Exercises — Acceleration of rolling objects on inclines — comparison

Shape factor defined hai $\beta =I/(m{R}^2)$ ke roop mein — tum literally $m{R}^2$ ke aage wala number padh lete ho. Yahan $I=\frac{2}{5}m{R}^2$ hai, isliye $\beta =\frac{\frac{2}{5}m{R}^2}{m{R}^2}=\frac{2}{5}=\mathrm{0.40.}$ $2/5$ ka $\beta$ solid sphere hai. ✔

Acceleration hai $a=\frac{g\sin \theta }{1+\beta }$. Bada $\beta$ denominator mein hota hai, isliye $a$ ko chota karta hai. Sphere ka $\beta$ chota hai ($2/5<1$), isliye $a$ bada, isliye woh jeetta hai. Solid sphere pehle. ✔

Solid cylinder $\Rightarrow \beta =\frac{1}{2}$, isliye $1+\beta =\frac{3}{2}$. $a=\frac{g\sin \theta }{1+\beta }=\frac{9.8\times \sin 25^{\circ }}{1.5}.$ Ab $\sin 25^{\circ }=0.4226$ hai, isliye numerator hai $9.8\times 0.4226=4.141$. $a=\frac{4.141}{1.5}=2.76{\text{m/s}}^2.$ Size check: yeh free-slide value $g\sin \theta =4.14$ se kam hona chahiye — aur $2.76<4.14$ hai. ✔

Upar engine (b) mein build kiya gaya energy result use karo: $h$ drop karne par $mgh$ release hoti hai, jo split hoti hai forward motion $\frac{1}{2}m{v}^2$ mein aur spin $\frac{1}{2}I{\omega }^2=\frac{1}{2}\beta m{v}^2$ mein, jisse $mgh=\frac{1}{2}m{v}^2(1+\beta )$ milta hai aur hence $v=\sqrt{\frac{2gh}{1+\beta }}.$ Solid sphere $\Rightarrow \beta =\frac{2}{5}$, isliye $1+\beta =\frac{7}{5}$. $v=\sqrt{\frac{2(9.8)(2.0)}{1.4}}=\sqrt{\frac{39.2}{1.4}}=\sqrt{28.0}=5.29\text{m/s}.$ Sanity check: ek frictionless sliding block $\sqrt{2gh}=\sqrt{39.2}=6.26$ m/s tak pahunchta — roller slower hai kyunki kuch energy spin mein gayi ($(1+\beta )$ tax). $5.29<6.26$. ✔

Humein friction $a$ ke terms mein chahiye. Friction rim par act karta hai (centre se $R$ distance par), isliye uska torque $fR$ hai; Newton ka rotational law usse $I\alpha$ ke equal set karta hai — spin-inertia $I$ times angular acceleration $\alpha$ (upar define kiya: $\alpha =a/R$, spin-rate change jo centre ki acceleration se juda hai). $I=\beta m{R}^2$ aur $\alpha =a/R$ substitute karo, bilkul engine (a) ki tarah: $fR=I\alpha =(\beta m{R}^2)\frac{a}{R}=\beta mRa\Rightarrow f=\beta ma.$ Pehle cylinder ke liye $a$ nikalo ($\beta =\frac{1}{2}$): $a=\frac{g\sin \theta }{1+\beta }=\frac{9.8\times 0.5}{1.5}=\frac{4.9}{1.5}=3.267{\text{m/s}}^2.$ Phir $f=\beta ma=\frac{1}{2}(3.0)(3.267)=4.90\text{N}.$ Figure s01 dekho: yellow $f$ slope ke upar point karta hai jabki body neeche slide karta hai — kyunki friction ka kaam woh red torque arc ($fR$) supply karna hai jo cylinder ko spin up karta hai. ✔

Static friction zyada se zyada $\mu N$ supply kar sakta hai, jahan $N=mg\cos \theta$ normal force hai jo is page ke upar define ki gayi hai — slope ki perpendicular push, figure s01 mein blue arrow ke roop mein. Rolling tab tak hoti hai jab tak required friction $f=\beta ma$ yeh cap cross na kare: $\beta ma\le \mu mg\cos \theta .$ Mass cancel ho jaata hai. $a=\frac{g\sin \theta }{1+\beta }$ se rearrange karo:

Dono ki constant acceleration $a=\frac{g\sin \theta }{1+\beta }$ hai, rest se start karte hain, isliye distance $s=\frac{1}{2}a{t}^2$ se milti hai. Accelerations compute karo. Common factor $g\sin 20^{\circ }=9.8\times 0.3420=3.352$.

• Solid sphere $\beta =\frac{2}{5}$, $1+\beta =1.4$: $a_1=3.352/1.4=2.394{\text{m/s}}^2$.
• Hollow sphere $\beta =\frac{2}{3}$, $1+\beta =1.6667$: $a_2=3.352/1.6667=2.011{\text{m/s}}^2$.

Time for solid sphere ko $L=4.0$ m cover karne ka: $L=\frac{1}{2}{a}_1{t}^2\Rightarrow t=\sqrt{\frac{2L}{a_1}}=\sqrt{\frac{8.0}{2.394}}=\sqrt{3.342}=1.828\text{s}.$ Usi instant par hollow sphere ne travel kiya hua distance: $s_2=\frac{1}{2}{a}_2{t}^2=\frac{1}{2}(2.011)(1.828^2)=\frac{1}{2}(2.011)(3.342)=3.361\text{m}.$ Toh peeche ka gap hai: $\Delta =L-s_2=4.0-3.361=0.639\approx 0.64\text{m}.$ Hollow sphere lagbhag 0.64 m peeche hai jab solid sphere finish karta hai. ✔

$L=\frac{1}{2}a{t}^2$ se $L$ fixed leke: $t=\sqrt{2L/a}\propto \sqrt{1/a}\propto \sqrt{1+\beta }$ (kyunki $a\propto 1/(1+\beta )$). Isliye

Energy result $v=\sqrt{\frac{2gh}{1+\beta }}$ (engine (b)) ko invert karo. Dono sides square karo: $v^2=\frac{2gh}{1+\beta }\Rightarrow 1+\beta =\frac{2gh}{v^2}.$ Plug in karo: $2gh=2(9.8)(1.5)=29.4$ aur $v^2=4.2^2=17.64$. $1+\beta =\frac{29.4}{17.64}=1.6667\Rightarrow \beta =0.6667=\frac{2}{3}.$ $2/3$ ka $\beta$ hollow sphere (thin shell) hai. ✔

Pure rolling tab tak chalti hai jab tak surface jo friction de sakta hai ($\mu$) woh friction chahiye ($\frac{\beta \tan \theta }{1+\beta }$) se zyada ya equal ho: $\mu \ge \frac{\beta \tan \theta }{1+\beta }$. Sabse steep allowed angle tab hota hai jab dono equal hon: $\mu =\frac{\beta \tan {\theta }_{\max }}{1+\beta }\Rightarrow \tan {\theta }_{\max }=\frac{\mu (1+\beta )}{\beta }.$ Ring ke liye $\beta =1$, $1+\beta =2$: $\tan {\theta }_{\max }=\frac{0.25\times 2}{1}=0.5\Rightarrow {\theta }_{\max }=\arctan (0.5)=26.57^{\circ }.$ $\approx 26.6^{\circ }$ se zyada steep hone par ring skid karne lagti hai. ✔

$a=\frac{g\sin \theta }{1+\beta }$ mein $\beta \to 0$ lo: ${\lim }_{\beta \to 0}a=\frac{g\sin \theta }{1+0}=g\sin \theta .$ Yeh exactly frictionless incline par sliding block hai. Physically $\beta =I/m{R}^2\to 0$ ka matlab hai koi rotational inertia nahi — ek point mass, ya saara mass axis par — isliye kuch spin up nahi karna padta aur saari gravity translation mein jaati hai. Similarly $v=\sqrt{2gh/(1+\beta )}\to \sqrt{2gh}$, free-fall/slide speed. Rolling law apne zero-spin corner mein sliding law ko contain karta hai. ✔

• Roller par friction, $a$ ke terms mein? ::: $f=\beta ma$
• $\alpha$ kya hai, aur yeh $a$ se kaise link hota hai? ::: angular acceleration; $\alpha =a/R$
• Incline par $N$ kya hai? ::: normal force $=mg\cos \theta$
• Minimum $\mu$ for pure rolling? ::: ${\mu }_{\min }=\frac{\beta \tan \theta }{1+\beta }$
• Sabse steep rolling angle? ::: ${\theta }_{\max }=\arctan (\mu (1+\beta )/\beta )$
• Measured bottom speed se $\beta$ recover karo? ::: $1+\beta =\frac{2gh}{v^2}$
• Do shapes ka descent-time ratio? ::: $\frac{t_2}{t_1}=\sqrt{\frac{1+{\beta }_2}{1+{\beta }_1}}$