Motion of System of Particles and Rigid Body: JEE Main PYQ, 2020–2025

Answer: (A)

For a thin lamina, the perpendicular axis theorem says I_z = I_x + I_y, where I_x and I_y are moments about two perpendicular axes in the plane. For a square of side a, I_z through the centre is M(a²+a²)/12 = Ma²/6. By symmetry, I_x = I_y, so each in-plane axis through the centre, parallel to a side, has I = Ma²/12.

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: (D)

The kinetic energy of rotation is K = ½Iω² = ½ × 2 × 10² = 100 J. It is the rotational analogue of ½mv², with I in place of m and ω in place of v.

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: (A)

For a rigid body in rotational equilibrium, the net torque about any point must be zero. Choose the hinge A as the pivot so the hinge force has no torque. The rod's weight acts at its centre of mass, 1 m from A, and produces a clockwise torque of 40 N x 1 m. The rope tension acts at B, 2 m from A, and produces an anticlockwise torque T x 2 m. Equating the two gives T = 20 N.

Answer: (C)

M a_cm = F_ext: the centre of mass of a system moves as if all its mass were concentrated there and all the external forces were applied there. Internal forces, which occur in equal and opposite pairs, do not affect it.

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: (B)

A couple is a pair of equal and opposite forces with different lines of action. Its net force is zero, but its net torque is not. The torque of a couple is τ = F × d, where d is the perpendicular distance between the two lines of action, and this value is the same about every point. Here τ = 5 N × 0.4 m = 2 N m. Note that the answer does not depend on where the reference point is chosen.