Systems of Particles and Rotational Motion Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1For any body of uniform mass density and geometric symmetry, the centre of mass lies at its centre of symmetry.
2. 2(i) Sphere: at its centre.
3. 3(ii) Cylinder: at the midpoint of its axis.
4. 4(iii) Ring: at its geometric centre — which is inside the hole, not on the material.
5. 5(iv) Cube: at its geometric centre.
6. 6No — the centre of mass need not lie inside the body, as the ring shows (its CM is at its centre, in empty space).

Step-by-step solution

1. 1Take the H nucleus at x = 0 and the Cl nucleus at x = 1.27 Å.
2. 2Let m_H = m; then m_Cl = 35.5m.
3. 3x_CM = (m_H · 0 + m_Cl × 1.27)/(m_H + m_Cl) = 35.5m × 1.27/(36.5m).
4. 4x_CM = (35.5 × 1.27)/36.5 ≈ 1.235 Å from the hydrogen nucleus.
5. 5Equivalently, it is about 0.035 Å from the chlorine nucleus, which is why we say the CM lies very close to the chlorine atom.

Step-by-step solution

1. 1The floor is smooth, so no external horizontal force acts on the trolley + child system.
2. 2With zero net external force, the total momentum, and hence the velocity of the centre of mass, is unchanged.
3. 3Internal forces (the child pushing on the trolley) cannot change the CM velocity.
4. 4Hence the CM continues to move with speed V, whatever the child does.

Step-by-step solution

1. 1Let a and b be drawn from a common vertex, with angle θ between them.
2. 2Area of the triangle = ½ × base × height = ½ |a| × |b| sin θ.
3. 3By definition of the cross product, |a × b| = |a||b| sin θ.
4. 4Therefore area = ½ |a × b|, as required.

Step-by-step solution

1. 1Consider the parallelepiped with edges a, b and c.
2. 2|b × c| equals the area of the base parallelogram, with normal along (b × c).
3. 3The height is a·(b × c)/|b × c| — the projection of a along the normal to the base.
4. 4Volume = base area × height = |b × c| × [a·(b × c)/|b × c|] = a·(b × c).
5. 5So the magnitude of the scalar triple product equals the volume of the parallelepiped.

Step-by-step solution

1. 1l = r × p = (x î + y ĵ + z k̂) × (pₓ î + pᵧ ĵ + p_z k̂).
2. 2Expanding with î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (cyclic):
3. 3lₓ = y p_z − z pᵧ; lᵧ = z pₓ − x p_z; l_z = x pᵧ − y pₓ.
4. 4If motion is confined to the x-y plane, z = 0 and p_z = 0 for all time.
5. 5Then lₓ = 0, lᵧ = 0, and only l_z = x pᵧ − y pₓ survives — the angular momentum has only a z-component.

Step-by-step solution

1. 1Let particle 1 move along the line y = 0 and particle 2 along the parallel line y = d (with appropriate velocities).
2. 2Take an arbitrary reference point O. The horizontal positions of the two particles must match to keep them abreast, so the lever arms about O for both particles coincide.
3. 3Let the common lever arm from O to the vertical plane of the particles be x₀; about O, particle 1's angular momentum is m v x₀ and particle 2's is −m v x₀.
4. 4The two contributions cancel in x₀ but combine through the separation d: the total angular momentum is L = m v d k̂ (perpendicular to the plane).
5. 5Since m, v and d are fixed, the result is independent of the choice of O.

Step-by-step solution

1. 1Use the exact triangle ratios: sin 36.9° = cos 53.1° = 0.6 and cos 36.9° = sin 53.1° = 0.8.
2. 2Let T₁ be the tension at the left end (36.9° to the vertical) and T₂ at the right (53.1°).
3. 3Horizontal balance: T₁ sin 36.9° = T₂ sin 53.1°, i.e. 0.6 T₁ = 0.8 T₂ ⇒ T₁ = (4/3)T₂.
4. 4Vertical balance: T₁ cos 36.9° + T₂ cos 53.1° = W, i.e. 0.8 T₁ + 0.6 T₂ = W.
5. 5Substitute: 0.8 × (4/3)T₂ + 0.6 T₂ = W ⇒ (1.067 + 0.6)T₂ = W ⇒ T₂ = 0.6 W.
6. 6Torques about the left end: W d = (T₂ cos 53.1°)(2 m) = 0.6 × 0.6 W × 2.
7. 7d = 0.72 m.

Step-by-step solution

1. 1mg = 1800 × 9.8 = 17,640 N. Let R_f be the total reaction on the front axle and R_b on the back axle.
2. 2Vertical balance: R_f + R_b = 17,640 N.
3. 3Torques about the front axle: R_b × 1.8 = 17,640 × 1.05.
4. 4R_b = (17,640 × 1.05)/1.8 = 10,290 N.
5. 5R_f = 17,640 − 10,290 = 7350 N.
6. 6Per wheel (two wheels per axle): front = 7350/2 = 3675 N; back = 10,290/2 = 5145 N.

Step-by-step solution

1. 1I of hollow cylinder = MR²; I of solid sphere = (2/5)MR².
2. 2For the same torque, α = τ/I — the body with the smaller moment of inertia gets the larger angular acceleration.
3. 3After a given time, ω = αt (starting from rest).
4. 4Since (2/5)MR² < MR², the solid sphere has the larger α and hence the greater angular speed.

Step-by-step solution

1. 1I = ½MR² = ½ × 20 × (0.25)² = 0.625 kg m².
2. 2Rotational KE = ½Iω² = ½ × 0.625 × (100)² = 3125 J.
3. 3Angular momentum L = Iω = 0.625 × 100 = 62.5 kg m² s⁻¹.

Step-by-step solution

1. 1(a) No external torque acts, so angular momentum is conserved: I₁ω₁ = I₂ω₂.
2. 2I₂ = (2/5)I₁, so ω₂ = (I₁/I₂)ω₁ = (5/2) × 40 = 100 rev/min.
3. 3(b) KE_rot = ½Iω². Initial KE₁ = ½I₁ω₁².
4. 4New KE₂ = ½(I₂)(ω₂)² = ½(2/5 I₁)(5/2 ω₁)² = ½I₁ × (2/5)(25/4)ω₁² = (5/2)½I₁ω₁² = 2.5 KE₁.
5. 5The child does muscular (internal) work in pulling his arms in; this internal energy is converted into rotational kinetic energy.

Step-by-step solution

1. 1I of a hollow cylinder = MR² = 3 × (0.4)² = 0.48 kg m².
2. 2Torque τ = FR = 30 × 0.4 = 12 N m.
3. 3Angular acceleration α = τ/I = 12/0.48 = 25 rad s⁻².
4. 4Linear acceleration of the rope a = αR = 25 × 0.4 = 10 m s⁻².

Step-by-step solution

1. 1Power P = τω (torque × angular speed).
2. 2P = 180 × 200 = 36,000 W.
3. 3P = 36 kW.

Step-by-step solution

1. 1Treat the hole as a negative mass. Mass of the full disk m = σπR²; mass of the hole m_h = σπ(R/2)² = m/4.
2. 2Take the origin at the centre of the disk, with the hole centred at x = R/2.
3. 3x_CM = (m × 0 − m_h × (R/2))/(m − m_h).
4. 4x_CM = −(m/4)(R/2)/(3m/4) = −(mR/8)(4/3m) = −R/6.
5. 5The centre of gravity lies on the line joining the hole to the centre, at R/6 from the centre on the side opposite the hole.

Step-by-step solution

1. 1The two coins together have mass 10 g, placed at the 12.0 cm mark.
2. 2The new balance point is 45.0 cm, so the CG of the stick alone (at 50.0 cm) is 5.0 cm to one side of it, and the coins are 33.0 cm to the other side.
3. 3Torque balance about the 45.0 cm point: M × 5.0 = 10 × 33.
4. 4M = 330/5 = 66 g.

Step-by-step solution

1. 1K_trans = ½mv² = ½ × 5.30 × 10⁻²⁶ × (500)² = 6.625 × 10⁻²¹ J.
2. 2K_rot = (2/3)K_trans = ½Iω².
3. 3ω² = [2 × (2/3) × 6.625 × 10⁻²¹]/1.94 × 10⁻⁴⁶ = (8.833 × 10⁻²¹)/(1.94 × 10⁻⁴⁶).
4. 4ω² ≈ 4.55 × 10²⁵.
5. 5ω ≈ 6.75 × 10¹² rad s⁻¹.