Exercises — Conservation of angular momentum — conditions

Har baar ek test apply karo: kya is axis ke baare mein net external torque hai?

• (a) Conserved. Muscular pull internal hai; ice frictionless hai isliye koi external twist nahi. $L$ fixed. ✅
• (b) NOT conserved. Bacche ka push off-axis apply hone wali ek external force hai → external torque → $L$ badhta hai. ❌
• (c) Conserved. Gravity ek central force hai jo straight Sun ki taraf point karti hai, isliye $\vec{F}\parallel \vec{r}$; cross-product zero-rule se $\vec{\tau }=\vec{r}\times \vec{F}=0$. $L$ fixed — yeh Kepler's 2nd law hai. ✅
• (d) NOT conserved. Contact par friction off-axis act karta hai aur spin oppose karta hai → $L$ ke opposite sign ka external torque → $L$ decay hokar zero ho jaata hai. ❌

Agar $\vec{L}$ ka time-derivative zero hai, toh $\frac{d\vec{L}}{dt}={\vec{\tau }}_{ext,net}$ se net external torque zero hai, aur $\vec{L}$ magnitude aur direction dono mein constant hai (space mein ek fixed arrow hai).

Condition: frictionless, internal pull → $L$ spin axis ke baare mein conserved (fixed vertical axis, isliye $L=I\omega$ apply hota hai). Woh poore time ek hi taraf spin karti hai, isliye saare terms positive hain aur signs out ho jaate hain. $I_1{\omega }_1=I_2{\omega }_2\Rightarrow {\omega }_2=\frac{8\times 1.5}{3}=4\text{rad/s}.$ Kinetic energy (toolkit se) $KE=\frac{1}{2}I{\omega }^2$. Kyunki $I\omega$ fixed hai, likho $KE=\frac{1}{2}(I\omega )\omega =\frac{1}{2}L\omega$, isliye $KE\propto \omega$: $\frac{K{E}_2}{K{E}_1}=\frac{{\omega }_2}{{\omega }_1}=\frac{4}{1.5}=\frac{8}{3}\approx \mathrm{2.67.}$ Energy badh gayi — uske muscles ne work kiya. Dekho Rotational kinetic energy.

Condition: gravity axis ke parallel hai (uspar koi torque nahi); impact internal hai → axis ke baare mein $L$ conserved. Disc aur putty aakhir mein same direction mein ghoomte hain → saare terms positive. Point-mass addition: $m{r}^2=0.4\times 0.6^2=0.144{\text{kg⋅m}}^2$. $I_d{\omega }_0=(I_d+m{r}^2){\omega }_f\Rightarrow {\omega }_f=\frac{1.2\times 6}{1.2+0.144}=\frac{7.2}{1.344}\approx 5.36\text{rad/s}.$

Gravity central hai → $\vec{F}\parallel \vec{r}$ → $\vec{\tau }=\vec{r}\times \vec{F}=0$ → $L=mvr$ conserved (apsides par $\vec{v}\perp \vec{r}$, isliye poora $vr$ count hota hai). Orbit ek hi direction mein circulate karta hai, isliye dono sides same sign rakhte hain. $v_1{r}_1=v_2{r}_2\Rightarrow v_2=\frac{54\times 0.6}{5.4}=6\text{km/s}.$ Nau guna door → ek-nouwaan speed. Slow aur door, fast aur paas: numbers mein Kepler's 2nd law.

Torque $\vec{\tau }=\vec{r}\times \vec{F}$ hai, aur iski size $rF\sin \theta$ hai jahan $\theta$, $\vec{r}$ aur $\vec{F}$ ke beech ka angle hai.

• Vertical axis (figure mein red weight arrow): weight $\vec{F}=m\vec{g}$ seedha neeche, vertical axis ke parallel point karta hai. Us axis ke baare mein kisi bhi ghoomne mein $\vec{r}$ flat pata hai, aur $\vec{r}\times \vec{F}$ ka vertical component zero hota hai kyunki force ka axis ke around koi sideways push nahi hai. Dekho kaise red arrow gray axis arrow ke parallel chalti hai — parallel matlab $\sin \theta =0$ → zero vertical torque → vertical $L$ conserved (condition 3: component-wise conservation).
• Horizontal rim axis: ab wohi red weight arrow us axis se horizontally door sit karti hai (figure mein black $\vec{r}$) aur uske perpendicular point karti hai. Yahan $\theta =90^{\circ }$, $\sin \theta =1$, isliye $\vec{r}\times \vec{F}\ne 0$ → wahan torque hai → horizontal axis ke baare mein $L$ protected nahi hai. Conservation axis-by-axis hoti hai, kabhi automatically "har cheez ke baare mein" nahi.

Condition: jo friction unhe couple karti hai woh do-disc system ke liye internal hai; koi external axial torque nahi → $L$ conserved. Dono same direction mein spin karte hain (B rest par start karta hai), isliye saare terms positive hain. $I_A{\omega }_A+I_B{\omega }_B=(I_A+I_B){\omega }_f\Rightarrow {\omega }_f=\frac{2\times 9+0}{6}=3\text{rad/s}.$ $K{E}_i=\frac{1}{2}(2)(9{)}^2=81\text{J}$. $K{E}_f=\frac{1}{2}(6)(3{)}^2=27\text{J}$. Lost fraction $=\frac{81-27}{81}=\frac{54}{81}=\frac{2}{3}\approx 0.667$. Do-tihaayi friction heat mein gayi — phir bhi $L$ exactly conserved hai.

Condition (figure se padho): red tension arrow $\vec{T}$ exactly black position arrow $\vec{r}$ ke along lie karta hai (dono central hole ki taraf point karte hain). Parallel arrows → $\sin \theta =0$ → $\vec{\tau }=\vec{r}\times \vec{T}=0$ → $L=mvr$ conserved. $v_1{r}_1=v_2{r}_2\Rightarrow v_2=\frac{3\times 0.8}{0.4}=6\text{m/s}.$ Pull radial hai isliye koi torque exert nahi karta, lekin jaise puck inward spiral karta hai woh hole ki taraf move karta hai, yaani tension ke along, isliye tension work karta hai. Toolkit se work–energy theorem $W=\Delta KE$ use karte hue: $W=K{E}_2-K{E}_1=\frac{1}{2}(0.5)(6^2-3^2)=0.25\times 27=6.75\text{J}.$ $L$ constant, tumhare pulling se energy badh gayi — skater jaisa hi lesson. (Figure mein notice karo ki $\vec{v}$ ko $\vec{r}$ ke perpendicular draw kiya gaya hai: yahi perpendicularity hai jo $L=mvr$ ko full $v$ ke saath banati hai.)

Origin choice: axis ko hinge par rakho. Hinge impact ke dauran ek bada unknown reaction force exert karta hai, lekin woh force hinge par act karta hai jahan $\vec{r}=0$ hai, isliye uska torque $\vec{r}\times \vec{F}=0\times \vec{F}=0$ hai — hinge ke baare mein zero torque. Gravity bahut chhote collision time par act karti hai → negligible impulse-torque. Isliye $L$ hinge ke baare mein conserved hai (parent note se "smart origin choose karo" rule). Take the bullet's incoming swing sense as $+$. Bullet ka incoming $L=mvr=0.01\times 400\times 0.9=3.6{\text{kg⋅m}}^2/\text{s}$ (positive). Embed hone ke baad, total $I=I_{rod}+m{r}^2=1.6+0.01\times 0.9^2=1.6081{\text{kg⋅m}}^2$. $\omega =\frac{L}{I}=\frac{3.6}{1.6081}\approx 2.239\text{rad/s}.$ Positive $\omega$ → rod us hi direction mein swing karti hai jis direction mein bullet ja rahi thi, as expected.

$K{E}_i=\frac{1}{2}m{v}^2=\frac{1}{2}(0.01)(400{)}^2=800\text{J}$. $K{E}_f=\frac{1}{2}I{\omega }^2=\frac{1}{2}(1.6081)(2.239{)}^2\approx 4.03\text{J}$.

• Linear momentum: conserved nahi — hinge ek external horizontal force impulse deliver karta hai. (Dekho Conservation of linear momentum kyun hinge ise toot deta hai.)
• Angular momentum about hinge: conserved — hinge force ka wahan zero moment arm hai.
• Kinetic energy: conserved nahi — embedding inelastic hai, $\approx 796\text{J}$ became heat. Yahi is topic ka poora point hai: teen conservation questions ke teen independent answers hain, jo teen alag conditions se decide hote hain.

Inelastic spin collision — kaun si quantity survive karti hai? ::: Angular momentum ($L_i=L_f$); kinetic energy lost ho jaati hai. Radial/central force torque aur energy ke saath kya karta hai? ::: Zero torque (isliye $L$ constant) lekin woh phir bhi work kar sakta hai (energy $W=\Delta KE$ se change hoti hai). Impact ke dauran hinge/pivot par axis kyun choose karte hain? ::: Wahan reaction force ka zero moment arm hota hai, isliye zero torque, isliye us axis ke baare mein $L$ conserved hota hai. Skater/puck speed up hoti hai — kya energy conserved hai? ::: Nahi; muscles/tumhare pull se work hota hai. $L$ conserved hai, energy nahi. Cross product $\vec{r}\times \vec{F}$ kab zero hota hai? ::: Jab $\vec{r}$ aur $\vec{F}$ parallel hon ($\sin \theta =0$), yaani central force ya axis par apply hone wali force. Same axis par do angular momenta kaise add karte hain? ::: Signed numbers ki tarah (right-hand rule): same sense adds, opposite sense subtracts.