System of Particles and Rotational Motion — NCERT Class 11 Physics

All these bodies have uniform mass distribution and symmetric geometry, so the centre of mass is at their geometric centre in each case: (a) Sphere — at the centre of the sphere (b) Cylinder — at the midpoint of its axis (c) Ring — at the centre of the ring (d) Cube — at the intersection of its diagonals

Let the Cl atom be at origin. H atom is at distance 1.27 Å from Cl. m_H = m, m_Cl = 35.5m

x_CM = (m_Cl × 0 + m_H × 1.27)/(m_Cl + m_H) = (m × 1.27)/(35.5m + m) = 1.27/36.5 = 0.0348 Å from the Cl atom

There is no external horizontal force on the system (child + trolley). By conservation of momentum, the centre of mass of the system continues to move with the same velocity. When the child runs in one direction, the trolley moves in the opposite direction to conserve momentum. However, the speed of the centre of mass remains unchanged. The trolley's speed changes depending on the child's motion, but the system's overall motion (CM velocity) stays the same.

L_total = Σ(rᵢ × mᵢvᵢ)

Let rᵢ = R_CM + rᵢ' (where rᵢ' is position relative to CM) vᵢ = V_CM + vᵢ'

L_total = Σ(R_CM + rᵢ') × mᵢ(V_CM + vᵢ') = R_CM × M V_CM + R_CM × Σmᵢvᵢ' + Σmᵢrᵢ' × V_CM + Σrᵢ' × mᵢvᵢ'

Since Σmᵢrᵢ' = 0 (CM frame) and Σmᵢvᵢ' = 0: L_total = R_CM × MV_CM + Σrᵢ' × mᵢvᵢ' = L_CM + L_about CM

(a) About the centre: I = m(l²) + m(l²) = 2ml²

(b) About one end: I = m(0²) + m((2l)²) = 4ml²

I_CM = ½MR² (about centre, perpendicular to plane)

For an axis in the plane of the disc through the centre: I_diameter = ¼MR² (by perpendicular axes theorem)

Using parallel axes theorem, shifting by distance R (to the tangent): I = I_diameter + MR² = ¼MR² + MR² = 5MR²/4

For rolling without slipping: a = Rα

Translational: Ma = Mg sin θ − f (friction) Rotational: Iα = fR → (½MR²)(a/R) = fR → f = ½Ma

Substituting: Ma = Mg sin θ − ½Ma → 3Ma/2 = Mg sin θ a = (2/3)g sin θ = (2g sin θ)/3

(See intext Q1 above.) All are at the geometric centre.

(See intext Q2 above.) 0.0348 Å from the Cl atom (toward the H atom).

No effect. The centre of mass velocity remains unchanged because there is no external horizontal force acting on the system.

The area of the triangle formed by a and b as two sides is:

Proved.

The component of a along the direction of (b × c) gives the height

(Where φ is the angle between a and the normal to the b-c plane.)

ω₀ = 0, ω = 50 rad s⁻¹, t = 10 s

α = (ω − ω₀)/t = (50 − 0)/10 = 5 rad s⁻²

r = 0.1 m, I = 0.2 kg m², t = 5 s

ω₁ = 120 × 2π/60 = 4π rad s⁻¹ ω₂ = 300 × 2π/60 = 10π rad s⁻¹

α = (ω₂ − ω₁)/t = (10π − 4π)/5 = 6π/5 rad s⁻²

Torque τ = Iα = 0.2 × 6π/5 = 0.24π N m

F = τ/r = 0.24π/0.1 = 2.4π = 7.54 N

m = 20 kg, R = 0.25 m, ω = 100 rad s⁻¹ I = ½MR² = ½ × 20 × (0.25)² = ½ × 20 × 0.0625 = 0.625 kg m²

KE_rot = ½Iω² = ½ × 0.625 × 10000 = 3125 J

L = Iω = 0.625 × 100 = 62.5 kg m² s⁻¹

By conservation of angular momentum: I₁ω₁ = I₂ω₂

When the child folds his arms, the mass distribution shifts closer to the axis, decreasing the moment of inertia. Since L = Iω = constant, a decrease in I causes an increase in ω.

I₁/I₂ = ω₂/ω₁ = 45/30 = 1.5

The moment of inertia decreases by a factor of 1.5, i.e., I₂ = (2/3)I₁

m = 1800 kg, d = 1.8 m, x = 1.05 m (CG from front axle)

Taking moments about the front axle: R_front × 0 + R_rear × 1.8 = mg × 1.05 R_rear = 1800 × 9.8 × 1.05/1.8 = 18522/1.8 = 10290 N (total on both rear wheels) Force per rear wheel = 10290/2 = 5145 N

R_front = mg − R_rear = 17640 − 10290 = 7350 N (total on both front wheels) Force per front wheel = 7350/2 = 3675 N

For rolling without slipping: a = g sin θ/(1 + I/(MR²))

• Ring: I = MR² → a = g sin θ/2
• Disc: I = ½MR² → a = 2g sin θ/3
• Sphere: I = (2/5)MR² → a = 5g sin θ/7

Since 5/7 > 2/3 > 1/2, the sphere has the greatest acceleration and reaches the bottom first, followed by the disc, and then the ring.

m = 5 kg, v = 5 m s⁻¹, I = (2/5)MR², v = Rω → ω = v/R

KE_total = KE_trans + KE_rot = ½mv² + ½Iω² = ½mv² + ½(2/5)MR² × (v/R)² = ½mv² + ⅕mv² = ½(5)(25) + ⅕(5)(25) = 62.5 + 25 = 87.5 J

a = g sin θ/(1 + I/(MR²)) = g sin θ/(1 + ½) = (2/3)g sin θ

a = (2/3)(9.8)(0.5) = 3.27 m s⁻² (deceleration going up)

v² = u² + 2as → 0 = 25 + 2(−3.27)s s = 25/6.54 = 3.82 m

Alternatively, using energy conservation: ½mv²(1 + I/(MR²)) = mgs sin θ ½v²(1 + ½) = gs sin θ s = 3v²/(4g sin θ) = 3(25)/(4 × 9.8 × 0.5) = 75/19.6 = 3.82 m

By conservation of angular momentum: I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω

Since discs are identical: I₁ = I₂ = I I(ω₀) + I(2ω₀) = 2Iω 3Iω₀ = 2Iω ω = 3ω₀/2

m_bullet = 0.01 kg, v = 500 m s⁻¹ Door: M = 12 kg, width = 1 m, I_door = (1/3)ML² = (1/3)(12)(1²) = 4 kg m²

Bullet embeds at centre (r = 0.5 m from hinge) I_bullet = m_bullet × r² = 0.01 × 0.25 = 0.0025 kg m²

Angular momentum before: L = m_bullet × v × r = 0.01 × 500 × 0.5 = 2.5 kg m² s⁻¹ Total I after = I_door + I_bullet = 4 + 0.0025 = 4.0025 kg m²

ω = L/I = 2.5/4.0025 = 0.625 rad s⁻¹