🔄 Chapter 7: System of Particles and Rotational Motion (Class 11) 🔩
📌 Introduction
Ab tak humne point objects (ya particles) ke motion ko study kiya (translational motion). Lekin real world mein objects finite size ke hote hain (system of particles / rigid bodies). Ab hum aisi bodies ke motion ko study karenge, especially their **rotational motion** (ghurnan gati).
Ab tak humne bindu vastuon (point objects) ki gati (translational motion) ko study kiya. Lekin asli duniya mein vastuon ka size hota hai (kanon ka nikay / dridh pind). Ab hum aisi bodies ke motion ko study karenge, khaaskar unki **ghurnan gati (rotational motion)**.- System of Particles: A collection of a large number of particles interacting with each other.
- Rigid Body: A body with a perfectly definite and unchanging shape. The distances between all pairs of particles of such a body do not change during motion. (Idealization). Dridh Pind (Rigid Body): Ek aisi vastu jiska aakar bilkul nishchit aur aparivartansheel ho. Aisi vastu ke kanon ke sabhi jodon ke beech ki dooriyan gati ke dauran nahi badalti hain. (Ek adarsh kalpana).
🎯 Centre of Mass (Dravyaman Kendra – CM)
Centre of Mass (CM): A point associated with a system of particles (or a rigid body) where the entire mass of the system can be assumed to be concentrated for describing its translational motion. It moves as if the total external force were applied at this point.
Dravyaman Kendra (CM): Kanon ke nikay (ya dridh pind) se juda ek aisa bindu jahan uske translational motion ka varnan karne ke liye nikay ka saara dravyaman kendrit maana ja sakta hai. Yeh aise chalta hai jaise saara bahari bal isi bindu par lag raha ho.🧑🤝🧑 CM of a Two-Particle System
Consider two particles of masses m₁ and m₂ with position vectors r₁ and r₂.
Position Vector of Centre of Mass (RCM):
RCM = m₁r₁ + m₂r₂m₁ + m₂
- If origin is chosen such that particles are on x-axis at x₁ and x₂, then
XCM = (m₁x₁ + m₂x₂) / (m₁ + m₂). - If the origin is AT the centre of mass, then
m₁r₁ + m₂r₂ = 0. - For a system of ‘n’ particles:
RCM = (Σ miri) / (Σ mi) = (1/M) Σ miri(where M is total mass). - Coordinates:
XCM = (1/M) Σ mixi,YCM = (1/M) Σ miyi,ZCM = (1/M) Σ mizi.
✅ Solved Example (CM of 2 Particles)
Problem 1: Find the centre of mass of two particles of masses 1 kg and 2 kg situated at x=0 and x=3 m respectively on the x-axis.
Given: m₁ = 1 kg, x₁ = 0 m; m₂ = 2 kg, x₂ = 3 m.
Since motion is along x-axis only, we find XCM.
XCM = m₁x₁ + m₂x₂m₁ + m₂
XCM = (1)(0) + (2)(3)1 + 2
XCM = 0 + 63 = 63 = 2 m.
Answer: Centre of mass is at x = 2 m.
⏩ Momentum Conservation and Centre of Mass Motion
- Total linear momentum of a system of particles (Psys) = Vector sum of individual momenta (
Σpi). - Velocity of Centre of Mass (VCM):
VCM = dRCMdt = Σ miviM = PsysM. - Therefore, Total Momentum
Psys = M VCM. (Total mass × Velocity of CM). - Newton’s 2nd Law for the System:
Fext = dPsysdt = d(MVCM)dt. - If mass M is constant,
Fext = M dVCMdt = M ACM(where ACM is acceleration of CM). - Conclusion: The centre of mass of a system moves as if the entire mass of the system were concentrated at the CM and the total external force were applied at that point. 🎯 System ka centre of mass aise chalta hai jaise poora mass CM par concentrate ho aur saara external force usi par lag raha ho.
- If Fext = 0: Then ACM = 0, which means VCM = constant. The centre of mass moves with constant velocity (or stays at rest if initially at rest). This relates to conservation of momentum.
🔩 Centre of Mass of a Rigid Body
- A rigid body is a system of closely packed particles with fixed relative distances.
- The Centre of Mass (CM) of a rigid body is a fixed point relative to the body.
- For bodies with uniform mass distribution and symmetry, the CM coincides with the geometrical centre. Symmetric bodies (jaise ring, disc, sphere, rod) jinka mass uniformly distributed hai, unka CM unke geometrical centre par hota hai.
- Example: CM of a uniform ring/disc is at its center. CM of a sphere is at its center.
- Centre of Mass of a Uniform Rod: Located at its midpoint (geometrical centre). Ek Saman Chhad (Uniform Rod) ka CM: Uske **madhyabindu (midpoint)** par hota hai.
✅ Solved Example (CM Motion)
Problem 2: Two bodies of masses 1 kg and 3 kg are moving towards each other with speeds 3 m/s and 1 m/s respectively. Find the velocity of their centre of mass.
Let m₁=1kg, m₂=3kg. Assume they move along x-axis.
Let direction of m₁ be positive. u₁ = +3 m/s.
m₂ moves towards m₁, so its velocity is opposite. u₂ = -1 m/s.
Total mass M = m₁ + m₂ = 1 + 3 = 4 kg.
Total initial momentum Psys = m₁u₁ + m₂u₂
Psys = (1)(+3) + (3)(-1) = 3 - 3 = 0 kg m/s.
Velocity of CM, VCM = Psys / M.
VCM = 0 / 4 = 0 m/s.
(Since net external force is likely zero here, CM velocity remains constant. If initially zero, it remains zero).
Answer: Velocity of Centre of Mass = 0 m/s
🔄 Rotational Motion Basics
Now we consider motion where the body rotates about an axis.
🔩➡️ Moment of Force (Torque – τ)
Torque (or Moment of Force): The turning effect of a force about an axis of rotation. It is the rotational analogue of force.
Bal Aaghoorn (Torque): Kisi ghoornan aksh (axis of rotation) ke paritah ek bal ka ghurnan prabhav. Yeh force ka rotational equivalent hai.Torque (τ) due to force F acting at position r from the axis/pivot:
τ = r × F (Vector/Cross Product)
Magnitude: τ = r F sin θ
(where θ is the angle between r and F).
Alternatively: τ = Force × Perpendicular distance of line of action of force from axis (Lever arm).
- Torque is a vector quantity. Its direction is perpendicular to the plane of r and F (given by Right-Hand Rule). (Torque ki disha r aur F ke plane ke perpendicular hoti hai).
- SI Unit: newton-metre (N m).
- Dimensional Formula: [Torque] = [r][F] = [L][MLT⁻²] = [ML²T⁻²] (same as work/energy, but represents turning effect).
- Torque causes angular acceleration.
Example: Opening a Door 🚪
You apply force at the handle (far from hinges – axis of rotation). Perpendicular distance (lever arm) is large, so less force needed to produce turning effect (torque). Pushing near hinges is harder because lever arm is small, needing more force for same torque.
💫 Angular Momentum (L)
Angular Momentum (L): The rotational analogue of linear momentum. It measures the quantity of rotational motion.
Koniya Samveg (L): Raikhik samveg ka rotational equivalent. Yeh ghurnan gati ki matra ko maapta hai.For a particle with linear momentum p at position r from the axis:
L = r × p (Vector/Cross Product)
Magnitude: L = r p sin φ = rmv sin φ
(where φ is the angle between r and p).
- Angular Momentum is a vector quantity. Its direction is perpendicular to the plane of r and p (Right-Hand Rule).
- SI Unit: kg m²/s.
- Dimensional Formula: [L] [MLT⁻¹] = [ML²T⁻¹].
Relation between Torque and Angular Momentum:
- Similar to Newton’s 2nd law (F = dp/dt), the rotational equivalent is:
Net External Torque (τext) = Rate of change of Angular Momentum
τext = dLdt
🔒 Law of Conservation of Angular Momentum
Statement: If the net external torque acting on a system is zero, then the total angular momentum (L) of the system remains constant (conserved).
If τext = 0, then L = Constant
Applications:
- Diver/Skater Spinning: When a diver tucks in or skater brings arms closer, their moment of inertia (I) decreases. To conserve L=Iω, their angular velocity (ω) increases, making them spin faster 🤸❄️. Extending limbs increases I, decreasing ω.
- Planetary Motion: Planets orbit sun under gravity (a central force). Torque due to gravity about sun is zero. Hence, angular momentum (L = mvr sin(90°) = mvr for near circular orbit) is conserved. Actually, areal velocity (dA/dt = L/2m) is constant (Kepler’s 2nd Law).
— End of Part 1: Notes & Concepts —
Part 2 (Equilibrium, Rotational Motion, MOI, Q&A, Quiz) will follow your confirmation.
Chapter 7 (Part 2): Equilibrium, Rotation & MOI
(Continuing from Centre of Mass, Torque, Angular Momentum)
⚖️ Equilibrium of Rigid Bodies (Dridh Pindo Ka Santulan)
For a rigid body to be in equilibrium, it must satisfy two conditions simultaneously:
Ek rigid body ke equilibrium mein hone ke liye, do shartein ek saath poori honi chahiye:- 1. Translational Equilibrium: The net external force acting on the body must be zero. Sthanantariy Santulan: Vastu par lagne wala net bahari bal zero hona chahiye.
- 2. Rotational Equilibrium: The net external torque acting on the body about any point must be zero. Ghurnan Santulan: Vastu par kisi bhi bindu ke sapeksh lagne wala net bahari torque zero hona chahiye.
Σ Fext = 0
(This ensures the Centre of Mass has zero acceleration: ACM = 0).
Σ τext = 0
(This ensures the body has zero angular acceleration: α = 0).
If both conditions are met, the body is in complete equilibrium (neither translating nor rotating, or moving with constant linear AND angular velocity).
Agar dono conditions poori hoti hain, toh body poorn santulan mein hai.✅ Solved Example (Equilibrium)
Problem 1: A uniform rod of weight W is supported horizontally by two vertical strings attached at its ends. Find the tension in each string.
Let the tensions be T₁ and T₂ acting upwards at the ends. Length of rod = L.
Weight W acts downwards at the Centre of Mass (midpoint, L/2 from ends).
Translational Equilibrium (Vertical forces):
Sum of upward forces = Sum of downward forces.
T₁ + T₂ = W — (Eq 1)
Rotational Equilibrium (Torques about one end, say where T₁ acts):
Net torque about this end must be zero.
Torque due to T₁ = 0 (passes through pivot).
Torque due to W = W × (L/2) (clockwise, let’s say negative).
Torque due to T₂ = T₂ × L (counter-clockwise, let’s say positive).
Στ = 0 ⇒ T₂ × L - W × (L/2) = 0.
T₂ L = W L / 2
Cancel L: T₂ = W / 2.
Substitute T₂ in Eq 1: T₁ + W/2 = W.
T₁ = W - W/2 = W / 2.
Answer: Tension in each string is W/2.
🔄 Rigid Body Rotation and Equations
When a rigid body rotates about a fixed axis, every particle of the body moves in a circle whose centre lies on the axis and radius is the perpendicular distance from the axis.
Jab ek rigid body ek sthir axis ke chaaron taraf ghoomti hai, toh uska har kan ek circle mein ghoomta hai jiska kendra axis par hota hai.Rotational Analogues of Linear Motion Quantities:
- Linear Displacement (s) ↔️ Angular Displacement (θ) [unit: radian]
- Linear Velocity (v = ds/dt) ↔️ Angular Velocity (ω = dθ/dt) [unit: rad/s]
- Linear Acceleration (a = dv/dt) ↔️ Angular Acceleration (α = dω/dt) [unit: rad/s²]
- Mass (m – inertia for linear) ↔️ Moment of Inertia (I – inertia for rotation)
- Force (F = ma) ↔️ Torque (τ = Iα)
- Linear Momentum (p = mv) ↔️ Angular Momentum (L = Iω – for rotation about fixed axis)
Equations of Rotational Motion (For CONSTANT Angular Acceleration α):
These are analogous to linear equations of motion.
Let ω₀ = Initial Angular Velocity, ω = Final Angular Velocity, θ = Angular Displacement, t = Time, α = Constant Angular Acceleration.
1st Eqn: ω = ω₀ + αt (Analogue of v = u + at)
2nd Eqn: θ = ω₀t + 12αt² (Analogue of s = ut + ½at²)
3rd Eqn: ω² = ω₀² + 2αθ (Analogue of v² = u² + 2as)
Relation between Linear and Angular Variables (for particle at distance r from axis):
- Linear speed
v = ωr - Tangential acceleration
aT = αr - Centripetal acceleration
ac = v²/r = ω²r
🏋️♀️🔄 Moment of Inertia (Jadatva Aaghoorn – I)
Moment of Inertia (I): The rotational analogue of mass. It represents the resistance of a rigid body to change its state of rotational motion. It depends not only on mass but also on how the mass is distributed relative to the axis of rotation.
Jadatva Aaghoorn (I): Dravyamaan ka rotational equivalent. Yeh ek rigid body ki apni ghurnan gati ki avastha mein parivartan ke prati **pratirodh (resistance)** ko darshata hai. Yeh sirf mass par hi nahi, balki is par bhi nirbhar karta hai ki mass ghurnan aksh ke sapeksh kaise vitarit hai.For a single particle of mass ‘m’ at perpendicular distance ‘r’ from axis:
I = mr²
For a system of n particles:
I = Σ miri² = m₁r₁² + m₂r₂² + ... + mnrn²
- Moment of Inertia (MOI) is a scalar quantity.
- SI Unit: kg m².
- Dimensional Formula: [ML²].
- It depends on: Mass of body, Size & shape of body, Distribution of mass about axis, Position & orientation of axis of rotation.
📏 Radius of Gyration (Ghoornan Trija – K)
Radius of Gyration (K): The perpendicular distance from the axis of rotation to a point where the entire mass (M) of the body can be assumed to be concentrated, so that its moment of inertia about the axis is the same as the actual MOI of the body.
Ghoornan Trija (K): Ghurnan aksh se us bindu tak ki lambvat doori jahan vastu ka poora dravyaman (M) kendrit maana ja sakta hai, taaki uska us aksh ke paritah jadatva aaghoorn vastu ke vastavik MOI ke barabar ho.Moment of Inertia I = MK²
Radius of Gyration K = √(I/M)
- SI Unit of K: metre (m).
📐 Values of Moments of Inertia (MOI) for Simple Geometrical Objects (No Derivation Required for Board Level):
- Thin Uniform Rod, Axis through centre ⟂ length:
I = 112 ML² - Thin Uniform Rod, Axis through one end ⟂ length:
I = 13 ML² - Circular Ring, Axis through centre ⟂ plane:
I = MR² - Circular Ring, Axis along diameter:
I = 12 MR² - Circular Disc, Axis through centre ⟂ plane:
I = 12 MR² - Circular Disc, Axis along diameter:
I = 14 MR² - Hollow Cylinder, Axis of cylinder:
I = MR² - Solid Cylinder, Axis of cylinder:
I = 12 MR² - Solid Sphere, Axis through centre (diameter):
I = 25 MR² - Hollow Sphere, Axis through centre (diameter):
I = 23 MR²
(M = Total Mass, R = Radius, L = Length)
Theorems for MOI (Names only needed for JEE/NEET context usually):
- Theorem of Parallel Axes: I = ICM + Md² (I about any axis = I about parallel axis through CM + Mass × (distance between axes)²)
- Theorem of Perpendicular Axes (For planar bodies only): Iz = Ix + Iy (MOI about axis ⟂ plane = Sum of MOIs about two ⟂ axes in the plane intersecting at the same point).
💡 Solved Practice Problems: Rotation & MOI
Problem 6: A wheel starting from rest rotates with a constant angular acceleration of 2.0 rad/s². What is its angular velocity after 5 seconds?
Given: Initial angular velocity ω₀ = 0 (rest).
Angular acceleration α = 2.0 rad/s².
Time t = 5 s.
Use 1st equation of rotational motion: ω = ω₀ + αt.
ω = 0 + (2.0)(5) = 10 rad/s.
Answer: 10 rad/s
Problem 7: Find the moment of inertia of a solid sphere of mass 5 kg and radius 0.1 m about its diameter.
Given: Mass M = 5 kg, Radius R = 0.1 m.
Object: Solid Sphere, Axis: Diameter.
Formula for MOI of solid sphere about diameter: I = 25 MR².
I = 25 × (5) × (0.1)²
I = 2 × (0.01) = 0.02 kg m².
Answer: 0.02 kg m²
❓ Sawal Jawab (Questions & Answers)
🤏 Very Short Answer Questions
1. What is Centre of Mass (CM)?
A point where the entire mass of a system is assumed to be concentrated for translational motion.
2. Where is the CM of a uniform rod located?
At its midpoint (geometrical centre).
3. Write the relation between total momentum (P) and velocity of CM (VCM).
P = M VCM (where M is total mass).
4. What happens to the motion of CM if net external force is zero?
Its velocity remains constant (VCM = constant).
5. Define Torque (Moment of Force).
The turning effect of a force about an axis.
6. What is the SI unit of Torque?
Newton-metre (N m).
7. Is Torque a scalar or vector?
Vector.
8. Define Angular Momentum (L).
Moment of linear momentum (L = r × p). Rotational analogue of momentum.
9. State the Law of Conservation of Angular Momentum.
If net external torque is zero, total angular momentum is conserved (constant).
10. Give an application of conservation of angular momentum.
Diver tucking in, Skater spinning faster, Planetary motion.
11. What are the conditions for equilibrium of a rigid body?
Net external force = 0 AND Net external torque = 0.
12. Define Moment of Inertia (I).
Measure of resistance to change in rotational motion (Rotational Inertia).
13. On what factors does Moment of Inertia depend?
Mass, Shape/Size, Mass Distribution, Axis of Rotation.
14. Write the SI unit of Moment of Inertia.
kg m².
15. Define Radius of Gyration (K).
Distance from axis where entire mass could be concentrated to give same MOI (I = MK²).
16. What is the MOI of a thin rod about axis through centre, perpendicular to length?
ML²/12.
17. What is the MOI of a solid sphere about its diameter?
(2/5)MR².
18. Write the rotational analogue of F=ma.
τ = Iα.
19. Write one equation of rotational motion for constant angular acceleration.
ω = ω₀ + αt (OR others).
20. What is the relation between linear velocity (v) and angular velocity (ω)?
v = ωr.
📝 Short Answer Questions
1. Find the CM of three particles at vertices of an equilateral triangle (side L), masses m, m, m.
- By symmetry, CM lies at the centroid.
- Centroid divides median in 2:1 ratio. Altitude = L√3/2. Distance from vertex to centroid = (2/3) * Altitude.
- Alternatively, use coordinates: A(0,0), B(L,0), C(L/2, L√3/2).
- XCM = (m*0 + m*L + m*L/2)/(m+m+m) = (3mL/2)/(3m) = L/2.
- YCM = (m*0 + m*0 + m*L√3/2)/(3m) = (mL√3/2)/(3m) = L√3/6 = L/(2√3).
- Answer: At centroid (L/2, L/(2√3)).
2. Explain motion of centre of mass under gravity (e.g., projectile explosion).
- Consider projectile exploding mid-air. Forces during explosion are internal. Gravity is the only external force (neglecting air resistance).
- Since external force (gravity) acts on the system, the total momentum is NOT conserved, but the motion of CM *is* affected only by gravity.
- Before explosion, CM follows parabolic path.
- After explosion, individual fragments follow different paths, but their **Centre of Mass continues to follow the original parabolic path** as if the explosion never occurred.
3. Define torque. When is torque maximum and minimum?
- Torque (τ) = Turning effect of force = r × F.
- Magnitude τ = r F sin θ.
- Max Torque: When sin θ is max (=1), i.e., θ = 90°. Force applied perpendicular to position vector (r). Max τ = rF.
- Min Torque: When sin θ is min (=0), i.e., θ = 0° or 180°. Force applied along or opposite to position vector (passes through axis). Min τ = 0.
4. State the relation between torque (τ) and angular momentum (L). What is its significance?
- Relation: τext = dL/dt (Net external torque = rate of change of angular momentum).
- Significance: It’s the rotational analogue of Newton’s 2nd Law (F=dp/dt). It states that torque is required to change the angular momentum of a system. If torque is zero, angular momentum remains constant (conserved).
5. Give two examples explaining conservation of angular momentum.
- Skater/Dancer: Pulling arms/legs in decreases Moment of Inertia (I). Since no external torque (τext≈0), Angular Momentum (L=Iω) must be conserved. To keep L constant, Angular Velocity (ω) must increase (spins faster).
- Planet orbiting Sun: Gravitational force by Sun acts towards Sun (along position vector r). Torque (τ=r×F) is zero because r and F are anti-parallel (θ=180°, sin180°=0). Thus, angular momentum of planet is conserved.
6. Write down the equations of rotational motion analogous to linear equations.
- Linear -> Rotational
- v = u + at -> ω = ω₀ + αt
- s = ut + ½at² -> θ = ω₀t + ½αt²
- v² = u² + 2as -> ω² = ω₀² + 2αθ
7. What is Radius of Gyration (K)? How is it related to Moment of Inertia (I)?
- K is the distance from axis where the entire mass (M) could be concentrated to give the same MOI.
- Relation: I = MK².
- K depends on the body’s shape, mass distribution, and axis of rotation.
8. State the theorems of Parallel Axes and Perpendicular Axes for Moment of Inertia.
- Parallel Axes: I = ICM + Md² (MOI about any axis = MOI about parallel axis through CM + Mass × (distance)²).
- Perpendicular Axes (Planar bodies): Iz = Ix + Iy (MOI about axis ⟂ plane = Sum of MOIs about two ⟂ axes in the plane, intersecting at the same point).
9. A wheel accelerates from rest to 120 rpm in 20 seconds. Find angular acceleration (α) and angular displacement (θ).
- ω₀ = 0 (rest). ω = 120 rpm = 120 × (2π/60) rad/s = 4π rad/s. t = 20 s.
- Use ω = ω₀ + αt => 4π = 0 + α(20) => α = 4π / 20 = π/5 rad/s².
- Use θ = ω₀t + ½αt² => θ = 0 + ½(π/5)(20)² = ½(π/5)(400) = 40π radians.
- Answer: α = π/5 rad/s², θ = 40π rad.
10. Calculate the Moment of Inertia of a circular disc (Mass 2kg, Radius 0.5m) about an axis passing through its edge and perpendicular to its plane.
- Use Parallel Axis Theorem: I = ICM + Md².
- Axis through edge is parallel to axis through CM (perpendicular to plane). Distance d = Radius R = 0.5 m.
- ICM for disc about axis ⟂ plane = ½MR².
- ICM = ½ (2 kg) (0.5 m)² = 1 × 0.25 = 0.25 kg m².
- I = ICM + MR² = 0.25 + (2)(0.5)² = 0.25 + 2(0.25) = 0.25 + 0.5 = 0.75 kg m².
- Answer: 0.75 kg m².
11. Find the torque of a force F = (-3i + j + 5k)N about the origin, acting on a particle whose position vector is r = (7i + 3j + k)m.
Torque τ = r × F.
Use determinant method for cross product:
τ = i j k
7 3 1
-3 1 5
= i[(3)(5) - (1)(1)] - j[(7)(5) - (1)(-3)] + k[(7)(1) - (3)(-3)]
= i[15 - 1] - j[35 - (-3)] + k[7 - (-9)]
= i[14] - j[38] + k[16]
Answer: τ = (14i - 38j + 16k) Nm
12. Why does a diver pull their limbs in to make faster somersaults?
- By pulling limbs inwards, the diver decreases their distribution of mass from the axis of rotation.
- This decreases their Moment of Inertia (I).
- Since no significant external torque acts mid-air, their Angular Momentum (L = Iω) is conserved.
- To keep L constant, if I decreases, Angular Velocity (ω) must increase.
- Therefore, they spin faster.
13. Find the CM of a system consisting of two particles m₁=2g at (1,1) cm and m₂=3g at (1,-4) cm.
- XCM = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (2*1 + 3*1)/(2+3) = (2+3)/5 = 5/5 = 1 cm.
- YCM = (m₁y₁ + m₂y₂)/(m₁ + m₂) = (2*1 + 3*(-4))/(2+3) = (2 – 12)/5 = -10/5 = -2 cm.
- Answer: CM is at (1, -2) cm.
14. When is work done by a force equal to zero?
Work (W = Fd cos θ) is zero when:
- Force F = 0.
- Displacement d = 0.
- Angle θ between F and d is 90° (Force is perpendicular to displacement).
15. Can centre of mass lie outside the body?
- Yes.
- For bodies that are not solid or have non-uniform shapes/mass distribution.
- Example: CM of a hollow ring or hollow sphere lies at its geometrical center, which is outside the material of the body. CM of an L-shaped object might also lie outside it.
16. Calculate angular momentum (L=Iω) of Earth about its rotation axis. (Assume Earth is uniform sphere M=6×10²⁴ kg, R=6.4×10⁶ m).
- Earth rotates once in T=24 hrs = 24×3600 s = 86400 s.
- Angular velocity ω = 2π/T = 2π / 86400 ≈ 7.27 × 10⁻⁵ rad/s.
- MOI of solid sphere I = (2/5)MR².
- I = (2/5)(6×10²⁴)(6.4×10⁶)² = 0.4 × 6×10²⁴ × (40.96×10¹²)
- I ≈ 2.4 × 10²⁴ × 41 × 10¹² ≈ 98.4 × 10³⁶ kg m².
- L = Iω ≈ (98.4 × 10³⁶) × (7.27 × 10⁻⁵) ≈ 715 × 10³¹ kg m²/s.
- Answer: Approx 7.15 × 10³³ kg m²/s.
📜 Long Answer Questions
1. Explain Centre of Mass for a two-particle system and n-particle system. Discuss the motion of the centre of mass.
- Definition: Point where system’s mass seems concentrated for translational motion.
- Two Particles (m₁, r₁; m₂, r₂): RCM = (m₁r₁ + m₂r₂) / (m₁ + m₂).
- N Particles: RCM = (Σ miri) / M.
- CM Motion: Its velocity VCM = Psys / M. Its acceleration ACM = Fext / M. The CM moves as if total mass M is at CM and total external force Fext acts on it. If Fext=0, VCM is constant.
2. Define Torque and Angular Momentum. Derive the relation between them.
- Torque (τ): Turning effect of force. τ = r × F. Vector. Unit: Nm.
- Angular Momentum (L): Rotational analogue of momentum. Measures quantity of rotation. For a particle, L = r × p. Vector. Unit: kg m²/s.
- Relation: Start with L = r × p. Differentiate wrt time: dL/dt = d/dt (r × p) = (dr/dt × p) + (r × dp/dt) [Using product rule for cross product]. dr/dt = v. p = mv. So, v × mv = m(v×v) = 0 (cross product of parallel vectors). dp/dt = Fext. Thus, dL/dt = 0 + (r × Fext) = τext. Relation: **τext = dL/dt**.
3. State and explain the law of conservation of angular momentum with two suitable examples.
Statement: If the net external torque on a system is zero, its total angular momentum (L) remains constant.
Explanation: Since τext = dL/dt, if τext=0, then dL/dt = 0, implying L is constant. For rotation about fixed axis, L=Iω. So, if τext=0, Iω = constant.
Examples:
- Ice Skater Spin: Skater starts spinning slowly with arms outstretched (high I). Pulling arms in decreases I. To conserve L (=Iω), angular speed ω increases.
- Diver/Gymnast: Tucking body into ball shape reduces I, increases ω for faster somersaults. Stretching out increases I, slows rotation before entering water/landing.
- Planet Motion: Torque of sun’s gravity on planet about sun is zero. So, planet’s angular momentum is conserved.
4. Discuss the equilibrium of a rigid body. State the conditions necessary for it.
A rigid body is in **equilibrium** if it has zero linear acceleration (ACM=0) AND zero angular acceleration (α=0).
Conditions for Equilibrium:
- Translational Equilibrium: The vector sum of all **external forces** acting on the body must be zero.
ΣFext = 0.
This ensures the centre of mass does not accelerate (VCM is constant, could be zero). In component form: ΣFx=0, ΣFy=0, ΣFz=0. - Rotational Equilibrium: The vector sum of all **external torques** acting on the body *about any arbitrary point* must be zero.
Στext = 0.
This ensures the body does not have angular acceleration (ω is constant, could be zero). In component form: Στx=0, Στy=0, Στz=0.
If both conditions are met, the body is in complete (mechanical) equilibrium.
5. Define Moment of Inertia and Radius of Gyration. Write MOI formulas for a ring and disc about an axis through centre perpendicular to plane.
Moment of Inertia (I): Rotational analogue of mass; measures resistance to change in rotational motion. Depends on mass and its distribution relative to axis. Formula for system of particles: I = Σmiri². Unit: kg m².
Radius of Gyration (K): Distance from axis where entire mass (M) seems concentrated for MOI purposes. Formula: I = MK² or K = √(I/M). Unit: m.
MOI Formulas:
- Thin Ring (Axis ⟂ plane, through centre): I = MR²
- Uniform Disc (Axis ⟂ plane, through centre): I = ½ MR²
6. State and explain the theorem of parallel axes for moment of inertia.
Statement: The moment of inertia (I) of a rigid body about any axis is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass (ICM) and the product of its total mass (M) and the square of the perpendicular distance (d) between the two parallel axes.
I = ICM + Md²
Explanation:
- This theorem relates the MOI about an axis through the CM to the MOI about any other axis *parallel* to it.
- ICM is the MOI about the axis passing through the centre of mass.
- ‘d’ is the perpendicular distance separating the two parallel axes.
- It shows that MOI is minimum about an axis passing through the centre of mass.
- Useful for finding MOI about axes that do not pass through the CM, if ICM is known. Ex: MOI of rod about end = MOI about center (ML²/12) + M(L/2)² = ML²/12 + ML²/4 = ML²/3.
7. State and explain the theorem of perpendicular axes for moment of inertia.
Statement: The moment of inertia of a *planar* rigid body (lamina) about an axis perpendicular to its plane (say, z-axis) is equal to the sum of its moments of inertia about two mutually perpendicular axes lying *in the plane* of the body and intersecting at the point where the perpendicular axis passes through it (say, x-axis and y-axis).
Iz = Ix + Iy
Explanation:
- This theorem is applicable **only to planar bodies** (like a disc, ring, thin plate).
- It relates the MOI about three mutually perpendicular axes, where two (x, y) lie in the plane of the body and the third (z) is perpendicular to the plane, all intersecting at a common point.
- Useful for finding MOI about an axis in the plane if MOI about perpendicular axis is known, or vice-versa. Ex: For a disc, Iz (axis ⟂ plane) = ½MR². By symmetry, Ix = Iy (about diameters). Using theorem, Iz = Ix + Iy = 2Ix. So, Ix (diameter) = Iz / 2 = (½MR²)/2 = ¼MR².
8. Compare Linear Motion and Rotational Motion parameters.
Many rotational quantities have direct analogues in linear motion:
- Displacement (s) ↔️ Angular Displacement (θ)
- Velocity (v = ds/dt) ↔️ Angular Velocity (ω = dθ/dt)
- Acceleration (a = dv/dt) ↔️ Angular Acceleration (α = dω/dt)
- Mass (m – Inertia) ↔️ Moment of Inertia (I – Rotational Inertia)
- Force (F) ↔️ Torque (τ)
- Newton’s 2nd Law (F = ma) ↔️ Newton’s 2nd Law for Rotation (τ = Iα)
- Linear Momentum (p = mv) ↔️ Angular Momentum (L = Iω)
- Work (W = Fs) ↔️ Work (W = τθ)
- Kinetic Energy (K = ½mv²) ↔️ Rotational Kinetic Energy (KR = ½Iω²)
9. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/s. The radius of the cylinder is 0.25 m. Calculate its rotational kinetic energy and angular momentum.
- Given: M=20kg, ω=100 rad/s, R=0.25m.
- 1. Find Moment of Inertia (I) for Solid Cylinder: I = ½MR².
I = ½ (20 kg) (0.25 m)² = 10 × (0.0625) = 0.625 kg m². - 2. Calculate Rotational Kinetic Energy (KR): KR = ½Iω².
KR = ½ (0.625) (100)² = ½ × 0.625 × 10000 = 0.625 × 5000 = 3125 J. - 3. Calculate Angular Momentum (L): L = Iω.
L = (0.625) × (100) = 62.5 kg m²/s.
Answer: KR = 3125 J, L = 62.5 kg m²/s.
10. Explain how a cat manages to land on its feet when dropped upside down (related to angular momentum).
This involves conservation of angular momentum and changing moment of inertia:
- Initially, when dropped, the cat has zero angular momentum (L=0) about its center of mass, assuming no initial rotation is given.
- Since gravity acts through the CM (no external torque about CM), the cat’s **total angular momentum must remain zero** throughout the fall.
- The cat cleverly changes its **shape** and **moment of inertia** during the fall.
- It first tucks its front legs in and extends its hind legs, rotating the front part of its body. This causes the back part to rotate slightly in the opposite direction (to keep total L=0), but by a smaller angle due to different MOI of front/back parts in this configuration.
- Then, it extends its front legs and tucks its hind legs, rotating its back part further. Again, the front part rotates slightly opposite to conserve L=0.
- By repeating such coordinated movements (twisting its spine and adjusting limbs), it can change its orientation significantly (turn 180°) while maintaining zero total angular momentum.
- This allows it to orient its feet downwards before landing.
11. What is meant by equilibrium? Discuss stable, unstable, and neutral equilibrium with examples.
Equilibrium means Net Force = 0 and Net Torque = 0.
Types based on stability after slight displacement:
- Stable Equilibrium: If slightly displaced, the body tends to **return** to its original equilibrium position. Potential energy is **minimum** here. Ex: Cone resting on its base, Ball in a bowl.
- Unstable Equilibrium: If slightly displaced, the body tends to move **further away** from its equilibrium position. Potential energy is **maximum** here. Ex: Cone balanced on its vertex, Ball on top of an inverted bowl.
- Neutral Equilibrium: If slightly displaced, the body stays in the new position (remains in equilibrium). Potential energy is **constant**. Ex: Cone resting on its side, Ball on a flat horizontal surface.
12. Define linear momentum and angular momentum. State the law of conservation for both.
- Linear Momentum (p): Product of mass and velocity (p=mv). Vector. Measures quantity of translational motion.
Conservation Law: If Net External Force is zero, total linear momentum is conserved. - Angular Momentum (L): Rotational analogue. Moment of linear momentum (L=r×p). Vector. Measures quantity of rotational motion.
Conservation Law: If Net External Torque is zero, total angular momentum is conserved.
13. Why are handles placed at the edge of a door?
- Opening/Closing a door requires applying a **Torque** about the hinges (axis of rotation).
- Torque τ = r F sin θ, where r is the distance from the axis to the point where force F is applied.
- To produce a certain torque needed to rotate the door, if we increase the distance ‘r’ (by placing the handle far from the hinges at the edge), the force ‘F’ required is reduced (τ = r⬆️ F⬇️ sin 90°).
- Placing the handle at the edge maximizes the lever arm ‘r’, making it easier (requiring less force) to open or close the door.
14. Compare Linear and Rotational Motion across various physical quantities and laws.
- Displacement (s) ↔️ Angular Displacement (θ)
- Velocity (v) ↔️ Angular Velocity (ω)
- Acceleration (a) ↔️ Angular Acceleration (α)
- Mass (m) ↔️ Moment of Inertia (I)
- Force (F) ↔️ Torque (τ)
- F = ma ↔️ τ = Iα
- Linear Momentum (p = mv) ↔️ Angular Momentum (L = Iω)
- Work (Fs) ↔️ Work (τθ)
- KE (½mv²) ↔️ Rotational KE (½Iω²)
- Conservation of p (if Fext=0) ↔️ Conservation of L (if τext=0)
15. Find the moment of inertia of a thin ring about its diameter.
- Let Iz be the MOI about the axis through the centre and perpendicular to the plane of the ring. We know Iz = MR².
- Let Ix and Iy be the MOIs about two perpendicular diameters lying in the plane of the ring.
- By symmetry, the MOI about any diameter is the same, so Ix = Iy.
- Apply the Theorem of Perpendicular Axes (applicable for this planar ring):
Iz = Ix + Iy - Since Ix = Iy, we have Iz = Ix + Ix = 2Ix.
- Therefore, Ix (MOI about diameter) = Iz / 2.
- Idiameter = (MR²) / 2 = ½ MR².
