Exercises — Circular motion — centripetal acceleration derivation

Hume $v$ aur $r$ diye gaye hain, isliye in dono se bani form use karo: $a_c=\frac{v^2}{r}=\frac{6^2}{2}=\frac{36}{2}=18{\text{m/s}}^2.$ Humne kya kiya: speed ko square kiya, radius se divide kiya. Yeh form kyun: $v$ aur $r$ exactly wahi hain jo hume diye gaye hain — $\omega$ ke zariye detour lene ki zaroorat nahi. Answer: $a_c=18{\text{m/s}}^2$, centre ki taraf directed.

$\omega$ aur $r$ diye gaye hain, seedha $a_c={\omega }^{2}r$ use karo: $a_c={\omega }^{2}r=(4{)}^2(0.25)=16\times 0.25=4{\text{m/s}}^2.$ Yeh form kyun: angular speed diya gaya hai, isliye ${\omega }^{2}r$ pehle $v$ calculate karne se bachata hai. Answer: $a_c=4{\text{m/s}}^2$ centre ki taraf.

Yeh circle ke centre ki taraf point karta hai, hamesha velocity ke perpendicular. Kyunki yeh $\vec{v}$ ke perpendicular hai, yeh sirf velocity ki direction badalta hai aur speed (magnitude) kabhi nahi badalta. Neeche figure dekho — red arrow hamesha black velocity se $90^{\circ }$ par hai.

Hume period diya gaya hai, $v$ ya $\omega$ nahi. Pehle convert karo: ek poora loop $2\pi$ radians ka hota hai, time $T$ mein, isliye $\omega =\frac{2\pi }{T}=\frac{2\pi }{0.5}=4\pi \text{rad/s}\approx 12.566\text{rad/s}.$ Yeh step kyun: $\omega$ timing se acceleration ka bridge hai. Ab: $a_c={\omega }^{2}r=(4\pi {)}^2(0.8)=16{\pi }^2\times 0.8=12.8{\pi }^2\approx 126.3{\text{m/s}}^2.$ Answer: $a_c\approx 126.3{\text{m/s}}^2$. (Equivalent roop mein $a_c=\frac{4{\pi }^{2}r}{T^2}$.)

Friction centripetal acceleration supply karta hai, isliye car tab tak safe hai jab tak required $a_c=v^2/r$ friction ki di hui limit se zyada na ho. Limit par: $\frac{v^2}{r}=6\Rightarrow v^2=6r=6\times 45=270\Rightarrow v=\sqrt{270}\approx 16.43\text{m/s}.$ Humne kya kiya: $a_c=v^2/r$ ko $v$ ke liye rearrange kiya. Kyun: hume acceleration aur radius pata hai, speed chahiye. Answer: $v_{\max }\approx 16.43\text{m/s}$ (lagbhag $59\text{km/h}$).

"Revolutions per minute" ko radians per second banana padega. Har revolution $2\pi$ rad ka hota hai, aur ek minute $60$ s ka: $\omega =600\times \frac{2\pi }{60}=600\times \frac{2\pi }{60}=20\pi \text{rad/s}\approx 62.83\text{rad/s}.$ Yeh step kyun: formula sirf rad/s mein $\omega$ accept karta hai, rpm mein kabhi nahi. $a_c={\omega }^{2}r=(20\pi {)}^2(0.3)=400{\pi }^2\times 0.3=120{\pi }^2\approx 1184.4{\text{m/s}}^2.$ Answer: $a_c\approx 1184.4{\text{m/s}}^2$ — $120g$ se zyada! Isliye tez ghoomne wale parts fly apart hote hain.

Centripetal force $F_c=\frac{m{v}^2}{r}$ hai. $m$ aur $r$ fixed hain, isliye $F_c\propto v^2$. Speed teeni karne par: $\frac{F_{\text{new}}}{F_{\text{old}}}=\frac{(3v{)}^2}{v^2}=9.$ Humne kya kiya: constants ko fixed rakha aur sirf varying part $v^2$ dekha. Kyun: ratio method har constant ko cancel kar deta hai, isliye actual numbers ki zaroorat nahi. Answer: force $9\times$ bada ho jaata hai.

Same turntable ⇒ same $\omega$. $\omega$ form use karo kyunki $\omega$ shared hai: $a_c={\omega }^{2}r\Rightarrow \frac{a_B}{a_A}=\frac{{\omega }^2(2r)}{{\omega }^{2}r}=2.$ $\omega$ form kyun: $\omega$ common hone par $a_c$ directly $r$ ke proportional hai, aur comparison instant ho jaata hai. Agar $v^2/r$ use karte toh pehle note karna padta ki $v=\omega r$ har ek ke liye alag hai — zyada kaam, same answer. Answer: particle B (outer) ka centripetal acceleration double hai.

Ab $v$ shared hai, isliye $v$ form use karo: $a_c=\frac{v^2}{r}\Rightarrow \frac{a_B}{a_A}=\frac{v^2/(2r)}{v^2/r}=\frac{1}{2}.$ Switch kyun: jo quantity constant rakhi jaati hai wahi decide karti hai kaunsa form sabse clean comparison deta hai. $v$ fixed hone par $a_c\propto 1/r$, isliye bada radius chota acceleration deta hai — Q3.2 ka ulta. Answer: particle A (inner) ka acceleration B se double hai. Same objects, ulta conclusion — kyunki kya fixed hai alag hai.

(a) $v$ aur $r$ diye gaye hain: $a_c=\frac{v^2}{r}=\frac{3^2}{0.6}=\frac{9}{0.6}=15{\text{m/s}}^2.$ (b) Newton's Second Law se, net inward force $F_c=m{a}_c$ hai. Yahan string tension hi woh inward force hai (yeh extra force nahi hai — yeh hi centripetal force hai): $T=m{a}_c=0.2\times 15=3\text{N}.$ Kyun: hum real force identify karte hain jo inward pull supply karta hai, phir use $m{v}^2/r$ ke barabar set karte hain. Answer: $a_c=15{\text{m/s}}^2$, tension $T=3\text{N}$.

Hill ke top par, gravity ($mg$, downward, centre ki taraf) aur normal force ($N$, upward, centre se door) dono centre ki vertical line ke saath act karte hain. Net downward = inward force centripetal requirement supply karta hai: $mg-N=\frac{m{v}^2}{r}.$ Car "just loses contact" jab road push nahi kar sakti — woh $N=0$ hai: $mg=\frac{m{v}^2}{r}\Rightarrow v^2=gr=9.8\times 40=392\Rightarrow v=\sqrt{392}\approx 19.8\text{m/s}.$ Humne kya kiya: radial (vertical) direction mein Newton's 2nd law likhi, phir lift-off condition $N=0$ impose ki. $N=0$ kyun: contact exactly tab khat hota hai jab surface push karna band kare — mass $m$ cancel ho jaata hai, isliye car kitni bhi bhaari ho, fark nahi padta. Answer: $v\approx 19.8\text{m/s}$ (lagbhag $71\text{km/h}$).

Do real forces act karte hain: tension $T$ string ke along, aur gravity $mg$ neeche. Tension ke components lo (figure dekho):

• Vertical: $T\cos \theta =mg$ (bob upar ya neeche nahi jaata).
• Horizontal: $T\sin \theta =\frac{m{v}^2}{R}$ (yeh horizontal part hi centripetal force hai).

Circle ka radius $R=L\sin \theta$ hai (axis se bob tak horizontal doori). $T$ cancel karne ke liye horizontal equation ko vertical se divide karo: $\tan \theta =\frac{v^2}{Rg}\Rightarrow v^2=Rg\tan \theta =L\sin \theta g\tan \theta .$ Divide kyun kiya: $T$ pata nahi aur chahiye bhi nahi — divide karne se eliminate ho jaata hai, yeh ek classic trick hai. $L=1$, $\theta =30^{\circ }$ ($\sin 30^{\circ }=0.5$, $\tan 30^{\circ }=0.5774$), $g=9.8$ plug in karo: $v^2=1\times 0.5\times 9.8\times 0.5774\approx 2.829\Rightarrow v\approx 1.682\text{m/s}.$ Answer: $v\approx 1.68\text{m/s}$.

(a) Yahan gravity hi centripetal force hai — Gravitation — orbital motion dekho. Isliye us height par gravitational acceleration centripetal acceleration ke barabar hai: $g_{\text{orbit}}=\frac{v^2}{r}\Rightarrow v^2=g_{\text{orbit}}r=8.0\times 7.0\times 10^6=5.6\times 10^7.$ $v=\sqrt{5.6\times 10^7}\approx 7483\text{m/s}\approx 7.48\text{km/s}.$ Dono kyun equate kiye: koi aur satellite ko inward nahi kheench raha, isliye gravity akele puri $v^2/r$ supply karni hai. Mass $m$ cancel ho jaata hai — orbital speed satellite ki mass par depend nahi karti.

(b) Period circumference divided by speed hai (Angular velocity and period use karte hue): $T=\frac{2\pi r}{v}=\frac{2\pi (7.0\times 10^6)}{7483}\approx \frac{4.398\times 10^7}{7483}\approx 5877\text{s}.$ Minutes mein convert karo: $5877/60\approx 97.9\text{min}$. Answer: $v\approx 7.48\text{km/s}$, period $T\approx 97.9\text{minutes}$ — ek realistic low-Earth-orbit value (lagbhag 90–100 min).

Loop ke bottom par, centre seedha bead ke upar hai. Isliye "centre ki taraf" matlab upar hai. Normal force $N$ upar (inward) push karta hai; gravity $mg$ neeche (outward) kheechti hai. Radial (vertical) direction mein Newton's 2nd law, inward ko positive lete hue: $N-mg=\frac{m{v}^2}{r}.$ $N$ ke liye solve karo: $N=mg+\frac{m{v}^2}{r}=0.1\times 9.8+\frac{0.1\times 4^2}{0.5}=0.98+\frac{1.6}{0.5}=0.98+3.2=4.18\text{N}.$ Humne kya kiya: bottom par "inward" kis taraf point karta hai identify kiya, phir us line ke saath real forces ka sum kiya. $N>mg$ kyun: bottom par wire ko bead ko upar rakhna bhi hai aur path ko upar ki taraf bend karna bhi hai, isliye yeh akele weight se zyada push karta hai. Answer: $N\approx 4.18\text{N}$, upar ki taraf directed (centre ki taraf).

• Kya tum diye gaye quantities ke basis par $v^2/r$ vs ${\omega }^{2}r$ choose kar sakte ho? (L1–L2)
• Kya tum rpm aur period ko rad/s mein convert kar sakte ho? (L2)
• Kya tumhe pata hai ki $a_c\propto r$ (fixed $\omega$) lekin $a_c\propto 1/r$ (fixed $v$)? (L3)
• Kya tum ek real force ko $m{v}^2/r$ ke barabar set karte ho, naya invent karne ki jagah? (L4)
• Kya tum vertical loop par har position ke liye "inward" ko re-point karte ho? (L5)