Motion of System of Particles and Rigid Body: JEE Advanced PYQ, 2007–2026

Answer: (A)

There is no external torque about the axis, so angular momentum is conserved. Platform: I = ½(100)(2²) = 200 kg m². Person at the rim: 50 × 2² = 200 kg m². Initial I = 400 kg m². At the centre the person contributes about zero, so final I = 200 kg m². Then 400 × 2 = 200 × ω, so ω = 4 rad s⁻¹.

Answer: (A)

The rod is acted on by three forces: the tension in each string and its weight. A rigid body in equilibrium under three non-parallel coplanar forces needs their lines of action to meet at one point. The two strings meet at P, so the vertical line through the centre of gravity must also pass through P. Let P be at height h above the rod. The horizontal distance from P to A is h/tan30° = h√3, and from P to B is h/tan60° = h/√3. Hence L = h(√3 + 1/√3) = 4h/√3, and the distance of the centre of gravity from A is h√3 = (√3 · √3/4) L = 3L/4. Check by torques about A. Horizontal balance gives T_A cos30° = T_B cos60°. Vertical balance gives T_A sin30° + T_B sin60° = W. Together they give T_B = (√3/2)W. Torque balance about A gives T_B sin60° · L = W·x, so (3W/4)L = W·x and x = 3L/4.

Answer: (A)

L = r × p = m(r × v). For motion in the xy-plane only the z component survives: L_z = m(x v_y − y v_x) = 2[(1)(−1) − (2)(3)] = 2(−7) = −14 kg m² s⁻¹. So L = −14 k̂ kg m² s⁻¹, which is clockwise when viewed from +z.

Answer: (A)

Use the subtraction method, with the mass of the intact disc M. Full disc about the axis: I₁ = MR²/2. The hole has mass m = M·(π(R/2)²)/(πR²) = M/4. Its moment of inertia about its own centre is ½·(M/4)(R/2)² = MR²/32. The axis is R/2 from the hole's centre, so by the parallel-axis theorem the hole contributes MR²/32 + (M/4)(R/2)² = MR²/32 + MR²/16 = 3MR²/32. Remaining part: I = MR²/2 − 3MR²/32 = 16MR²/32 − 3MR²/32 = 13MR²/32.

Answer: (C)

For a continuous distribution, x_cm = (∫x dm)/(∫dm), with dm = λ dx = kx² dx. The numerator is ∫₀ᴸ kx³ dx = kL⁴/4. The denominator, the total mass, is ∫₀ᴸ kx² dx = kL³/3. So x_cm = (kL⁴/4)/(kL³/3) = 3L/4. The centre of mass lies beyond the midpoint because more mass sits at the far end.

Answer: (D)

There is no external torque about the vertical axis, so the total angular momentum stays zero. The platform has I_p = ½MR² = ½ × 100 × 4 = 200 kg m². The man has I_m = mR² = 40 × 4 = 160 kg m². Suppose the platform turns at ω in one direction. The man's velocity relative to the platform is u, so his angular velocity relative to the ground is u/R − ω = 0.75 − ω. Then 160(0.75 − ω) − 200ω = 0, so 120 = 360ω and ω = 1/3 ≈ 0.33 rad/s. The key point is that angular momentum must be calculated with velocities measured relative to the ground, not relative to the platform.

Answer: (A)

Torque about a point is τ = r × F. Both vectors lie in the xy-plane, so the torque points along z with τ_z = xF_y − yF_x. Substituting x = 2, y = 3, F_x = 4, F_y = −1 gives τ_z = (2)(−1) − (3)(4) = −2 − 12 = −14 N m. So τ = −14 k N m, i.e. clockwise when viewed from +z. Using F × r flips the sign, the dot product gives a scalar, and a sign error in the determinant gives 10.

Answer: (C)

For a system of particles, the angular momentum about any point is L = L_cm + R_cm × M V_cm. Here L_cm is the angular momentum about the centre of mass. The centre of mass is at height R above the chosen floor point and moves horizontally, so the orbital term has magnitude M v R = 2 × 3 × 0.2 = 1.2 kg m²/s. The spin term is I_cm ω with I_cm = ½MR² = 0.04 kg m² and ω = v/R = 15 rad/s, giving 0.6 kg m²/s. For rolling without slipping, the sense of rotation is the same as the sense of r × p of the centre of mass about the floor point, so the two terms add to 1.8 kg m²/s. As a check, this equals I_contact·ω = (3/2)MR²·ω = 1.8.

Answer: (A)

The parallel axis theorem I = I_cm + Md² holds only with respect to an axis through the centre of mass. Step 1: from axis A, 5 = I_cm + 2(1)², so I_cm = 3 kg m². Step 2: for axis B, I_B = 3 + 2(2)² = 11 kg m². Applying the theorem between A and B directly, or forgetting to square the distance, gives the wrong options.

Answer: (D)

About the point of suspension, r (length l) is always perpendicular to the velocity, so |L| = m v l = 0.5 × 3 × 2 = 3 kg m²/s. L is perpendicular to r, and r makes angle θ with the vertical, so L makes angle θ with the horizontal. Its vertical component is L sin θ and its horizontal component is L cos θ. The tension passes through the pivot and has zero torque, but gravity gives a horizontal torque about the pivot, so L changes direction and rotates about the vertical with the bob. After half a revolution the horizontal component has reversed and the vertical component is unchanged, so ΔL = 2L cos θ = 2 × 3 × 0.8 = 4.8 kg m²/s.