Class 11 Physics · Chapter 6
Short answer:
Class 11 Physics Chapter 6 — System of Particles and Rotational Motion ka matlab hai: ab tak jo bhi motion padha (Chapter 2 se Chapter 5 tak — motion in a straight line, motion in a plane, laws of motion, work-energy-power) sab mein body ko ek point particle maan liya gaya tha. Is chapter se body ka size matter karna shuru karta hai — extended rigid bodies, unka centre of mass (CM), torque, angular momentum, moment of inertia aur rotational equilibrium.
Rationalised 2026-27 syllabus mein ye Chapter 6 hai (Physical World standalone chapter hata diya gaya hai, isliye Units and Measurements se ek number peeche shift hua hai). Official NCERT textbook PDF (keph106.pdf, ncert.nic.in) directly verify kiya gaya hai is spec ke liye — is edition mein rolling motion ka poora section (purani book ka 6.13) hata diya gaya hai, aur parallel axis theorem / perpendicular axis theorem bhi syllabus se bahar hain. Table 6.1 mein moment of inertia ki values sirf di gayi hain, derive nahi karayi jaati — khud NCERT ki line hai "derivations of these expressions are beyond the scope of this textbook". Chapter mein 12 solved Examples aur 17 end-chapter Exercises hain — koi "Additional Exercises" section nahi (purani non-rationalised book mein tha).
Is chapter ke saare important formulas — CM, torque τ = r × F, angular momentum L = Iω, aur K = ½Iω² — neeche formula table mein diye hain.
Chapter 5 tak humne har body ko ek point mass ki tarah treat kiya — jaise ek car ko bhi ek dot maan ke uski motion nikaal li. Ye approximation kaafi jagah chal jaati hai, lekin ek spinning top, ek ghoomta hua ceiling fan, ya ek rolling cylinder ko sirf ek point maan ke nahi samajh sakte — inki motion mein translation ke saath-saath rotation bhi hota hai. Chapter 6 "System of Particles and Rotational Motion" yahi gap bharta hai: ye batata hai ki jab body ka size matter karta hai (rigid body — jiski shape fix rehti hai), to uski motion ko kaise describe karein.
Chapter do bade hisson mein bant jaata hai. Pehla hissa centre of mass (CM) ke around hai — kaise ek complicated system ki poori translational motion ek single point (CM) se capture ho jaati hai, chahe system ke andar kuch bhi ho raha ho (explosion, internal forces, whatever). Doosra hissa rotation ke around hai — vector product, angular velocity, torque, angular momentum, rigid-body equilibrium, moment of inertia, aur rotational kinematics/dynamics ke equations, jo linear motion (jo Chapter 2 "Motion in a Straight Line" mein padha tha) ke bilkul parallel hain (x → θ, v → ω, m → I, F → τ).
Ye chapter conceptually Chapter 4 (Laws of Motion) aur Chapter 5 (Work, Energy and Power) par directly build karta hai — F=ma ka rotational version τ=Iα yahin milta hai, aur conservation of angular momentum, conservation of linear momentum ka hi cousin hai jo Laws of Motion mein aaya tha. Aage Chapter 8 (Gravitation) mein bhi angular momentum conservation dobara kaam aata hai (satellite orbits mein), aur is poori "mechanics" series (Motion in a Straight Line → Motion in a Plane → Laws of Motion → Work Energy Power → System of Particles and Rotational Motion → Gravitation) ke baad book Part II mein Mechanical Properties of Solids/Fluids, Thermodynamics, Kinetic Theory, Oscillations aur Waves ki taraf move karti hai.
Numerically ye ek calculation-heavy chapter hai — CM location, vector/cross products, torque-balance (equilibrium) problems, aur moment-of-inertia based dynamics — isliye har solved example aur exercise neeche full step-by-step working ke saath diya gaya hai, CBSE marking-scheme style mein.
Chapter 6 Summary — 5 Minute Revision
Chapter 6 ka core summary:
- Rigid body ki motion ya to pure translation hoti hai, ya pure rotation (fixed axis ke around), ya dono ka combination (jaise rolling — jiska sirf intro-level mention hai, formal treatment/derivation is chapter se hata di gayi hai).
- Centre of mass ek n-particle system ka wo point hai jahan poore system ka mass concentrated maan ke motion nikaali ja sakti hai: R = Σmᵢrᵢ / M. CM ki velocity V = P/M, aur agar total external force zero ho to CM uniform velocity se move karta hai (linear momentum conservation).
- Vector (cross) product a × b ek naya tool hai jiski zaroorat torque aur angular momentum define karne ke liye padti hai. Iska magnitude ab sinθ hai, direction right-hand rule se, aur a × b ≠ b × a (non-commutative).
- Angular velocity aur linear velocity ka relation v = ω × r hai; fixed axis ke around har particle ka ω same hota hai chahe uska r alag ho.
- Torque τ = r × F force ka rotational analogue hai, aur angular momentum l = r × p linear momentum ka rotational analogue. dL/dt = τ_ext — ye poore chapter ka sabse important result hai (Newton's second law ka rotational version, system of particles ke liye).
- Agar τ_ext = 0, to angular momentum conserved rehta hai — isi se ice-skater ka spin fast hona (arms fold karne par) explain hota hai.
- Rigid body equilibrium ke liye do conditions chahiye: ΣF = 0 (translational) aur Στ = 0 (rotational). Centre of gravity wo point hai jahan total gravitational torque zero hota hai — uniform g mein ye CM ke saath coincide karta hai.
- Moment of inertia I = Σmᵢrᵢ² rotational motion mein mass ka analogue hai — ye axis, shape aur mass-distribution par depend karta hai (Table 6.1 ki standard values yaad rakhni padti hain, is rationalised syllabus mein derive nahi karni).
- Rotational kinematics (ω = ω₀+αt, θ = ω₀t+½αt², ω²=ω₀²+2αθ) exactly linear kinematics jaisi hai, sirf x→θ, v→ω, a→α.
- Rotational dynamics: τ = Iα (Newton's 2nd law ka rotational version), K = ½Iω² (rotational KE), P = τω (power), aur L = Iω (symmetric bodies, fixed axis).
In-Text Questions — Solutions
Udaharan 6.1 (NCERT Example 6.1): Ek equilateral triangle ke teeno vertices par teen particles rakhe hain jinka mass 100 g, 150 g aur 200 g hai. Triangle ka har side 0.5 m hai. In teen particles ka centre of mass (CM) nikaalo.
Solution: Axes is tarah choose karte hain ki O(0,0) par 100 g, A(0.5, 0) par 150 g, aur B(0.25, 0.25√3) par 200 g ho (equilateral triangle ki geometry se).
X = [100(0) + 150(0.5) + 200(0.25)] / (100+150+200) = (0+75+50)/450 = 125/450 = 5/18 m ≈ 0.28 m
Y = [100(0) + 150(0) + 200(0.25√3)] / 450 = 50√3/450 = 1/(3√3) ≈ 0.19 m
Answer: CM point O se X ≈ 0.28 m aur Y ≈ 0.19 m par hai. Notice — ye triangle ka geometric centre (centroid) nahi hai, kyunki masses unequal hain.
Udaharan 6.2 (NCERT Example 6.2): Ek triangular lamina (thin flat plate) ka centre of mass dhoondo.
Solution (conceptual proof): Lamina ∆LMN ko base MN ke parallel bahut saari narrow strips mein baant lo. Symmetry se har strip ka CM uske apne midpoint par hoga. Sab strips ke midpoints ko jodne par median LP milta hai — isliye poore triangle ka CM is median LP par kahin lie karega. Yahi argument doosri do medians MQ aur NR ke liye bhi lagta hai.
Answer: Teeno medians ka common point hi centroid G hai, isliye triangular lamina ka CM uski centroid par hota hai.
Udaharan 6.3 (NCERT Example 6.3): Ek uniform L-shaped lamina ka mass 3 kg hai, jo teen 1m × 1m squares se bani hai (har square 1 kg). Iska centre of mass nikaalo.
Solution: Teeno 1 kg squares ke geometric centres (jo unke apne CM hain) coordinates hain: C1(0.5, 0.5), C2(1.5, 0.5), C3(0.5, 1.5).
X = [1(0.5)+1(1.5)+1(0.5)] / (1+1+1) = 2.5/3 = 5/6 m ≈ 0.83 m
Y = [1(0.5)+1(0.5)+1(1.5)] / 3 = 2.5/3 = 5/6 m ≈ 0.83 m
Answer: CM (X, Y) = (5/6 m, 5/6 m) ≈ (0.83 m, 0.83 m), jo L-shape ki symmetry-diagonal OD par lie karta hai.
Udaharan 6.4 (NCERT Example 6.4): Do vectors a = 3î − 4ĵ + 5k̂ aur b = −2î + ĵ − 3k̂ ke scalar (dot) product aur vector (cross) product nikaalo.
Scalar product:
a·b = (3)(−2) + (−4)(1) + (5)(−3) = −6 − 4 − 15 = −25
Vector product (determinant rule se):
a × b = | î ĵ k̂ ; 3 −4 5 ; −2 1 −3 |
= î[(−4)(−3) − (5)(1)] − ĵ[(3)(−3) − (5)(−2)] + k̂[(3)(1) − (−4)(−2)]
= î(12−5) − ĵ(−9+10) + k̂(3−8) = 7î − ĵ − 5k̂
Answer: a·b = −25 ; a×b = 7î − ĵ − 5k̂ (aur b×a = −7î + ĵ + 5k̂, kyunki cross product commutative nahi hota).
Udaharan 6.5 (NCERT Example 6.5): Origin ke around ek force F = 7î + 3ĵ − 5k̂ ka torque nikaalo jo ek particle par act kar rahi hai jiska position vector r = î − ĵ + k̂ hai.
Solution: τ = r × F, determinant rule se:
τ = | î ĵ k̂ ; 1 −1 1 ; 7 3 −5 |
= î[(−1)(−5) − (1)(3)] − ĵ[(1)(−5) − (1)(7)] + k̂[(1)(3) − (−1)(7)]
= î(5−3) − ĵ(−5−7) + k̂(3+7) = 2î + 12ĵ + 10k̂
Answer: τ = (2î + 12ĵ + 10k̂) N m.
Udaharan 6.6 (NCERT Example 6.6): Dikhao ki constant velocity se move kar rahe ek single particle ka angular momentum, kisi bhi point ke around, poori motion mein constant rehta hai.
Solution: Maano particle velocity v se point P par hai, aur arbitrary origin O hai. Angular momentum l = r × mv, jiska magnitude l = mvr sinθ hai, jahan θ, r aur v ke beech ka angle hai.
r sinθ = OM, jo O se line-of-v (v ki direction wali line) ki perpendicular distance hai. Chunki particle sirf apni straight-line trajectory par move karta hai, iski velocity ki direction wali line hamesha same rehti hai — isliye OM (perpendicular distance) time ke saath change nahi hoti, ek constant hai.
Iske alawa, l ki direction bhi fixed hai (r aur v ke plane ke perpendicular, page ke andar ki taraf) — ye bhi change nahi hoti.
Answer: Chunki magnitude (mv × OM) aur direction dono constant hain, l poori motion mein constant (conserved) rehta hai. Wajah: is particle par koi external torque origin O ke around act nahi kar raha (chunki v constant hai, koi force hi nahi hai).
Udaharan 6.7 (NCERT Example 6.7): Dikhao ki ek couple (do equal-magnitude, opposite-direction forces jinki lines of action alag hain) ka moment, us point par depend nahi karta jiske around moments liye ja rahe hain.
Solution: Ek couple mein force F point B par aur −F point A par act karti hai. In dono ke position vectors origin O se r₂ aur r₁ hain. O ke around dono forces ka total moment:
Moment = r₁ × (−F) + r₂ × F = (r₂ − r₁) × F
Ab AB = r₂ − r₁ (A se B tak ka vector), jo sirf A aur B ki relative position par depend karta hai — origin O kahan hai, us par nahi.
Moment of couple = AB × F
Answer: Chunki final expression mein origin O bilkul present hi nahi hai, moment of a couple kisi bhi point ke around same rehta hai — origin-independent hota hai.
Udaharan 6.8 (NCERT Example 6.8): Ek 4.00 kg, 70 cm lambi metal bar dono ends se 10 cm door do knife-edges par supported hai. Ek 6.00 kg load ek end se 30 cm par latka hai. Knife-edges par reactions nikaalo (bar uniform aur homogeneous hai).
Solution: Bar AB=70 cm, G (uska apna CG) center par, AG=35 cm. Load P par AP=30 cm, isliye PG=5 cm. Knife edges K1, K2: AK1=BK2=10 cm, isliye K1G=K2G=25 cm.
Translational equilibrium: R1+R2 = W+W1 = (4.00+6.00)g = 10.00 × 9.8 = 98.0 N ... (i)
Rotational equilibrium (moments about G): −R1(25) + W1(5) + R2(25) = 0, aur W1=6.00×9.8=58.8 N
R1 − R2 = W1(5)/25 = 58.8/5 = 11.76 N ... (ii)
(i) aur (ii) solve karne par: R1 = (98.0+11.76)/2 = 54.88 N, R2 = (98.0−11.76)/2 = 43.12 N
Answer: Knife edge K1 par reaction ≈ 55 N, K2 par ≈ 43 N.
Udaharan 6.9 (NCERT Example 6.9): 20 kg ki 3 m lambi ladder frictionless wall par leaning hai, uske pair floor par wall se 1 m par hain. Wall aur floor ki reaction forces nikaalo.
Solution: Foot A wall se AC=1 m par; Pythagoras se BC = √(3²−1²) = √8 = 2√2 m. Wall frictionless, isliye wall ki reaction F1 sirf horizontal hai. Floor ki reaction do parts mein: normal N (vertical) aur friction F (horizontal).
Vertical equilibrium: N = W = 20×9.8 = 196.0 N
Rotational equilibrium (A ke around moments): F1(2√2) = W(1/2) → F1 = W/(4√2) = 196/(4×1.414) = 34.6 N
Horizontal equilibrium: F = F1 = 34.6 N
Floor ki total reaction: F2 = √(N²+F²) = √(196² + 34.6²) ≈ 199.0 N, jo horizontal se angle α = tan⁻¹(4√2) ≈ 80° par hai.
Answer: Wall ki reaction F1 ≈ 34.6 N (horizontal); Floor ki reaction F2 ≈ 199 N, horizontal se ≈80° par.
Udaharan 6.10 (NCERT Example 6.10): First principles se derive karo: ω = ω₀ + αt (uniform angular acceleration ke liye).
Solution: Uniform angular acceleration ka matlab: dω/dt = α = constant.
Dono side integrate karne par: ω = ∫α dt + c = αt + c (chunki α constant hai)
t=0 par ω=ω₀ (diya gaya hai), isliye c = ω₀.
Answer: ω = αt + ω₀, jo required equation hai. (Eq. 6.37 aur 6.38 similarly ω=dθ/dt ko integrate karke nikalte hain — isko exercise ki tarah chhoda gaya hai NCERT mein.)
Udaharan 6.11 (NCERT Example 6.11): Motor wheel ki angular speed 1200 rpm se 3120 rpm ho jaati hai 16 seconds mein. (i) Angular acceleration nikaalo (uniform maan kar). (ii) Is time mein kitni revolutions hui?
Solution:
ω₀ = 2π(1200)/60 = 40π rad/s ; ω = 2π(3120)/60 = 104π rad/s
(i) α = (ω−ω₀)/t = (104π−40π)/16 = 64π/16 = 4π rad/s² ≈ 12.57 rad/s²
(ii) θ = ω₀t + ½αt² = 40π(16) + ½(4π)(16²) = 640π + 512π = 1152π rad
Revolutions = θ/2π = 1152π/2π = 576
Answer: α = 4π rad/s² ≈ 12.6 rad/s² ; total 576 revolutions.
Udaharan 6.12 (NCERT Example 6.12): Ek flywheel (mass 20 kg, radius 20 cm) ki rim par ek cord ghoomi hai, jis par 25 N ka pull lagaya jaata hai (frictionless bearings). (a) Angular acceleration nikaalo. (b) 2 m cord unwind hone par pull se hua work nikaalo. (c) Us point par wheel ki KE nikaalo. (d) (b) aur (c) compare karo.
Solution: τ = FR = 25×0.20 = 5.0 N m. Flywheel ki I (disc, axis se) = MR²/2 = 20(0.2)²/2 = 0.4 kg m².
(a) α = τ/I = 5.0/0.4 = 12.5 rad/s²
(b) Work by pull = F × distance = 25 × 2 = 50 J
(c) θ = length unwound/R = 2/0.2 = 10 rad. ω² = 2αθ = 2(12.5)(10) = 250 (rest se start).
KE = ½Iω² = ½(0.4)(250) = 50 J
Answer: α = 12.5 rad/s² ; Work = 50 J ; KE = 50 J — dono equal hain, kyunki friction absent hone se pull ka poora work rotational KE mein convert ho jaata hai (energy conservation).
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q17)
6.1 (i) Sphere, (ii) cylinder, (iii) ring, aur (iv) cube — in sabki, uniform mass density maan kar, centre of mass ki location batao. Kya kisi body ka CM zaroori hai body ke andar hi ho?
(i) Sphere: uska geometric centre, jo body ke andar hai.
(ii) Cylinder: uski symmetry axis par, height ke bilkul beech mein — body ke andar (ya axis line par).
(iii) Ring: uska geometric centre — jo ring ke material (wire) mein nahi, balki ring ke beech ki khaali jagah mein hai.
(iv) Cube: uska geometric centre — body ke andar.
Answer (second part): Nahi, CM hamesha body ke andar hona zaroori nahi hai. Ring iska sabse achha example hai — uska CM material ke bahar (khaali space) mein hota hai. (L-shaped lamina jaisi irregular shapes mein bhi CM material se bahar aa sakta hai.)
6.2 HCl molecule mein dono nuclei ke beech separation ~1.27 Å hai (1 Å = 10⁻¹⁰ m). Molecule ke CM ki approximate location nikaalo, agar chlorine atom hydrogen atom se ~35.5 guna heavy hai aur poora mass nucleus mein concentrated hai.
Solution: H atom origin par (mass m), Cl atom 1.27 Å par (mass 35.5m).
X_cm = [m(0) + 35.5m(1.27)] / (m + 35.5m) = 35.5(1.27)/36.5 = 45.085/36.5 ≈ 1.235 Å
Answer: CM, H atom se ≈1.235 Å door hai (yaani Cl atom se sirf ≈0.035 Å) — matlab CM Cl atom (heavier atom) ke bahut paas hai, dono nuclei ko jodne wali line par.
6.3 Ek bachcha ek lambi trolley ke ek end par baitha hai, jo smooth horizontal floor par constant speed V se chal rahi hai. Agar bachcha uthkar trolley par kisi bhi tarah idhar-udhar bhaage, to (trolley + bachcha) system ke CM ki speed kya hogi?
Solution: Bachcha ka trolley par daudna sirf internal forces generate karta hai (bachche ke pair aur trolley ke beech). Floor smooth hai, isliye koi external horizontal force system par act nahi kar rahi.
Answer: Chunki total external force zero hai, CM ki velocity change nahi hogi — CM ki speed V hi rahegi, bachcha kaise bhi bhaage.
6.4 Dikhao ki vectors a aur b se bane triangle ka area, a × b ke magnitude ka aadha hota hai.
Solution: Maano O origin hai, OA = a aur OB = b. Parallelogram OACB (a aur b adjacent sides) ka area = |a × b| (ye cross-product ki standard geometric property hai — base × height).
Diagonal AB is parallelogram ko do congruent triangles mein baant deta hai — triangle OAB aur triangle ACB.
Answer: Isliye Area(triangle OAB) = ½ × Area(parallelogram) = ½|a × b|.
6.5 Dikhao ki a·(b × c), a, b aur c se bane parallelepiped ke volume ke barabar (magnitude mein) hota hai.
Solution: b × c ek vector hai jiska magnitude = |b||c| sinθ (θ, b aur c ke beech ka angle) — ye exactly b aur c se bane base parallelogram ka area hai. Iski direction base-plane ke perpendicular hai.
a·(b×c) = |a| |b×c| cosφ, jahan φ, a aur (b×c) ke beech ka angle hai. |a| cosφ, a ka (b×c) ki direction mein component hai — yehi parallelepiped ki height hai (base-plane se perpendicular).
Answer: a·(b×c) = (base area) × (height) = Volume of parallelepiped. (Magnitude mein — sign, orientation par depend karta hai.)
6.6 Ek particle ka position vector r (components x, y, z) aur momentum p (components pₓ, p_y, p_z) hai. Angular momentum l ke x, y, z axes ke along components nikaalo. Dikhao ki agar particle sirf x-y plane mein move kare, to angular momentum ka sirf z-component hota hai.
Solution: l = r × p, determinant rule se components:
lₓ = y·p_z − z·p_y
l_y = z·pₓ − x·p_z
l_z = x·p_y − y·pₓ
Agar particle sirf x-y plane mein move kare, to z = 0 aur p_z = 0 (velocity ka z-component hi nahi hai).
lₓ = y(0) − 0(p_y) = 0 ; l_y = 0(pₓ) − x(0) = 0 ; l_z = x·p_y − y·pₓ (generally non-zero)
Answer: lₓ = l_y = 0, sirf l_z bacha rehta hai — isliye angular momentum ka sirf z-component hota hai.
6.7 Do particles, har ek ka mass m aur speed v, opposite directions mein do parallel lines (distance d apart) ke along travel kar rahe hain. Dikhao ki dono-particle system ka angular momentum vector, kisi bhi point ke around liya jaaye, same rehta hai.
Solution (general property): Kisi bhi origin O ke around L_O = Σrᵢ×pᵢ. Agar O′ ek doosra origin hai jiski position O se a hai (rᵢ = a + rᵢ′), to:
L_O = Σ(a+rᵢ′)×pᵢ = a×(Σpᵢ) + Σrᵢ′×pᵢ = a×P + L_O′
Yahan P = total linear momentum = p₁+p₂ = mv + m(−v) = 0 (particles opposite directions mein equal speed se ja rahe hain).
Answer: Chunki P=0, L_O = L_O′ hamesha — matlab angular momentum origin ke choice se independent hai. (Iska magnitude, ek line par origin lekar nikala ja sakta hai: us particle ka contribution zero hoga jiska r uski apni velocity ke parallel hai, aur doosre particle ka contribution = mvd — isliye total |L| = mvd, hamesha same.)
6.8 Ek non-uniform bar (weight W) do strings se latki hai jo vertical se 36.9° aur 53.1° ke angles banati hain. Bar 2 m lambi hai. Bar ke left end se uske centre of gravity ki distance d nikaalo (Fig. 6.33).
Solution: sin36.9°≈0.6, cos36.9°≈0.8; sin53.1°≈0.8, cos53.1°≈0.6.
Horizontal equilibrium: T1 sin36.9° = T2 sin53.1° → 0.6T1 = 0.8T2 → T1 = (4/3)T2
Vertical equilibrium: T1 cos36.9° + T2 cos53.1° = W → 0.8T1 + 0.6T2 = W
0.8(4/3)T2 + 0.6T2 = W → (16/15)T2 + 0.6T2 = (5/3)T2 = W → T2 = 0.6W ; T1 = 0.8W
Left end (A) ke around moments (sirf vertical components ka moment lagta hai): T2 cos53.1° × 2 = W × d
(0.6W)(0.6)(2) = Wd → 0.72W = Wd → d = 0.72 m
Answer: d = 0.72 m (left end se, jahan 36.9° wali string bandhi hai).
6.9 Ek car ka weight 1800 kg hai. Front aur back axles ke beech distance 1.8 m hai. Uska centre of gravity front axle se 1.05 m peeche hai. Har front wheel aur har back wheel par ground ki force nikaalo.
Solution: W = 1800×9.8 = 17640 N. Rf (dono front wheels) + Rb (dono back wheels) = W
Front axle ke around moments: Rb × 1.8 = W × 1.05
Rb = (17640×1.05)/1.8 = 18522/1.8 = 10290 N
Rf = W − Rb = 17640 − 10290 = 7350 N
Answer: Har back wheel = 10290/2 = 5145 N; har front wheel = 7350/2 = 3675 N.
6.10 Same magnitude ke torques ek hollow cylinder aur ek solid sphere par lagaye jaate hain — dono ka mass aur radius same hai. Cylinder apni symmetry axis ke around, aur sphere apne centre se guzarne wali axis ke around free hai. Given time ke baad kisme zyada angular speed hogi?
Solution: Hollow cylinder: I_cyl = MR². Solid sphere: I_sph = (2/5)MR². Same M, R ke liye I_sph < I_cyl.
α = τ/I → same τ ke liye, chota I zyada α deta hai → α_sph > α_cyl
Dono rest se start karte hain (assume), to ω = αt — jitna zyada α, utni zyada ω same t par.
Answer: Solid sphere zyada angular speed acquire karega (kyunki uska moment of inertia kam hai).
6.11 Ek 20 kg ka solid cylinder apni axis ke around 100 rad/s angular speed se rotate kar raha hai. Cylinder ka radius 0.25 m hai. Rotation se judi kinetic energy nikaalo. Cylinder ke angular momentum ka magnitude bhi nikaalo (axis ke around).
Solution: I (solid cylinder, axis) = MR²/2 = 20(0.25)²/2 = 20(0.0625)/2 = 0.625 kg m²
K = ½Iω² = ½(0.625)(100)² = ½(0.625)(10000) = 3125 J
L = Iω = 0.625 × 100 = 62.5 kg m²/s
Answer: K = 3125 J ; L = 62.5 kg m²/s.
6.12 (a) Ek bachcha turntable ke centre par apne arms outstretched karke khada hai, turntable 40 rev/min se rotate ho rahi hai. Agar bachcha apne hands fold kar le aur apna moment of inertia initial value ka 2/5 kar de, to uski nayi angular speed kya hogi (turntable friction-free maano)? (b) Dikhao ki bachche ki nayi rotational KE initial KE se zyada hai. Ye extra energy kahan se aayi?
Solution (a): Koi external torque nahi hai (frictionless), isliye angular momentum conserved: I₁ω₁ = I₂ω₂
I₁(40) = (2/5)I₁ × ω₂ → ω₂ = 40 × 5/2 = 100 rev/min
Solution (b):
KE₂/KE₁ = (I₂ω₂²)/(I₁ω₁²) = (2/5)(100/40)² = (2/5)(2.5)² = (2/5)(6.25) = 2.5
Answer: ω₂ = 100 rev/min. KE₂ = 2.5 × KE₁, yaani nayi KE zyada hai. Ye extra energy bachche ke apni muscular (internal biochemical) energy se aayi hai — jab wo apne hands andar khinchta hai, uske muscles internal work karte hain jo rotational KE mein convert ho jaata hai (angular momentum conserved rehta hai, KE nahi).
6.13 Ek negligible-mass rope, 3 kg mass aur 40 cm radius wale ek hollow cylinder par ghoomi hai. Agar rope ko 30 N ke force se khincha jaaye, to cylinder ka angular acceleration kya hoga? Rope ka linear acceleration kya hoga (no slipping maan kar)?
Solution: I (hollow cylinder) = MR² = 3(0.4)² = 3(0.16) = 0.48 kg m²
τ = FR = 30 × 0.4 = 12 N m
α = τ/I = 12/0.48 = 25 rad/s²
No slipping ki condition se, rope ki linear acceleration = rim ki tangential acceleration:
a = αR = 25 × 0.4 = 10 m/s²
Answer: α = 25 rad/s² ; a = 10 m/s².
6.14 Ek rotor ko uniform angular speed 200 rad/s par maintain karne ke liye engine ko 180 N m ka torque transmit karna padta hai. Engine ko kitni power chahiye (engine 100% efficient maano)?
Solution:
P = τω = 180 × 200 = 36000 W = 36 kW
Answer: Engine ko 36 kW power chahiye. (Note: uniform ω ka matlab hai zero net torque zaroori hota, lekin friction ko counter karne ke liye applied torque zaroori hai — isliye engine ko ye power denii padti hai.)
6.15 Ek uniform disc (radius R) mein se ek circular hole (radius R/2) kaata gaya hai. Hole ka centre original disc ke centre se R/2 door hai. Bachi hui flat body ka centre of gravity locate karo.
Solution (negative-mass method): Original poore disc ka mass M maano (radius R, mass per unit area σ = M/πR²). Hole ka mass = σ × π(R/2)² = M/4.
Origin O = original disc ka centre. Hole ka centre, O se distance R/2 par (say x-axis par).
Bachi hui body = (poora disc, mass M, O par) + (hole ka disc, mass −M/4, R/2 par):
X_cm = [M(0) + (−M/4)(R/2)] / (M − M/4) = (−MR/8) / (3M/4) = −R/6
Answer: Bachi hui body ka CG, original centre O se R/6 door hai, hole ki opposite direction mein (yaani hole se door ki taraf).
6.16 Ek metre stick knife edge par uske centre par balance hai. Jab 5 g ke do coins (ek doosre ke upar) 12.0 cm mark par rakhe jaate hain, to stick 45.0 cm mark par balance hoti hai. Metre stick ka mass kya hai?
Solution: Chunki stick akele (bina coins ke) 50 cm mark (centre) par balance hoti thi, iska matlab stick ki apni CG 50 cm mark par hai.
Naya pivot 45.0 cm par hai. Coins (10 g total) 12 cm par hain — pivot se distance = 45−12 = 33 cm. Stick ka weight 50 cm par hai — pivot se distance = 50−45 = 5 cm.
Torque balance (masses se, kyunki g dono taraf cancel hota hai):
10 g × 33 cm = M_stick × 5 cm → M_stick = 330/5 = 66 g
Answer: Metre stick ka mass = 66 g.
6.17 Oxygen molecule ka mass 5.30×10⁻²⁶ kg hai aur uska moment of inertia (centre se guzarne wali, dono atoms ko jodne wali line ke perpendicular axis ke around) 1.94×10⁻⁴⁶ kg m² hai. Agar gas mein aise molecule ki mean speed 500 m/s hai aur uski rotational KE, translational KE ki 2/3 hai, to molecule ki average angular velocity nikaalo.
Solution:
KE_trans = ½mv² = ½(5.30×10⁻²⁶)(500)² = ½(5.30×10⁻²⁶)(2.5×10⁵) = 6.625×10⁻²¹ J
KE_rot = (2/3)KE_trans = (2/3)(6.625×10⁻²¹) = 4.4167×10⁻²¹ J
KE_rot = ½Iω² se:
ω² = 2(KE_rot)/I = 2(4.4167×10⁻²¹)/(1.94×10⁻⁴⁶) = 8.833×10⁻²¹/1.94×10⁻⁴⁶ ≈ 4.553×10²⁵
ω = √(4.553×10²⁵) ≈ 6.75×10¹² rad/s
Answer: ω ≈ 6.75×10¹² rad/s.
Important Equations — Ek Nazar Me
| Concept | Formula | Notes |
|---|---|---|
| Centre of mass (n particles) | R = Σmᵢrᵢ / M, jahan M = Σmᵢ | X = Σmᵢxᵢ/M (aur Y, Z similarly) |
| Velocity / acceleration of CM | V = P/M ; M A = Fext | Fext=0 par CM uniform velocity se move karta hai |
| Linear momentum conservation | Agar Fext=0, to P = constant | System of particles ka Newton's 2nd law: dP/dt = Fext |
| Vector (cross) product | |a × b| = ab sinθ, direction right-hand rule se | a × b = −(b × a); non-commutative |
| Linear-angular velocity relation | v = ω × r ; v = ωr (magnitude) | Fixed axis ke around har particle ka ω same hota hai |
| Angular acceleration | α = dω/dt | Linear motion ke a = dv/dt ka analogue |
| Torque (moment of force) | τ = r × F ; τ = rF sinθ | SI unit: N m (dimensionally kaam/energy jaisa, par vector hai) |
| Angular momentum (particle) | l = r × p ; l = rp sinθ | Linear momentum ka rotational analogue |
| Torque–angular momentum relation | dL/dt = τext | System of particles ke liye rotational Newton's 2nd law |
| Conservation of angular momentum | Agar τext=0, to L = constant | Ice-skater / turntable examples yahi use karte hain |
| Rigid body equilibrium | ΣF = 0 (translational) aur Στ = 0 (rotational) | Dono conditions ek saath chahiye |
| Principle of moments (lever) | F₁d₁ = F₂d₂ | Load arm × load = effort arm × effort |
| Centre of gravity | Σmᵢrᵢ (CG ke around) = 0 | Uniform g mein CG = CM |
| Moment of inertia | I = Σmᵢrᵢ² | Rotational motion mein mass ka analogue; axis-dependent |
| Radius of gyration | I = Mk² | k = geometric property (axis + shape par depend) |
| MI — thin ring (perp. axis, centre) | I = MR² | Table 6.1 ki standard values (derivation syllabus se bahar hai) |
| MI — ring (diameter) | I = MR²/2 | |
| MI — thin rod (perp. axis, midpoint) | I = ML²/12 | |
| MI — disc (perp. axis, centre) | I = MR²/2 | |
| MI — disc (diameter) | I = MR²/4 | |
| MI — hollow cylinder (axis) | I = MR² | |
| MI — solid cylinder (axis) | I = MR²/2 | |
| MI — solid sphere (diameter) | I = (2/5)MR² | |
| Rotational kinematics (uniform α) | ω = ω₀ + αt ; θ = ω₀t + ½αt² ; ω² = ω₀² + 2αθ | Linear kinematics ka exact analogue (x→θ, v→ω, a→α) |
| Work done by torque | W = τθ ; dW = τdθ | Linear W = Fs ka analogue |
| Power (rotational) | P = τω | Linear P = Fv ka analogue |
| Newton's 2nd law (rotational) | τ = Iα | Linear F = ma ka analogue |
| Kinetic energy of rotation | K = ½Iω² | Linear K = ½mv² ka analogue |
| Angular momentum (fixed-axis, symmetric body) | L = Iω | Linear p = mv ka analogue |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Centre of mass (CM) aur centre of gravity (CG) ko hamesha ek hi maan lena — dono sirf uniform gravitational field mein coincide karte hain; CM sirf mass-distribution par depend karta hai, gravity par bilkul nahi.
- v = rω use karte waqt vector nature bhool jaana — sahi relation v = ω × r hai, aur r wahi lena chahiye jo rotation axis se perpendicular ho, poora position vector nahi.
- Torque nikalte waqt bas τ = rF laga dena, sinθ (angle between r aur F) bhool jaana — force agar r ke parallel/anti-parallel ho to torque zero hota hai, ye bhool jaate hain.
- Table 6.1 se galat moment-of-inertia formula pick karna — jaise disc ke liye MR²/2 (perpendicular axis, centre) use kar lena jab question diameter ke around MI (MR²/4) maang raha ho. Axis kaunsi hai, ye pehle confirm karo.
- Cross product ka order badal dena — a × b aur b × a ka magnitude same hai lekin direction opposite; equilibrium/torque calculations mein sign galat aane se poora answer ulta ho jaata hai.
- Numerical problems mein rpm (rev/min) ko seedha equations mein daal dena bina rad/s mein convert kiye — ω hamesha rad/s mein hona chahiye kinematics/dynamics equations lagane se pehle (1 rev/min = 2π/60 rad/s).
Board-Style Important Questions
- Conceptual: Ek couple ka net force zero hota hai lekin net torque nahi — is statement ko justify karo ek example ke saath.
- Short Answer: Kisi rigid body ke liye do conditions likho jo mechanical equilibrium ke liye zaroori hain, aur inhe symbolic form mein express karo.
- Derivation: Derive karo ki uniform angular acceleration ke case mein θ = ω₀t + ½αt² hota hai, first principles (calculus) se.
- Short Answer: Ek ring aur ek solid sphere, same mass aur radius ke saath, same axis ke around rotate ho rahe hain. Dono ke moment of inertia compare karo aur bataao same angular velocity par kiski rotational KE zyada hogi.
- Long Answer: Angular momentum conservation ka principle likho aur ise use karke samjhao ki jab ek ice-skater apne arms andar khinchti hai to uski spin speed kyun badh jaati hai. Ek numerical example ke saath illustrate karo.
- Long Answer: Ek non-uniform bar do strings se latki hai jo vertical se different angles banati hain. Bar ke centre of gravity ki position nikalne ka general method derive karo, aur ek numerical example solve karo.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics mein System of Particles and Rotational Motion chapter number kya hai — Chapter 6 ya Chapter 7?
Rationalised 2026-27 syllabus mein ye Chapter 6 hai. Purani (pre-rationalisation) NCERT book mein ye Chapter 7 hota tha, kyunki tab "Physical World" ek standalone Chapter 1 hota tha. Ab wo chapter hata diya gaya hai, isliye Units and Measurements Chapter 1 ban gaya hai aur baaki sab chapters ek number peeche shift ho gaye — is wajah se sabhi class 11 physics ncert solutions ki chapter-numbering bhi update ho gayi hai across websites, isliye number confirm karte waqt latest edition dekhna zaroori hai.
Kya rolling motion is chapter mein padhaya jaata hai (2026-27 syllabus mein)?
Nahi. Rationalised syllabus mein rolling motion ka poora formal section (purani book ka Section 6.13, jisme rolling ke KE aur velocity ki derivation hoti thi) hata diya gaya hai. Chapter sirf introduction (Section 6.1) mein rolling ka mention motivation ke taur par karta hai — 'rolling motion translation + rotation ka combination hai' — lekin uski poori dynamics is chapter se bahar hai. Ye class 11 physics deleted syllabus 2026-27 ka ek important part hai jo students aksar miss kar dete hain.
Kya parallel axis theorem aur perpendicular axis theorem ab bhi is chapter mein hain?
Nahi, ye dono theorems bhi current rationalised edition se hata diye gaye hain. Table 6.1 mein sirf standard bodies (ring, disc, rod, cylinder, sphere) ki moment of inertia values di gayi hain — unki derivation, aur ek axis se doosre axis par shift karne ke rules (parallel/perpendicular axis theorem), textbook khud kehta hai ki ye 'beyond the scope of this textbook' hain (higher classes mein padhaya jaayega).
Class 11 Physics ke saare chapters ek jagah kahan milenge (2026-27 session ke liye)?
Rationalised syllabus mein Class 11 Physics ab 14 chapters ka hai, do parts mein publish hota hai (Part I: mechanics-heavy chapters jaise Motion in a Straight Line, Motion in a Plane, Laws of Motion, Work Energy Power, System of Particles and Rotational Motion, Gravitation; Part II: Mechanical Properties of Solids/Fluids, Thermodynamics, Kinetic Theory of Gases, Oscillations, Waves). Agar class 11 physics all chapters pdf 2026-27 chahiye, to official ncert.nic.in textbook page hi sabse reliable source hai — third-party PDFs mein purani (non-rationalised) content reh jaana common issue hai.
Ye chapter pehle padhe chapters se kaise connect hota hai?
System of Particles and Rotational Motion, chapter 2 motion in a straight line aur chapter 3 motion in a plane ki linear kinematics ko rotational version deta hai (x→θ, v→ω), aur laws of motion class 11 ke F=ma ko τ=Iα mein extend karta hai. Angular momentum conservation ka concept aage gravitation class 11 numericals with solutions mein bhi kaam aata hai (satellite/planetary motion mein), isliye ye chapter is series ki ek core building-block hai.
Is chapter ke baad book mein aage kya aata hai?
System of Particles and Rotational Motion (aur Gravitation) ke baad Part II shuru hota hai: Mechanical Properties of Solids aur Fluids (mechanical properties of fluids class 11 numericals wale chapters), phir Thermodynamics (thermodynamics class 11 physics notes), Kinetic Theory of Gases (kinetic theory of gases class 11 derivation), Oscillations (oscillations class 11 physics important questions), aur last mein Waves (waves class 11 physics doppler effect formula). Har chapter apni tarah ka calculation-heavy hai, isliye is chapter mein torque/angular-momentum ki practice zaroor karo — baaki chapters mein bhi similar rigor chahiye hogi.
Class 11 Physics — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Physics ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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