Exercises — Centripetal force — what provides it in various situations

Answer: string mein tension. String sirf apni length ke saath, aapke haath ki taraf (centre ki taraf) pull kar sakti hai. Woh inward pull exactly wahi kaam karta hai jo centripetal requirement maangti hai. Kisi aur cheez ka inward horizontal component nahi hota (gravity seedha neeche point karti hai, air drag negligible hota hai). Isliye $T=\frac{m{v}^2}{r}$.

Answer: gravity — Earth ki gravitational pull Moon par. Yeh Moon se Earth ki taraf (orbit ke centre ki taraf) point karti hai. Hum ise requirement ke barabar rakhte hain: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$.

Answer: tyres aur road ke beech static friction, jo horizontally inward (roundabout ke centre ki taraf) point karti hai. Yeh static friction hai, sliding friction nahi, kyunki tyre ka contact patch sideways skid nahi kar raha — woh grip karta hai. Woh sideways grip hi ek maatra available horizontal force hai, isliye yeh $\frac{m{v}^2}{r}$ supply karta hai.

Sirf tension inward kaam karta hai, isliye $T=\frac{m{v}^2}{r}$. $T=\frac{0.20\times 3.0^2}{0.50}=\frac{0.20\times 9.0}{0.50}=\frac{1.8}{0.50}=3.6\text{N}$ Tension $=3.6\text{N}$, ball se haath ki taraf directed.

Friction zyada se zyada ${\mu }_{s}mg$ supply kar sakta hai. Ise requirement ke barabar karo: ${\mu }_{s}mg=\frac{m{v}_{\max }^2}{r}\Rightarrow v_{\max }=\sqrt{{\mu }_{s}gr}$ Mass $m$ cancel ho jaata hai — heavy aur light dono cars ki same limit hoti hai. $v_{\max }=\sqrt{0.80\times 9.8\times 50}=\sqrt{392}\approx 19.8\text{m/s}$ $v_{\max }\approx 19.8\text{m/s}$ (approximately $71\text{km/h}$).

Magnetic force $qvB$ hamesha velocity ke perpendicular hoti hai — yeh electron ko kabhi speed up nahi karti, sirf bend karti hai, jo exactly wahi hai jo centripetal force karta hai. $qvB=\frac{m{v}^2}{r}$ set karo aur solve karo: $r=\frac{mv}{qB}=\frac{9.11\times 10^{-31}\times 2.0\times 10^6}{1.60\times 10^{-19}\times 0.010}$ $r=\frac{1.822\times 10^{-24}}{1.60\times 10^{-21}}\approx 1.14\times 10^{-3}\text{m}\approx 1.1\text{mm}$

Figure dekho: road surface tilt karti hai, isliye normal force $N$ (surface ke perpendicular) bhi tilt karti hai. Yeh ek vertical slice $N\cos \theta$ aur ek horizontal slice $N\sin \theta$ (centre ki taraf pointing) mein split ho jaati hai.

• Vertical (koi up/down acceleration nahi): $N\cos \theta =mg$.
• Horizontal (yeh centripetal requirement hai): $N\sin \theta =\frac{m{v}^2}{r}$.

Doosre ko pehle se divide karo — $N$ aur $m$ dono cancel ho jaate hain, ek clean relation milta hai: $\tan \theta =\frac{v^2}{rg}\Rightarrow v=\sqrt{rg\tan \theta }$ $v=\sqrt{80\times 9.8\times \tan 20^{\circ }}=\sqrt{80\times 9.8\times 0.3640}\approx \sqrt{285.3}\approx 16.9\text{m/s}$ $v\approx 16.9\text{m/s}$.

String tension $T$ tilt karti hai, bilkul upar normal force ki tarah:

• Vertical: $T\cos \theta =mg$, isliye $T=\frac{mg}{\cos \theta }=\frac{0.30\times 9.8}{\cos 30^{\circ }}=\frac{2.94}{0.8660}\approx 3.39\text{N}$.
• Horizontal: $T\sin \theta =\frac{m{v}^2}{r}$, aur circle ka radius hai $r=L\sin \theta =1.2\times 0.5=0.60\text{m}$.

$v$ ke liye solve karo: $v=\sqrt{\frac{T\sin \theta \cdot r}{m}}=\sqrt{\frac{3.39\times 0.5\times 0.60}{0.30}}=\sqrt{3.39}\approx 1.84\text{m/s}$ $T\approx 3.39\text{N}$, $v\approx 1.84\text{m/s}$.

(a) Top. Figure ke top ko dekho: gravity ($mg$, neeche) aur tension ($T$, bhi neeche centre ki taraf) dono inward point karte hain. Isliye $T+mg=\frac{m{v}_{\text{top}}^2}{r}.$ String pull kar sakti hai lekin push nahi, isliye woh kam se kam $T=0$ kar sakti hai. Tab gravity akele poori requirement supply karni chahiye: $mg=\frac{m{v}_{\min }^2}{r}\Rightarrow v_{\min }=\sqrt{gr}=\sqrt{9.8\times 1.0}\approx 3.13\text{m/s}.$

(b) Bottom. Energy conservation use karo. Bottom top se $2r$ neeche hai, isliye ball $mg(2r)$ ke barabar kinetic energy gain karti hai: $\frac{1}{2}m{v}_{\text{bot}}^2=\frac{1}{2}m{v}_{\text{top}}^2+mg(2r)$ $v_{\text{bot}}^2=v_{\text{top}}^2+4gr=gr+4gr=5gr=5\times 9.8\times 1.0=49$ $v_{\text{bot}}=7.0\text{m/s}.$

(c) Bottom par Tension. Bottom par, tension upar (inward) point karta hai lekin gravity neeche (outward) point karti hai. Net inward hai $T-mg$: $T-mg=\frac{m{v}_{\text{bot}}^2}{r}\Rightarrow T=mg+\frac{m{v}_{\text{bot}}^2}{r}$ $T=0.50\times 9.8+\frac{0.50\times 49}{1.0}=4.9+24.5=29.4\text{N}.$ $v_{\min }\approx 3.13$ m/s, $v_{\text{bot}}=7.0$ m/s, $T_{\text{bot}}=29.4$ N.

Gravity centripetal force provide karti hai: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$. $m$ aur ek $r$ cancel karo: $v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{3.986\times 10^{14}}{7.0\times 10^6}}=\sqrt{5.694\times 10^7}\approx 7546\text{m/s}.$ Period circumference divided by speed hai: $T=\frac{2\pi r}{v}=\frac{2\pi \times 7.0\times 10^6}{7546}\approx 5828\text{s}\approx 97\text{min}.$

Maximum speed par car outward/up the slope slide karne ki koshish karti hai, isliye friction $f={\mu }_{s}N$ down the slope point karta hai (impending slide ko oppose karte hue). Figure ki tarah $N$ aur $f$ ko vertical aur horizontal mein resolve karo.

Vertical (koi vertical acceleration nahi): $N\cos \theta =mg+f\sin \theta =mg+{\mu }_{s}N\sin \theta .$ Horizontal (centripetal): normal ka horizontal slice aur friction ka horizontal slice dono inward point karte hain: $N\sin \theta +{\mu }_{s}N\cos \theta =\frac{m{v}^2}{r}.$ Pehle se $N$ solve karo: $N(\cos \theta -{\mu }_s\sin \theta )=mg$. Doosre mein substitute karo aur $m$ cancel karo: $v_{\max }=\sqrt{rg\frac{\sin \theta +{\mu }_s\cos \theta }{\cos \theta -{\mu }_s\sin \theta }}.$ Plug in karo ($\sin 15^{\circ }=0.2588,\cos 15^{\circ }=0.9659$): $v_{\max }=\sqrt{100\times 9.8\times \frac{0.2588+0.40\times 0.9659}{0.9659-0.40\times 0.2588}}=\sqrt{980\times \frac{0.6452}{0.8624}}$ $=\sqrt{980\times 0.7481}\approx \sqrt{733.1}\approx 27.1\text{m/s}.$ $v_{\max }\approx 27.1\text{m/s}$ (approximately $98\text{km/h}$).

Pehle geometry. Har string hypotenuse hai ($1.25\text{m}$); horizontal reach $r=1.0\text{m}$ hai, isliye vertical reach $\sqrt{1.25^2-1.0^2}=\sqrt{0.5625}=0.75\text{m}$ hai — $1.5\text{m}$ separation ke consistent ($2\times 0.75$). Isliye $\cos \theta =r/L=1.0/1.25=0.80$ (horizontal) aur $\sin \theta =0.75/1.25=0.60$ (vertical). Upper string inward aur upar pull karti hai; lower string inward aur neeche pull karti hai.

Vertical balance (upar positive): upper string ka up-part minus lower string ka down-part minus weight $=0$: $T_1(0.60)-T_2(0.60)-mg=0\Rightarrow T_1-T_2=\frac{mg}{0.60}=\frac{2.0\times 9.8}{0.60}=32.67\text{N}.(\star )$ Horizontal (centripetal): dono inward horizontal parts add hote hain: $T_1(0.80)+T_2(0.80)=\frac{m{v}^2}{r}=\frac{2.0\times 36}{1.0}=72\text{N}\Rightarrow T_1+T_2=\frac{72}{0.80}=90\text{N}.(\star \star )$ $(\star )$ aur $(\star \star )$ add karo: $2T_1=122.67\Rightarrow T_1\approx 61.3\text{N}$. Phir $T_2=90-61.3\approx 28.7\text{N}$. $T_1\approx 61.3\text{N}$ (upper), $T_2\approx 28.7\text{N}$ (lower). Dono positive → dono strings genuinely taut hain. ✓

L1 ne tumhe provider ko naam dena sikhaya ::: tension / gravity / friction / normal / electromagnetic. L2 ne tumhe $F_{\text{provider}}=m{v}^2/r$ sikhaya ::: ek unknown ke liye solve karo, aur mass aksar cancel hota hai. L3 ne tumhe tilted force ko resolve karna sikhaya ::: vertical part weight balance karta hai, horizontal part centripetal hai. L4 ne tumhe sikhaya ki position matter karti hai ::: top par gravity inward add hoti hai, bottom par woh subtract hoti hai. L5 ne tumhe sikhaya kya pehle tootega / kis taraf slide hoga ::: friction direction aur taut-string checks answer decide karte hain.