Parallel axis theorem — I = I_CM + Md² — proof: Exercises

1.5.6 · D4 · Physics › Rotational Mechanics › Parallel axis theorem — I = I_CM + Md² — proof

Level 1 — Recognition

Goal: situation padho aur values seedha $I = I_{C M} + M d^{2}$ mein daal do.

L1.1

Ek thin rod jiska mass $M$ aur length $L$ hai, apne midpoint se guzarne wale axis ke baare mein spin karta hai, rod ke perpendicular. $I$ likho. Kya theorem ki zarurat bhi hai?

Recall Solution

Axis kya hai: yeh CM se guzarti hai (uniform rod ka midpoint). Theorem kyun nahi: agar axis already CM se guzarti hai, toh $d = 0$ , isliye $M d^{2} = 0$ . $I = I_{C M} + M (0)^{2} = \frac{1}{12} M L^{2}$ Theorem ki zarurat nahi — lekin phir bhi agree karta hai. Yeh sanity check hai: CM par correction vanish ho jaata hai.

L1.2

Ek solid disc jiska mass $M$ , radius $R$ hai, disc ke perpendicular ek axis ke baare mein spin karta hai jo uske rim par ek point se guzarti hai. $I$ nikalo.

Recall Solution

$d$ identify karo: rim centre se perpendicular distance $R$ par hai, isliye $d = R$ . $I = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}$

L1.3

Ek thin ring (hoop) jiska mass $M$ , radius $R$ hai, apne plane ke perpendicular ek axis ke baare mein spin karta hai jo ring par hi ek point se guzarti hai. $I$ nikalo.

Recall Solution

$d = R$ fir se (ring ka edge uske centre se $R$ door hai). $I = M R^{2} + M R^{2} = 2 M R^{2}$

Level 2 — Application

Goal: $d$ khud compute karo, phir theorem apply karo.

L2.1

Ek thin rod, mass $M$ , length $L$ , rod ke perpendicular ek axis ke baare mein spin karta hai jo ek end se one-quarter of the way par ek point se guzarti hai. $I$ nikalo.

Figure — Parallel axis theorem — I = I_CM + Md² — proof

Recall Solution

$d$ nikalo (figure dekho). CM midpoint par hai, kisi bhi end se $L /2$ door. Nayi axis end se $L /4$ par hai. Toh axis aur CM ke beech gap hai: $d = \frac{L}{2} - \frac{L}{4} = \frac{L}{4}$ Subtract kyun karein: $d$ CM-se-nayi-axis distance hai, aur dono points ek hi end se measure kiye gaye hain. $I = \frac{1}{12} M L^{2} + M (\frac{L}{4})^{2} = \frac{1}{12} M L^{2} + \frac{1}{16} M L^{2}$ Common denominator $48$ : $\frac{4}{48} M L^{2} + \frac{3}{48} M L^{2} = \frac{7}{48} M L^{2}$ . $I = \frac{7}{48} M L^{2}$

L2.2

Ek solid sphere jiska mass $M$ , radius $R$ hai, ek axis ke baare mein spin karta hai jo uski surface ke tangent hai (bahar se sirf touch karti hai). $I$ nikalo.

Recall Solution

Tangent line surface ko touch karti hai, isliye yeh centre se exactly ek radius door hai: $d = R$ . $I = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2}$

L2.3

Ek thin rectangular plate, mass $M$ , sides $a$ aur $b$ , plate ke perpendicular ek axis ke baare mein spin karta hai jo ek corner se guzarti hai. $I$ nikalo.

Recall Solution

$d$ nikalo. CM rectangle ke centre par hai. Ek corner is distance par hai: $d = (\frac{a}{2})^{2} + (\frac{b}{2})^{2} = \frac{1}{2} a^{2} + b^{2}$ Square root kyun: corner dono $x$ aur $y$ mein offset hai, aur straight-line distance hypotenuse hoti hai — yeh parallel axis ke liye perpendicular distance hai. $I = \frac{1}{12} M (a^{2} + b^{2}) + M \cdot \frac{1}{4} (a^{2} + b^{2}) = (\frac{1}{12} + \frac{3}{12}) M (a^{2} + b^{2}) = \frac{1}{3} M (a^{2} + b^{2})$

Level 3 — Analysis

Goal: sirf plug in nahi, theorem ki structure ke baare mein reason karo.

L3.1

Do parallel axes, A aur B, ek body ke CM ke opposite sides par hain, ek hi line mein. Axis A, CM se $d_{A} = 3 R$ door hai; axis B, CM se $d_{B} = 2 R$ door hai, doosri taraf. Body ek solid sphere hai, mass $M$ , radius $R$ . $I_{A} - I_{B}$ nikalo.

Recall Solution

KEY insight (parent ka Example 3 trap): tum A aur B ki separation directly use nahi kar sakte. Har ek ko CM se route karo. $I_{A} = \frac{2}{5} M R^{2} + M (3 R)^{2} = \frac{2}{5} M R^{2} + 9 M R^{2}$ $I_{B} = \frac{2}{5} M R^{2} + M (2 R)^{2} = \frac{2}{5} M R^{2} + 4 M R^{2}$ Subtraction par $I_{C M}$ terms cancel ho jaate hain: $I_{A} - I_{B} = 9 M R^{2} - 4 M R^{2} = 5 M R^{2}$ Side matter kyun nahi karta: $d$ , $d^{2}$ ke roop mein aata hai, isliye CM ke "left" ya "right" hona irrelevant hai — sirf distance ki magnitude count karti hai.

L3.2

Ek body ka moment of inertia uske CM se $2 m$ door ek axis ke baare mein $I_{1}$ hai, aur CM se $5 m$ door parallel axis ke baare mein $I_{2}$ hai. Diya hua hai $I_{2} - I_{1} = 84 kg\cdotm^{2}$ , mass $M$ nikalo.

Recall Solution

CM se dono setup karo: $I_{1} = I_{C M} + M (2)^{2}, I_{2} = I_{C M} + M (5)^{2}$ Subtract karo unknown $I_{C M}$ ko khatam karne ke liye: $I_{2} - I_{1} = M (25 - 4) = 21 M$ Subtract kyun karein: hum $I_{C M}$ nahi jaante aur zarurat bhi nahi — differencing ise eliminate kar deta hai, bilkul wahi algebra jo proof rely karta hai. $21 M = 84 \Rightarrow M = 4 kg$

L3.3

Dikhao ki ek fixed direction ke PARALLEL SAARE axes mein, CM se guzarne wala axis minimum moment of inertia deta hai, aur koi bhi doosra usse kitna zyada deta hai.

Recall Solution

Kisi bhi parallel axis ke liye $I (d) = I_{C M} + M d^{2}$ . Ise $d$ ki function ki tarah padho: $M d^{2} \geq 0$ hamesha (mass $M > 0$ , square $\geq 0$ ), aur yeh $0$ tabhi hota hai jab $d = 0$ . Isliye $I (d) \geq I_{C M}$ har $d$ ke liye, equality sirf CM par hai. Toh CM-axis unique minimum hai, aur koi bhi doosra parallel axis isse exactly $M d^{2}$ zyada hai. Dekho Radius of gyration: yeh minimum $I_{C M}$ us direction ke liye sabse chhhota possible gyration radius fix karta hai.

Level 4 — Synthesis

Goal: parallel axis theorem ko kisi doosre idea ke saath combine karo.

L4.1 (perpendicular axis theorem ke saath)

Ek thin uniform disc, mass $M$ , radius $R$ , ek plane mein pada hai. Rim point se guzarne wale diameter ke baare mein $I$ nikalo — yaani ek in-plane axis jo ek tangent-jaisi diameter direction mein hai lekin rim par ek point se guzarti hai. Practically: pehle centre se guzarne wale in-plane axis ke baare mein $I$ nikalo, phir rim par parallel in-plane axis ke baare mein.

Recall Solution

Step 1 — Perpendicular axis theorem se in-plane central axis. Flat disc ke liye, $I_{z} = I_{x} + I_{y}$ jahan $z$ disc ke perpendicular hai aur $x, y$ do in-plane diameters hain. Symmetry se $I_{x} = I_{y}$ , aur $I_{z} = \frac{1}{2} M R^{2}$ . Toh: $2 I_{x} = \frac{1}{2} M R^{2} \Rightarrow I_{x} = I_{diameter} = \frac{1}{4} M R^{2}$ Step 2 — rim par shift (parallel axis). Rim axis is diameter ke parallel hai, centre se $d = R$ door: $I = \frac{1}{4} M R^{2} + M R^{2} = \frac{5}{4} M R^{2}$ Dono tools kyun: perpendicular axis ne hume central diameter inertia diya (directly integrate karna mushkil hai); parallel axis ne phir use rim par move kiya. Ek theorem $I_{C M}$ banata hai, doosra use relocate karta hai.

L4.2 (rotational kinetic energy ke saath)

L4.1 ki disc (mass $M$ , radius $R$ ) rim axis perpendicular to disc ( $I = \frac{3}{2} M R^{2}$ L1.2 se) ke baare mein angular speed $ω$ par spin kari. Uski Rotational kinetic energy likho, aur ise central perpendicular axis ke baare mein usi $ω$ par spin karne se compare karo.

Recall Solution

Rotational kinetic energy hai $K E = \frac{1}{2} I ω^{2}$ . Rim axis: $K E_{rim} = \frac{1}{2} (\frac{3}{2} M R^{2}) ω^{2} = \frac{3}{4} M R^{2} ω^{2}$ . Central axis: $K E_{cm} = \frac{1}{2} (\frac{1}{2} M R^{2}) ω^{2} = \frac{1}{4} M R^{2} ω^{2}$ . Ratio: $K E_{rim} / K E_{cm} = 3$ . Rim ke baare mein spin karne mein usi $ω$ par teen guna energy lagti hai — extra $M d^{2} = M R^{2}$ directly extra kinetic energy ke roop mein dikhta hai. Yeh parent ka "bahar jaane ke liye $M d^{2}$ pay karo" physical roop mein hai.

Level 5 — Mastery

Goal: multi-step chain banao aur har link justify karo.

L5.1 (physical pendulum period)

Ek uniform rod, mass $M$ , length $L$ , ek end par pivoted hai aur gravity $g$ ke under Physical pendulum ki tarah swing karta hai. Small-oscillation period hai: $T = 2 π \frac{I _{pivot}}{M g d _{cm}}$ jahan $I_{pivot}$ pivot ke baare mein inertia hai aur $d_{cm}$ pivot-to-CM distance hai. $L = 1.2 m$ , $g = 9.8 m/s^{2}$ ke liye $T$ nikalo.

Recall Solution

Step 1 — parallel axis theorem se $I_{pivot}$ nikalo. Pivot end par hai, CM middle mein, $d = L /2$ : $I_{pivot} = \frac{1}{12} M L^{2} + M (\frac{L}{2})^{2} = \frac{1}{3} M L^{2}$ Step 2 — $d_{cm}$ identify karo. CM pivot ke $L /2$ neeche hang karta hai, isliye $d_{cm} = L /2$ . Step 3 — substitute karo. Mass $M$ cancel ho jaata hai (numerator aur denominator dono mein aata hai): $T = 2 π \frac{\frac{1}{3} M L ^{2}}{M g \cdot \frac{L}{2}} = 2 π \frac{2 L}{3 g}$ $M$ kyun cancel hota hai: inertia mass ke saath badhti hai aur restoring torque bhi; yeh trade off karte hain, isliye period sirf geometry par depend karta hai. Numbers: $T = 2 π \frac{2 ( 1.2 )}{3 ( 9.8 )} = 2 π 0.081633 \approx 2 π (0.28572) \approx 1.795 s$ . $T \approx 1.80 s$

L5.2 (composite body — dumbbell ki inertia)

Ek "dumbbell": ek thin rod jiska mass $m_{r}$ aur length $L$ hai, aur har end par ek point mass $m_{p}$ fixed hai. Rod ke centre se guzarne wale perpendicular axis ke baare mein $I$ nikalo, phir ek end se guzarne wale axis ke baare mein, sab rod ke perpendicular. $m_{r} = 2 kg$ , $m_{p} = 1 kg$ , $L = 2 m$ use karo.

Recall Solution

Inertia ki additivity: total $I$ = har piece ka $I$ usi axis ke baare mein, sab ka sum. Centre axis.

Rod apne centre ke baare mein: $\frac{1}{12} m_{r} L^{2}$ .
Har end mass $L /2$ distance par: $m_{p} (L /2)^{2}$ ; dono milake: $2 m_{p} (L /2)^{2} = \frac{1}{2} m_{p} L^{2}$ . $I_{centre} = \frac{1}{12} m_{r} L^{2} + \frac{1}{2} m_{p} L^{2}$ Numbers: $\frac{1}{12} (2) (4) + \frac{1}{2} (1) (4) = 0.6667 + 2 = 2.6667 kg\cdotm^{2}$ . End axis. Ab poore system ko shift karo. Pehle total mass aur CM nikalo.
Total mass $M = m_{r} + 2 m_{p} = 2 + 2 = 4 kg$ .
Symmetry se composite CM abhi bhi rod ka centre hai. Toh end axis poore dumbbell ke CM se $d = L /2 = 1 m$ door hai.
Parallel axis theorem total mass use karta hai: $I_{end} = I_{centre} + M (L /2)^{2}$ . $I_{end} = 2.6667 + 4 (1)^{2} = 6.6667 kg\cdotm^{2}$ Total mass kyun, koi piece nahi: theorem ka $M$ woh mass hai jo sab kuch nayi axis par move ho raha hai; poora rigid dumbbell saath shift hota hai.