Exercises — Rolling without slipping — v = Rω condition

Kya chahiye: spin $\omega$ ko centre speed $v$ se link karna. Yeh velocity form hai. Yeh form kyun: humein $\omega$ diya hai aur $v$ chahiye — koi time-change nahi, koi distance nahi, toh use karo $v=R\omega$. $v=R\omega =0.5\times 4=2\text{ m/s}.$

Kya chahiye: point-speed rule "bottom zero, centre vee, top twice". Kyun: rolling, translation ($+v$ har jagah) aur rotation ko combine karta hai (bottom par backward $R\omega$, top par forward $R\omega$). Kyunki $R\omega =v$:

• Top $=v+R\omega =v+v=2v=6\text{ m/s}$.
• Contact $=v-R\omega =v-v=0\text{ m/s}$.

Kya chahiye: acceleration form, kyunki dono quantities speed ke rates-of-change hain. Yeh form kyun: $\alpha$ diya hai, $a$ chahiye — constraint ko ek baar differentiate karo ⇒ $a=R\alpha$. $a=R\alpha =0.30\times 6=1.8{\text{ m/s}}^2.$

(a) Spin nikalne ke liye $v=R\omega$ ko rearrange karo: $\omega =\frac{v}{R}=\frac{5}{0.35}=14.29\text{ rad/s}.$ Divide kyun karein: humein $v$ aur $R$ pata hai, $\omega$ chahiye; $R$, $\omega$ ko multiply karta hai, toh divide karke undo karo. (b) Top speed $=2v=10\text{ m/s}$ (translation $v$ plus forward rotation $R\omega =v$).

Kya chahiye: displacement form $x_{cm}=R\theta$, kyunki humein angle turned diya gaya hai aur distance chahiye. Radians kyun: constraint mein $\theta$ radians mein hota hai. Ek revolution $=2\pi$ rad, toh $\theta =2\times 2\pi =4\pi \text{ rad}.$ $x_{cm}=R\theta =0.20\times 4\pi =2.513\text{ m}.$ Sanity check: distance $=$ turns ki sankhya $\times$ circumference $=2\times 2\pi R=2\times 2\pi (0.20)=2.513\text{ m}$. ✓ Same answer, jaise hona chahiye.

Step 1 — linear acceleration: $a=\frac{\Delta v}{t}=\frac{4-0}{5}=0.8{\text{ m/s}}^2$. Kyun: uniform acceleration ka matlab hai $a$ constant hai $=$ (speed mein change)$/$(time). Step 2 — constraint se convert karo: kyunki kabhi slip nahi karta, $a=R\alpha$, toh $\alpha =\frac{a}{R}=\frac{0.8}{0.10}=8{\text{ rad/s}}^2.$

Kya karte hain: 3 o'clock point par dono velocities ko vectors ke roop mein add karo — yeh exactly wahi amber right-triangle hai jo figure mein bana hai.

• Translation (cyan arrow, right ki taraf): $+v$, horizontal (aage).
• Centre ke baare mein Rotation (white arrow, upar ki taraf): rim point centre se $R$ door hai; uski rotational speed $R\omega =v$ hai. 3 o'clock point par, rotation use seedha upar le jaata hai (wahan circle ki tangent).

Perpendicular kyun: rotational velocity hamesha circle ki tangent hoti hai, yaani centre se point tak ki line ke perpendicular. 3 o'clock par woh line horizontal hai, toh tangent vertical hai — isliye cyan aur white arrows figure mein right angle par milte hain.

Combine karo: dono arrows right triangle ki do sides hain; amber arrow uska hypotenuse hai (resultant). Pythagoras se: $v_{3\text{ o’clock}}=\sqrt{v^2+v^2}=v\sqrt{2}\approx 1.414v,$ horizontal se $45^{\circ }$ upar aage ki taraf point karta hua — amber arrow ka slope.

Yeh kyun matter karta hai: yeh dikhata hai ki point speed smoothly vary karti hai $0$ (bottom) se $v\sqrt{2}$ (side) se $2v$ (top) tak — wheel ek rigid block nahi hai jo ek hi speed par chal raha ho.

Trick: kyunki contact point instantaneously rest par hai, pure wheel ko us contact point ke baare mein pure rotation maana ja sakta hai (Instantaneous axis of rotation) — figure mein bottom par woh amber pivot dot. Allowed kyun: ek rotating rigid body ke point ki speed $=\omega \times (\text{axis se distance})$. Contact point ki zero speed hai, toh woh us instant ke liye axis hai.

• Centre: contact axis se distance $=R$ (figure mein white double-arrow), toh speed $=\omega R=v$. ✓ (jo hum jaante hain woh match karta hai) — short white velocity arrow se dikhaya gaya.
• Top: contact axis se distance $=2R$ (taller cyan double-arrow — top, contact point ke bilkul diametrically opposite hai). Toh speed $=\omega (2R)=2\omega R=2v$. ✓ — longer cyan velocity arrow se dikhaya gaya, exactly white wale se double length.

Figure "twice the distance ⇒ twice the speed" reasoning ko visible banata hai: same $\omega$, double radius arm.

Kya chahiye: rolling KE $=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$ — translational plus rotational. Constraint kyun aata hai: humein $v$ pata hai, $\omega$ nahi, lekin no-slip deta hai $\omega =v/R$, toh sab kuch $v$ mein likh sakte hain. $KE=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\left(\frac{v}{R}\right)}^2=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2.$ $R$ cancel kyun hota hai: $I$ mein $R^2$ hai jo ${\omega }^2$ se $1/R^2$ se milta hai — rolling problems ki pehchaan. $KE=\frac{7}{10}(2)(3^2)=0.7\times 2\times 9=12.6\text{ J}.$

(a) $v^2=u^2+2ad$ use karo start speed $u=0$ ke saath: $a=\frac{v^2}{2d}=\frac{6^2}{2\times 9}=\frac{36}{18}=2{\text{ m/s}}^2.$ Yeh kinematic equation kyun: yeh speed, distance aur acceleration ko bina time ke link karta hai — exactly woh teen quantities jo hamare paas hain. (b) No slipping, toh differentiated constraint laagu hota hai: $\alpha =\frac{a}{R}=\frac{2}{0.30}=6.667{\text{ rad/s}}^2.$

(a) Constraint test karo: agar rolling hai, toh $v$ hona chahiye $R\omega =0.30\times 30=9\text{ m/s}$ ke barabar. Lekin car sirf $6\text{ m/s}$ par move karti hai. Kyunki $6\ne 9$, yeh slipping hai — $v=R\omega$ laagu nahi hota. (b) Contact-point velocity $=v_{cm}-R\omega$ (translation aage, rotation peeche drag karta hai): $v_{contact}=6-9=-3\text{ m/s}.$ Minus sign ka matlab hai contact patch ground ke relative peeche $3\text{ m/s}$ par move kar raha hai — tyre road se faster spin kar raha hai, toh peeche skid karta hai (ab kinetic friction laagu hota hai, Static vs kinetic friction ke mutabik).

(a) $\alpha =\frac{{\omega }_1-{\omega }_0}{t}=\frac{10-2}{4}=2{\text{ rad/s}}^2.$ (b) No slip ⇒ $a=R\alpha =0.4\times 2=0.8{\text{ m/s}}^2.$ (c) Total angle turned: kyunki $\alpha$ constant hai, angular speed ${\omega }_0$ se ${\omega }_1$ tak seedhi line mein badhti hai. Kisi bhi quantity ke liye jo constant rate se change hoti hai, average value simply start aur end ka midpoint hota hai, $\bar{\omega }=\frac{{\omega }_0+{\omega }_1}{2}$, aur covered distance $=$ average rate $\times$ time. Isliye constant angular acceleration ke liye hum $\theta =\bar{\omega }t$ use kar sakte hain (yeh $\omega$-vs-$t$ graph ke neeche straight-line ka exact area hai — ek trapezium): $\theta =\frac{{\omega }_0+{\omega }_1}{2}t=\frac{2+10}{2}\times 4=24\text{ rad}.$ $x_{cm}=R\theta =0.4\times 24=9.6\text{ m}.$ Linear speeds se cross-check: $v_0=R{\omega }_0=0.8\text{ m/s}$, $v_1=R{\omega }_1=4\text{ m/s}$, average $=2.4\text{ m/s}$, times $4\text{ s}=9.6\text{ m}$. ✓

Har ek ki energy set up karo $\omega =v/R$ use karke taaki $\frac{1}{2}I{\omega }^2=\frac{1}{2}I{v}^2/R^2$.

• Hoop: rotational $=\frac{1}{2}(m{R}^2)(v^2/R^2)=\frac{1}{2}m{v}^2$. Translational $=\frac{1}{2}m{v}^2$. Total $=m{v}^2$. Rotational fraction $=\frac{\frac{1}{2}m{v}^2}{m{v}^2}=\frac{1}{2}=50\%$.
• Disc: rotational $=\frac{1}{2}(\frac{1}{2}m{R}^2)(v^2/R^2)=\frac{1}{4}m{v}^2$. Translational $=\frac{1}{2}m{v}^2$. Total $=\frac{3}{4}m{v}^2$. Rotational fraction $=\frac{\frac{1}{4}m{v}^2}{\frac{3}{4}m{v}^2}=\frac{1}{3}\approx 33.3\%$. Answer: hoop zyaada fraction ($50\%$ vs $33.3\%$) rotation mein store karta hai, kyunki uska mass axis se door hota hai, jisse bada Moment of inertia milta hai.