c$ bilkul dominate karta hai — high spin par centre-ki-taraf pull, speed-up ko bilkul dabaata hai. Neeche figure dekho: total (green) arrow radial direction se thoda sa bhi nahi tilt hota.

Pehle geometry set up karo. Total acceleration upar figure mein green arrow hai: yeh ek right triangle ka diagonal hai jiske do legs hain radial leg $a_c$ (inward point karta hai) aur tangential leg $a_t$ (motion ke saath point karta hai). Total arrow aur radial (inward) leg ke beech ke angle ko $\varphi$ bolte hain — wahi tilt jo humse pucha gaya hai. Is triangle par, $a_t$ woh side hai jo $\varphi$ ke opposite hai aur $a_c$ woh side hai jo $\varphi$ ke adjacent hai, isliye "opposite over adjacent" $\varphi$ ka tangent deta hai: $\tan \varphi =\frac{a_t}{a_c}=\frac{2.5}{200}=0.0125$ Tangent ko undo karne ke liye (yani "kis angle ka tan yeh hai?" dhundhne ke liye) hum arctan use karte hain: $\varphi =\arctan (0.0125)\approx 0.716°$ Barely radial se off — jaise expect kiya tha jab $a_c\gg a_t$.

"Steady speed" matlab speed change nahi ho rahi, isliye $a_t=0$. Lekin direction change ho rahi hai, isliye centripetal acceleration hai. Yahan hume seedha $v$ diya gaya hai, isliye handy form $a_c=v^2/r$ hai: $a_c=\frac{v^2}{r}=\frac{15^2}{50}=\frac{225}{50}=4.5{\text{m/s}}^2$ Total acceleration $=4.5{\text{m/s}}^2$, centre ki taraf point karta hua. (Is acceleration ko supply karne wali force ke liye Centripetal force and circular motion dekho.)

(a) Bridge $v=r\omega$ ko invert karke $\omega =v/r$ nikalo: $\omega =\frac{v}{r}=\frac{10}{0.25}=40\text{rad/s}$ (b) KYUN yeh formula. Hume $\omega$ mein change diya gaya hai ek angle turned $\theta$ ke upar (time ke upar nahi), isliye hume ek aise relation ki zaroorat hai jo $\omega$, $\alpha$ aur $\theta$ ko time eliminate karke link kare. Constant $\alpha$ ke liye woh relation hai ${\omega }^2={\omega }_0^2+2\alpha \theta$. Yeh kahan se aata hai? Yeh linear kinematics formula $v^2=v_0^2+2as$ ka exact angular twin hai: distance $s$ ko angle $\theta$ se replace karo, speed $v$ ko $\omega$ se, aur linear acceleration $a$ ko $\alpha$ se. (Tum ise $\omega ={\omega }_0+\alpha t$ lekar, $t$ solve karke, aur $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$ mein substitute karke derive kar sakte ho.) ${\omega }_0=0$ ke saath (rest se shuru): ${\omega }^2=2\alpha \theta \Rightarrow \alpha =\frac{{\omega }^2}{2\theta }=\frac{40^2}{2\times 8}=\frac{1600}{16}=100{\text{rad/s}}^2$ (c) Linear tangential accel tak bridge: $a_t=r\alpha =0.25\times 100=25{\text{m/s}}^2$

(a) $a_c=r{\omega }^2$ se, $\omega$ solve karo: $\omega =\sqrt{\frac{a_c}{r}}=\sqrt{\frac{90}{0.10}}=\sqrt{900}=30\text{rad/s}$ (b) $a_t=r\alpha$ se: $\alpha =\frac{a_t}{r}=\frac{30}{0.10}=300{\text{rad/s}}^2$ (c) Speed ke liye bridge: $v=r\omega =0.10\times 30=3\text{m/s}$

(a) Speed uniformly $0$ se $12\text{m/s}$ tak $3\text{s}$ mein grow karti hai, isliye $a_t=\frac{\Delta v}{\Delta t}=\frac{12}{3}=4{\text{m/s}}^2$ (b) $a_t=r\alpha$ se: $\alpha =\frac{a_t}{r}=\frac{4}{0.8}=5{\text{rad/s}}^2$ (c) $t=3\text{s}$ par, $v=12$, isliye $a_c=v^2/r$ use karo: $a_c=\frac{12^2}{0.8}=\frac{144}{0.8}=180{\text{m/s}}^2$ (d) Perpendicular combine: $a=\sqrt{a_c^2+a_t^2}=\sqrt{180^2+4^2}=\sqrt{32416}\approx 180.044{\text{m/s}}^2$

(a) Rolling without slipping centre ki speed ko spin se tie karta hai: $v_{\text{cm}}=R\omega$, isliye $\omega =\frac{v_{\text{cm}}}{R}=\frac{7}{0.35}=20\text{rad/s}$ (b) Top point mein rolling velocity ($v_{\text{cm}}$ forward) plus centre ke baare mein rim ki rotational velocity (top par $R\omega$ forward) hoti hai. Yeh add hote hain: $v_{\text{top}}=v_{\text{cm}}+R\omega =7+7=14\text{m/s}$ (c) Contact point par rotational velocity ($R\omega$ backward) forward $v_{\text{cm}}$ ko cancel karta hai: $v_{\text{bottom}}=v_{\text{cm}}-R\omega =7-7=0\text{m/s}$ Bottom point instantaneously rest par hai — yahi "without slipping" ka matlab hai. (Zyada Rolling motion mein.)

Total-acceleration arrow ka radial direction se tilt $\tan \varphi =a_t/a_c=r\alpha /(r{\omega }^2)=\alpha /{\omega }^2$ hai. Jaise $\omega \to \infty$, $\alpha /{\omega }^2\to 0$, isliye $\varphi \to 0$: arrow centre ki taraf snap hota hai. Yeh confirm karta hai ki fast spinners almost purely inward pull feel karte hain. Woh equal hain jab $a_c=a_t$: $r{\omega }^2=r\alpha \Rightarrow {\omega }^2=\alpha \Rightarrow \omega =\sqrt{\alpha }=\sqrt{10}\approx 3.162\text{rad/s}$ (Neat: equal-point sirf $\alpha$ par depend karta hai, $r$ par nahi, kyunki $r$ cancel ho jaata hai.)

Har bridge formula mein ek factor $r$ hota hai, aur yahan $r=0$ hai: $v=r\omega =0,a_t=r\alpha =0,a_c=r{\omega }^2=0$ Pivot point bilkul move nahi karta — yeh axis par baitha hai. Angular quantities $\omega ,\alpha$ bade hain, lekin linear quantities sab vanish ho jaate hain kyunki trace karne ke liye koi circle nahi hai. Yeh boundary case dikhata hai ki axis kyun special hai: poori disk $\omega$ share karti hai, phir bhi axis point ki speed zero hai.

(a) $a_c=r{\omega }^2$ se, $\omega$ solve karo: $\omega =\sqrt{\frac{a_c}{r}}=\sqrt{\frac{500}{0.20}}=\sqrt{2500}=50\text{rad/s}$ (b) Rim speed: $v=r\omega =0.20\times 50=10\text{m/s}$ (c) $2.5\text{s}$ mein rest se $50\text{rad/s}$ tak pahunchne ke liye, $\alpha =\Delta \omega /\Delta t$ se: $\alpha =\frac{50-0}{2.5}=20{\text{rad/s}}^2,a_t=r\alpha =0.20\times 20=4{\text{m/s}}^2$

$v=r\omega$ (ek baar multiply karo) ::: spin se linear speed $a_t=r\alpha$ (tangential, $\alpha$ use karta hai; sign $\alpha$ follow karta hai) ::: speed kaise change hoti hai $a_c=r{\omega }^2=v^2/r$ (radial, hamesha inward, kabhi reverse nahi) ::: direction kaise change hoti hai $a_t$ aur $a_c$ combine karo ::: $a=\sqrt{a_t^2+a_c^2}$ kyunki yeh perpendicular hain rolling without slipping ::: $v_{\text{cm}}=R\omega$, bottom point rest par