Systems of Particles & Rotational Motion

1

The centre of mass of a system of two particles divides the line joining them in the ratio of:Easy

A. Their masses

B. Inverse of their masses

C. Square of their masses

D. Their velocities

✅ Solution

The CM divides the line joining two masses in the inverse ratio of their masses.X = (m₁x₁ + m₂x₂)/(m₁ + m₂)So it is closer to the heavier mass — i.e., divides in inverse ratio of masses.

Answer: B — Inverse of their masses

2

For two particles of equal mass, the centre of mass lies:Easy

A. At the heavier particle

B. At the lighter particle

C. Exactly midway between them

D. Outside the system

✅ Solution

X = (mx₁ + mx₂)/(2m) = (x₁ + x₂)/2For equal masses, the CM is simply the average of their positions — exactly midway.

Answer: C

3

A rigid body is said to be in mechanical equilibrium when:Easy

A. Net force = 0 only

B. Net torque = 0 only

C. Both net force = 0 and net torque = 0

D. Net velocity = 0

✅ Solution

For mechanical equilibrium, both conditions must hold: ΣF = 0 (translational equilibrium) AND Στ = 0 (rotational equilibrium)A body can satisfy one without the other — but both are needed for full equilibrium.

Answer: C

4

Torque is defined as:Easy

A. τ = r · F (dot product)

B. τ = r × F (cross product)

C. τ = F / r

D. τ = F × r (F cross r)

✅ Solution

τ = r × F, magnitude: τ = rF sinθNote: τ = r × F (not F × r). Since cross product is anti-commutative, F × r = −r × F.

Answer: B — τ = r × F

5

The SI unit of moment of inertia is:Easy

A. kg·m

B. kg·m²

C. kg/m²

D. N·m²

✅ Solution

I = Σmᵢrᵢ² → units = kg × m² = kg·m²Moment of inertia is the rotational analogue of mass. Dimensions: [ML²].

Answer: B — kg·m²

6

The moment of inertia of a thin circular ring of mass M and radius R about an axis through its centre, perpendicular to its plane is:Easy

A. MR²/2

B. MR²

C. 2MR²

D. MR²/4

✅ Solution

For a thin ring, every particle is at distance R from the axis. I = ΣmᵢR² = MR²This is the maximum MI for a given mass and radius (all mass at the rim).

Answer: B — MR²

7

The moment of inertia of a solid disc of mass M and radius R about an axis through its centre perpendicular to its plane is:Moderate

A. MR²

B. MR²/4

C. MR²/2

D. 2MR²/5

✅ Solution

I_disc (perpendicular axis) = MR²/2This is less than MR² (ring) because disc mass is distributed at various distances from centre, not all at R.

Answer: C — MR²/2

8

Angular momentum of a particle is defined as:Easy

A. l = p × r

B. l = r · p

C. l = r × p

D. l = m × ω

✅ Solution

l = r × p = r × mvIt is the rotational analogue of linear momentum. Its magnitude: l = rp sinθ.

Answer: C — l = r × p

9

For a rotating rigid body with symmetric mass distribution about its axis, the relation between angular momentum L and angular velocity ω is:Moderate

A. L = Iω²

B. L = Iω

C. L = I/ω

D. L = ½Iω

✅ Solution

L = Iω (for symmetric bodies rotating about fixed axis)Compare with linear momentum p = mv. Here I replaces m, and ω replaces v.

Answer: B — L = Iω

10

Newton's second law for rotational motion about a fixed axis is:Moderate

A. τ = Iω

B. τ = Iα

C. τ = Iω²

D. τ = I/α

✅ Solution

τ = Iα (analogous to F = ma)Torque τ produces angular acceleration α inversely proportional to moment of inertia I.

Answer: B — τ = Iα

11

If the total external torque on a system is zero, then:Moderate

A. Linear momentum is conserved

B. Angular momentum is conserved

C. Kinetic energy is conserved

D. Potential energy is conserved

✅ Solution

From dL/dt = τ_ext. If τ_ext = 0: dL/dt = 0 → L = Iω = constantThis is the Law of Conservation of Angular Momentum.

Answer: B — Angular momentum is conserved

12

A skater pulls her arms in while spinning. Her angular speed:Moderate

A. Decreases

B. Increases

C. Remains unchanged

D. Becomes zero

✅ Solution

By conservation of angular momentum (no external torque): I₁ω₁ = I₂ω₂Pulling arms in decreases I (moment of inertia), so ω must increase to conserve Iω.

Answer: B — Increases

13

The kinetic energy of a body rotating with angular velocity ω and moment of inertia I is:Easy

A. Iω

B. Iω²

C. ½Iω²

D. 2Iω²

✅ Solution

KE_rotation = ½Iω²Analogous to KE_translation = ½mv². Here I replaces m, ω replaces v.

Answer: C — ½Iω²

14

The velocity of the centre of mass of a system of particles is related to total linear momentum P and total mass M as:Easy

A. V = M/P

B. V = P × M

C. V = P/M

D. V = P²/M

✅ Solution

P = MV → V = P/MThe total momentum of the system equals the total mass times the velocity of the CM.

Answer: C — V = P/M

15

A couple produces:Moderate

A. Translational motion only

B. Rotation without translation

C. Translation without rotation

D. No motion

✅ Solution

A couple consists of two equal and opposite forces with different lines of action. Net force = 0 (no translation), but net torque ≠ 0 (causes rotation). Example: opening a bottle cap.

Answer: B — Rotation without translation

16

The moment of inertia of a solid sphere of mass M and radius R about its diameter is:Moderate

A. MR²

B. 2MR²/3

C. 2MR²/5

D. MR²/2

✅ Solution

From NCERT Table 6.1: I_solid sphere (diameter) = 2MR²/5This is one of the standard results that must be memorised for exam purposes.

Answer: C — 2MR²/5

17

The relation between linear velocity v and angular velocity ω for a particle at distance r from the axis is:Easy

A. v = ω/r

B. v = ωr

C. v = ω²r

D. v = r/ω

✅ Solution

v = ωr (in vector form: v = ω × r)All particles of a rigid body have the same angular velocity ω but different linear speeds depending on their distance from the axis.

Answer: B — v = ωr

18

The cross product (vector product) of two parallel vectors is:Easy

A. Their product ab

B. Maximum (ab)

C. Zero (null vector)

D. ab/2

✅ Solution

For parallel vectors, θ = 0°. |a × b| = ab sin0° = 0 → null vectorAlso: î×î = ĵ×ĵ = k̂×k̂ = 0

Answer: C — Zero (null vector)

19

A projectile explodes in mid-air. The centre of mass of fragments:Moderate

A. Falls vertically down

B. Continues on the same parabolic path

C. Moves in a straight line

D. Stops at the point of explosion

✅ Solution

Explosion forces are internal — they don't affect CM motion. The only external force is gravity (unchanged). So: MA = F_ext (gravity) → CM follows same parabola

Answer: B — Continues on the same parabolic path

20

The radius of gyration k of a body is defined by:Moderate

A. I = Mk

B. I = Mk²

C. I = M/k²

D. I = Mk³

✅ Solution

I = Mk² → k = √(I/M)k (radius of gyration) is the RMS distance of the body's particles from the axis. It is a geometric property of the body and the axis.

Answer: B — I = Mk²

21

In pure translational motion of a rigid body, at any instant:Easy

A. All particles have different velocities

B. All particles have the same velocity

C. All particles have the same angular velocity

D. Particles near the axis are stationary

✅ Solution

By definition: in pure translational motion, all particles of the body move with the same velocity at any given instant. (In rotation, all particles have same ω but different v.)

Answer: B

22

Power delivered by a torque τ acting on a body rotating with angular velocity ω is:Moderate

A. P = τ/ω

B. P = τ²ω

C. P = τω

D. P = τ + ω

✅ Solution

P = τω (analogous to P = Fv in linear motion)This follows from: dW = τ dθ → P = dW/dt = τ(dθ/dt) = τω.

Answer: C — P = τω

23

The centre of mass of a uniform triangular lamina coincides with its:Moderate

A. Incentre

B. Circumcentre

C. Centroid

D. Orthocentre

✅ Solution

The CM of a uniform triangular lamina lies at the intersection of its three medians, which is the centroid G. The centroid divides each median in 2:1 ratio from vertex.

Answer: C — Centroid

24

The first equation of rotational kinematics (uniform angular acceleration) is:Moderate

A. ω = ω₀ + αt²

B. ω = ω₀ + αt

C. θ = ω₀t + ½αt²

D. ω² = ω₀² − 2αθ

✅ Solution

The three rotational kinematic equations are: ω = ω₀ + αt (1st equation)θ = θ₀ + ω₀t + ½αt² (2nd equation)ω² = ω₀² + 2α(θ−θ₀) (3rd equation)

Answer: B — ω = ω₀ + αt

25

Given equal torques are applied to a hollow cylinder (I = MR²) and a solid sphere (I = 2MR²/5) of same mass and radius. Which gains greater angular speed after time t?Hard

A. Hollow cylinder

B. Solid sphere

C. Both gain equal speed

D. Depends on mass

✅ Solution

α = τ/I. Greater I → smaller α → smaller ω after time t. I_cylinder = MR² (larger)I_sphere = 2MR²/5 = 0.4MR² (smaller)α_sphere > α_cylinder → ω_sphere > ω_cylinder

Answer: B — Solid sphere gains greater angular speed

1

Three particles of masses 100 g, 150 g and 200 g are located at the vertices of an equilateral triangle of side 0.5 m. Find the (X, Y) coordinates of the centre of mass. (Take 100 g at origin, 150 g at (0.5, 0) m, 200 g at (0.25, 0.25√3) m) Moderate3 Marks

✅ Solution

Total mass M = 100 + 150 + 200 = 450 g X = (100×0 + 150×0.5 + 200×0.25)/450 = (0+75+50)/450 = 125/450 = 5/18 m ≈ 0.278 m Y = (100×0 + 150×0 + 200×0.25√3)/450 = (50√3)/450 = √3/9 = 1/(3√3) m ≈ 0.192 m

CM = (5/18 m, 1/3√3 m) ≈ (0.278 m, 0.192 m)

2

In a HCl molecule, the nuclei are separated by 1.27 Å. A chlorine atom is 35.5 times as massive as a hydrogen atom. Find the location of the centre of mass from the hydrogen nucleus. Moderate3 Marks

✅ Solution

Let m_H = m, m_Cl = 35.5m. Place H at x=0, Cl at x = 1.27 Å. X_cm = (m×0 + 35.5m×1.27) / (m + 35.5m) = (35.5×1.27)/36.5 X_cm = 45.085/36.5 ≈ 1.235 Å from H nucleus The CM is very close to the Cl nucleus (heavier atom).

CM is ≈ 1.235 Å from H nucleus (≈ 0.035 Å from Cl)

3

Find the scalar and vector products of vectors a = (3î − 4ĵ + 5k̂) and b = (−2î + ĵ − 3k̂). Moderate3 Marks

✅ Solution

Scalar product: a·b = (3)(−2)+(−4)(1)+(5)(−3) = −6−4−15 = −25 Vector product a × b using determinant: a × b = î[(−4)(−3)−(5)(1)] − ĵ[(3)(−3)−(5)(−2)] + k̂[(3)(1)−(−4)(−2)] = î[12−5] − ĵ[−9+10] + k̂[3−8] = 7î − ĵ − 5k̂

a·b = −25; a×b = 7î − ĵ − 5k̂

4

Find the torque of force F = 7î + 3ĵ − 5k̂ about the origin, acting on a particle at position r = î − ĵ + k̂. Moderate3 Marks

✅ Solution

τ = r × F, computed via the determinant method: τ = | î ĵ k̂ |
| 1 −1 1 |
| 7 3 −5 | = î[(−1)(−5)−(1)(3)] − ĵ[(1)(−5)−(1)(7)] + k̂[(1)(3)−(−1)(7)] = î[5−3] − ĵ[−5−7] + k̂[3+7] = 2î + 12ĵ + 10k̂

τ = 2î + 12ĵ + 10k̂ N·m

5

A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/s. Radius = 0.25 m. Find: (a) Kinetic energy of rotation and (b) Angular momentum about its axis. Moderate3 Marks

✅ Solution

I = MR²/2 = (20 × 0.0625)/2 = 0.625 kg·m² (a) KE = ½Iω² = ½ × 0.625 × 10000 = 3125 J (b) L = Iω = 0.625 × 100 = 62.5 kg·m²/s

(a) KE = 3125 J; (b) L = 62.5 kg·m²/s

6

A child (arms outstretched) sits on a turntable rotating at 40 rev/min. He folds his arms and reduces his moment of inertia to 2/5 of the original. Find the new angular speed. Also find the ratio of new KE to initial KE. Moderate3 Marks

✅ Solution

By conservation of angular momentum: I₁ω₁ = I₂ω₂ I₁ × 40 = (2/5)I₁ × ω₂ → ω₂ = (5/2) × 40 = 100 rev/min KE ratio: KE₂/KE₁ = ½I₂ω₂² / ½I₁ω₁² = (I₂/I₁)(ω₂/ω₁)² = (2/5) × (100/40)² = (2/5) × 6.25 = 2.5 KE increases (energy comes from work done by muscles).

New angular speed = 100 rev/min; KE ratio = 2.5 (KE increases 2.5×)

7

A rope of negligible mass is wound around a hollow cylinder of mass 3 kg and radius 40 cm. The rope is pulled with a force of 30 N. Find (a) angular acceleration of the cylinder and (b) linear acceleration of the rope. Moderate3 Marks

✅ Solution

I = MR² = 3 × (0.4)² = 0.48 kg·m² Torque τ = F × R = 30 × 0.4 = 12 N·m (a) α = τ/I = 12/0.48 = 25 rad/s² (b) a = αR = 25 × 0.4 = 10 m/s²

(a) α = 25 rad/s²; (b) a = 10 m/s²

8

An engine maintains a rotor at uniform angular speed of 200 rad/s by transmitting a torque of 180 N·m. What power is required? (Assume 100% efficiency.) Easy3 Marks

✅ Solution

P = τω = 180 × 200 = 36,000 W = 36 kW At uniform angular speed, torque is needed only to counter friction (not to accelerate). At 100% efficiency, all electrical power = mechanical power.

Power required = 36,000 W = 36 kW

9

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. Find: (a) angular acceleration and (b) number of revolutions made during this time. Moderate3 Marks

✅ Solution

ω₀ = 2π×1200/60 = 40π rad/s; ω = 2π×3120/60 = 104π rad/s (a) α = (ω−ω₀)/t = (104π−40π)/16 = 64π/16 = 4π rad/s² Angular displacement: θ = ω₀t + ½αt² = 40π×16 + ½×4π×256 θ = 640π + 512π = 1152π rad (b) Revolutions = θ/(2π) = 1152π/2π = 576 revolutions

(a) α = 4π rad/s² ≈ 12.57 rad/s²; (b) 576 revolutions

10

A flywheel of mass 20 kg and radius 20 cm has a cord wound around its rim. A steady pull of 25 N is applied. Starting from rest, find: (a) angular acceleration, (b) work done when 2 m of cord unwinds, (c) KE gained. Moderate3 Marks

✅ Solution

I = MR²/2 = 20×(0.2)²/2 = 0.4 kg·m²; τ = 25×0.2 = 5 N·m (a) α = τ/I = 5/0.4 = 12.5 rad/s² (b) W = F×d = 25×2 = 50 J θ = 2m/0.2m = 10 rad; ω² = 2αθ = 2×12.5×10 = 250 (rad/s)² (c) KE = ½Iω² = ½×0.4×250 = 50 J ✓ (equals work done)

(a) 12.5 rad/s²; (b) 50 J; (c) 50 J — energy is conserved

11

A metal bar 70 cm long, mass 4 kg, is supported on two knife edges 10 cm from each end. A 6 kg load is suspended 30 cm from one end. Find reactions R₁ and R₂ at the knife edges. (g = 9.8 m/s²) Moderate3 Marks

✅ Solution

W = 4×9.8 = 39.2 N at centre (35 cm); W₁ = 6×9.8 = 58.8 N at 30 cm from A K₁ at 10 cm, K₂ at 60 cm from A. Taking moments about G (35 cm from A): R₁+R₂ = W+W₁ = 39.2+58.8 = 98 N ... (i) −R₁(25) + W₁(5) + R₂(25) = 0 → R₁−R₂ = 58.8×5/25 = 11.76 N ... (ii) Solving: R₁ = 54.88 N ≈ 55 N; R₂ = 43.12 N ≈ 43 N

R₁ ≈ 54.9 N; R₂ ≈ 43.1 N

12

A uniform disc of radius R has a circular hole of radius R/2 cut from it, with the hole's centre at R/2 from the disc's centre. Find the centre of gravity of the remaining flat body. Hard3 Marks

✅ Solution

Let surface mass density = σ. Full disc mass M = σπR². Removed disc: mass m = σπ(R/2)² = M/4 Remaining mass = M − M/4 = 3M/4 CM of full disc at origin (0). CM of removed disc at x = R/2. X_cm = (M×0 − (M/4)×(R/2)) / (3M/4) = (−MR/8) / (3M/4) = −R/6 The negative sign means CM is R/6 to the left of original centre.

CG is at R/6 from the centre, away from the hole

13

A metre stick is balanced at its centre. Two coins of mass 5 g each are placed at the 12 cm mark. The stick balances at 45 cm. Find the mass of the metre stick. Moderate3 Marks

✅ Solution

New pivot is at 45 cm. Coins (10 g) are at 12 cm, i.e., 33 cm from pivot (to the left). CM of metre stick = 50 cm mark, which is 5 cm to the right of new pivot (45 cm). Taking moments about new pivot (45 cm): 10g × 33 = M × 5 M = 330/5 = 66 g

Mass of metre stick = 66 g

14

A 3 m ladder (20 kg) leans against a frictionless wall with its foot 1 m from the wall on the floor. Find the reaction forces of the wall (F₁) and floor (normal N, friction F). (g = 9.8 m/s²) Hard3 Marks

✅ Solution

W = 20×9.8 = 196 N; BC = √(3²−1²) = √8 = 2√2 m (height) Taking moments about foot A: 2√2 × F₁ = (1/2) × W F₁ = 196/(4√2) = 196/5.657 ≈ 34.6 N (wall reaction, horizontal) Vertical: N = W = 196 N; Horizontal: F = F₁ = 34.6 N F₂ = √(N² + F²) = √(196² + 34.6²) ≈ 199 N

F₁(wall) ≈ 34.6 N; N = 196 N; F(friction) ≈ 34.6 N; Floor reaction F₂ ≈ 199 N

15

An oxygen molecule (mass 5.30×10⁻²⁶ kg, I = 1.94×10⁻⁴⁶ kg·m²) has mean translational speed 500 m/s. Its rotational KE = (2/3) × translational KE. Find average angular velocity ω of the molecule. Hard3 Marks

✅ Solution

KE_trans = ½mv² = ½ × 5.30×10⁻²⁶ × (500)² = ½ × 5.30×10⁻²⁶ × 250000 KE_trans = 6.625 × 10⁻²¹ J KE_rot = (2/3) × 6.625×10⁻²¹ = 4.417 × 10⁻²¹ J ½Iω² = 4.417×10⁻²¹ → ω² = 2×4.417×10⁻²¹ / (1.94×10⁻⁴⁶) ω² = 8.834×10⁻²¹ / 1.94×10⁻⁴⁶ ≈ 4.55×10²⁵ ω ≈ 6.75 × 10¹² rad/s

Average angular velocity ω ≈ 6.75 × 10¹² rad/s

📋 General Instructions

Section A — Multiple Choice Questions

Section B — Numerical Problems