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

1.5.3 · Physics › Rotational Mechanics

WHY karte hain hum ek bridge ki zaroorat?

Setup karna: arc length hi sab kuch ki neev hai

WHY radians matter karte hain: $s=r\theta$ ka relation tab hi kaam karta hai jab $\theta$ radians mein ho. Agar degrees use karo, toh $\pi /180$ ke ugly factors aate. Radians is tarah banaye gaye hain ki arc length = radius × angle, cleanly.

$v=r\omega$ derive karna scratch se

Derivation (HOW): Arc length $s=r\theta$ se shuru karo. Rigid body ke liye, $r$ time mein constant rehta hai (point axis ke kareeb ya door drift nahi karta). Time ke saath differentiate karo: $\frac{ds}{dt}=r\frac{d\theta }{dt}$ Ab $\frac{ds}{dt}=v$ (speed = path ke saath distance ke change ki rate) aur $\frac{d\theta }{dt}=\omega$ (angular velocity). Isliye: $\boxed{v=r\omega }$

• Yeh step kyun? Hum differentiate kar sake kyunki $r$ constant hai — derivative se bahar nikaalo. Agar $r$ change hota, toh ek extra term aata.
• WHAT iska matlab hai: $v$ tangential speed hai — yeh circle ke saath, radius ke perpendicular point karta hai.

$a_t=r\alpha$ derive karna (tangential acceleration)

Derivation (HOW): $v=r\omega$ lo aur dobara differentiate karo (rigid body $\Rightarrow r$ constant): $\frac{dv}{dt}=r\frac{d\omega }{dt}\Rightarrow a_t=r\alpha$ jahan $\alpha =d\omega /dt$ angular acceleration hai.

• Yeh step kyun? $a_t$ speed ke change ki rate hai. Yeh tabhi exist karta hai jab rotation speed up ya slow down ho raha ho. Agar $\omega$ constant hai, toh $\alpha =0$ aur $a_t=0$.
• WHAT yeh karta hai: $a_t$ velocity ki magnitude change karta hai, uski direction nahi.

$a_c=r{\omega }^2$ derive karna (centripetal acceleration)

Derivation (HOW) — first principles: Constant speed par bhi, circle par ek point accelerate karta hai kyunki uski velocity ki direction continuously change hoti rehti hai. Position vector likho: $\vec{r}=r(\cos \theta ,\sin \theta ),\theta =\omega t$ Velocity (differentiate karo, $\omega$ constant): $\vec{v}=r\omega (-\sin \theta ,\cos \theta )$ Acceleration (dobara differentiate karo): $\vec{a}=r{\omega }^2(-\cos \theta ,-\sin \theta )=-{\omega }^2\vec{r}$ Minus sign ka matlab hai $\vec{a}$ center ki taraf point karta hai ($\vec{r}$ ke opposite). Iska magnitude: $\boxed{a_c=r{\omega }^2}$ $v=r\omega \Rightarrow \omega =v/r$ use karo, substitute karo: $a_c=r(v/r{)}^2=v^2/r$. ✓

• Yeh step kyun? Humne vector differentiate kiya, sirf speed nahi — isi tarah humne direction change capture kiya, jo centripetal acceleration ka poora source hai.

Dual Coding: geometry

Diagram mein ek rotating disk par ek point dikhaya gaya hai jisme $\vec{v}$ tangent hai, ${\vec{a}}_t$ tangent hai ($v$ ke saath ya uske against), aur ${\vec{a}}_c$ center ki taraf point karta hai. Bada $r$ → same $\omega$ ke liye lamba $\vec{v}$ arrow.

Worked Examples

Forecast-then-Verify

Common Mistakes (Steel-manned)

Mnemonic

Feynman: ek 12-saal ke bacche ko explain karo

Flashcards

Connections

• Angular velocity and angular acceleration — source quantities $\omega ,\alpha$.
• Radian measure and arc length — kyun $s=r\theta$ is sab ki neev hai.
• Centripetal force and circular motion — $F_c=mr{\omega }^2$ mein $a_c$ use hota hai.
• Uniform vs non-uniform circular motion — $a_t=0$ vs $a_t\ne 0$.
• Moment of inertia — kinetic energy $\frac{1}{2}m{v}^2$ ko $v=r\omega$ se sum karne par $\frac{1}{2}I{\omega }^2$ milta hai.
• Rolling motion — contact point par $v=r\omega$ combine karta hai.

Concept Map