JEE Physics System Of Particles and Rotational Motion MCQs Set 30

Answer: (b) \( I_0 + \dfrac{ML^2}{4} \)

Explanation:
1. By the parallel axes theorem, I = I₀ + Md² with d = L/2.
2. I = I₀ + ML²/4.

Teacher's Note:
1. Use the parallel axes theorem.
2. The end is L/2 from the centre.

Answer: (d) increase by a factor of 4

Explanation:
1. The tension passes through the centre, so L = mvr is conserved: v′ = 2v when r′ = r/2.
2. K = ½mv², so K′ = ½m(2v)² = 4K.

Teacher's Note:
1. Conserve angular momentum.
2. K is proportional to v².

Answer: (c) F₁ + F₂

Explanation:
1. The perpendicular distance from O to each side is the same, say d (the inradius).
2. F₁ (along A to B) and F₂ (along B to C) both turn the triangle anticlockwise about O, each with torque Fd; F₃ (along A to C) gives a clockwise torque F₃d.
3. Zero net torque: F₁d + F₂d = F₃d, so F₃ = F₁ + F₂.

Teacher's Note:
1. All three lines of action are at the same distance from O.
2. Check the sense of rotation of each force.

Answer: (d) 0.6 m

Explanation:
1. Total kinetic energy of the rolling cylinder = ¾mv² = ¾ × 3 × 16 = 36 J.
2. All of it goes into the spring: ½kx² = 36, so x² = 72/200 = 0.36 and x = 0.6 m.

Teacher's Note:
1. Use K = (3/4)mv² for a rolling solid cylinder.
2. Equate to ½kx².

Answer: (a) 2π s

Explanation:
1. Angular momentum is conserved: man's angular momentum + platform's angular momentum = 0.
2. mvR = Iω_p gives ω_p = 50 × 1 × 2/200 = 0.5 rad/s (opposite sense).
3. The man's angular velocity relative to the ground is v/R = 0.5 rad/s, so relative to the platform it is 0.5 + 0.5 = 1 rad/s.
4. Time for one revolution relative to the platform = 2π/1 = 2π s.

Teacher's Note:
1. The platform turns the opposite way to the man.
2. Add the angular speeds to get the relative angular speed.

Answer: (d) 157 N

Explanation:
1. α = 2 rev/s² = 4π rad/s².
2. TR = Iα = ½MR²α, so T = ½MRα = ½ × 50 × 0.5 × 4π = 50π.
3. T ≈ 157 N.

Teacher's Note:
1. Convert revolutions to radians.
2. Torque of the tension = TR.

Answer: (a) 5 : 7

Explanation:
1. Rolling: a = g sinθ/(1 + 2/5) = (5/7)g sinθ.
2. Slipping (smooth incline): a = g sinθ.
3. Ratio = 5 : 7.

Teacher's Note:
1. Use the rolling-acceleration formula.
2. A sliding body accelerates at g sinθ.

Answer: (b) \( 2mv_0^2 \)

Explanation:
1. The tension acts through the hole, so angular momentum is conserved: mv₀R₀ = mv(R₀/2), so v = 2v₀.
2. K = ½m(2v₀)² = 2mv₀².

Teacher's Note:
1. Conserve angular momentum about the hole.
2. Then find K from the new speed.

Answer: (c) \( \dfrac{W(d - x)}{d} \)

Explanation:
1. Take torques about B: N_A × d = W × (d − x).
2. N_A = W(d − x)/d.

Teacher's Note:
1. Take torques about the other knife edge.
2. The weight acts at distance d − x from B.

Answer: (b) 6.66 kg m²/s²

Explanation:
1. v = 54 km/h = 15 m/s, so ω = v/R = 15/0.45 = 33.3 rad/s.
2. Angular momentum lost: ΔL = Iω = 3 × 33.3 = 100 kg m²/s.
3. Average torque = ΔL/t = 100/15 ≈ 6.66 kg m²/s² (N m).

Teacher's Note:
1. Convert 54 km/h to m/s.
2. Torque = change in angular momentum / time.

Answer: (a) \( x = \dfrac{m_2L}{m_1 + m_2} \)

Explanation:
1. Work = ½Iω₀², so it is minimum when I is minimum.
2. I = m₁x² + m₂(L − x)². Setting dI/dx = 2m₁x − 2m₂(L − x) = 0 gives x = m₂L/(m₁ + m₂).
3. This is the centre of mass (second derivative is positive, so it is a minimum).

Teacher's Note:
1. Work done equals the rotational kinetic energy gained.
2. Minimise I with respect to x.

Answer: (b) −1

Explanation:
1. Angular momentum is conserved when the torque τ = r × F is zero, that is, when F is parallel to r.
2. r = 2 i − 6 j − 12 k = −2(−i + 3 j + 6 k), so F must be proportional to (−i + 3 j + 6 k).
3. Comparing the j and k components (3 and 6) gives F = (−1) i + 3 j + 6 k, so α = −1.

Teacher's Note:
1. Zero torque needs parallel vectors.
2. Match the ratio of the components.

Answer: (b) \( 13MR^2/32 \)

Explanation:
1. The hole has radius R/2, so its mass is M/4 and its centre is R/2 from the centre of the disc.
2. I(hole about the centre) = ½(M/4)(R/2)² + (M/4)(R/2)² = 3MR²/32.
3. I(remaining) = ½MR² − 3MR²/32 = 13MR²/32.

Teacher's Note:
1. Mass of a part is proportional to its area.
2. Use the parallel axes theorem for the hole.

Answer: (a) 8.0

Explanation:
1. ω = αt = 4 rad/s. Tangential acceleration a_t = αR = 2 × 0.5 = 1 m/s².
2. Centripetal acceleration a_c = ω²R = 16 × 0.5 = 8 m/s².
3. Net a = √(1² + 8²) ≈ 8.06 ≈ 8.0 m/s².

Teacher's Note:
1. Find both the tangential and the radial components.
2. Combine them as perpendicular vectors.

Answer: (a) L_B > L_A

Explanation:
1. K = L²/(2I), so L = √(2IK).
2. For equal K, L is proportional to √I. Since I_B > I_A, L_B > L_A.

Teacher's Note:
1. Express L in terms of K and I.
2. The masses given do not matter.

Answer: (d) 1 : 5

Explanation:
1. E_sphere = ½(2/5 mR²)ω² = (1/5)mR²ω².
2. E_cylinder = ½(½mR²)(2ω)² = mR²ω².
3. E_sphere/E_cylinder = 1/5.

Teacher's Note:
1. Use I = (2/5)mR² for the sphere and ½mR² for the cylinder.
2. Square the angular speed ratio.

Answer: (c) \( \dfrac{m_1m_2}{m_1 + m_2}\ell^2 \)

Explanation:
1. The centre of mass is at distance m₂ℓ/(m₁ + m₂) from m₁ and m₁ℓ/(m₁ + m₂) from m₂.
2. I = m₁[m₂ℓ/(m₁ + m₂)]² + m₂[m₁ℓ/(m₁ + m₂)]² = m₁m₂ℓ²(m₂ + m₁)/(m₁ + m₂)² = m₁m₂ℓ²/(m₁ + m₂).

Teacher's Note:
1. Locate the centre of mass first.
2. The result is the reduced mass times ℓ².

Answer: (b) 25 rad/s²

Explanation:
1. Torque τ = FR = 30 × 0.4 = 12 N m.
2. I = MR² = 3 × 0.16 = 0.48 kg m².
3. α = τ/I = 12/0.48 = 25 rad/s².

Teacher's Note:
1. Hollow cylinder: I = MR².
2. Angular acceleration is measured in rad/s².

Answer: (a) \( \dfrac{1}{4}I(\omega_1 - \omega_2)^2 \)

Explanation:
1. Angular momentum is conserved: Iω₁ + Iω₂ = 2Iω, so ω = (ω₁ + ω₂)/2.
2. Loss = ½Iω₁² + ½Iω₂² − ½(2I)ω² = ½I(ω₁² + ω₂²) − ¼I(ω₁ + ω₂)².
3. Loss = ¼I(ω₁ − ω₂)².

Teacher's Note:
1. Find the common angular velocity first.
2. Subtract the final energy from the initial energy.

Answer: (a) \( W_C > W_B > W_A \)

Explanation:
1. Work needed = initial rotational kinetic energy = ½Iω².
2. I(sphere) = (2/5)MR², I(disc) = ½MR², I(ring) = MR².
3. So W_C > W_B > W_A.

Teacher's Note:
1. Work to stop equals the rotational kinetic energy.
2. Compare the moments of inertia.

Answer: (d) \( -7\hat i - 4\hat j - 8\hat k \)

Explanation:
1. Position of the point of application relative to the reference point: r = (2 − 2, 0 + 2, −3 + 2) = (0, 2, −1).
2. τ = r × F = [2(−6) − (−1)(5)] i + [(−1)(4) − 0(−6)] j + [0(5) − 2(4)] k.
3. τ = (−12 + 5) i − 4 j − 8 k = −7 i − 4 j − 8 k.

Teacher's Note:
1. Use the vector from the reference point to the point of application.
2. Evaluate the determinant of r × F.