Exercises — Vertical circular motion — minimum speed conditions

Top par minimum tab hoti hai jab tension $T=0$ ho aur sirf gravity centripetal pull deti hai: $mg=\frac{m{v}_{\text{top}}^2}{r}\Rightarrow v_{\text{top}}=\sqrt{gr}.$ $v_{\text{top}}=\sqrt{9.8\times 2.0}=\sqrt{19.6}=4.43\text{m/s}.$

$v_{\text{bot}}=\sqrt{5gr}=\sqrt{5\times 9.8\times 2.0}=\sqrt{98}=9.90\text{m/s}.$

Minimum-speed loop ke liye bottom tension $6mg$ hoti hai: $T_{\text{bot}}=6mg=6\times 0.50\times 9.8=29.4\text{N}.$

(a) $v_{\text{top}}=\sqrt{gr}=\sqrt{9.8\times 1.2}=\sqrt{11.76}=3.43\text{m/s}.$ (b) Estimate use karte hue $T=\frac{2\pi r}{v}=\frac{2\pi (1.2)}{3.43}=\frac{7.540}{3.43}=2.20\text{s}.$ (Real period shorter hogi kyunki ball bottom par zyada fast hoti hai — yeh sirf ek top-speed estimate hai.)

(a) $v_{\text{bot}}=\sqrt{5gr}=\sqrt{5\times 9.8\times 6.0}=\sqrt{294}=17.15\text{m/s}.$ (b) Side par (height $r$): $v_{\text{side}}=\sqrt{3gr}=\sqrt{3\times 9.8\times 6.0}=\sqrt{176.4}=13.28\text{m/s}.$

Top par gravity aur tension dono neeche point karte hain (toward center), isliye: $T+mg=\frac{m{v}^2}{r}\Rightarrow T=\frac{m{v}^2}{r}-mg.$ $T=\frac{0.30\times 6.0^2}{0.80}-0.30\times 9.8=\frac{10.8}{0.80}-2.94=13.5-2.94=10.56\text{N}.$

Top par speed energy conservation se, height gained $=2r=10\text{m}$: $v_{\text{top}}^2=v_{\text{bot}}^2-4gr=18^2-4(9.8)(5.0)=324-196=128,$ $v_{\text{top}}=\sqrt{128}=11.31\text{m/s}.$ Top par normal force (dono $N$ aur $mg$ center ki taraf neeche point karte hain): $N+mg=\frac{m{v}_{\text{top}}^2}{r}\Rightarrow N=\frac{m{v}_{\text{top}}^2}{r}-mg.$ $N=\frac{600\times 128}{5.0}-600\times 9.8=15360-5880=9480\text{N}.$

Force equations: $T_{\text{bot}}-mg=\frac{m{v}_{\text{bot}}^2}{r},T_{\text{top}}+mg=\frac{m{v}_{\text{top}}^2}{r}.$ Subtract karo: $T_{\text{bot}}-T_{\text{top}}-2mg=\frac{m}{r}\left(v_{\text{bot}}^2-v_{\text{top}}^2\right).$ Energy conservation height $2r$ par: $v_{\text{bot}}^2-v_{\text{top}}^2=4gr$. Substitute karo: $T_{\text{bot}}-T_{\text{top}}-2mg=\frac{m}{r}(4gr)=4mg\Rightarrow T_{\text{bot}}-T_{\text{top}}=6mg.$ Yeh har speed ke liye hold karta hai, sirf minimum ke liye nahi — $v$ dependence cancel ho gayi.

Loop complete karne ke liye loop bottom par $v_{\text{bot}}^2=5gr$ chahiye. Release height $h$ se energy: $mgh=\frac{1}{2}m{v}_{\text{bot}}^2=\frac{1}{2}m(5gr)\Rightarrow h=\frac{5r}{2}=2.5r.$ Ek neat, mass- aur $g$-free result: kam se kam do aur aadhe radii ki height se shuru karo.

(a) Rod push kar sakti hai, isliye top condition sirf $v_{\text{top}}\ge 0$ hai; minimum $v_{\text{top}}=0$ hai. Phir: $v_{\text{bot}}^2=v_{\text{top}}^2+4gr=0+4gr=4gr,$ $v_{\text{bot}}=\sqrt{4gr}=2\sqrt{gr}=2\sqrt{9.8\times 1.5}=2\sqrt{14.7}=7.67\text{m/s}.$ (b) Top par $v_{\text{top}}=0$ ke saath, centripetal requirement $m{v}^2/r=0$ hai. Force equation (neeche = toward center positive maante hue) $F_{\text{rod}}+mg=0$ hai, isliye $F_{\text{rod}}=-mg$. Negative sign ka matlab hai rod force upar point karta hai — yaani rod ball ko outward (upar, center se door) push karti hai taaki use hold kiya ja sake. $|F_{\text{rod}}|=mg=0.40\times 9.8=3.92\text{N}\text{(ek push)}.$

Hill ke top par, center neeche hota hai aur ground upar push karta hai (center se door), gravity neeche (center ki taraf). Centripetal neeche ki taraf hai, isliye: $mg-N=\frac{m{v}^2}{r}.$ Skier ground chhod deta hai jab $N=0$: $mg=\frac{m{v}^2}{r}\Rightarrow v=\sqrt{gr}=\sqrt{9.8\times 20}=\sqrt{196}=14.0\text{m/s}.$ Same $\sqrt{gr}$ form — lekin ab yeh ground par rehne ke liye maximum allowed speed hai, minimum nahi. Geometry (center neeche) identical hai; sirf yeh flip hota hai ki kaunsi force "extra" hai.

(a) $v_{\text{bot}}=\sqrt{5gr}=\sqrt{5\times 9.8\times 0.90}=\sqrt{44.1}=6.64\text{m/s}.$ (b) $v_{\text{side}}=\sqrt{3gr}=\sqrt{3\times 9.8\times 0.90}=\sqrt{26.46}=5.14\text{m/s}.$ (c) Side par center horizontal hota hai, isliye sirf horizontal component matter karta hai. Tension poori tarah horizontal hai (toward center) aur gravity poori tarah vertical hai (horizontal/centripetal direction mein kuch contribute nahi karti): $T_{\text{side}}=\frac{m{v}_{\text{side}}^2}{r}=\frac{0.25\times 26.46}{0.90}=\frac{6.615}{0.90}=7.35\text{N}.$ (Note karo $T_{\text{side}}=3mg=3\times 0.25\times 9.8=7.35\text{N}$ — match karta hai, kyunki $v_{\text{side}}^2=3gr$.)

Geometry setup karo. $\theta$ wo angle hai jo top se sweep hua hai. Ball ki height top se neeche $r(1-\cos \theta )$ hai, isliye bottom se uski height $r(1+\cos \theta )$ hai. Top se energy ($v_{\text{top}}^2=gr$): $v^2=v_{\text{top}}^2+2g[r(1-\cos \theta )]=gr+2gr(1-\cos \theta )=gr(3-2\cos \theta ).$

Angle $\theta$ par radial equation. Gravity ka component center ki taraf $mg\cos \theta$ hai (jab string vertical-up direction se $\theta$ angle banati hai). Isliye: $T+mg\cos \theta =\frac{m{v}^2}{r}=mg(3-2\cos \theta ).$ $T=mg(3-2\cos \theta )-mg\cos \theta =mg(3-3\cos \theta )=3mg(1-\cos \theta ).$

$T=mg$ impose karo: $3mg(1-\cos \theta )=mg\Rightarrow 1-\cos \theta =\frac{1}{3}\Rightarrow \cos \theta =\frac{2}{3}.$ $\theta =\arccos \left(\frac{2}{3}\right)=48.19^{\circ }.$ Endpoints check karo: $\theta =0$ par (top) $T=0$ ✓; $\theta =180^{\circ }$ par (bottom) $T=3mg(1-(-1))=6mg$ ✓. Formula dono known cases reproduce karta hai.

String ko $v_{\text{bot}}^2=5gr$ chahiye; rod ko $v_{\text{bot}}^2=4gr$ chahiye. Kinetic energy $\frac{1}{2}m{v}^2$ (same $m$, same $r$): $\frac{K{E}_{\text{string}}}{K{E}_{\text{rod}}}=\frac{5gr}{4gr}=\frac{5}{4}=\mathrm{1.25.}$ String version ko $25\%$ zyada launch energy chahiye, purely isliye kyunki use poore top tak string taut rakhni padti hai.

Wire par bead rod ki tarah hai: wire bead par dono inward aur outward push kar sakti hai (wo bead ke around wrap karti hai), isliye koi "slack" condition nahi hai. Loop complete karne ke liye sirf yahi zaroori hai ki bead top par move kare, yaani $v_{\text{top}}\ge 0$; minimum $v_{\text{top}}=0$ hai. $v_0^2=v_{\text{top}}^2+4gr=4gr\Rightarrow v_{0,\min }=2\sqrt{gr}.$ Comparison: rod ke same ($2\sqrt{gr}$), string ($\sqrt{5gr}$) se kam. Ek string sirf inward pull kar sakti hai, isliye use itni speed chahiye ki required inward force kabhi zero na ho — yahi extra demand poori wajah hai ki $\sqrt{5gr}>2\sqrt{gr}$.

$\sqrt{5gr}$ speed chahiye, aur top se zyada kyun? Bottom. Use top par $\sqrt{gr}$ ke saath pahunchna hai, aur height $2r$ chadhne se $v^2$ mein $4gr$ add hota hai: $v_{\text{bot}}^2=gr+4gr=5gr$.

$r$ radius ka loop complete karne ke liye minimum release height (frictionless slide)? $h=2.5r$ — $mgh=\frac{1}{2}m(5gr)$ se.

$\theta$ par vertical circle par general tension formula (minimum-speed loop)? $T=3mg(1-\cos \theta )$: top par $0$ aur bottom par $6mg$ deta hai.