Exercises — Angular momentum L = Iω (fixed axis), L = r × p (general)

• Single particle: $\vec{L}=\vec{r}\times \vec{p}$, magnitude $L=rp\sin \theta =p{r}_{\perp }$.
• Rigid body, fixed/symmetry axis: $L=I\omega$, with $I=\sum m_i{r}_i^2$.
• No external torque ⇒ $L$ conserved: $I_1{\omega }_1=I_2{\omega }_2$.

WHAT we ask: $L=rp\sin \theta$. WHY: sirf motion ka "ghoomne wala" part count karta hai. Velocity us line ke along point kar rahi hai jo ball ko $O$ se jodhti hai, isliye $\vec{r}$ aur $\vec{p}$ parallel hain: $\theta =0^{\circ }$, hence $\sin \theta =0$. $L=rp\sin 0^{\circ }=0.$ Meaning: origin ki taraf (ya usse door) seedhi motion ka zero lever arm hota hai $r_{\perp }=r\sin \theta =0$ — kuch bhi $O$ ke around swirl nahi karta.

WHY $L=I\omega$? Rigid body, fixed symmetry axis — exactly woh case jahan sum $\sum m_i{r}_i^2\omega$ collapse hokar $I\omega$ ban jaata hai. $L=I\omega =(0.5)(10)=5\mathrm{kg}{\mathrm{m}}^2/s.$

Step 1 — momentum: $\vec{p}=m\vec{v}=(0,1,0)\mathrm{kg}m/s$. Step 2 — cross product (dekho Cross Product). $\vec{r}=(r_x,r_y,0)$ aur $\vec{p}=(p_x,p_y,0)$ ke liye sirf $z$-component bachta hai: $L_z=r_x{p}_y-r_y{p}_x=(3)(1)-(4)(0)=3.$ Toh $\vec{L}=(0,0,3)\mathrm{kg}{\mathrm{m}}^2/s$, magnitude $|\vec{L}|=3\mathrm{kg}{\mathrm{m}}^2/s$, right-hand rule se page ke bahar (+$z$) point karta hai. Lever arm se check: $r=\sqrt{3^2+4^2}=5$, $p=1$, aur $\sin \theta$ is tarah work out hota hai ki $r_{\perp }p=3$. Consistent. ✓

Step 1 — chosen axis ke baare mein $I$ (dekho Moment of Inertia): $I=\frac{1}{3}M{\ell }^2=\frac{1}{3}(3)(2^2)=4\mathrm{kg}{\mathrm{m}}^2.$ Step 2 — $L=I\omega$ apply karo: $L=I\omega =(4)(4)=16\mathrm{kg}{\mathrm{m}}^2/s.$ End axis kyun matter karta hai: wohi rod apne centre ke baare mein $I=\frac{1}{12}M{\ell }^2=1\mathrm{kg}{\mathrm{m}}^2$ rakhti hai — char guna chhota. Axis choice $I$ ko change karta hai, isliye $L$ bhi change hota hai.

(a) $L$ conserve karo (koi external torque nahi — sirf internal muscular pull, dekho Conservation of Angular Momentum): $I_1{\omega }_1=I_2{\omega }_2\Rightarrow {\omega }_2=\frac{(6)(3)}{2}=9rad/s.$ (b) $KE=\frac{1}{2}I{\omega }^2$ use karke (dekho Rotational Kinetic Energy): $K{E}_1=\frac{1}{2}(6)(3^2)=27\mathrm{J},K{E}_2=\frac{1}{2}(2)(9^2)=81\mathrm{J}.$ (c) Extra $54\mathrm{J}$ skater ki muscles se aayi jo mass ko andar kheenchte waqt outward "centrifugal" tendency ke against kaam karti hain. $L$ conserved hai; $KE$ nahi — energy internally add ki gayi.

The tool: ${\vec{\tau }}_{\text{ext}}=d\vec{L}/dt$ ($\vec{L}=\vec{r}\times \vec{p}$ differentiate karo; dekho Torque). Koi force nahi ⇒ $\vec{F}=0$ ⇒ torque $\vec{\tau }=\vec{r}\times \vec{F}=0$ ⇒ $d\vec{L}/dt=0$, toh $\vec{L}$ constant hai. Geometric picture: $L=p{r}_{\perp }$. Jaise ball glide karti hai, $\vec{r}$ swing aur lengthen karta hai, lekin $O$ se motion ki (fixed) line tak perpendicular distance $r_{\perp }$ kabhi nahi badlta, aur $p$ constant hai. Do constants multiply karo ⇒ constant $L$. Lever arm, na ki $r$, woh cheez hai jo nature yahan fixed rakhti hai.

(a) Gravity ek central force hai: $\vec{F}$, $\vec{r}$ ke antiparallel hai, toh $\vec{\tau }=\vec{r}\times \vec{F}=0$. Isliye $d\vec{L}/dt=0$ aur $\vec{L}$ conserved hai (yahi hai Kepler's Second Law — equal time mein equal areas). (b) Perihelion aur aphelion par velocity $\vec{r}$ ke perpendicular hai, toh $\theta =90^{\circ }$ aur $L=mvr$. $L$ conserve karo (mass cancel ho jaata hai): $m{v}_p{r}_p=m{v}_a{r}_a\Rightarrow v_a=\frac{v_p{r}_p}{r_a}=\frac{(6.0\times 10^4)(1.0\times 10^{11})}{1.5\times 10^{11}}=4.0\times 10^{4}m/s.$ Meaning: bahar jaao ⇒ slower, exactly "equal time mein equal areas sweep" ke matching.

(a) Unke beech sirf internal forces hain (contact par friction) ⇒ common axis ke baare mein koi external torque nahi ⇒ $L$ conserved: $I_A{\omega }_A=(I_A+I_B){\omega }_f\Rightarrow {\omega }_f=\frac{(2)(12)}{2+4}=4rad/s.$ (b) $K{E}_i=\frac{1}{2}{I}_A{\omega }_A^2=\frac{1}{2}(2)(144)=144\mathrm{J}.$ $K{E}_f=\frac{1}{2}(6)(4^2)=48\mathrm{J}.$ Fraction lost $=\frac{144-48}{144}=\frac{96}{144}=\frac{2}{3}\approx 0.667$. Meaning: bilkul perfectly inelastic linear collision ki tarah — $L$ (rotational "momentum") conserved hai, lekin sliding friction $\frac{2}{3}$ energy heat ke roop mein burn kar deta hai.

(a) Point mass at radius $R$: $I_{\text{beetle}}=m{R}^2=(0.10)(0.50^2)=0.025\mathrm{kg}{\mathrm{m}}^2$. (b) $L$ conserve karo (beetle ki legs disc ko push karti hain lekin woh internal torque hai; koi bhi cheez system ko externally twist nahi karti):

• Initial: beetle centre par contribute karta hai $0$, toh $L_i=I_d{\omega }_0=(0.20)(6.0)=1.2\mathrm{kg}{\mathrm{m}}^2/s$.
• Final: $L_f=(I_d+m{R}^2){\omega }_1=(0.225){\omega }_1$. ${\omega }_1=\frac{1.2}{0.225}=5.333rad/s\approx 5.33rad/s.$ (c) $K{E}_i=\frac{1}{2}(0.20)(6.0^2)=3.6\mathrm{J}$; $K{E}_f=\frac{1}{2}(0.225)(5.333^2)=3.2\mathrm{J}$. Energy girta hai $0.4\mathrm{J}$ se: beetle ki muscles net negative work absorb karti hain (use bahar jaate waqt rim ki motion ke against brake lagana padta hai), toh $KE$ decrease hota hai chahe kuch bhi externally act nahi kiya.

(a) $\vec{p}=(1,3,0)$; $L_z=r_x{p}_y-r_y{p}_x=(2)(3)-(0)(1)=6$. Toh $\vec{L}=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. (b) $\vec{r}$ $+x$ ke along point karta hai, toh radial velocity hai $v_x=1m/s$ (seedha bahar — kuch contribute nahi karta, $\theta =0$). Tangential velocity hai $v_y=3m/s$, $\vec{r}$ ke perpendicular. Sirf wahi swirl karta hai: $L=r(m{v}_y)=2\times (1\times 3)=6\mathrm{kg}{\mathrm{m}}^2/s.✓$ Radial $v_x$ particle ko $O$ ki taraf/door move karta hai — zero lever arm, zero $L$. (c) $L=I\omega$ ke liye chahiye ek rigid body about a fixed axis jiska ek well-defined single $\omega$ ho. Ek akela particle jo straight/general path par drift kar raha ho uski koi fixed rotation axis nahi hoti aur koi rigid $I$ nahi hota — tumhe general $\vec{L}=\vec{r}\times \vec{p}$ use karna padega.

• Origin ki taraf seedhi motion ka $L=0$ hota hai ::: kyunki $\theta =0\Rightarrow r_{\perp }=0$
• Central force ⇒ $L$ conserved ::: torque $\vec{r}\times \vec{F}=0$ jab $\vec{F}\parallel \vec{r}$
• Sticking/inelastic rotation ::: $L$ conserve karo, $KE$ loss expect karo
• Insaan turntable par bahar jaaye ::: $\omega$ girta hai (larger $I$), $KE$ change hoti hai (woh kaam karta hai)
• $L=I\omega$ ke liye chahiye ::: rigid body, fixed/symmetry axis — warna $\vec{r}\times \vec{p}$ use karo