Exercises — Moment of inertia of - rod (about end, centre), disk, ring, sphere (solid, hollow), cylinder

KAUNSI body, KAUNSA axis? Ek ring apne central perpendicular axis ke baare mein. Saara mass axis se ek hi perpendicular distance $R$ par baitha hai, isliye sabke liye $r=R$ hai — yahi exactly woh case hai $I_{\text{ring}}=M{R}^2$. $I=M{R}^2=(2)(0.5{)}^2=2\times 0.25=0.5\mathrm{kg}{\mathrm{m}}^2$

Dono spheres mein se kaun sa? Solid, isliye "Full Five" → $I=\frac{2}{5}M{R}^2$. $I=\frac{2}{5}(3)(0.2{)}^2=0.4\times 3\times 0.04=0.048\mathrm{kg}{\mathrm{m}}^2$

End axis → $I=\frac{1}{3}M{L}^2$ (mass centre se zyada, end se average mein zyada door baitha hai). $I=\frac{1}{3}(1.2)(1{)}^2=0.4\mathrm{kg}{\mathrm{m}}^2$

Step 1 — centre value. $I_{\text{cm}}=\frac{1}{12}M{L}^2=\frac{1}{12}(1.2)(1{)}^2=0.1\mathrm{kg}{\mathrm{m}}^2.$ Step 2 — theorem se shift karo. Parallel-axis theorem kehta hai $I=I_{\text{cm}}+M{d}^2$, jahan $d$ do parallel axes ke beech ka gap hai. Centre se end tak $d=L/2=0.5\mathrm{m}$ hai. $I_{\text{end}}=0.1+(1.2)(0.5{)}^2=0.1+0.3=0.4\mathrm{kg}{\mathrm{m}}^2$ Yeh Solution 1.3 se exactly match karta hai. ✓ Theorem $M{d}^2$ add karta hai kyunki centre of mass se axis door karna hamesha body ko spin karna mushkil banata hai.

Step 1 — $I$. Central axis ke baare mein disk: $I=\frac{1}{2}M{R}^2=\frac{1}{2}(0.8)(0.3{)}^2=0.036\mathrm{kg}{\mathrm{m}}^2.$ Step 2 — energy. Rotational KE, $\frac{1}{2}m{v}^2$ jaisi hai lekin $m$ ki jagah $I$ aur $v$ ki jagah $\omega$ aata hai: $KE=\frac{1}{2}I{\omega }^2=\frac{1}{2}(0.036)(10{)}^2=0.5\times 0.036\times 100=1.8\mathrm{J}$

Theorem. Kisi bhi flat (planar) body ke liye, $I_z=I_x+I_y$, jahan $z$ perpendicular axis hai aur $x,y$ do in-plane axes hain ek hi point se. Figure dekhte hain: dono diameters $x$ aur $y$ symmetry se interchangeable hain, isliye $I_x=I_y\equiv I_d$.

Solve karo. Hum jaante hain $I_z=\frac{1}{2}M{R}^2$ (central perpendicular axis). Tab $I_z=I_x+I_y=2I_d\Rightarrow I_d=\frac{1}{2}{I}_z=\frac{1}{4}M{R}^2$ Toh diameter ke baare mein disk ka $I$ $\boxed{\frac{1}{4}M{R}^2}$ hai — perpendicular-axis value ka half.

Additivity. Moment of inertia add hota hai: total $I$ = rod ka $I$ + har point mass ka $I$, sabhi ek hi axis ke baare mein.

• Rod: $I_{\text{rod}}=\frac{1}{12}M{L}^2=\frac{1}{12}(1)(1.2{)}^2=\frac{1.44}{12}=0.12\mathrm{kg}{\mathrm{m}}^2.$
• Har point mass $r=L/2=0.6\mathrm{m}$ par baitha hai: contribution $m{r}^2=0.5(0.6{)}^2=0.18\mathrm{kg}{\mathrm{m}}^2$. Dono milake: $0.36$. $I_{\text{total}}=0.12+0.36=0.48\mathrm{kg}{\mathrm{m}}^2$

Formulas compare karo. Hollow $=\frac{2}{3}M{R}^2$, solid $=\frac{2}{5}M{R}^2$. Kyunki $\frac{2}{3}>\frac{2}{5}$, hollow shell spin karna mushkil hai — uska saara mass outer radius $R$ par baitha hai, jo maximum possible average $r^2$ deta hai; solid sphere apna zyaadatar mass centre ke paas chhupaata hai jahan $r<R$ hai. Percentage. $\frac{I_{\text{hollow}}}{I_{\text{solid}}}=\frac{2/3}{2/5}=\frac{5}{3}\approx 1.667$ Toh hollow sphere ka $I$, $\frac{5}{3}-1=\frac{2}{3}\approx 66.7\%$ zyada hai.

Definition. Radius of gyration $k$ woh single distance hai jahan aap saara mass rakh sakte ho aur real $I$ reproduce kar sako: $I=M{k}^2\Rightarrow k=\sqrt{I/M}$.

• Cylinder: $M{k}^2=\frac{1}{2}M{R}^2\Rightarrow k^2=\frac{1}{2}{R}^2\Rightarrow k/R=\frac{1}{\sqrt{2}}\approx \mathrm{0.707.}$
• Ring: $M{k}^2=M{R}^2\Rightarrow k/R=1.$ Ring ka mass effectively sab rim par hai ($k=R$); cylinder ka andar ki taraf ghusa hua hai ($k\approx 0.71R$), isliye woh roll-down race jeet jaata hai.

Direct integration. Ring ko $xy$-plane mein rakho, radius $R$, aur $x$-axis ko apna rotation axis chunno. Angle $\varphi$ par ek mass element position $(R\cos \varphi ,R\sin \varphi )$ par hai. $x$-axis tak uski perpendicular distance $y$-coordinate hai: $r=R\sin \varphi$.

$dm=\frac{M}{2\pi }d\varphi$ ke saath (mass poore circle mein evenly spread hai): $I_d={\int }_0^{2\pi }(R\sin \varphi {)}^2\frac{M}{2\pi }d\varphi =\frac{M{R}^2}{2\pi }{\int }_0^{2\pi }{\sin }^2\varphi d\varphi$ ${\int }_0^{2\pi }{\sin }^2\varphi d\varphi =\pi$ use karke: $I_d=\frac{M{R}^2}{2\pi }\cdot \pi =\frac{1}{2}M{R}^2$ Perpendicular-axis theorem se check karo. Flat ring ke liye, $I_z=I_x+I_y=2I_d$. Kyunki $I_z=M{R}^2$, hum paate hain $I_d=\frac{1}{2}M{R}^2$. ✓ Dono agree karte hain.

Step 1 — central value. Disk, central perpendicular axis: $I_{\text{cm}}=\frac{1}{2}M{R}^2=\frac{1}{2}(1)(0.2{)}^2=0.02\mathrm{kg}{\mathrm{m}}^2.$ Step 2 — rim tak shift karo. Tangent axis central axis ke parallel hai, $d=R=0.2\mathrm{m}$ offset ke saath. Parallel-axis theorem: $I=I_{\text{cm}}+M{d}^2=0.02+(1)(0.2{)}^2=0.02+0.04=0.06\mathrm{kg}{\mathrm{m}}^2$ Equivalently $I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$ — yaad rakhne layak ek neat closed form.

Insight. Energy $=\frac{1}{2}I{\omega }^2$ hai, aur $M,R,\omega$ fixed hain, toh winner woh hai jiska $I$ zyada hai. Ring har gram ko $R$ par rakhta hai (maximum $r^2$), isliye woh zaroor jeetega.

• Disk: $I=\frac{1}{2}M{R}^2=\frac{1}{2}(4)(0.25{)}^2=0.125\mathrm{kg}{\mathrm{m}}^2.$ $E_{\text{disk}}=\frac{1}{2}(0.125)(50{)}^2=\frac{1}{2}(0.125)(2500)=156.25\mathrm{J}$
• Ring: $I=M{R}^2=(4)(0.25{)}^2=0.25\mathrm{kg}{\mathrm{m}}^2.$ $E_{\text{ring}}=\frac{1}{2}(0.25)(2500)=312.5\mathrm{J}$ Ring exactly double energy store karta hai. Design note: real flywheels mass ko rim par push karte hain, lekin material strength limit karti hai ki high $\omega$ par kitna rim stress survive kar sako — yeh pure $I$ se aage ek genuine engineering trade-off hai.

Rod part (apne end axis ke baare mein): $I_{\text{rod}}=\frac{1}{3}{M}_r{L}^2=\frac{1}{3}(1)(0.6{)}^2=\frac{0.36}{3}=0.12\mathrm{kg}{\mathrm{m}}^2.$ Sphere part. Sphere ek door axis ke baare mein spin kar raha hai, isliye parallel-axis use karo: apna $I_{\text{cm}}=\frac{2}{5}{M}_s{R}^2$ plus $M_s{d}^2$ jahan $d=L=0.6\mathrm{m}$. $I_{\text{cm}}=\frac{2}{5}(2)(0.1{)}^2=0.4\times 2\times 0.01=0.008\mathrm{kg}{\mathrm{m}}^2$ $M_s{d}^2=(2)(0.6{)}^2=0.72\mathrm{kg}{\mathrm{m}}^2\Rightarrow I_{\text{sphere}}=0.008+0.72=0.728$ Total. $I_{\text{total}}=0.12+0.728=0.848\mathrm{kg}{\mathrm{m}}^2$ Dhyan do ki sphere ka khud ka spin term ($0.008$) $M_s{d}^2$ transport term ($0.72$) ke aage kitna chhota hai — jab ek compact body axis se bahut door ho, woh almost ek point mass ki tarah behave karta hai.

Parallel-axis hamesha $I$ badhata hai ::: kyunki $M{d}^2\ge 0$; CM se door jaana kabhi help nahi karta. Perpendicular-axis sirf kaam karta hai ::: thin flat (planar) bodies ke liye, kyunki ise $z=0$ chahiye. Ring vs disk flywheel energy ratio ::: $2:1$ (ring double store karta hai, saara mass $R$ par). Radius of gyration $k$ ::: woh distance jahan saara mass rakh ke same $I$ milta: $I=M{k}^2$.