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

$I$ chosen axis par depend karta hai: same rod apne centre ke baare mein $\frac{1}{12}M{L}^2$ hai lekin end ke baare mein $\frac{1}{3}M{L}^2$ hai. Mass ek number hoti hai; $I$ ek number hota hai per axis.

$I$ is baat par depend karta hai ki mass kahan hai, sirf kitni hai nahi. Ek ring ($M{R}^2$) same $M,R$ wali disk ($\frac{1}{2}M{R}^2$) se zyada hoti hai kyunki uski mass rim tak push out hoti hai.

$I=\int r^{2}dm$ squared distances ko positive mass elements se multiply karke sum karta hai, isliye har term $\ge 0$ hai. Yeh zero tabhi hoga jab saari mass bilkul axis par ho.

$r^2$ ki wajah se, $r$ double karne par us element ka contribution chaar guna ho jaata hai. Isliye "door wali mass itni mehengi padti hai."

$M$ aur $R$ wale ek solid cylinder aur ek thin disk ka long axis ke baare mein same $I$ hota hai. ::: True. Dono $\frac{1}{2}M{R}^2$ hote hain. Ek cylinder bas disks ki stack hai, aur length puri tarah cancel ho jaati hai.

$I=\frac{2}{3}M{R}^2$ hota hai. ::: False. Shell cylinder ki saari mass radius $R$ par hoti hai, isliye woh $M{R}^2$ hai (ring ki tarah). Sirf hollow sphere $\frac{2}{3}M{R}^2$ hai, kyunki uski mass $r=0$ (poles) se $r=R$ (equator) tak range karti hai.

$M,R$), ring ka $I$ sabse bada hota hai. ::: True. Mean-square distance $⟨r^2⟩=I/(M{R}^2)$ equals $1$ (ring), $\frac{2}{3}$ (hollow sphere), $\frac{1}{2}$ (disk), $\frac{2}{5}$ (solid sphere). Kyunki koi bhi mass distance $R$ se zyada nahi ho sakti, sabse bada possible $⟨r^2⟩$ $1$ hai — ring, jo saari mass ko bilkul $R$ par rakhti hai. Figure 2 mein yeh mass-distributions side by side dikhaye gaye hain.

$M{R}^2$ hai, jo tab achieve hota hai jab saari mass sabse door allowed distance $R$ par ho — yahi ring/thin-shell-cylinder ka case hai.

$r$ har point se disk ke centre tak ki distance hai." ::: Galat. Diameter axis ke liye, $r$ us line tak perpendicular drop hai, isliye diameter par wale points ka $r=0$ hai. Distance-to-centre tabhi $r$ ke barabar hoti hai jab axis centre se hokar plane ke perpendicular guzrti ho.

$I_{\text{end}}=I_{\text{cm}}+M{d}^2$ mein $d=L$ lene se rod-about-end ki value milti hai." ::: Galat $d$ hai. Parallel-axis mein $d$ do parallel axes ke beech ka gap hai. Centre-of-mass axis rod ke midpoint par hoti hai, aur end axis tip par; ek uniform rod ka midpoint uski length ke beech mein hota hai, isliye gap $d=L/2$ hai. Isse milta hai $\frac{1}{12}M{L}^2+M(L/2{)}^2=\frac{1}{3}M{L}^2$.

$I$ zyada hona chahiye." ::: Galat. Long axis ke baare mein har slice ek identical disk hai jiska $\frac{1}{2}dm{R}^2$ hai; jab sum karo, sirf total mass aur $R$ bachte hain. Length irrelevant hai: $\frac{1}{2}M{R}^2$.

$I$ derive karne ke liye flat rings use nahi kar sakte — sirf spherical shells valid hain; rings use karna galti hai." ::: Dono valid hain. Flat rings mein slice karke $\int r^{2}dm$ integrate karne se sahi $\frac{2}{5}M{R}^2$ milta hai; bas isme do-variable (radius aur height) geometry setup chahiye. Spherical-shell route zyada quick hai kyunki har shell pehle se known $\frac{2}{3}{r}^{2}dm$ contribute karti hai — yeh shortcut hai, correctness ka issue nahi.

$I$ nikalne ke liye ek mushkil integral chahiye tha kyunki mass ek circle ke around spread hai." ::: Galat feeling. Ring ka har point same distance $R$ par hai, isliye $R^2$ seedha $\int dm$ se bahar aa jaata hai, aur bina kisi real integration ke $M{R}^2$ milta hai.

$I_{\text{diameter}}=2I_z$ hota hai." ::: Ulta hai. Flat body ke liye, $I_z=I_x+I_y=2I_{\text{diameter}}$, isliye $I_{\text{diameter}}=\frac{1}{2}{I}_z=\frac{1}{4}M{R}^2$, double nahi. Dekho Perpendicular Axis Theorem.

$r$ squared kyun appear karta hai, pehli power mein kyun nahi? ::: Ek particle ki speed $v=\omega r$ hai, aur uski kinetic energy $\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2{r}^2$ hai. $I=\sum m{r}^2$ define karne se rotational KE $\frac{1}{2}I{\omega }^2$ ban jaati hai, jo $\frac{1}{2}m{v}^2$ ko mirror karti hai. Dekho Rotational Kinetic Energy.

$\frac{2}{5}$ hollow sphere ke $\frac{2}{3}$ se chhota kyun hai? ::: Ek solid sphere mein zyaatar mass centre ke paas hoti hai ($r<R$), jo average $r^2$ ko kam karta hai. Ek hollow sphere saari mass surface par rakhti hai, jo average par zyada door hai.

half kyun hai? ::: Ek ring ki saari mass $R$ par hai; ek disk same mass ko $0\to R$ tak andar ki taraf spread karti hai, isliye average $r^2$ kam ho jaata hai. Exact integral ${\int }_0^R{r}^{3}dr$ factor $\frac{1}{2}$ produce karta hai.

$I$ (per unit $M{R}^2$) matlab zyada energy spin mein jaati hai rather than sliding mein, isliye object slower accelerate karta hai — ring solid sphere se harti hai. Dekho Rolling Motion.

$I$ ko "mass ka rotational analogue" kyun kaha jaata hai? ::: $\tau =I\alpha$ mein (dekho Torque and Angular Acceleration), $I$ bilkul wahi role play karta hai jo $m$ $F=ma$ mein karta hai: yeh angular acceleration ke against resistance measure karta hai.

par baitha ek point mass ka $I$ kya hoga? ::: Zero. Uski perpendicular distance $r=0$ hai, isliye $r^{2}dm=0$. Axis par koi mass kuch contribute nahi karta.

$I$ us axis ke baare mein kya hoga jo uski apni length ke along chalti ho? ::: Bilkul zero. Ek one-dimensional rod ka har point axis par hota hai, isliye har jagah $r=0$ hai aur $\int r^{2}dm=0$. Jaana-pehchana $\frac{1}{12}M{L}^2$ sirf us axis ke liye apply hota hai jo rod ke perpendicular ho.

$I$ kya approach karta hai? ::: Yeh $M{R}^2$ approach karta hai, ring ke identical — kyunki thin shell ki saari mass single radius $R$ par hoti hai.

$M$ fixed), toh $I$ ka kya hoga? ::: $I=M{R}^2\to 0$. Saari mass axis par collapse ho jaati hai, isliye rotation resist karne ke liye kuch nahi bachta — yeh axis par ek point ban jaata hai.

$\varphi$ banati ho. Jab $\varphi$ $0$ se $90^{\circ }$ tak jaata hai toh $I$ kaise badlta hai? ::: Sirf har element ka perpendicular part count karta hai, isliye rod ke along $x$ distance par ek element $r=x\sin \varphi$ par hota hai, jo deta hai $I(\varphi )=\frac{1}{12}M{L}^2{\sin }^2\varphi$. $\varphi =0$ par (axis rod ke along) $I=0$; $\varphi =90^{\circ }$ par (perpendicular) $I=\frac{1}{12}M{L}^2$. Figure 3 mein same centre-point axis teen angles se tilted dikhaya gaya hai — same $M$, same centre, alag $I$.

$I$ same hoga? ::: Zarori nahi. Parallel-axis (dekho Parallel Axis Theorem) sirf parallel axes compare karta hai; alag directions mass ko alag tarah se slice kar sakti hain aur alag $I$ de sakti hain (upar tilted-rod case dekho). Radius of gyration (dekho Radius of Gyration) yeh per axis capture karta hai.

$I$ se kam ho sakta hai? ::: Nahi, parallel axes ke liye. Parallel-axis kehta hai $I=I_{\text{cm}}+M{d}^2\ge I_{\text{cm}}$ kyunki $M{d}^2\ge 0$ hai. Centre-of-mass axis hamesha uske parallel axes mein minimum deta hai.