Exercises — Angular displacement θ, angular velocity ω, angular acceleration α

Unit kya bata raha hai. Radians angle ki duniya mein hote hain. "Per second" ka matlab hai "time ke saath change ki rate." "Per second squared" ka matlab hai "rate of change of a rate."

• $\text{rad}$ akela → ek angle → $\theta$.
• $\text{rad/s}$ → angle per time → $\omega$.
• ${\text{rad/s}}^2$ → (angle per time) per time → $\alpha$, angular acceleration.

Toh $4.5{\text{rad/s}}^2$ ek angular acceleration hai: iska matlab hai ki spin rate $\omega$ har second $4.5\text{rad/s}$ badhti hai.

Radians kyun. Ek full turn mein poori circumference $2\pi r$ jitna arc sweep hota hai, aur $\theta =s/r=2\pi r/r=2\pi$ rad. Toh ek revolution $=2\pi$ rad — idea ke liye koi calculator nahi chahiye, bas multiply karo. $\theta =3\times 2\pi =6\pi \approx 18.85\text{rad}.$

Kya share hota hai aur kya scale hota hai. Kisi bhi rigid body ke liye, har point same time mein same angle se ghumta hai — isliye humne angular language invent ki.

• Angle: dono cheetiyan $\frac{\pi }{2}$ rad sweep karti hain — barabar.
• Arc (distance): $s=r\theta$, toh cheenti B $s_B=0.3\times \frac{\pi }{2}\approx 0.471$ m chalti hai jabki cheenti A $s_A=0.1\times \frac{\pi }{2}\approx 0.157$ m chalti hai.

Cheenti B 3 guna zyada arc cover karti hai, sirf isliye kyunki woh 3 guna zyada door hai. (Neeche figure dekho — same wedge angle, longer outer arc.)

Yeh equation kyun. Hum jaante hain start ${\omega }_0=0$, end $\omega$, aur time $t$ — yeh exactly woh trio hai jo $\omega ={\omega }_0+\alpha t$ mein hai. $\alpha =\frac{\omega -{\omega }_0}{t}=\frac{12-0}{3}=4{\text{rad/s}}^2.$ Meaning check: $+4$ matlab $\omega$ har second $4\text{rad/s}$ badhta hai: $0\to 4\to 8\to 12$. ✓

$v=r\omega$ aur $a_t=r\alpha$ kyun. Yeh do "bridges" hain angular world se linear world mein, radius $r$ par baithe kisi point ke liye. $v=r\omega =0.25\times 12=3\text{m/s}.$ $a_t=r\alpha =0.25\times 4=1.0{\text{m/s}}^2.$ $a_t$ acceleration ka woh part hai jo path ke saath hai — yahi rim ki speed ko badhata hai.

Displacement equation kyun. Hum jaante hain ${\omega }_0=0$, $\alpha =4$, $t=3$ aur chahiye $\theta$ — yeh hai $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$. $\theta =0+\frac{1}{2}(4)(3^2)=\frac{1}{2}(4)(9)=18\text{rad}.$ Turns: $18/(2\pi )\approx 2.86$ revolutions.

Teesri equation kyun. Hum jaante hain ${\omega }_0$, final $\omega =0$, aur $\theta$, lekin $t$ nahi. Bina $t$ wali sirf ek kinematic equation hai: ${\omega }^2={\omega }_0^2+2\alpha \theta$. $0=40^2+2\alpha (100)\Rightarrow 0=1600+200\alpha .$ $\alpha =-\frac{1600}{200}=-8{\text{rad/s}}^2.$ Minus sign physics hai: $\alpha$ ka direction $\omega$ ke opposite hai, yaani yeh deceleration hai.

Do accelerations kyun. Rim ka ek point ek saath do kaam karta hai: uski speed badhti hai (tangential, path ke along) aur uski direction bend hoti hai (centripetal, centre ki taraf). Yeh dono perpendicular hain, toh hum Pythagoras se combine karte hain. $a_t=r\alpha =0.2\times 6=1.2{\text{m/s}}^2(\text{path ke along}).$ $a_c={\omega }^{2}r=10^2\times 0.2=20{\text{m/s}}^2(\text{centre ki taraf}).$ $a=\sqrt{a_t^2+a_c^2}=\sqrt{1.2^2+20^2}=\sqrt{1.44+400}=\sqrt{401.44}\approx 20.04{\text{m/s}}^2.$ Dhyan do $a_c\gg a_t$: zyada spin par turning hi speeding-up par dominant hoti hai. Figure mein do perpendicular arrows aur unka resultant dikhaya gaya hai.

(b) ke liye differentiate kyun. Average kehta hai "total angle ÷ total time"; instantaneous kehta hai "rate abhi iske moment", jo derivative hai $\omega =d\theta /dt$.

• (a) Average: $\theta (2)=2(8)=16$, $\theta (1)=2(1)=2$. ${\omega }_{avg}=\frac{16-2}{2-1}=14\text{rad/s}.$
• (b) Instantaneous: $\omega =\frac{d}{dt}(2t^3)=6t^2$, toh $t=2$ par: $\omega =6(4)=24\text{rad/s}.$

Alag kyun hain: motor accelerate kar raha hai ($\alpha =d\omega /dt=12t\ne 0$), toh end par rate ($24$) interval pe average rate ($14$) se zyada hai. Constant $\alpha$ ke liye dono sirf interval ke midpoint par equal hote.

Pehle angular mein convert kyun. Rolling without slipping road-distance $s_{road}$ ko wheel-angle se link karta hai: har turn mein car ek circumference aage badhti hai. Toh car ki forward speed $v$, wheel ke spin $\omega$ se $v=r\omega$ ke through, forward acceleration $a$ se $a=r\alpha$ ke through, aur road distance $s_{road}$ se $s_{road}=r\theta$ ke through connect hoti hai.

Step 1 — angular speeds: ${\omega }_0=\frac{v_0}{r}=\frac{5}{0.35}\approx 14.286\text{rad/s},\omega =\frac{20}{0.35}\approx 57.143\text{rad/s}.$ Step 2 — ghuma hua angle: wheel $150$ m roll karta hai, toh $\theta =s_{road}/r=150/0.35\approx 428.57$ rad. Step 3 — α (time-free, time nahi pata): ${\omega }^2={\omega }_0^2+2\alpha \theta \Rightarrow 57.143^2=14.286^2+2\alpha (428.57).$ $3265.3=204.1+857.14\alpha \Rightarrow \alpha =\frac{3061.2}{857.14}\approx 3.57{\text{rad/s}}^2.$ Linear kinematics se cross-check: $a=(v^2-v_0^2)/(2s)=(400-25)/300=1.25{\text{m/s}}^2$, aur $a/r=1.25/0.35\approx 3.57{\text{rad/s}}^2$. ✓ Same answer, do alag languages. Revolutions: $428.57/(2\pi )\approx 68.2$ rev.

Phases mein kyun todna. Constant-$\alpha$ kinematics ki equations sirf us phase ke andar apply hoti hain jahan $\alpha$ constant ho. Do alag $\alpha$ values → do alag calculations, phir add karo.

Phase 1 (accelerating, $0$–$5$ s): ${\omega }_1={\omega }_0+{\alpha }_{1}t=0+8(5)=40\text{rad/s}.$ ${\theta }_1=\frac{1}{2}{\alpha }_1{t}^2=\frac{1}{2}(8)(25)=100\text{rad}.$ Phase 2 (constant speed, $5$–$10$ s): $\alpha =0$, toh ${\theta }_2={\omega }_{1}t=40\times 5=200$ rad. Total: $\theta ={\theta }_1+{\theta }_2=100+200=300\text{rad}.$

Sign kyun dekhna. $\omega$ aur $\alpha$ pehle opposite directions mein hain (wheel slow ho raha hai), lekin $\alpha$ kabhi nahi rukta — toh rukne ke baad $\omega$ negative ho jaata hai (reverse spin). Equations yeh sab automatically handle karte hain agar tum signs rakho.

(a) Momentary stop: $\omega =0$ set karo $\omega ={\omega }_0+\alpha t$ mein: $0=6+(-3)t\Rightarrow t=2\text{s}.$ (b) $t=4$ s par displacement: $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2=6(4)+\frac{1}{2}(-3)(16)=24-24=0\text{rad}.$ Net angle zero hai — wheel aage gaya, ruka, phir wapas apne start par aa gaya. (c) $t=4$ s par: $\omega =6+(-3)(4)=-6\text{rad/s}$: yeh ab backward spin kar raha hai jitni speed se forward shuru kiya tha. (θ-vs-t parabola dekho — yeh utha, $t=2$ par peak kiya, aur $t=4$ par $0$ par wapas aa gaya.)

Distance ≠ displacement kyun. Displacement net hoti hai (sign ke saath); total angle turned forward aur backward dono parts ko positive amounts mein add karta hai, kyunki wheel physically dono taraf chala.

Forward stretch (0 → 2 s): $t=2$ par peak angle: ${\theta }_{peak}=6(2)+\frac{1}{2}(-3)(4)=12-6=6\text{rad}.$ Backward stretch (2 → 4 s): $+6$ rad se $0$ rad tak wapas → aur $6$ rad ka turning (reverse mein). Total angle turned: $|{\theta }_{forward}|+|{\theta }_{back}|=6+6=12\text{rad}.$ Toh wheel ne $12$ rad ka arc sweep kiya, bhawahe uska net displacement $0$ hai.

Limit test kyun. Ek acchi general formula ko simpler special case mein collapse karna chahiye jab extra ingredient (yahan $\alpha$) switch off ho — yeh poore framework ka sanity check hai, aur exactly batata hai ki general one ke andar kaunsi familiar motion baithi hai.

Step 1 — displacement equation mein $\alpha$ term hatao: $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2\stackrel{\alpha \to 0}{\to }\theta ={\omega }_{0}t.$ $\frac{1}{2}\alpha t^2$ term disappear ho jaata hai kyunki yeh $\alpha =0$ se multiply hai. Jo bachta hai, $\theta ={\omega }_{0}t$, woh angle hai jo time ke saath linearly badhta hai — constant spin rate ki pehchaan.

Step 2 — check karo ki baaki do equations bhi agree karti hain: $\omega ={\omega }_0+\alpha t\stackrel{\alpha \to 0}{\to }\omega ={\omega }_0(\text{constant spin}),$ ${\omega }^2={\omega }_0^2+2\alpha \theta \stackrel{\alpha \to 0}{\to }{\omega }^2={\omega }_0^2(\text{speed kabhi nahi badlti}).$ Teeno consistently collapse karte hain — framework self-consistent hai.

Step 3 — physical picture (Uniform Circular Motion se connect karo). $\alpha =0$ ke saath tangential acceleration $a_t=r\alpha \to 0$, toh rim ki speed kabhi nahi badlti. Jo bachta hai woh centripetal acceleration $a_c={\omega }^{2}r$ hai, jo sirf direction bend karta hai. Constant speed se circle trace karta hua ek body jiska velocity continuously turn karta ho, woh exactly uniform circular motion hai. Toh uniform circular motion koi alag theory nahi — yeh angular kinematics ka $\alpha \to 0$ corner hai, usi tarah jaise constant-velocity motion linear kinematics ka $a\to 0$ corner hai.

Kya ab tum bina dekhe ye kar sakte ho:

• bata sako ki koi unit θ/ω/α mein se kiska hai? ::: Haan — "per seconds" gino: koi nahi→θ, ek→ω, do→α.
• jab $t$ missing ho toh time-free equation chuno? ::: Haan — ${\omega }^2={\omega }_0^2+2\alpha \theta$ use karo.
• jahan α badlta hai wahan two-phase spin split karo? ::: Haan — har phase solve karo, ω aage le jao, θ add karo.
• signs rakho taaki reversing wheel mein net θ=0 mile lekin path 12 rad? ::: Haan — ek positive direction fix karo aur kabhi minus mat chodo.