Exercises — Moment of inertia I = Σmᵢrᵢ² — concept

Axis kya hai: $y$-axis woh vertical line hai jahan $x=0$ hota hai. Figure dekho

Mass axis par hi hai, isliye uski perpendicular distance $r=0$ hai. $I=m{r}^2=4\times 0^2=0{\text{kg m}}^2.$ Jo mass axis par hi baith jaati hai woh kuch bhi contribute nahi karti — tum use us line ke baare mein circle mein nahi ghuma sakte jis par woh already baithi hai.

Har mass ko uski squared distance se multiply karke sum karo: $I=1(0.2{)}^2+2(0.4{)}^2+3(0.6{)}^2=0.04+0.32+1.08=1.44{\text{kg m}}^2.$ Dhyan do ki $m_3$ (sabse door) $1.08$ contribute karta hai — total ka teen-chautha — kyunki uski distance square hoti hai.

Geometry kya hai: figure dekho

Kyun hum $\frac{1}{2}I{v}^2$ seedha nahi likh sakte: har corner mass apne circle mein move karta hai, isliye chaaon particles ki ordinary kinetic energy $\tfrac12 m v^2$ add karte hain aur dekhte hain $I$ kaise aata hai. Step 1 — har corner ki speed. Rigid body mein har particle ek hi $\omega$ share karta hai, aur angular velocity se har ek ko speed milti hai $v_i=\omega r_i$. Yahan chaaon ki $r_i=r=0.7071\text{m}$ hai, isliye har ek $v=\omega r=3\times 0.7071=2.121\text{m/s}$ se move karta hai. Step 2 — linear KE sum karo, phir factor karo. $K{E}_{\text{rot}}={\sum }_i\frac{1}{2}{m}_i{v}_i^2={\sum }_i\frac{1}{2}{m}_i(\omega r_i{)}^2=\frac{1}{2}(\underset{=I}{\underbrace{\sum _i{m}_i{r}_i^2}}){\omega }^2=\frac{1}{2}I{\omega }^2.$ Bracket bilkul exactly moment of inertia hai — yahi wajah hai ki $\frac{1}{2}\sum m(\omega r{)}^2$ collapse hokar $\frac{1}{2}I{\omega }^2$ ban jaata hai: $\omega$ sab mein common hai, isliye factor out ho jaata hai aur pure geometry sum $I$ bach jaata hai. Step 3 — plug in karo. $K{E}_{\text{rot}}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 4\times 3^2=18\text{J}.$

(a) $m_1$ se guzarne wala axis: toh $r_1=0$, $r_2=L=2\text{m}$. $I_a=3(0{)}^2+5(2{)}^2=20{\text{kg m}}^2.$ (b) Midpoint se guzarne wala axis: har mass axis se $L/2=1\text{m}$ door hai. $I_b=3(1{)}^2+5(1{)}^2=3+5=8{\text{kg m}}^2.$ (c) $I_a=20>I_b=8$. End axis jeet jaata hai kyunki woh poore bhaari $5\text{kg}$ mass ko full distance $2\text{m}$ (squared to $4$) par force karta hai, jabki midpoint sab ko sirf $1\text{m}$ par rakhta hai. Distance dominate karta hai yahan: same masses, sirf axis ke relative position badlne se $I$ $2.5$ ke factor se badal jaata hai.

$I$ tab minimize hota hai jab axis centre of mass se guzre, kyunki woh balance point hai jo mass-weighted spread ko sabse chhota rakhta hai. Ise locate karo (distance $x$ $m_1$ se $2\text{m}$ rod ke along): $x=\frac{m_{2}L}{m_1+m_2}=\frac{5\times 2}{3+5}=1.25\text{m}.$ Tab $r_1=1.25$, $r_2=2-1.25=0.75$: $I_{\min }=3(1.25{)}^2+5(0.75{)}^2=4.6875+2.8125=7.5{\text{kg m}}^2.$ Waakai $7.5<8<20$ — centre-of-mass axis midpoint aur end dono ko beat karta hai. Bhaari mass is axis ke karib hai (sirf $0.75\text{m}$), jise $r^2$ weighting reward karti hai.

Theorem geometrically kya kehta hai: figure dekho

(a) $L$ ka conservation: $I_1{\omega }_1=I_2{\omega }_2$, isliye ${\omega }_2=\frac{I_1{\omega }_1}{I_2}=\frac{6\times 2}{2}=6\text{rad/s}.$ $I$ ko 3 ke factor se shrink karne par woh 3 ke factor se speed up hoti hai. (b) Nayi energy: $K{E}_2=\frac{1}{2}{I}_2{\omega }_2^2=\frac{1}{2}\times 2\times 6^2=36\text{J}.$ Purani energy: $K{E}_1=\frac{1}{2}\times 6\times 2^2=12\text{J}$. Energy teen guna ho gayi. Kahan se aayi? Uski muscles ne arms ko andar kheenchne mein outward (centrifugal) tendency ke against kaam kiya. Momentum conserved hai (koi external torque nahi), lekin energy nahi — skater actively $24\text{J}$ add karti hai.

(a) Chhhon masses $r=R=0.4\text{m}$ share karte hain: $I=6m{R}^2=6\times 0.5\times 0.4^2=6\times 0.5\times 0.16=0.48{\text{kg m}}^2.$ (b) Rotation ka Newton's second law, $\tau =I\alpha$: $\alpha =\frac{\tau }{I}=\frac{1.2}{0.48}=2.5{\text{rad/s}}^2.$ (c) Rest se constant $\alpha$ ke saath: $\omega =\alpha t=2.5\times 5=12.5\text{rad/s}.$ (d) Kinematics se energy: $KE=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 0.48\times 12.5^2=0.24\times 156.25=37.5\text{J}.$ Torque dwara kiye kaam se energy, jahan $\theta =\frac{1}{2}\alpha t^2=\frac{1}{2}\times 2.5\times 25=31.25\text{rad}$: $W=\tau \theta =1.2\times 31.25=37.5\text{J}.✓$ Dono routes $37.5\text{J}$ dete hain — rotation ka work–energy theorem internally consistent hai.

Chain mein peeche se kaam karo. Pehle $KE=\frac{1}{2}I{\omega }^2$ se required $I$: $I=\frac{2KE}{{\omega }^2}=\frac{2\times 25}{10^2}=0.5{\text{kg m}}^2.$ Do masses mein se har ek $m{r}^2$ contribute karta hai, isliye $I=2m{r}^2$: $0.5=2\times 1\times r^2\Rightarrow r^2=0.25\Rightarrow r=0.5\text{m}.$ Har mass ko axis se $0.5\text{m}$ door rakho.

$\sum m_i{r}_i^2$ se continuous integral $\int r^{2}dm$ tak Yahan har exercise ne discrete point masses use kiye, isliye $I={\sum }_i{m}_i{r}_i^2$ ek finite sum tha. Solid object ka kya — ek rod, ek disk, ek real skater — jahan mass continuously phaili ho? Bridge: object ko bahut saare tiny chunks mein kaato. Maano $i$-th chunk ka ek tiny mass $\Delta m_i$ hai perpendicular distance $r_i$ par. Tab $I={\sum }_i{r}_i^2\Delta m_i.$ Ab chunks ko chhhota aur zyada numerous banao bina limit ke. Sum ek integral ban jaata hai — woh tool jo yeh jawab deta hai "infinitely many infinitesimal pieces ke sum ka limit kya hai?": $\boxed{I=\int r^{2}dm.}$ Integral kyun aur bada sum nahi? Kyunki ek continuous body mein uncountably many points hain; koi finite list of $m_i$ use describe nahi kar sakti. Integral $\int r^{2}dm$ us hi recipe ka exact continuous version hai — still "mass ka har tukda times uski distance-squared, sab add up." Hamare L2.2 square par sanity check: chaar discrete corners ne sum finite rakha ($I=4{\text{kg m}}^2$). Usi mass wala ek real solid square plate $\int r^{2}dm$ use karega, aur uski mass, average par centre ke karib phaili hogi, isliye ek chhota $I$ degi — parent note se wahi "far mass dominates" logic, ab continuous form mein.