System of Particles and Rotational Motion: Class 11 Physics Chapter 6 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

120 min
1 simulation
24 practice questions
39 MCQs

Is chapter mein aap kya seekhenge

Rigid body ki motion mein pure translation, fixed axis ke around rotation aur dono ke combination mein farq pehchano
Particle systems aur symmetric va composite bodies ka centre of mass dhoondho, aur iski madad se batao ki external forces ke under system kaise move karta hai
Right-hand rule aur components se vector products nikalo, aur unse velocity, torque aur angular momentum pata karo
Systems of particles par linear aur angular momentum ka conservation lagao
Equilibrium ki conditions se levers, beams aur ladders ke problems solve karo, aur centres of gravity dhoondho
Moment of inertia aur radius of gyration define karo, aur common bodies ke standard values use karo
Rotational kinematic equations aur torque, work, power va angular momentum ke relations se fixed-axis rotation ke problems solve karo

NCERT Physics Class 11, Chapter 6, pp. 92–126 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

6.1 Introduction

Ab tak humne cricket ball, car aur yahan tak ki Earth ko bhi ek single point maan liya jismein saara mass hai. Yeh particle model tab theek hai jab sirf path matter karta hai. Lekin yeh nahi samjha sakta ki ghoomta hua lattu seedha kyun khada rehta hai, mechanic lamba spanner kyun uthata hai, ya Kathak dancer apni baahen andar kheench kar aur tez kaise ghoomti hai. Asli bodies ka size hota hai, aur size hi unhe ghoomne deta hai.

Is chapter mein hum ek extended body ko system of particles maante hain aur do sawaal poochhte hain. System poora ek saath kaise move karta hai? Iska jawaab ek khaas point par tika hai, centre of mass. Aur body ghoomti kaise hai? Iske liye humein teen nayi quantities chahiye: torque, angular momentum aur moment of inertia, aur har ek kisi aisi cheez ka rotational partner hai jo tum pehle se jaante ho.

Ghoomne ke problems tab aasaan ho jaate hain jab body apna shape bana kar rakhti hai. Rigid body (दृढ़ पिंड) ek idealised body hai jiske har pair of particles ke beech ki distance fixed rehti hai, chahe koi bhi forces lagein. Asli bodies bas iske kareeb hoti hain: truck ke neeche steel girder thoda jhuk jaata hai aur bounce karte waqt rubber ball chapti ho jaati hai. Phir bhi cycle ka wheel, darwaaza, lattu ya satellite normal motion mein itna kam shape badalte hain ki unhe rigid maanne se accuracy mein lagbhag koi farq nahi padta.

Definition

Rigid body

Ek ideal body jiske saare particles ek doosre se fixed distance par rehte hain, isliye kisi bhi force se uska size aur shape kabhi nahi badalta.

ExampleHinges par ghoomta darwaaza lagbhag rigid hai; goondha ja raha aata nahi.

6.1.1 Rigid body kis tarah ki motion kar sakti hai?

Floors ke beech upar jaata lift cabin dekho, ya desk se bahar khiskta drawer. Us par do dots lagao, ek top ke paas aur ek corner ke paas: har instant par dono dots same velocity se move karte hain, isliye woh bilkul ek jaise, parallel paths banate hain. Yeh pure translation hai: kisi bhi instant par body ke har particle ki velocity same hoti hai.

Ab sadak par rolling karta cycle ka front wheel dekho. Wheel cycle ke saath aage jaata hai, to translation hai, lekin uske points ab ek speed se nahi chalte. Tyre ka top aage bhaagta hai, hub cycle ki speed se chalta hai, aur sadak ko chhoota point us pal rest par hota hai agar tyre skid na kare. Rolling mein translation ke saath ek doosri tarah ki motion judi hoti hai, aur Fig. 6.1 dono ko alag karke dikhata hai.

3/3Top par $2 v$ , centre par $v$ , contact point par zero.

Fig. 6.1Bina slipping ke rolling = centre ka translation + centre ke around rotation.NCERT Fig. 6.2 · p. 93

Level ground par ek wheel, jiske top, centre aur contact point par arrows hain. Step 1, translation: teeno arrows aage ki taraf hain aur sabki length v hai. Step 2, centre ke around rotation: top arrow aage aur bottom wala peeche, dono ki length v, aur centre par koi arrow nahi. Step 3, rolling: top arrow 2v, centre v, aur contact point ki koi velocity nahi.

Woh doosri motion kya hai? Body ko aise pakdo ki woh bilkul translate na kar sake, jaise hinges darwaaze ko ya axle sewing-machine ke wheel ko pakadta hai. Tab woh sirf ghoom sakti hai, aur woh fixed line uska axis of rotation hai. Chakki (haath se chalne wala pathar), revolving door aur wind turbine sab fixed axes ke around ghoomte hain.

Jab body fixed axis ke around ghoomti hai, to uska har particle ek circle mein chalta hai jiska plane axis ke right angle par hai aur centre axis par (Fig. 6.2). Alag heights ke particles alag, parallel planes mein ghoomte hain, aur axis se door wala particle utne hi time mein bada circle poora karta hai. Axis par pada particle bilkul move nahi karta.

4/4C axis par hai aur bilkul move nahi karta.

Fig. 6.2Fixed-axis rotation mein har particle apne perpendicular plane mein axis ke chaaron taraf ghoomta hai.NCERT Fig. 6.4 · p. 94

Vertical dashed axis ke around ghoomta ek drum. Step 1 drum aur axis dikhata hai. Step 2 particle A jodta hai, neeche aur axis ke paas, axis par centred ek chhote horizontal circle par. Step 3 particle B jodta hai, upar aur rim ke paas, ek bade circle par. Step 4 axis par particle C mark karta hai, jo move nahi karta.

Axis ka ek jagah tike rehna zaroori nahi. Ghoomta lattu apni nok se guzarne wale axis ke around ghoomta hai, aur ghoomte waqt woh axis dheere-dheere vertical ke chaaron taraf ghoomta hai aur ek cone banata hai. Axis ke is dheeme ghoomne ko precession kehte hain. Kandhe se ghoomti tumhari baanh ek aur example hai: axis joint se guzarta hai lekin kai directions mein point kar sakta hai. Dono mein body ka ek point fixed hai, poori line nahi. Is chapter mein hum fixed axis wale simple case par hi rahenge.

Try karke dekho

Pehle khiskao, phir ghumao

Tumhe chahiye

ek 30 cm scale
chart paper ki ek sheet
do pencils
ek dost

Pehle guess karo

Tum scale ko ek curve ke along bina ghumaye khiskate ho. Dono dots ke paths ek doosre se kaise milenge?

Paths bilkul same honge, bas khisake hueEnd wale dot ka path lamba hogaBeech wale dot ka path lamba hoga

30 cm scale par do dots lagao: dot A ek end par aur dot B 15 cm ke mark par.
Scale ko chart paper par rakho. Ise ek halke curve ke along khiskao, paper ke edge ke parallel rakhte hue, aur dost se kaho ki pencils se A aur B ke paths trace kare.
Phir se shuru karo, lekin is baar scale ko waise hi curve ke along chalte hue quarter turn ghoomne do. Dono dots phir se trace karo.

6.2 Centre of mass

Cricket bat ko ek ungli par balance karo. Balance point beech mein nahi balki bhaari blade ke kaafi paas hota hai: jahan zyada mass hai, balance point udhar jhukta hai. Centre of mass (द्रव्यमान केंद्र) is roz ke idea ko ek formula mein badal deta hai.

Shuru karo $x$ -axis par do particles se jinke masses $m_{1}$ aur $m_{2}$ hain, origin O se $x_{1}$ aur $x_{2}$ par. Unka centre of mass yahan hota hai

Do particles ka centre of mass

X = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}

(6.1)

Yeh dono positions ka mass-weighted average hai. Barabar masses ke liye yeh simple average, yaani midpoint hai. Bhaari particle $X$ ko apni taraf kheenchta hai: agar $m_{2} = 3 m_{1}$ , to centre of mass $m_{1}$ se $m_{2}$ ki taraf teen-chauthai raaste par hota hai (Fig. 6.3).

3/3Centre of mass par support dene se rod balance hoti hai: $m_{1} d_{1} = m_{2} d_{2}$ .

Fig. 6.3Centre of mass bhaari particle ke paas hota hai, jahan

m_{1} d_{1} = m_{2} d_{2}

.NCERT Fig. 6.7 · p. 95

Ek light rod par do balls aur unke centre of mass ka marker. Step 1: barabar masses, marker midpoint par. Step 2: right wali ball teen guna bhaari hai aur marker three-quarter point par chala jaata hai. Step 3: marker ke neeche support rod ko balance karta hai, distances d1 aur d2 mark hain, d1 d2 ka teen guna.

Line par $n$ particles ke liye bhi rule same hai: har position ko uske mass se multiply karo, jodo, aur total mass $M = \sum m_{i}$ se divide karo.

\begin{aligned} X & = \frac{\sum_{i = 1}^{n} m_{i} x_{i}}{\sum_{i = 1}^{n} m_{i}} \\ = \frac{1}{M} \sum_{i = 1}^{n} m_{i} x_{i} \end{aligned}

(6.2)

Agar particles ek plane mein ya space mein faile hain, to har coordinate par alag se yahi weighted average lagao:

\begin{aligned} X & = \frac{1}{M} \sum m_{i} x_{i} \\ Y & = \frac{1}{M} \sum m_{i} y_{i} \\ Z & = \frac{1}{M} \sum m_{i} z_{i} \end{aligned}

(6.4)

Teen barabar masses ke liye yeh teeno corners ka average deta hai, jo unke triangle ka centroid hai. Position vectors $𝐫_{i} = x_{i} \hat{𝐢} + y_{i} \hat{𝐣} + z_{i} \hat{𝐤}$ ke saath teeno equations ek vector equation mein simat jaati hain:

Centre of mass (vector form)

𝐑 = \frac{1}{M} \sum_{i = 1}^{n} m_{i} 𝐫_{i}

(6.4d)

Ek nateeja yaad rakhne layak hai: agar origin centre of mass par rakho, to $\sum m_{i} 𝐫_{i} = 𝟎$ . Hum ise Section 6.8.2 mein use karenge.

Ek steel scale ya flywheel mein lagbhag $10^{25}$ atoms hote hain, ek-ek karke jodne ke liye bahut zyada. Iske bajaye socho ki body ko mass $Δ m$ ke chhote pieces mein kaata gaya, pieces par $Δ m 𝐫$ jodo, aur pieces ko bina limit chhota hone do. Sums integrals ban jaate hain:

𝐑 = \frac{1}{M} \int 𝐫 d m

(6.5b)

jahan

$d m$: position $𝐫$ par mass ka ek chhota element
$M = \int d m$: total mass

Centre of symmetry wali uniform bodies ke liye integral ki zaroorat nahi. Patli uniform rod ke beech mein origin rakho: $x$ par har chhote piece $d m$ ka $- x$ par barabar mass ka jodidaar hai, isliye products $x d m$ pairs mein cancel hote hain aur $\int x d m = 0$ . $(x, y, z)$ ko $(- x, - y, - z)$ ke saath jodne wali yahi trick uniform ring, disc ya sphere, ya uniform cross-section wali moti rod ke liye bhi chalti hai, isliye unka centre of mass geometric centre hai. Dhyaan do ki centre of mass ka material mein hona zaroori nahi: ring ka centre of mass uske khaali beech mein hai.

Example 6.1NCERT Example 6.1 · p. 97

Set-square par teen brass weights

$2.0 k g$ , $3.0 k g$ aur $5.0 k g$ ke brass weights ek right-angled triangle ke corners par rakhe hain. Right-angle wale corner ko origin lo, jahan $2.0 k g$ weight hai, $3.0 k g$ weight $(0.40 m, 0)$ par aur $5.0 k g$ weight $(0, 0.30 m)$ par. Centre of mass dhoondho. Kya yeh triangle ke centroid par hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.2NCERT Example 6.2 · p. 98

Triangular sign kahan balance hota hai?

Dikhao ki uniform triangular plate, jaise plywood ka road sign, ka centre of mass woh point hai jahan uske teeno medians milte hain.

3/3CA ke parallel strips ke saath dohraao: medians G par katte hain, jo centroid hai.

Fig. 6.4Har strip apne midpoint par balance hoti hai, aur saare midpoints ek median par hain.NCERT Fig. 6.10 · p. 98

Triangle ABC. Step 1: ise BC ke parallel patli strips mein kaata gaya hai. Step 2: dots har strip ka midpoint dikhate hain, aur yeh A se nikalne wale median par hain. Step 3: B se median joda gaya hai, aur dono medians G par katte hain, jo centroid hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.3NCERT Example 6.3 · p. 98

T-shape ka name board

Ek name board $10 k g$ per square metre wali uniform sheet metal se kaata gaya hai. Uska top bar $80 c m$ chauda aur $20 c m$ ooncha hai; bar ke beech se $20 c m$ chauda aur $60 c m$ ooncha stem latakta hai (Fig. 6.5). Iska centre of mass dhoondho.

Fig. 6.5T ko do rectangles mein todo aur har ek ko uske centre par ek point mass maano.NCERT p. 98

Axes par T-shape ki plate, origin uske 20 cm by 60 cm stem ke neeche, aur upar 80 cm by 20 cm ka bar. Dono rectangles ke centres C1 aur C2, aur unke beech centre of mass G, vertical line of symmetry par hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveCentre of mass lab

Metres mein ek square grid jismein teen rangeen point masses A, B aur C hain jinhe tum drag kar sakte ho. Neeche ke sliders har mass ko 1 se 10 kilogram tak set karte hain. Ek cross centre of mass dikhata hai aur ek dashed triangle teeno masses ko jodta hai, comparison ke liye uske centroid par ek chhota ring hai. Readouts total mass aur centre of mass ke coordinates X aur Y dikhate hain. Ek button teeno masses ko barabar kar deta hai, taaki tum centre of mass ko centroid par koodte dekh sako.

Fig. 6.6Teeno masses ko drag karo aur unke size badlo; cross centre of mass dikhata hai aur readout uske coordinates batata hai.

Quick check

3 mein se 1

2.0 k g

ball

x = 0

par hai aur ek

6.0 k g

ball

x = 0.80 m

par. Unka centre of mass kahan hai?

Answer dekho

Answer 0.6 m

X = (6.0 \times 0.80) / 8.0 = 0.60 m

, bhaari ball ki taraf teen-chauthai raaste par.

6.3 Centre of mass ki motion

Centre of mass itna important kyun hai? Kyunki yeh bahut simple tareeke se move karta hai, chahe parts ki motion kitni bhi complicated ho.

DerivationCentre of mass ke liye Newton ka second law

Definition ko $M 𝐑 = \sum m_{i} 𝐫_{i}$ likho aur har mass ko constant rakhte hue time ke respect mein differentiate karo:

M 𝐕 = \sum_{i} m_{i} 𝐯_{i}

(6.8)

Yahan $𝐕 = d 𝐑 / d t$ centre of mass ki velocity hai. Ek baar aur differentiate karne se uska acceleration $𝐀$ milta hai:

\begin{aligned} M 𝐀 & = \sum_{i} m_{i} 𝐚_{i} \\ = \sum_{i} 𝐅_{i} \end{aligned}

(6.9)

Newton ke second law se har $m_{i} 𝐚_{i}$ particle $i$ par lagne wala total force $𝐅_{i}$ hai. Is force ke do parts hain: system ke bahar ki bodies se aane wale external forces, aur doosre particles se aane wale internal forces. Internal forces action–reaction pairs mein aate hain (third law), isliye sum mein do-do karke cancel ho jaate hain. Sirf external forces bachte hain:

Centre of mass ki motion

M 𝐀 = 𝐅_{ext}

(6.11)

(6.11) ko ek vaade ki tarah padho. System ke parts chahe jaise dhakka-mukki karein, ghoomein ya bikhar jaayein, ek point, centre of mass, Newton ke second law ko aise follow karta hai jaise woh mass $M$ ka akela particle ho jise total external force dhakel raha ho.

Isse do nateeje nikalte hain. Pehla, centre of mass ki motion predict karne ke liye internal forces jaanne ki zaroorat nahi: na explosion ka push, na muscles ka pull, na do trolleys ke beech ka spring. Doosra, yeh pichhle chapters ko sahi thehraata hai, jahan humne car ya block ko point maana aur uske rotation aur internal motion ko chupchaap ignore kiya. Ab hum jaante hain ki uska centre of mass sach mein point ki tarah move karta hai, baaki kuch bhi ho raha ho.

Socho ek festival shell apni udaan ke top par phat jaata hai (Fig. 6.7). Phatne se pehle gravity shell ko kheenchti hai; baad mein har piece ko kheenchti hai, aur pieces ke weights jud kar phir bhi utne hi total bante hain. Explosion sirf pieces ko ek doosre ke khilaaf dhakelta hai, isliye (6.11) nahi badalta, aur uska answer bhi nahi: pieces ki mass-weighted average position shell ki original parabola par chalti rehti hai. Yeh tab tak sahi hai jab tak koi piece zameen par na gire aur ground system ko push karna shuru na kare. (Hum poore time air drag ignore karte hain.)

3/3Pieces ka centre of mass original parabola par hi rehta hai.

Fig. 6.7Burst ke internal forces centre of mass ko uski parabola se nahi hata sakte.NCERT Fig. 6.12 · p. 100

Left mein launcher se ek dashed parabola shuru hoti hai. Step 1: shell uske along upar jaata hai. Step 2: shell top par phatta hai, pieces bahar ki taraf udte hain. Step 3: paanch pieces apne-apne paths par bikhre hain, aur unke centre of mass ka cross original parabola par hai.

Quick check

3 mein se 1

Parabola par udta ek shell hawa mein do barabar halves mein phat jaata hai. Air resistance ignore karo; kisi bhi half ke zameen par girne se pehle, dono halves ka centre of mass

burst point se seedha neeche girta haioriginal parabola par chalta rehta haiconstant velocity se seedhi line mein chalta haiburst point par hi ruka rehta hai

Answer dekho

Answer (B) original parabola par chalta rehta hai

Explosion ke forces internal hain. Ab bhi sirf gravity hi external force hai, isliye centre of mass usi parabola par rehta hai. Seedha neeche girne ke liye ek sideways external force chahiye hota jo uski horizontal velocity hata de.

6.4 System of particles ka linear momentum

Ek particle ka linear momentum $𝐩 = m 𝐯$ hai, aur Newton ka second law $𝐅 = d 𝐩 / d t$ hai. $n$ particles ke system ke liye hum total momentum ko vector sum define karte hain:

𝐏 = \sum_{i} 𝐩_{i} = \sum_{i} m_{i} 𝐯_{i}

(6.14)

(6.8) se compare karne par ek saaf result milta hai: total momentum = total mass guna centre of mass ki velocity.

𝐏 = M 𝐕

(6.15)

Differentiate karke, aur (6.11) use karke, poore system ke liye Newton ka second law milta hai:

System of particles ke liye second law

\frac{d 𝐏}{d t} = 𝐅_{ext}

(6.17)

LawLinear momentum ka conservation

Agar system par total external force zero hai, to uska total linear momentum constant rehta hai, aur uska centre of mass constant velocity se move karta hai.

Internal forces phir bhi alag-alag particles ko uljhe hue paths par bhej sakte hain. Sirf centre of mass ki guarantee hai ki woh free particle ki tarah steady speed se seedhi line mein chalega.

Radioactive decay dikhata hai ki yeh kyun matter karta hai. Rest par pada uranium-238 nucleus jab alpha particle emit karta hai, to pehle aur baad mein momentum zero hai, isliye bacha hua thorium-234 nucleus seedha peeche recoil karta hai; alpha particle se lagbhag 58 guna bhaari hone ki wajah se yeh lagbhag 58 guna dheere chalta hai. Agar uranium nucleus move kar raha hota, to laboratory ka observer fragments ko ek angle par udte dekhta, aur unka centre of mass bina ruke original line par chalta rehta. Peeth-se-peeth wali picture wahi event hai jo centre of mass ke saath chalne wale frame se dekha gaya, isiliye decays aur collisions aksar is centre-of-mass frame mein solve kiye jaate hain.

Yahi idea orbits ko bhi suljhaata hai. Gravity se bandhi do bodies, bina kisi bahari force ke, apne common centre of mass ke chaaron taraf ghoomti hain jabki woh point uniformly drift karta hai (Fig. 6.8). Earth aur Moon ke liye (Sun ka pull ignore karke), centre of mass, yaani barycentre, Earth ke centre se lagbhag $4700 k m$ door, Earth ke andar hai, kyunki Earth 81 guna bhaari hai. Earth iske chaaron taraf ek chhota monthly circle banati hai; Moon ek bada.

2/2Centre of mass rest par hai; dono bodies opposite sides par uske chaaron taraf ghoomti hain.

Fig. 6.8Tedhe-medhe paths = centre of mass ka steady drift + uske chaaron taraf circles.NCERT Fig. 6.14 · p. 101

Saath bandhi ek bhaari aur ek halki body. Step 1, laboratory frame: unke centre of mass ka cross seedhi line mein chalta hai; bhaari body thoda hilti hai aur halki body opposite side par bade loops banati hai. Step 2, centre-of-mass frame: cross ruka hua hai aur bodies ek chhote aur ek bade circle par chalti hain, hamesha ek doosre ke opposite.

Example

Naav par chalna

Ek $60 k g$ ka naavik shaant jheel par rest mein tair rahi $140 k g$ ki naav ke peechhe khada hai. Woh naav ke along $4.0 m$ aage chalta hai. Paani ka resistance ignore karo; naav kitna move karti hai, aur shore ke relative woh kitna move karta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

6.5 Do vectors ka vector product

Chapter 5 mein tumne force aur displacement ko scalar (dot) product se multiply kiya aur work mila, jo ek number hai. Rotation ke liye ek doosri tarah ka product chahiye, jiska answer ek vector ho jo dono factors ke perpendicular ho. Torque aur angular momentum dono isi tarah bante hain.

Definition

Vector product

Vector product (cross product) $𝐚 \times 𝐛$ ek vector hai jiska magnitude $a b \sin θ$ hai, jo $𝐚$ aur $𝐛$ dono ke perpendicular hai, aur jiska sense right-hand rule se milta hai.

Magnitude: $∣ 𝐚 \times 𝐛 ∣ = a b \sin θ$ , jahan $θ$ dono vectors ke beech ka chhota angle hai ( $0^{\circ}$ aur $180^{\circ}$ ke beech).
Direction: us line ke along jo $𝐚$ aur $𝐛$ wale plane ke perpendicular hai.
Sense: right hand ki ungliyon ko $𝐚$ se $𝐛$ ki taraf chhote angle se modo; tumhara angootha $𝐚 \times 𝐛$ ki direction dikhata hai. Isi tarah, $𝐚$ se $𝐛$ ki taraf ghumaya gaya right-handed screw $𝐚 \times 𝐛$ ke along aage badhta hai (Fig. 6.9).

2/4 $𝐚$ se $𝐛$ ki taraf ghumao: $𝐚 \times 𝐛$ upar point karta hai, size $a b \sin θ$ .

Fig. 6.9Pehle vector se doosre ki taraf modo: angootha cross product deta hai.NCERT Fig. 6.15 · p. 102

Ek jhuke hue plane mein ek point se bane vectors a aur b, beech mein angle theta. Step 2: a se b ki taraf ghumane par product a cross b plane se seedha upar point karta hai. Step 3: b se a ki taraf ghumane par product seedha neeche point karta hai. Step 4: b, a ke opposite point karta hai aur koi product arrow nahi dikhta.

Definition se kuch properties turant nikalti hain:

Order matter karta hai. $𝐛 \times 𝐚$ ka size $𝐚 \times 𝐛$ jitna hi hai lekin sense ulta: $𝐛 \times 𝐚 = - 𝐚 \times 𝐛$ . (Iske ulat, dot product ko order se farq nahi padta.)
Parallel vectors zero dete hain, kyunki $\sin 0^{\circ} = \sin 180^{\circ} = 0$ . Khaas kar $𝐚 \times 𝐚 = 𝟎$ , yaani null vector.
Distributive: $𝐚 \times (𝐛 + 𝐜) = 𝐚 \times 𝐛 + 𝐚 \times 𝐜$ , bilkul waise hi jaise $𝐚 \cdot (𝐛 + 𝐜) = 𝐚 \cdot 𝐛 + 𝐚 \cdot 𝐜$ .
Mirror behaviour: har coordinate ko reflect karne par ( $x \to - x$ , $y \to - y$ , $z \to - z$ ) ek ordinary vector ka sign ulat jaata hai, lekin $(- 𝐚) \times (- 𝐛) = 𝐚 \times 𝐛$ , isliye cross product ka sign nahi badalta.

Unit vectors ke liye $\hat{𝐢} \times \hat{𝐢} = \hat{𝐣} \times \hat{𝐣} = \hat{𝐤} \times \hat{𝐤} = 𝟎$ . Do alag unit vectors perpendicular hote hain, isliye unke product ka size 1 hai aur woh teesre axis ke along point karta hai:

\begin{aligned} \hat{𝐢} \times \hat{𝐣} & = \hat{𝐤} \\ \hat{𝐣} \times \hat{𝐤} & = \hat{𝐢} \\ \hat{𝐤} \times \hat{𝐢} & = \hat{𝐣} \end{aligned}

Yaad rakhne ki trick: cycle $i \to j \to k \to i$ ki direction mein chalne par plus sign milta hai, aur ulti direction mein minus sign, isliye $\hat{𝐣} \times \hat{𝐢} = - \hat{𝐤}$ aur $\hat{𝐢} \times \hat{𝐤} = - \hat{𝐣}$ . In rules se $(a_{x} \hat{𝐢} + a_{y} \hat{𝐣} + a_{z} \hat{𝐤}) \times (b_{x} \hat{𝐢} + b_{y} \hat{𝐣} + b_{z} \hat{𝐤})$ ko term by term multiply karne par milta hai

\begin{aligned} 𝐚 \times 𝐛 = \\ (a_{y} b_{z} - a_{z} b_{y}) \hat{𝐢} \\ + (a_{z} b_{x} - a_{x} b_{z}) \hat{𝐣} \\ + (a_{x} b_{y} - a_{y} b_{x}) \hat{𝐤} \end{aligned}

jise determinant ki tarah yaad rakhna sabse aasaan hai:

Components mein cross product

𝐚 \times 𝐛 = ∣ \begin{array}{ccc} \hat{𝐢} & \hat{𝐣} & \hat{𝐤} \\ a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \end{array} ∣

Example 6.4NCERT Example 6.4 · p. 103

Components mein dot aur cross products

$𝐚 = 2 \hat{𝐢} + 3 \hat{𝐣} - \hat{𝐤}$ aur $𝐛 = \hat{𝐢} - 2 \hat{𝐣} + 4 \hat{𝐤}$ ke liye $𝐚 \cdot 𝐛$ , $𝐚 \times 𝐛$ aur $𝐛 \times 𝐚$ nikalo. Apna cross product check karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

\hat{𝐤} \times \hat{𝐣}

barabar hai

\hat{𝐢}

𝟎

\hat{𝐣}

- \hat{𝐢}

Answer dekho

Answer (D) $- \hat{𝐢}$

\hat{𝐣} \times \hat{𝐤} = \hat{𝐢}

cycle follow karta hai;

\hat{𝐤} \times \hat{𝐣}

uske ulta jaata hai, isliye yeh

- \hat{𝐢}

hai. Yeh zero nahi hai, kyunki

\hat{𝐤}

aur

\hat{𝐣}

perpendicular hain.

6.6 Angular velocity aur linear velocity se uska relation

Fixed axis ke around ghoomti body par wapas aao, jise hum $z$ -axis lete hain. Axis se perpendicular distance $r_{⊥}$ par ek particle P usi radius ke circle par chalta hai, jiska centre axis par ek point C hai. Chhote time $Δ t$ mein radius CP ek angle $Δ θ$ ghoom jaata hai, jo particle ka angular displacement hai. Ratio $Δ θ / Δ t$ uski average angular velocity hai; $Δ t \to 0$ hone par yeh instantaneous angular velocity $ω = d θ / d t$ ban jaati hai.

Circular motion se tum jaante ho ki particle ki speed $v = ω r_{⊥}$ hai. Rigid body ki khaas baat yeh hai ki har particle same time mein same angle ghoomta hai, isliye ek hi $ω$ poori body ko describe karta hai. Bahar wale particles bas tez chalte hain:

v_{i} = ω r_{i}

(6.19)

jahan

$r_{i}$: axis se particle $i$ ki perpendicular distance

Axis par wale particles ka $r_{i} = 0$ hai aur woh ruke rehte hain, jaisa fixed axis ke liye zaroori hai. To pure rotation, pure translation ka jodidaar hai: translation mein har particle ki velocity same hoti hai; rotation mein har particle ki angular velocity same hoti hai.

Angular velocity ki bhi direction hoti hai, aur yeh vector ki tarah behave karti hai (ise prove karne ke liye yahan se zyada maths chahiye). Fixed axis ke liye $𝝎$ axis ke along point karta hai, us direction mein jismein body ke saath ghoomne wala right-handed screw aage badhta. Right hand ki ungliyon ko motion ke along modo aur angootha $𝝎$ deta hai. Deewar ghadi ki suiyan saamne se dekhne par clockwise ghoomti hain, isliye unka $𝝎$ deewar ke andar, tumse door point karta hai.

DerivationVelocity ek vector product ke roop mein

Origin O ko axis par rakho aur $𝐫 = 𝐎 𝐏$ ko particle ka position vector lo (Fig. 6.10). Ise particle ke circle ke centre C par todo:

\begin{aligned} 𝝎 \times 𝐫 & = 𝝎 \times 𝐎 𝐂 \\ + 𝝎 \times 𝐂 𝐏 \end{aligned}

Pehla term zero ho jaata hai kyunki $𝐎 𝐂$ axis ke along hai, $𝝎$ ke parallel. Doosra axis aur radius CP dono ke perpendicular hai, isliye yeh P par circle ke tangent ke along point karta hai. Iska size $ω r_{⊥}$ hai, kyunki $𝝎$ aur $𝐂 𝐏$ right angle par hain. Yahi size aur direction dono mein particle ki velocity hai:

𝐯 = 𝝎 \times 𝐫

(6.20)

(6.20) tab bhi sahi hai jab body ka sirf ek point fixed ho, jaise lattu ka, aur $𝐫$ usi point se naapa jaaye. Fixed axis ke liye $𝝎$ ki direction kabhi nahi badalti, haalaanki uska size badal sakta hai; zyada general rotation mein dono badal sakte hain.

3/3 $𝐯 = 𝝎 \times 𝐫$ circle ka tangent hai, size $ω r_{⊥}$ .

Fig. 6.10

𝐯 = 𝝎 \times 𝐫

circle ka tangent hai aur iska size

ω r_{⊥}

hai.NCERT Fig. 6.17 · p. 104

Ek vertical axis jis par upar ki taraf omega arrow hai, aur us par point C par centred ek horizontal circle, jis par particle P hai. Step 2 axis par origin O aur O se P tak position vector r jodta hai, jo axis ke along OC aur radius CP (length r perpendicular) se bana hai. Step 3 P par velocity v jodta hai, jo circle ka tangent hai.

6.6.1 Angular acceleration

Translation mein acceleration velocity ke badalne ki rate hai. Iska rotational jodidaar angular acceleration hai:

𝜶 = \frac{d 𝝎}{d t}

(6.21)

Fixed axis ke liye $𝝎$ ki direction fixed hai, isliye $𝜶$ bhi usi axis ke along hai aur hum sirf magnitudes ke saath kaam kar sakte hain:

α = \frac{d ω}{d t}

(6.22)

Iski SI unit $r a d / s^{2}$ hai. Axis se distance $r_{⊥}$ par ek particle ka tab apne circle ke along tangential acceleration $r_{⊥} α$ hota hai, axis ki taraf centripetal acceleration $ω^{2} r_{⊥}$ ke alawa.

Quick check

3 mein se 1

Ek ceiling fan

300

rpm par ghoomta hai. Axis se

0.60 m

door blade ki nok kitni tez chalti hai?

m/s

Answer dekho

Answer 18.85 m/s

ω = 300 \times 2 π / 60 = 31.4 r a d / s

, isliye

v = ω r = 31.4 \times 0.60 = 18.8 m / s

6.7 Torque aur angular momentum

Force linear momentum badalta hai. Body ka rotation kya badalta hai? Yeh section force aur momentum ke rotational partners define karta hai. Dono vector products hain, aur dono sabse zyada systems of particles aur rigid bodies ke liye kaam aate hain.

6.7.1 Moment of force (Torque)

Cycle repair shop par mechanic ek ziddi nut kholne ke liye sabse lamba spanner lagata hai aur uske door wale end par handle ke right angle par dhakka deta hai. Nut ke paas dhakka dena kam kaam karta hai, aur handle ke along nut ki taraf kheenchne se kuch nahi hota, chahe woh kitna bhi zor lagaye (Fig. 6.11). Ghumane ka asar force par, woh kahan lagta hai us par, aur uski direction par depend karta hai. Ghumane ke is asar ko moment of force, ya torque (बल आघूर्ण) kehte hain.

Origin O se position $𝐫$ par ek particle par lagne wale force $𝐅$ ke liye O ke about torque hai

Torque

𝝉 = 𝐫 \times 𝐅

(6.23)

jahan

$𝐫$: jahan $𝐅$ lagta hai us point ki position, O se naapi gayi

Iska magnitude, jahan $θ$ $𝐫$ aur $𝐅$ ke beech ka angle hai, do kaam ke tareekon se padha ja sakta hai:

τ = r F \sin θ

(6.24a)

τ = r_{⊥} F = r F_{⊥}

(6.24b)

jahan

$r_{⊥} = r \sin θ$: O se $𝐅$ ki line of action tak perpendicular distance (moment arm)
$F_{⊥} = F \sin θ$: $𝐅$ ka $𝐫$ ke perpendicular component

Seedha O ki taraf diya gaya dhakka, ya khud O par lagaya gaya force, O ke about kuch nahi ghuma sakta. Torque newton metres ( $N m$ ) mein naapa jaata hai, dimensions $[M L^{2} T^{- 2}]$ . Work ke dimensions bhi yahi hain, lekin yeh samaanta sirf upar-upar ki hai: work ek scalar hai jiski unit ko hum joule kehte hain, jabki torque ek vector hai, isliye torque hamesha $N m$ mein likhte hain, kabhi joules mein nahi.

2/4Moment arm ghat kar $r \sin θ$ ho jaata hai: $τ = r F \sin θ$ .

Fig. 6.11Torque = force × moment arm: O se line of action tak perpendicular distance.NCERT Fig. 6.18 · p. 106

O par nut par laga spanner. Step 1: force F door wale end par seedha dhakelta hai, isliye moment arm poori length r hai. Step 2: force angle theta par tircha hai, aur O se dashed perpendicular chhota moment arm r sin theta dikhata hai. Step 3: force handle ke along kheenchta hai aur uski line O se guzarti hai, isliye koi torque nahi. Step 4: nut ke paas seedhe force ka moment arm chhota hai.

Example 6.5NCERT Example 6.5 · p. 108

Components se force ka torque

Ek force $𝐅 = (3 \hat{𝐢} - 2 \hat{𝐣} + \hat{𝐤}) N$ origin se $𝐫 = (\hat{𝐢} + 2 \hat{𝐣} - \hat{𝐤}) m$ par ek bolt ke head par lagta hai. Origin ke about torque aur uska magnitude nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

6.7.2 Ek particle ka angular momentum

Torque force ka moment hai; angular momentum (कोणीय संवेग) linear momentum ka moment hai. O se position $𝐫$ par momentum $𝐩$ wale particle ke liye O ke about angular momentum hai

Ek particle ka angular momentum

𝐥 = 𝐫 \times 𝐩

(6.25a)

\begin{aligned} l & = r p \sin θ \\ = r_{⊥} p = r p_{⊥} \end{aligned}

(6.26)

Yahan $r_{⊥}$ O se us line tak ki perpendicular distance hai jiske along particle chal raha hai. Rest par pada particle, O par baitha particle, ya seedha O ki taraf ya O se door jaata particle, inka O ke about koi angular momentum nahi hota. SI unit $k g m^{2} / s$ hai, jo $J s$ ke barabar hai.

DerivationTorque angular momentum ke badalne ki rate hai

$𝐥 = 𝐫 \times 𝐩$ ko product rule se differentiate karo, factors ka order bana kar rakhte hue:

\begin{aligned} \frac{d 𝐥}{d t} & = \frac{d 𝐫}{d t} \times 𝐩 \\ + 𝐫 \times \frac{d 𝐩}{d t} \end{aligned}

Pehla term $𝐯 \times m 𝐯 = 𝟎$ hai, kyunki kisi vector ka parallel vector ke saath cross product zero hota hai. Doosre mein $d 𝐩 / d t = 𝐅$ . Isliye

\frac{d 𝐥}{d t} = 𝐫 \times 𝐅 = 𝝉

(6.27)

Yeh single particle ke liye $𝐅 = d 𝐩 / d t$ ka rotational form hai.

Example 6.6NCERT Example 6.6 · p. 109

Toll booth ke paas se guzarti car

Ek $1200 k g$ ki car seedhe highway par steady $20 m / s$ se chal rahi hai. Ek toll booth O sadak ki line se $50 m$ door hai (Fig. 6.12). O ke about car ka angular momentum nikalo (a) jab woh booth ke theek saamne ho aur (b) jab woh $120 m$ aage nikal jaaye. Kya O ke about koi torque hai?

Fig. 6.12O se car ki line of motion tak ki perpendicular distance kabhi nahi badalti.NCERT Fig. 6.19 · p. 109

Ek seedhi sadak jis par car do positions par hai, dono baar velocity v se right ki taraf chal rahi hai, aur sadak ke neeche booth O hai. Position 1 par car seedhe O ke upar hai, 50 m door. Position 2 par O se line 130 m lambi hai aur sadak ke saath angle theta banati hai, lekin perpendicular distance ab bhi 50 m hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Dost ke saath try karo

Ghoomta wheel jo girne se mana karta hai

Tumhe chahiye

cycle se nikala hua ek front wheel, jiske axle ke ends bahar nikle hon
lagbhag 1 m lambi mazboot rassi
ek dost

Pehle guess karo

Wheel apne axle ke ek end se latka hai aur tezi se ghoom raha hai. Jab tum doosra end chhodte ho, to wheel

neeche latak jaayega jab tak rassi ke neeche na aa jaayelagbhag level rahega jabki uska axle rassi ke chaaron taraf dheere-dheere ghoomegaaur tez ghoomne lagega

Rassi ko wheel ke axle ke ek end se mazbooti se baandho aur rassi ko haath failaakar upar pakdo.
Wheel ko bina ghumaye axle ka doosra end level pakdo, phir woh end chhod do. Dekho wheel kya karta hai.
Ab axle ko phir se level pakdo jab tumhara dost wheel ko jitna safe ho utna tez ghumaye.
Axle ka free end chhodo aur upar se dekho.

Spin ke bina wheel bas neeche latak jaata hai. Ghoomte hue yeh lagbhag level rehta hai aur uska axle dheere-dheere rassi ke chaaron taraf ghoomta hai. Wheel ke centre par lagne wale weight ka rassi ke about ek horizontal torque hai; $d 𝐋 = 𝝉 d t$ se $𝐋$ mein har chhota badlav horizontal aur $𝐋$ ke perpendicular hai, isliye $𝐋$ , aur uske saath axle, girne ke bajaye ghoom jaata hai. Torque ki direction nikalo aur check karo ki axle usi taraf precess karta hai jaisa $𝝉$ batata hai.

System of particles ke liye torque aur angular momentum

Kisi point ke about system ka total angular momentum uske particles ke angular momenta ka vector sum hai:

\begin{aligned} 𝐋 & = \sum_{i = 1}^{n} 𝐥_{i} \\ = \sum_{i = 1}^{n} 𝐫_{i} \times 𝐩_{i} \end{aligned}

(6.25b)

DerivationExternal torque total angular momentum badalta hai

(6.27) use karke term by term differentiate karo: $d 𝐋 / d t = \sum 𝝉_{i}$ , jahan $𝝉_{i} = 𝐫_{i} \times 𝐅_{i}$ particle $i$ par total force ka torque hai. Har force ko external aur internal part mein todo, taaki $𝝉 = 𝝉_{ext} + 𝝉_{int}$ .

Ab particles ke ek pair, $i$ aur $j$ , ko dekho. Third law se $j$ ki wajah se $i$ par force $𝐅_{i j} = - 𝐅_{j i}$ hai. Yeh bhi maano ki yeh forces particles ko jodne wali line ke along lagte hain. Tab unke torques cancel ho jaate hain:

\begin{aligned} 𝐫_{i} \times 𝐅_{i j} + 𝐫_{j} \times 𝐅_{j i} \\ = (𝐫_{i} - 𝐫_{j}) \times 𝐅_{i j} \\ = 𝟎 \end{aligned}

kyunki $𝐫_{i} - 𝐫_{j}$ , $𝐅_{i j}$ ke parallel hai. Har pair isi tarah cancel hota hai, isliye $𝝉_{int} = 𝟎$ aur

System ke liye torque aur angular momentum

\frac{d 𝐋}{d t} = 𝝉_{ext}

(6.28b)

Dekho iske liye kya chahiye tha: har particle ke liye second law, third law, aur yeh extra condition ki internal forces particles ko jodne wali lines ke along lagein. In sabke saath, kisi point ke about sirf external torque decide karta hai ki us point ke about angular momentum kitni tezi se badlega. (6.17) ki tarah, yeh kisi bhi system ke liye sahi hai, rigid ho ya nahi.

LawAngular momentum ka conservation

Jab kisi point ke about system par koi net external torque nahi lagta, to us point ke about system ka angular momentum badal nahi sakta.

6.8 Rigid body ka equilibrium

Ab se hum rigid bodies dekhenge, aur 'force' aur 'torque' se hamara matlab external wale hain. External forces do alag kaam karte hain. Unka vector sum body ka linear momentum badalta hai ((6.17)), aur unka total torque uska angular momentum badalta hai ((6.28b)). Rigid body mechanical equilibrium (यांत्रिक संतुलन) mein tab hoti hai jab dono mein se kuch nahi badal raha: na linear acceleration hai na angular acceleration. Iske liye do conditions chahiye:

Translational equilibrium

\sum_{i = 1}^{n} 𝐅_{i} = 𝟎

(6.30a)

Rotational equilibrium

\sum_{i = 1}^{n} 𝝉_{i} = 𝟎

(6.30b)

Har vector equation teen scalar equations ke barabar hai, isliye general case mein chhah conditions hain (Eqs. 6.31a, b): har axis ke along force components ka sum zero, aur torque components ka bhi. Aksar saare forces ek plane mein hote hain. Tab teen conditions kaafi hain: us plane mein do perpendicular directions ke along force components ka sum alag-alag zero ho, aur plane ke perpendicular kisi bhi axis ke about torques ka sum zero ho.

Ek kaam ki azaadi: jab forces ka sum zero ho, to total torque har point ke about same hota hai. (Origin ko $𝐬$ se khiskao: har $𝐫_{i}$ ban jaata hai $𝐫_{i} - 𝐬$ , aur total torque $- 𝐬 \times \sum 𝐅_{i} = 𝟎$ se badalta hai.) Isliye torques us point ke about lo jo sabse suvidhajanak ho, aksar woh jahan koi unknown force lagta hai, kyunki tab woh force hat jaata hai. Single particle ke liye sirf force wali condition chahiye, kyunki us par har force same point par lagta hai.

Dono conditions independent hain: body ek satisfy karke doosri fail kar sakti hai (Fig. 6.13). Carrom striker ko opposite edges par same direction mein do barabar dhakke do: centre ke about unke torques cancel ho jaate hain, lekin forces jud jaate hain, isliye woh bina ghoome khiskta hai. Dhakke opposite karo to forces cancel hote hain aur torques judte hain, isliye striker apni jagah par ghoomta hai.

2/2Net force zero, torques judte hain: yeh bina khiske apne centre ke about ghoomta hai.

Fig. 6.13Same direction ke dhakke striker ko translate karte hain; barabar opposite dhakke (couple) use ghumate hain.NCERT Fig. 6.20 · p. 110

Carrom board par ek carrom striker, upar se dekha gaya, uske centre par ek mark aur uske aar-paar ek line. Step 1: barabar forces uske top aur bottom edges ko ek hi taraf dhakelte hain, aur ghost outlines use bina ghoome khiskte dikhate hain. Step 2: forces opposite taraf dhakelte hain, aur ghost outlines marked line ko centre ke about ghoomte dikhate hain jabki striker apni jagah par rehta hai.

Alag lines of action wale do barabar aur opposite forces ek couple (बल युग्म) banate hain, jo body ke centre of mass ko hilaaye bina use ghumata hai. Jab tum ungli aur angoothe se tap ghumate ho, ya dono haathon se steering wheel, to tum couple lagate ho. Yahi cheez latakte bar magnet ko north ki taraf ghumati hai (Fig. 6.14): Earth ka field uske do poles ko barabar aur opposite forces se kheenchta hai, jo couple banate hain jab bhi magnet north–south line ke aar-paar ho. Jab woh north point karne lagta hai, dono forces ek hi line ke along lagte hain aur ghoomna ruk jaata hai.

SimulationCouple latakte magnet ko ghumata hai

Stage ke centre mein ek bar magnet dhaage se latka hai, upar se dekha gaya, top par north dikhane wala marker. Shuru mein yeh east se west point karta hai. Play dabane par yeh north ki taraf ghoomta hai, aage nikal jaata hai, wapas aata hai aur north point karte hue ruk jaata hai. Tum magnet ko kisi bhi starting angle par ghuma kar phir se chhod sakte ho.

Fig. 6.14Ek bar magnet Earth ke field mein dhaage se latka hai. Use chhodne ke liye Play dabao; use naye starting angle par ghumao aur phir se chhodo.

Try karo

Play dabao. Dono poles par forces barabar aur opposite hain, to magnet phir bhi kyun ghoomta hai?
Magnet ko itna ghumao ki uska north pole lagbhag seedha south ki taraf ho, phir Play dabao. Yeh itna dheere kyun shuru hota hai?
Kis starting angle se magnet sabse tezi se ghoomna shuru karta hai? Ise $τ = r_{⊥} F$ se jodo.

Example 6.7NCERT Example 6.7 · p. 111

Couple ka moment har point ke about same hota hai

Ek driver $0.20 m$ radius wale steering wheel ko dono haathon se ghumata hai: top wala haath $15 N$ se right dhakelta hai aur bottom wala $15 N$ se left. Centre ko origin lo, $x$ right ki taraf aur $y$ upar. Is couple ka moment nikalo (a) centre ke about, (b) bottom wale haath ke about, aur (c) kisi bhi point ke about. Dikhao ki result point par depend nahi karta.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

6.8.1 Principle of moments

Koi bhi rigid bar jo kisi fixed point ke about ghoom sake, lever hai, aur woh point uska fulcrum hai. Tum roz levers use karte ho: scissors, bottle opener, beam balance, supari kaatne wala sarota. Bar ko weightless maano, aur do parallel forces $F_{1}$ aur $F_{2}$ us par right angle par lagein, fulcrum ke opposite sides par distances $d_{1}$ aur $d_{2}$ par, jabki fulcrum reaction $R$ se wapas dhakelta hai (Fig. 6.15).

Fig. 6.15Equilibrium mein, load × load arm = effort × effort arm.NCERT Fig. 6.23 · p. 111

Ek bar triangular fulcrum par tika hai. Load F1 fulcrum ke left mein distance d1 par latka hai, effort F2 right mein badi distance d2 par neeche dhakelta hai, aur fulcrum ka reaction R upar point karta hai.

Forces balance hone chahiye, isliye $R = F_{1} + F_{2}$ . Torques ke liye fulcrum chunna samajhdaari hai: wahan unknown $R$ ka koi moment arm nahi aur woh hat jaata hai. Anticlockwise ko positive ginkar $d_{1} F_{1} - d_{2} F_{2} = 0$ , yaani

Principle of moments

d_{1} F_{1} = d_{2} F_{2}

(6.32a)

Yeh principle of moments (आघूर्णों का सिद्धांत) hai. Jab lever kuch uthata hai, to $F_{1}$ load hai aur $d_{1}$ load arm, jabki $F_{2}$ effort hai aur $d_{2}$ effort arm: load × load arm = effort × effort arm. Load aur effort ka ratio mechanical advantage hai:

M.A. = \frac{F_{1}}{F_{2}} = \frac{d_{2}}{d_{1}}

(6.32b)

Effort arm ko load arm ka paanch guna karo aur ek dhakka paanch guna bada load utha leta hai. Principle tab bhi chalta hai jab parallel forces lever ke perpendicular na hokar tirche hon: dono moment arms ek hi factor se chhote hote hain, isliye unka ratio nahi badalta.

Fig. 6.16Beam par weights latkao aur pivot khiskao. Beam tabhi level hota hai jab clockwise aur anticlockwise moments barabar hon.

6.8.2 Centre of gravity

Ek ruler tumhari ungli ki nok par sirf ek khaas point par balance hota hai, aur kahin nahi. Kyun? Ruler ke har chhote piece ka apna weight hai, aur har weight ka ungli ki nok ke about ek torque hai; ek side ke pieces ruler ko ek taraf ghumane ki koshish karte hain aur doosri side ke ulti taraf. Sirf ek point par yeh torques cancel hote hain, aur woh point ruler ka centre of gravity (CG) (गुरुत्व केंद्र) hai. Ungli ka upar ki taraf dhakka, jo total weight $M g$ ke barabar hai, force balance sambhaal leta hai.

DerivationUniform gravity mein centre of gravity

Har particle ki position $𝐫_{i}$ CG se naapo. CG woh point hai jiske about total gravitational torque zero hai:

\begin{aligned} 𝝉_{g} & = \sum_{i} 𝐫_{i} \times m_{i} 𝐠 \\ = 𝟎 \end{aligned}

(6.33)

Agar $𝐠$ har particle par same hai, to yeh sum se bahar aa jaata hai: $(\sum m_{i} 𝐫_{i}) \times 𝐠 = 𝟎$ . Body ko kaise bhi ghumao, woh apne CG par balance hoti hai, isliye yeh body ke relative $𝐠$ ki har direction ke liye sahi hona chahiye, jisse $\sum m_{i} 𝐫_{i} = 𝟎$ hona zaroori hai. Section 6.2 se, jis point ke about yeh sum zero hota hai woh centre of mass hai. Isliye uniform gravitational field mein centre of gravity hi centre of mass hai.

Phir bhi dono alag ideas hain. Centre of mass sirf is par depend karta hai ki mass kaise faila hai. Centre of gravity field par bhi depend karta hai, isliye itni badi body ke liye jismein $g$ badalta ho, jaise bahut ooncha tower, dono points thode alag ho jaate hain.

To chhoti bodies ke liye Section 6.2 ke methods CG bhi dhoondh dete hain. Irregular flat body ke liye plumb line yeh kaam karti hai, jaisa Fig. 6.17 aur neeche ki activity dikhati hai.

3/4Lines CG par milti hain; C se line bhi isi se guzarti hai.

Fig. 6.17CG suspension point ke neeche hota hai, isliye plumb lines usi par milti hain.NCERT Fig. 6.25 · p. 113

Aam ke shape ka cardboard jismein holes A, B aur C hain. Step 1: yeh A se guzarti pin par plumb line ke paas latka hai, aur A se vertical line kheenchi gayi hai. Step 2: yeh B se latka hai aur doosri line kheenchi gayi hai. Step 3: lines G par milti hain, aur C se line bhi G se guzarti hai. Step 4: card G ke neeche pencil ki nok par balance hota hai.

Try karke dekho

Cardboard cut-out ka CG dhoondho

Tumhe chahiye

mote cardboard ki ek sheet
scissors
ek pin aur ek cork board
ek dhaaga jisse ek chhota nut bandha ho
ek pencil aur ek ruler

Pehle guess karo

Tum holes A, B aur C se vertical lines kheenchte ho. Woh kahan milengi?

Teen alag points parZameen ke sabse paas wale hole parEk common point par

Mote cardboard se ek irregular shape kaato aur uske edge ke paas teen chhote holes, door-door, banao.
Hole A mein pin daal kar use cork board ya darwaaze ki frame mein lagao taaki card aazaadi se jhool sake. Plumb line ko bhi usi pin se latkao.
Jab sab ruk jaaye, dhaage ki line par do points mark karo aur unhe card ke aar-paar jodo.
Holes B aur C se dohraao, phir crossing point par card ko pencil ki nok par seedha balance karke dekho.

Example 6.8NCERT Example 6.8 · p. 113

Takhte par painter

$3.0 m$ lamba aur $15 k g$ mass ka uniform takhta do supports par tika hai: ek uske left end A par, aur ek uske right end se $0.50 m$ par. Ek $45 k g$ ki painter A se $1.0 m$ par khadi hai (Fig. 6.18). Har support kitna force lagata hai, nikalo. Takhte ke ulatne se pehle woh right end ki taraf kitni door chal sakti hai? $g = 9.8 m / s^{2}$ lo.

Fig. 6.18Takhte ka free-body diagram: do reactions upar, do weights neeche.NCERT Fig. 6.26 · p. 113

Ek 3.0 m ka takhta apne left end A par aur A se 2.5 m par supports par tika hai. Reactions R1 aur R2 upar point karte hain; painter ka weight 45g A se 1.0 m par neeche lagta hai aur takhte ka weight 15g uske beech mein. Ek scale distances metres mein dikhata hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.9NCERT Example 6.9 · p. 114

Smooth deewar se tiki seedhi

$15 k g$ mass ki ek uniform $5.0 m$ seedhi ek smooth safedi wali deewar se tiki hai, uska pair rough zameen par deewar se $3.0 m$ door hai (Fig. 6.19). Deewar aur zameen ke lagaye forces nikalo, aur friction ka sabse kam coefficient jo pair ko phisalne se roke. $g = 9.8 m / s^{2}$ lo.

Fig. 6.19Smooth deewar sirf apne right angle par dhakel sakti hai; baaki kaam pair par friction karta hai.NCERT Fig. 6.27 · p. 114

Ek seedhi AB jiska pair A deewar se 3.0 m door hai aur top B deewar par 4.0 m upar. Forces hain: midpoint par neeche weight W, B par deewar ka horizontal dhakka Fw, A par upar normal force N, aur A par deewar ki taraf friction f.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

6.9 Moment of inertia

Kise ghumana zyada mushkil hai: cycle ka wheel ya same mass aur radius ki steel disc? Sirf mass se faisla nahi hota; matter yeh karta hai ki mass axis ke relative kahan hai. Mass ka rotational partner dhoondhne ke liye fixed axis ke around ghoomti body ki kinetic energy dekho.

DerivationRotation ki kinetic energy

Axis se distance $r_{i}$ par mass $m_{i}$ ka particle $v_{i} = r_{i} ω$ se chalta hai ((6.19)), isliye uski kinetic energy $\frac{1}{2} m_{i} r_{i}^{2} ω^{2}$ hai. Saare particles ka $ω$ same hai, isliye poori body ke liye

\begin{aligned} K & = \sum_{i} \frac{1}{2} m_{i} v_{i}^{2} \\ = \frac{1}{2} ω^{2} \sum_{i} m_{i} r_{i}^{2} \end{aligned}

Sum $\sum m_{i} r_{i}^{2}$ sirf body aur axis par depend karta hai, is par nahi ki woh kitni tez ghoom rahi hai. Yahi moment of inertia $I$ hai:

Moment of inertia

I = \sum_{i = 1}^{n} m_{i} r_{i}^{2}

(6.34)

Rotational kinetic energy

K = \frac{1}{2} I ω^{2}

(6.35)

Translation ke $K = \frac{1}{2} M v^{2}$ se compare karo: $ω$ , $v$ ka role nibhata hai, aur $I$ mass ka.

Definition

Moment of inertia

Kisi diye gaye axis ke around ghoomti body ke liye $I = \sum m_{i} r_{i}^{2}$ , jahan $r_{i}$ us axis se particle $i$ ki perpendicular distance hai. SI unit $k g m^{2}$ ; dimensions $[M L^{2}]$ .

ExampleMass $M$ aur radius $R$ ke patle hoop ka saara mass uske central axis se distance $R$ par hai, isliye $I = M R^{2}$ .

Mass batata hai ki body apni linear motion ke badlav ka kitna virodh karti hai; moment of inertia (जड़त्व आघूर्ण) batata hai ki woh kisi diye gaye axis ke about apne rotation ke badlav ka kitna virodh karti hai. Mass ke ulat, yeh body ke liye fixed nahi hai. Fig. 6.20 side $a$ ke halke square frame par mass $m$ ke chaar barabar beads dikhata hai: teen alag axes ke about yeh $I = 2 m a^{2}$ , $2 m a^{2}$ aur $m a^{2}$ dete hain.

3/3Do beads axis par, do $a / \sqrt{2}$ par: $I = m a^{2}$ .

Fig. 6.20Wahi beads alag axes ke about alag moments of inertia rakhte hain.NCERT Fig. 6.28 · p. 115

Side a ke square frame ke corners par chaar barabar beads. Step 1: axis frame ke centre par uske perpendicular hai; har bead a over root 2 door hai, jisse 2ma squared milta hai. Step 2: axis ek side ke along hai; do beads us par hain aur do a door, jisse 2ma squared milta hai. Step 3: axis ek diagonal hai; do beads us par hain aur do a over root 2 door, jisse ma squared milta hai.

Continuous bodies ke calculations ke liye integration chahiye, jo tum aage ki classes mein padhoge. Table 6.1 kuch common uniform bodies ke results deta hai. Har entry (mass) × (length)² ke form mein hai, isliye hum hamesha likh sakte hain

I = M k^{2}

jahan

$k$: chune gaye axis ke about radius of gyration

Length $k$ radius of gyration (घूर्णन त्रिज्या) hai: axis se woh distance jahan ek single point mass $M$ ka moment of inertia body ke barabar hoga. Hoop ke liye uske central axis ke about $k = R$ , kyunki uska saara mass pehle se $R$ par hai. Solid sphere ke liye diameter ke about $k = \sqrt{2 / 5} R \approx 0.63 R$ , kyunki uska kaafi mass axis ke paas hai.

Table 6.1Mass

M

ki kuch uniform bodies ke moments of inertiaNCERT Table 6.1 · p. 116

Body	Axis	$I$	$k$
Patli rod, length $L$	Beech se, rod ke ⊥	$M L^{2} / 12$	$L / \sqrt{12}$
Patli ring (hoop), radius $R$	Centre se, uske plane ke ⊥	$M R^{2}$	$R$
Patli ring, radius $R$	Ek diameter	$M R^{2} / 2$	$R / \sqrt{2}$
Disc, radius $R$	Centre se, uske plane ke ⊥	$M R^{2} / 2$	$R / \sqrt{2}$
Disc, radius $R$	Ek diameter	$M R^{2} / 4$	$R / 2$
Patli deewar wala hollow cylinder, radius $R$	Uska apna axis	$M R^{2}$	$R$
Solid cylinder, radius $R$	Uska apna axis	$M R^{2} / 2$	$R / \sqrt{2}$
Solid sphere, radius $R$	Ek diameter	$2 M R^{2} / 5$	$R \sqrt{2 / 5}$

Short mein, $I$ sirf body ki property nahi hai: axis ko khiskao, ya mass ko uske paas ya usse door le jao, aur $I$ badal jaata hai.

Quick check

3 mein se 1

Ek patli ring aur ek uniform disc ka mass aur radius same hai. Centre se guzarne wale, plane ke perpendicular axis ke about,

ring ka moment of inertia zyada haidisc ka moment of inertia zyada haidono ke moments of inertia barabar hainyeh is par depend karta hai ki woh kitni tez ghoomte hain

Answer dekho

Answer (A) ring ka moment of inertia zyada hai

Ring ka saara mass distance

R

par hai (

I = M R^{2}

); disc ka kaafi mass axis ke zyada paas hai (

I = M R^{2} / 2

). Moment of inertia spin rate par depend nahi karta.

6.10 Fixed axis ke around rotational motion ki kinematics

Fixed axis ke around ghoomti body ki position fix karne ke liye bas ek number chahiye: angle $θ$ , jitna uske kisi ek particle tak ka radius us particle ke plane mein ek fixed reference line se ghooma hai. Ek variable, yaani ek degree of freedom, fixed-axis rotation ko seedhi line mein motion ka jodidaar banata hai, jismein $θ$ , $ω = d θ / d t$ aur $α = d ω / d t$ $x$ , $v$ aur $a$ ka role nibhate hain.

Line mein constant acceleration ke liye tum jaante ho $v = v_{0} + a t$ , $x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ aur $v^{2} = v_{0}^{2} + 2 a (x - x_{0})$ . Constant angular acceleration ke saath bilkul wahi algebra deta hai:

ω = ω_{0} + α t

(6.36)

θ = θ_{0} + ω_{0} t + \frac{1}{2} α t^{2}

(6.37)

ω^{2} = ω_{0}^{2} + 2 α (θ - θ_{0})

(6.38)

jahan

$θ_{0}, ω_{0}$: $t = 0$ par angular position aur angular velocity

DerivationIntegration se pehli do equations

Constant $α$ ka matlab $d ω / d t = α$ . $0$ se $t$ tak integrate karo, $t = 0$ par $ω = ω_{0}$ ke saath:

\begin{aligned} \int_{ω_{0}}^{ω} d ω & = \int_{0}^{t} α d t \\ ω & = ω_{0} + α t \end{aligned}

Ab $d θ / d t = ω_{0} + α t$ rakho aur phir integrate karo, $t = 0$ par $θ = θ_{0}$ ke saath:

\begin{aligned} θ - θ_{0} \\ = \int_{0}^{t} (ω_{0} + α t) d t \\ = ω_{0} t + \frac{1}{2} α t^{2} \end{aligned}

Graph se dekho to $θ - θ_{0}$ , $ω$ – $t$ line ke neeche ka area hai: ek rectangle $ω_{0} t$ plus ek triangle $\frac{1}{2} (α t) t$ (Fig. 6.21).

Fig. 6.21

ω

–

t

graph ke neeche ka area angular displacement hai.NCERT Fig. 6.29 · p. 117

Time ke against angular velocity ka straight-line graph, jo omega nought se time t par omega tak badhta hai. Uske neeche ka area ek rectangle, omega nought guna t, aur ek triangle, one half alpha t squared, mein bant jaata hai.

Example 6.10NCERT Example 6.10 · p. 117

Chain rule se teesri equation

$α = d ω / d t =$ constant se shuru karke, time use kiye bina (6.38) nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.11NCERT Example 6.11 · p. 117

Dheema padta ceiling fan

$330$ rpm par chalta ek ceiling fan band kiya jaata hai aur woh uniformly dheema hokar $90 s$ mein ruk jaata hai. Nikalo (i) uska angular acceleration, (ii) rukte hue woh kitne revolutions karta hai, aur (iii) band karne ke $30 s$ baad rpm mein uski speed.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek grinding wheel rest se

2.0 r a d / s^{2}

ke constant angular acceleration ke saath shuru hota hai.

5.0 s

mein woh kitna angle ghoomta hai?

rad

Answer dekho

Answer 25 rad

θ = \frac{1}{2} α t^{2} = \frac{1}{2} (2.0) (5.0)^{2} = 25 r a d

, lagbhag 4 revolutions.

6.11 Fixed axis ke around rotational motion ki dynamics

Humne $θ$ , $ω$ aur $α$ ko $x$ , $v$ aur $a$ ke saath, aur $I$ aur $τ$ ko mass aur force ke saath joda hai. Table 6.2 ke baaki jode, work, power, kinetic energy aur momentum ke liye, analogy se andaaza lagaye ja sakte hain. Lekin andaaza proof nahi hai, isliye is section mein hum unhe derive karte hain.

Table 6.2Translation aur fixed axis ke around rotation, saath-saathNCERT Table 6.2 · p. 119

	Translation	Fixed axis ke around rotation
Position	displacement $x$	angular displacement $θ$
Badalne ki rate	$v = d x / d t$	$ω = d θ / d t$
Uske badalne ki rate	$a = d v / d t$	$α = d ω / d t$
Inertia	mass $M$	moment of inertia $I$
Second law	$F = M a$	$τ = I α$
Work	$d W = F d s$	$d W = τ d θ$
Kinetic energy	$\frac{1}{2} M v^{2}$	$\frac{1}{2} I ω^{2}$
Power	$P = F v$	$P = τ ω$
Momentum	$p = M v$	$L = I ω$

Darwaaze ko uske handle par seedha neeche dabakar, ya hinges ki line ke along kheench kar ghumane ki koshish karo: hinges wapas dhakelte hain aur darwaaza wahin rehta hai. Torque ka sirf woh hissa jo hinge axis ke along point karta hai darwaaze ko ghuma sakta hai; koi bhi doosra hissa axis ko jhukane ki koshish karta hai, aur hinges ya bearings use rokne ke liye jitne forces chahiye utne de dete hain. Isliye fixed axis ke around rotation ke liye hum sirf torque ke axial component ke saath kaam karte hain, axis ke parallel force components ko ignore karte hain aur har moment arm uske perpendicular plane mein naapte hain.

DerivationTorque dwara kiya gaya work

Maan lo ek force $𝐅$ body ke ek point P par lagta hai, axis ke perpendicular plane mein, axis se distance $r$ par (Fig. 6.22). Jab body ek chhota angle $d θ$ ghoomti hai, to P apne circle ke along tangent ki direction mein $d s = r d θ$ chalta hai. Force ka sirf tangential hissa, $F \sin β$ (jahan $β$ , $𝐅$ aur P tak ke radius ke beech ka angle hai), work karta hai:

\begin{aligned} d W & = (F \sin β) d s \\ = (r F \sin β) d θ \\ = τ d θ \end{aligned}

kyunki $r F \sin β$ axis ke about torque hai. Kai forces hon to unke torques sab axis ke along hote hain aur algebraically judte hain, jabki $d θ$ har particle ke liye same hai. Isliye jab total external torque $τ$ ho, to kiya gaya work hai

d W = τ d θ

(6.39)

$d t$ se divide karne par di gayi power milti hai:

P = \frac{d W}{d t} = τ ω

(6.40)

2/3Wheel $d θ$ ghoomta hai; P $d s = r d θ$ chalta hai.

Fig. 6.22Wheel ghoomte samay force ka sirf tangential hissa work karta hai.NCERT Fig. 6.30 · p. 118

Centre C wala ek wheel, saamne se dekha gaya. Step 1: ek force F, P par radius se angle beta par lagta hai, jo radial part aur tangential part F sin beta mein bant jaata hai. Step 2: wheel d theta ghoomta hai aur P rim ke along P prime tak chalta hai, arc ds = r d theta. Step 3: kiya gaya work dW = tau d theta likha hai.

DerivationRotation ke liye Newton ka second law

Poori tarah rigid body ke andar kuch nahi khiskta ya mudta, isliye energy ke paas rotation ke alawa kahin jaane ki jagah nahi: torque dwara di gayi power us rate ke barabar hai jisse $\frac{1}{2} I ω^{2}$ badhta hai. $I$ ko constant lo (fixed mass, rigid body, axis body mein fixed) aur differentiate karo:

\begin{aligned} \frac{d}{d t} (\frac{1}{2} I ω^{2}) & = I ω \frac{d ω}{d t} \\ = I ω α \end{aligned}

Ise power $τ ω$ ke barabar rakho aur $ω$ cancel karo:

Rotation ke liye second law

τ = I α

(6.41)

Torque angular acceleration paida karta hai bilkul waise hi jaise force acceleration. Torque do guna karo to $α$ do guna; moment of inertia do guna karo to $α$ aadha.

Example 6.12NCERT Example 6.12 · p. 120

Ganne ke crusher ko ghumana

Ganne ka juice nikaalne wale crusher ka flywheel $30 k g$ mass aur $0.25 m$ radius ki uniform solid disc hai, jo frictionless axle par ghoomti hai. Uske rim par lipti halki rassi ko rest se shuru karke steady $40 N$ se kheencha jaata hai. Nikalo (a) angular acceleration, (b) $3.0 m$ rassi khulne tak pull dwara kiya gaya work, aur (c) tab flywheel ki kinetic energy. (d) (b) aur (c) ko compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

0.50 k g m^{2}

moment of inertia wale wheel par

4.0 N m

ka torque lagta hai. Uska angular acceleration kya hai?

rad/s²

Answer dekho

Answer 8 rad/s²

α = τ / I = 4.0 / 0.50 = 8.0 r a d / s^{2}

6.12 Fixed axis ke around rotation mein angular momentum

System of particles ke liye torque aur angular momentum mein humne dekha ki kisi point ke about external torque us point ke about angular momentum ka badlav chalata hai. Ab hum dhoondhte hain ki fixed axis ke around ghoomti rigid body ke liye woh angular momentum kya hai.

DerivationFixed axis ke about angular momentum

Origin O ko axis ( $z$ -axis) par rakho. C par centred circle par P par ek particle ke liye $𝐫 = 𝐎 𝐂 + 𝐂 𝐏$ likho, isliye $𝐥 = 𝐎 𝐂 \times m 𝐯 + 𝐂 𝐏 \times m 𝐯$ .

Doosre term mein $𝐂 𝐏$ aur $𝐯$ perpendicular hain, aur right-hand rule dikhata hai ki unka product axis ke along point karta hai. Iska size $r_{⊥} m v = m r_{⊥}^{2} ω$ hai. Pehla term axis ke perpendicular hai. Isliye $𝐥$ ka axis ke along wala hissa hai

𝐥_{z} = m r_{⊥}^{2} ω \hat{𝐤}

aur $𝐥$ ka ek extra sideways hissa bhi hai: single particle ke liye $𝐥$ ka $𝝎$ ke parallel hona zaroori nahi. Poori body par axial parts jodne se milta hai $\sum m_{i} r_{i}^{2} ω \hat{𝐤}$ , yaani

𝐋_{z} = I ω \hat{𝐤}

(6.42b)

Sideways parts, $\sum {𝐎 𝐂}_{i} \times m_{i} 𝐯_{i}$ , pairs mein cancel ho jaate hain jab body axis ke about symmetric ho: har particle ka usi circle par theek saamne ek jodidaar hota hai jiski velocity $- 𝐯_{i}$ hai. Aisi bodies ke liye

𝐋 = 𝐋_{z} = I 𝝎

(6.42d)

Axis ke about tedhi-medhi body, jaise cam ya rim par keechad ke lump wala wheel, ke liye $𝐋$ axis ke along nahi hota.

Kyunki $\hat{𝐤}$ fixed hai, $d 𝐋_{z} / d t = \frac{d}{d t} (I ω) \hat{𝐤}$ . Sirf axis ke along torque matter karta hai (baaki bearings sambhaal lete hain), isliye

\frac{d}{d t} (I ω) = τ

(6.43c)

jabki $𝐋$ ka perpendicular hissa constant rehta hai (Eq. 6.43b). Agar $I$ nahi badalta, to yeh phir se $τ = I α$ ((6.41)) hai, is baar energy ke bajaye angular momentum ke raaste se.

6.12.1 Angular momentum ka conservation

Agar axis ke about external torque zero hai, to (6.43c) deta hai

Angular momentum ka conservation (fixed axis)

I ω = constant

(6.44)

Ab $I$ badal sakta hai, jab tak body ya system bina external torque ke usi axis ke about ghoomta rahe. Agar $I$ ghatta hai to $ω$ usi proportion mein badhta hai, aur ulta bhi: $I_{1} ω_{1} = I_{2} ω_{2}$ (Fig. 6.23). Axis ke about symmetric bodies ke liye yahan $L_{z}$ hi poora $L$ hai.

Chakkar leti Kathak dancer baahen faila kar shuru karti hai aur unhe andar kheench kar aur tez ghoomti hai. Diver board se dheere ghoomte hue nikalta hai, tezi se somersault karne ke liye gend ki tarah simat jaata hai, phir paani mein jaane se pehle dheema hone ke liye khul jaata hai. Bade scale par, ek marta hua star jo neutron star mein collapse hota hai, ek second mein saikdon chakkar lagane lagta hai.

2/2Chhota $I$ , tez spin: $I_{1} ω_{1} = I_{2} ω_{2}$ .

Fig. 6.23Bina external torque ke

I ω

constant rehta hai: baahen andar kheenchne se spin tez hota hai.NCERT Fig. 6.32 · p. 122

Aazaadi se ghoomte stool par ek vyakti har haath mein paani ki ek bottle pakde hai. Step 1: baahen poori bahar, dheeme ghoomne ke liye ek curved arrow, label bada I, chhota omega. Step 2: bottles chhaati ki taraf kheenchi hui, tez ghoomne ke liye teen curved arrows, label chhota I, bada omega.

Try karke dekho

Ghoomti kursi

Tumhe chahiye

aazaadi se ghoomne wali swivel chair
do 1-litre paani ki bottles, bhari hui
ghadi wala ek dost

Pehle guess karo

Tum baahen bahar karke ghoom rahe ho aur phir bottles ko chhaati ki taraf kheench lete ho. Tumhara spin

dheema ho jaayegasame rahegatez ho jaayega

Swivel chair par pair zameen se upar karke baitho, har haath mein bhari paani ki bottle haath faila kar pakdo.
Dost se kaho ki chair ko halka sa ghumaye, phir chhod de.
Baahen bahar rakh kar 5 seconds mein kitne chakkar lagte hain gino. Phir bottles chhaati se lagao aur phir gino.
Bottles ko phir bahar le jao aur dekho kya hota hai.

Table 6.35 seconds mein chakkar

	5 s mein chakkar
Baahen bahar
Baahen andar

Tumhari entries apne aap save hoti hain.

Example

Kumhaar ke chaak par giri mitti

$0.40 k g m^{2}$ moment of inertia wala kumhaar ka chaak $6.0 r a d / s$ par aazaadi se ghoom raha hai. $2.0 k g$ ka mitti ka lump us par axis se $0.20 m$ door dheere se giraya jaata hai aur chipak jaata hai. Lump ko point mass maan kar aur friction ignore karke nayi angular speed aur khoyi hui kinetic energy nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 6.24Stool ghumane ke liye Play dabao, phir weights andar kheencho ya bahar dhakelo. Readouts dikhate hain ki kaun si quantity constant rehti hai.

Quick check

3 mein se 1

Ek skater

4.0 k g m^{2}

moment of inertia ke saath

2.0

rev/s par ghoom rahi hai. Woh baahen andar kheenchti hai aur uska moment of inertia ghat kar

1.6 k g m^{2}

ho jaata hai. Uska naya spin rate kya hai?

rev/s

Answer dekho

Answer 5 rev/s

I_{1} ω_{1} = I_{2} ω_{2}

ω_{2} = 4.0 \times 2.0 / 1.6 = 5.0

rev/s. Unit rev/s hi rakh sakte ho kyunki woh cancel ho jaati hai.

Recap

Summary aur key points

Rigid body apne particles ke beech ki distances fixed rakhti hai. Bina pivot ke yeh translate, ya translate aur rotate kar sakti hai; pivoted hone par sirf rotate kar sakti hai (Section 6.1.1).
Fixed axis ke about har particle axis ke right angle wale plane mein, axis par centred circle mein ghoomta hai; sabki angular velocity ek hi hai, $ω = d θ / d t$ (Section 6.6).
Centre of mass $𝐑 = \sum m_{i} 𝐫_{i} / M$ (ya $\int 𝐫 d m / M$ ) hai; symmetric uniform bodies ke liye yeh geometric centre hai, jo material ke bahar bhi ho sakta hai (Section 6.2).
Centre of mass mass $M$ ke particle ki tarah move karta hai jis par total external force lag raha ho: $M 𝐀 = 𝐅_{ext}$ . Internal forces iski motion nahi badal sakte (Section 6.3).
Total momentum $𝐏 = M 𝐕$ aur $d 𝐏 / d t = 𝐅_{ext}$ ; external force na ho to $𝐏$ conserve hota hai (Section 6.4).
$𝐚 \times 𝐛$ ka size $a b \sin θ$ hai, yeh dono vectors ke perpendicular hai aur sense right-hand rule se milta hai, aur $𝐛 \times 𝐚 = - 𝐚 \times 𝐛$ (Section 6.5).
Ghoomti body ke liye $𝐯 = 𝝎 \times 𝐫$ , aur angular acceleration $𝜶 = d 𝝎 / d t$ hai (Section 6.6).
Torque $𝝉 = 𝐫 \times 𝐅$ aur angular momentum $𝐥 = 𝐫 \times 𝐩$ ; system ke liye $d 𝐋 / d t = 𝝉_{ext}$ jab internal forces central hon, isliye $𝝉_{ext} = 𝟎$ hone par $𝐋$ conserve hota hai (Section 6.7).
Mechanical equilibrium ke liye $\sum 𝐅 = 𝟎$ aur $\sum 𝝉 = 𝟎$ chahiye; couple ka net force zero hai lekin torque har point ke about same hai (Section 6.8).
Principle of moments kehta hai: load × load arm = effort × effort arm; mechanical advantage $= d_{2} / d_{1}$ (Section 6.8.1).
Centre of gravity woh point hai jahan total gravitational torque zero ho jaata hai; $g$ uniform ho to yeh centre of mass par hi hota hai (Section 6.8.2).
Moment of inertia $I = \sum m_{i} r^{2} = M k^{2}$ axis par depend karta hai; rotational kinetic energy $\frac{1}{2} I ω^{2}$ hai (Section 6.9).
Constant $α$ ke saath: $ω = ω_{0} + α t$ , $θ - θ_{0} = ω_{0} t + \frac{1}{2} α t^{2}$ , $ω^{2} = ω_{0}^{2} + 2 α (θ - θ_{0})$ (Section 6.10).
Fixed axis ke about: $d W = τ d θ$ , $P = τ ω$ , $τ = I α$ , $L_{z} = I ω$ ; external torque na ho to $I ω$ constant rehta hai (Section 6.11, Section 6.12.1).

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Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 6.1	Practice Q1
NCERT Exercises 6.2	Practice Q2
NCERT Exercises 6.3	Practice Q3
NCERT Exercises 6.4	Practice Q4
NCERT Exercises 6.5	Practice Q5
NCERT Exercises 6.6	Practice Q6
NCERT Exercises 6.7	Practice Q7
NCERT Exercises 6.8	Practice Q8
NCERT Exercises 6.9	Practice Q9
NCERT Exercises 6.10	Practice Q10
NCERT Exercises 6.11	Practice Q11
NCERT Exercises 6.12	Practice Q12
NCERT Exercises 6.13	Practice Q13
NCERT Exercises 6.14	Practice Q14
NCERT Exercises 6.15	Practice Q15
NCERT Exercises 6.16	Practice Q16
NCERT Exercises 6.17	Practice Q17

Quick check: MCQs

6.1.1 Rigid body kis tarah ki motion kar sakti hai?3 questions
6.2 Centre of mass3 questions
6.3 Centre of mass ki motion3 questions
6.4 System of particles ka linear momentum3 questions
6.5 Do vectors ka vector product4 questions
6.6.1 Angular acceleration3 questions
System of particles ke liye torque aur angular momentum4 questions
6.8.2 Centre of gravity4 questions
6.9 Moment of inertia3 questions
6.10 Fixed axis ke around rotational motion ki kinematics3 questions
6.11 Fixed axis ke around rotational motion ki dynamics3 questions
6.12.1 Angular momentum ka conservation3 questions

Glossary · 13

Key terms (English + Hindi)

angular accelerationकोणीय त्वरणkoniya tvaran: Angular velocity ke badalne ki rate, $α = d ω / d t$ , $r a d / s^{2}$ mein.Chapter mein pehli baar yahan aaya →
angular momentumकोणीय संवेगkoniya sanveg: Linear momentum ka moment, $𝐥 = 𝐫 \times 𝐩$ ; apne axis ke about ghoomti symmetric body ke liye $L = I ω$ .Chapter mein pehli baar yahan aaya →
angular velocityकोणीय वेगkoniya veg: Angular position ke badalne ki rate, $ω = d θ / d t$ , rad/s mein; vector ke roop mein yeh right-hand rule se axis ke along point karti hai.Chapter mein pehli baar yahan aaya →

centre of gravityगुरुत्व केंद्रgurutva kendra: Woh point jiske about body par total gravitational torque zero hai; g uniform ho to yeh centre of mass hi hai.Chapter mein pehli baar yahan aaya →
centre of massद्रव्यमान केंद्रdravyaman kendra: System ki mass-weighted average position, $𝐑 = \sum m_{i} 𝐫_{i} / M$ . Yeh aise move karta hai jaise saara mass wahin ho aur saare external forces wahin lag rahe hon.Chapter mein pehli baar yahan aaya →
coupleबल युग्मbal yugm: Alag lines of action wale do barabar aur opposite forces. Yeh body ko translate kiye bina ghumata hai.Chapter mein pehli baar yahan aaya →

mechanical equilibriumयांत्रिक संतुलनyantrik santulan: Rigid body ki woh state jismein us par net external force aur net external torque dono zero hon.Chapter mein pehli baar yahan aaya →
moment of inertiaजड़त्व आघूर्णjadatva aaghurn: Kisi axis ke about body ka rotational inertia, $I = \sum m_{i} r_{i}^{2}$ , $k g m^{2}$ mein.Chapter mein pehli baar yahan aaya →

principle of momentsआघूर्णों का सिद्धांतaaghurnon ka siddhant: Equilibrium mein lever ke liye, load × load arm = effort × effort arm.Chapter mein pehli baar yahan aaya →

radius of gyrationघूर्णन त्रिज्याghurnan trijya: Axis se woh distance $k$ jahan poora mass rakhne par same moment of inertia mile: $I = M k^{2}$ .Chapter mein pehli baar yahan aaya →
rigid bodyदृढ़ पिंडdridh pind: Ek ideal body jismein kisi bhi force ke under har pair of particles ke beech ki distance fixed rehti hai.Chapter mein pehli baar yahan aaya →

torqueबल आघूर्णbal aaghurn: Kisi point ke about force ka ghumane wala asar, $𝝉 = 𝐫 \times 𝐅$ , size force × moment arm, N m mein.Chapter mein pehli baar yahan aaya →

vector productसदिश गुणनफलsadish gunanphal: Vector $𝐚 \times 𝐛$ jiska size $a b \sin θ$ hai, jo dono vectors ke perpendicular hai, aur jiska sense right-hand rule se milta hai.

System of Particles and Rotational Motion: Class 11 Physics Chapter 6 Hinglish mein

Is chapter mein aap kya seekhenge

6.1 Introduction

6.1.1 Rigid body kis tarah ki motion kar sakti hai?

6.2 Centre of mass

6.3 Centre of mass ki motion

6.4 System of particles ka linear momentum

6.5 Do vectors ka vector product

6.6 Angular velocity aur linear velocity se uska relation

6.6.1 Angular acceleration

6.7 Torque aur angular momentum

6.7.1 Moment of force (Torque)

6.7.2 Ek particle ka angular momentum

System of particles ke liye torque aur angular momentum

6.8 Rigid body ka equilibrium

6.8.1 Principle of moments

6.8.2 Centre of gravity

6.9 Moment of inertia

6.10 Fixed axis ke around rotational motion ki kinematics

6.11 Fixed axis ke around rotational motion ki dynamics

6.12 Fixed axis ke around rotation mein angular momentum

6.12.1 Angular momentum ka conservation

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

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Key terms (English + Hindi)

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