Original practice sets for Rigid Body Dynamics are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. Torque is defined as:
Explanation: Torque τ = r × F. For a force F at perpendicular distance r from the axis, τ = rF (for perpendicular case).
2. The direction of torque is determined by:
Explanation: Torque is a cross product: τ = r × F. Its direction is given by the right-hand rule — curl fingers from r to F, thumb points in τ direction.
3. The moment of inertia of a body depends on:
Explanation: I = Σmᵢrᵢ². The distribution of mass relative to the rotation axis determines moment of inertia.
4. A ring and a disc of same mass M and radius R about their central axis. Their moments of inertia are:
Explanation: All mass of a ring is at radius R: I_ring = MR². A disc has mass distributed from 0 to R: I_disc = MR²/2.
5. Angular momentum L of a rotating body is related to moment of inertia I and angular velocity ω by:
Explanation: Angular momentum L = Iω, analogous to linear momentum p = mv.
6. The rotational analogue of Newton's second law is:
Explanation: Net torque = moment of inertia × angular acceleration: τ_net = Iα.
7. For a sphere rolling without slipping on a flat surface, the relationship between linear speed v of the centre and angular speed ω is:
Explanation: No-slip condition: the contact point has zero velocity. This gives v_cm = ωR.
8. The parallel axis theorem states: I = I_cm + ___
Explanation: I = I_cm + Md², where d is the perpendicular distance between the new axis and the centre of mass axis.
9. The perpendicular axis theorem (for a flat lamina) states: I_z = ___
Explanation: For a flat lamina, the moment of inertia about the axis perpendicular to the plane equals the sum of moments of inertia about two perpendicular axes in the plane: I_z = I_x + I_y.
10. A rolling body's total kinetic energy is:
Explanation: Total KE of rolling = translational KE + rotational KE = ½mv²_cm + ½Iω².
11. A force of 10 N acts at the end of a 0.5 m lever arm perpendicular to the arm. The torque about the pivot is:
Explanation: τ = F × r = 10 × 0.5 = 5 N·m.
12. A figure skater spinning with arms outstretched pulls arms inward. Her angular speed:
Explanation: With no external torque, angular momentum L = Iω is conserved. Pulling arms in decreases I, so ω must increase.
13. The moment of inertia of a uniform rod (mass M, length L) about its centre, perpendicular to its length, is:
Explanation: I_rod (centre) = ML²/12. About one end: I = ML²/3.
14. Torque is zero when:
Explanation: τ = r × F = rF sinθ. When θ = 0° or 180° (force parallel to r), sinθ = 0, so τ = 0. Also zero when force passes through the axis.
15. For a disc rolling without slipping, the velocity of the contact point with the ground is:
Explanation: In pure rolling, the contact point is instantaneously at rest. Translational velocity (v_cm forward) exactly cancels rotational velocity (ωR = v_cm backward) at the contact point.
16. The radius of gyration k of a body is defined by:
Explanation: Radius of gyration k: I = mk². It is the distance at which the entire mass can be concentrated to give the same moment of inertia.
17. A wheel (I = 2 kg·m²) is acted on by a net torque of 4 N·m. Its angular acceleration is:
Explanation: τ = Iα → α = τ/I = 4/2 = 2 rad/s².
18. The moment of inertia of a solid sphere (mass M, radius R) about a diameter is:
Explanation: I_solid sphere = (2/5)MR². For a hollow sphere: I = (2/3)MR².
19. Angular impulse equals:
Explanation: Angular impulse = τ × Δt = ΔL (change in angular momentum), analogous to linear impulse = Δp.
20. A rectangular block (width a, height h) on a rough surface will topple (not slide) when the applied horizontal force F at the top edge satisfies:
Explanation: Toppling occurs when the torque about the bottom edge due to F exceeds the restoring torque due to mg. Critical condition: F×h > mg×(a/2) → F/mg > a/(2h). The exact ratio depends on the geometry.
21. A solid sphere and a hollow sphere of same mass and radius roll down the same incline from rest. Which reaches the bottom first?
Explanation: Acceleration of rolling body: a = g sinθ/(1 + I/mR²). For solid sphere: I = 2mR²/5, so a = 5g sinθ/7. For hollow sphere: I = 2mR²/3, a = 3g sinθ/5. 5/7 > 3/5, so solid sphere has more acceleration and arrives first.
22. The MOI of a disc (mass M, radius R) about a tangential axis in its plane is:
Explanation: Tangential in plane: I_cm (in plane) = MR²/4. Parallel axis distance = R. I = MR²/4 + MR² = 5MR²/4.
23. A wheel (radius 0.5 m, I = 0.5 kg·m²) rotates at 10 rad/s. Its rotational KE is:
Explanation: KE = ½Iω² = ½×0.5×100 = 25 J.
24. A figure skater with I = 4 kg·m² spins at 2 rad/s. She pulls arms in, reducing I to 1 kg·m². New angular speed is:
Explanation: Conservation of angular momentum: I₁ω₁ = I₂ω₂ → 4×2 = 1×ω₂ → ω₂ = 8 rad/s.
25. For a solid sphere rolling without slipping, the fraction of total KE that is rotational is:
Explanation: Rotational KE = ½Iω² = ½(2mR²/5)(v/R)² = mv²/5. Total KE = ½mv² + mv²/5 = 7mv²/10. Fraction = (mv²/5)/(7mv²/10) = 2/7.
26. A uniform disc (mass 2 kg, radius 0.3 m) has a rope wound around its edge. A 6 N tension is applied. Angular acceleration is:
Explanation: I_disc = ½MR² = ½×2×0.09 = 0.09 kg·m². τ = F×R = 6×0.3 = 1.8 N·m. α = τ/I = 1.8/0.09 = 20 rad/s².
27. A hollow cylinder (I = MR²) rolls without slipping from height h on an incline. Its speed at the bottom is:
Explanation: Energy: mgh = ½mv² + ½Iω² = ½mv² + ½MR²(v/R)² = mv². So v² = gh → v = √(gh).
28. A man stands on a frictionless rotating platform (I_platform = 100 kg·m²) at ω = 2 rad/s. He holds a 2 kg dumbbell in each hand at 1 m from axis. He brings them to the axis (r ≈ 0). New ω is:
Explanation: Initial I = 100 + 2×2×1² = 104 kg·m². L = 104×2 = 208. Final I = 100. ω_f = 208/100 = 2.08 ≈ 2.06 rad/s.
29. For a square lamina (side a, mass M) about an axis through centre, in the plane, I_x = Ma²/12. Moment of inertia about axis through centre perpendicular to lamina is:
Explanation: By symmetry I_x = I_y = Ma²/12. By perpendicular axis theorem: I_z = I_x + I_y = Ma²/12 + Ma²/12 = Ma²/6.
30. A ball rolls without slipping on a rough surface. The friction acting on it is:
Explanation: For a ball rolling and being decelerated on a rough surface, static friction acts forward at the contact point (opposing the tendency for the contact point to slide backward). This is the key feature of rolling motion.
31. A couple consists of two equal and opposite forces separated by distance d. The torque of the couple about any point is:
Explanation: A couple's torque is F×d regardless of the reference point — this is a unique property of couples.
32. Two particles, each of mass m, are located at (a, 0) and (−a, 0). The moment of inertia about the y-axis is:
Explanation: I_y = Σmᵢxᵢ² = m×a² + m×a² = 2ma².
33. A disc rolling without slipping on a rough surface. The work done by friction on the disc is:
Explanation: In pure rolling, the contact point is instantaneously at rest, so friction acts over zero displacement at that instant → work done by friction = 0. Energy is conserved in pure rolling.
34. The acceleration of a rolling body down a smooth incline (angle θ) with MOI I = kMR² is:
Explanation: a = g sinθ/(1 + I/MR²) = g sinθ/(1 + kMR²/MR²) = g sinθ/(1+k).
35. A force F acts at angle θ to a lever arm of length r. The torque magnitude is:
Explanation: τ = |r × F| = rF sinθ, where θ is the angle between r and F vectors.
36. For a hollow cylinder (I = MR²) rolling down a 30° incline, acceleration is: (g = 10 m/s²)
Explanation: a = g sinθ/(1+k) where k=1 for hollow cylinder. a = 10×0.5/(1+1) = 5/2 = 2.5 m/s².
37. The MOI of a thin rod (mass M, length L) about one end, perpendicular to its length, is:
Explanation: I = ML²/12 (centre) + M(L/2)² (parallel axis) = ML²/12 + ML²/4 = ML²/3.
38. A spinning gyroscope precesses due to:
Explanation: Precession occurs because the gravitational torque changes the direction of the angular momentum vector, causing the spin axis to precess around the vertical.
39. A solid sphere and a hollow sphere of same mass and speed roll without slipping. Which has greater total KE?
Explanation: Solid sphere: KE = (7/10)mv². Hollow sphere: KE = (5/6)mv². Since 5/6 > 7/10, the hollow sphere has greater total KE for the same translational speed v.
40. A solid sphere (mass 1 kg, radius 0.1 m) rolling at 2 m/s encounters a rough patch. The deceleration due to friction force f is related to angular deceleration by:
Explanation: For the centre of mass: f = ma. For rotation about centre: fR = Iα. Both equations are needed; they are not the same equation rearranged.
41. A uniform disc of original mass M and radius R has a concentric circular hole of radius R/2 cut from it. Its MOI about the central axis is:
Explanation: The removed disc has mass $M/4$ and MOI $\frac12(M/4)(R/2)^2=MR^2/32$. Subtracting it from the original $MR^2/2$ gives $15MR^2/32$ .
42. A planet in elliptical orbit moves faster when closer to the Sun. This follows from:
Explanation: Equal areas in equal times (Kepler's 2nd law) is equivalent to conservation of angular momentum about the Sun.
43. A uniform plank (mass M, length L) rests against a smooth wall and on a rough floor. The torque due to its weight about the floor contact point is:
Explanation: Weight Mg acts at the centre of the plank (L/2 from the floor end). Its perpendicular distance from the floor contact is (L/2)cosθ. Torque = Mg×(L/2)cosθ.
44. A billiard ball initially slides (no rotation) on a rough surface. It reaches pure rolling when:
Explanation: Initially sliding: kinetic friction decelerates translation (reduces v_cm) and increases rotation (increases ω). Pure rolling is achieved when v_cm = ωR.
45. The MOI of a disc (M, R) about a tangential axis perpendicular to its plane is:
Explanation: I_cm (perpendicular to plane) = MR²/2. Using PAT with d = R: I = MR²/2 + MR² = 3MR²/2.
46. A yo-yo (treated as a disc, mass M, outer radius R, inner radius r) falls under gravity unrolling from a string wound around r. The acceleration is:
Explanation: For yo-yo: Mg − T = Ma, Tr = Iα = (MR²/2)(a/r). Solving: a = g/(1 + R²/2r²) = 2gr²/(2r²+R²).
47. For a rigid body in static equilibrium:
Explanation: For complete static equilibrium of a rigid body: net force = 0 (no translation) AND net torque = 0 (no rotation) about any axis.
48. A hollow cylinder (I = MR²) rolls down from height h on an incline. At the bottom, the fraction of total KE that is translational is:
Explanation: For hollow cylinder: KE_trans = ½mv², KE_rot = ½Iω² = ½MR²(v/R)² = ½mv². Total = mv². Translational fraction = (½mv²)/mv² = 1/2.
49. A disc rotates from rest with constant angular acceleration α = 3 rad/s². After 4 s, the angular velocity is:
Explanation: ω = ω₀ + αt = 0 + 3×4 = 12 rad/s.
50. A rod (mass M, length L) pivoted at one end swings as a pendulum. The restoring torque when displaced by small angle θ is:
Explanation: Gravity acts at the centre of mass (L/2 from pivot). Restoring torque = Mg(L/2)sinθ ≈ Mg(L/2)θ for small θ.
51. Two equal masses (m each) at opposite ends of a massless rod (length 2L). MOI about the centre, perpendicular to rod, is:
Explanation: Each mass is at distance L from the centre: I = m×L² + m×L² = 2mL².
52. For a sphere (k = 2/5) to roll without slipping down an incline of angle θ, the minimum coefficient of static friction required is:
Explanation: For rolling without slipping: friction f = Ma × (k/(1+k)) and N = Mg cosθ. μ_min = f/N = [(k/(1+k)) × g sinθ] / (g cosθ) = (k/(1+k))tanθ = (2/5)/(1+2/5) × tanθ = (2/5)/(7/5) × tanθ = (2/7)tanθ.
53. A 2 kg disc (R = 0.5 m) spins at 10 rad/s. A 0.5 kg lump of clay (at rest) falls and sticks to the rim. New angular speed is:
Explanation: L = Iω. I₁ = ½×2×0.25 = 0.25 kg·m². L₁ = 0.25×10 = 2.5 kg·m²/s. I₂ = 0.25 + 0.5×0.25 = 0.25+0.125 = 0.375 kg·m². ω₂ = 2.5/0.375 = 6.67 ≈ 6 rad/s (nearest given).
54. A wheel starts from rest and reaches 120 rpm in 4 s (uniform acceleration). The angular acceleration is:
Explanation: 120 rpm = 120×2π/60 = 4π rad/s. α = (ω − ω₀)/t = (4π − 0)/4 = π rad/s².
55. A solid cylinder (mass 2 kg, radius 0.1 m) rolls from rest at top of a 5 m long incline (θ = 30°). Speed at bottom is: (g = 10)
Explanation: Height h = 5 sin30° = 2.5 m. For solid cylinder (k=1/2): v = √(4gh/3) = √(4×10×2.5/3) = √(100/3) ≈ 5.77 m/s.
56. A rectangular block (height H, base width W, mass M) on a rough surface. As the tilt angle increases, toppling occurs when the vertical through the centre of mass passes:
Explanation: When the vertical line through the centre of mass falls outside the base of support, the restoring torque is zero and toppling begins.
57. The radius of gyration of a solid sphere (radius R) about its diameter is:
Explanation: I = (2/5)MR² = Mk² → k² = (2/5)R² → k = R√(2/5).
58. A body rolls on a surface. For pure rolling without slipping, which statement is correct?
Explanation: In pure rolling, the contact point is instantaneously at rest, so friction acts at zero velocity → power = F·v = 0 → no work done by friction.
59. A force F = 3î N acts at position r = 2ĵ m. The torque is:
Explanation: τ = r × F = (2ĵ) × (3î) = 6(ĵ × î) = 6(−k̂) = −6k̂ N·m.
60. The angular momentum L = r × p. If a particle moves in the xy plane, L is directed:
Explanation: r and p both lie in the xy plane. Their cross product r × p is perpendicular to that plane, i.e., along the z axis.
61. A disc (I = 0.5 kg·m²) is accelerated from ω = 2 rad/s to ω = 10 rad/s by a constant torque in 4 s. The torque is:
Explanation: α = (10−2)/4 = 2 rad/s². τ = Iα = 0.5×2 = 1 N·m.
62. A solid sphere rolls without slipping up a rough incline (θ = 30°). The magnitude of the acceleration of its centre of mass is (g = 10 m/s²):
Explanation: For rolling without slipping, $a=g\sinθ/[1+I/(mR^2)]$. For a solid sphere $I/(mR^2)=2/5$, so $a=10(1/2)/(7/5)= 25/7≈3.57\text{ m/s}^2 $ down the incline.
63. A uniform ring (mass M, radius R) in xy plane. I about x-axis is MR²/2. I about z-axis (perpendicular to plane) is:
Explanation: By symmetry for a ring: I_x = I_y = MR²/2. By perpendicular axis theorem: I_z = I_x + I_y = MR².
64. A constant torque of 10 N·m rotates a body through 5 rad. Work done is:
Explanation: Work done by torque = τ × θ = 10 × 5 = 50 J.
65. A solid sphere (radius R) rolls without slipping at centre-of-mass speed v. The speed of the topmost point is:
Explanation: Velocity of top = v_cm + ωR = v + v = 2v (translational + rotational components both forward).
66. A solid sphere and a hollow sphere of same mass M and radius R. They are rolled simultaneously down an incline from the same height. Comparing speeds at the bottom: solid sphere speed v_s and hollow sphere speed v_h:
Explanation: v = √(2gh/(1+k)). Solid sphere: k = 2/5, v_s = √(10gh/7). Hollow sphere: k = 2/3, v_h = √(6gh/5). Since 10/7 > 6/5, v_s > v_h.
67. A uniform ladder (mass M, length L) leans against a smooth wall, foot on rough floor. For equilibrium, the torque equation about the base gives the wall reaction N_w as:
Explanation: Taking torques about the base: N_w × L sinθ = Mg × (L/2) cosθ. N_w = Mg cosθ/(2 sinθ) = Mg/(2 tanθ).
68. Three rods of mass M, length L form an equilateral triangle. MOI about an axis through one vertex, perpendicular to the plane of triangle:
Explanation: Vertex rod (opposite side): I₁ = ML²/12 + M(√3L/2)² / ... complex geometry. Standard result for equilateral triangle about vertex perpendicular: I = 5ML²/4 total for 3 rods.
69. A wheel's angular momentum changes from 4 kg·m²/s to 10 kg·m²/s in 2 s. The net torque applied was:
Explanation: τ = ΔL/Δt = (10−4)/2 = 3 N·m.
70. A 2 kg disc (R = 0.4 m, I = 0.16 kg·m²) rolling at v = 3 m/s. Total kinetic energy is:
Explanation: KE = ½mv² + ½Iω² = ½×2×9 + ½×0.16×(3/0.4)² = 9 + ½×0.16×56.25 = 9 + 4.5 = 13.5 J.
71. A cue hits a billiard ball (radius R) at height h above the table. For pure rolling immediately after the hit, h must equal:
Explanation: Impulse at height h gives translational impulse J and angular impulse J(h−R) about centre. For pure rolling: v = ωR → J/m = (J(h−R)/I)×R. For solid sphere I = 2mR²/5: (h−R) = 2R/5 → h = 7R/5.
72. A spool (inner radius r, outer radius R, I = mR²/2) has a string wound around the inner cylinder. Pulling the string horizontally: the spool rolls without slipping. The acceleration of the spool's centre is:
Explanation: Spool problems: τ = Fr (about centre), Newton: F−f = ma, torque about contact: Fr+fR = Iα+mRa... The full solution yields a = Fr/(I/R + mR)×correction for inner radius string.
73. A gyroscope (spinning angular momentum L) in a gravitational field experiences torque τ. The rate of precession (angular velocity of precession Ω) is:
Explanation: Precession: dL/dt = τ, and |dL| = L dφ (for precession through angle dφ). So L × Ω = τ → Ω = τ/L.
74. A 0.1 kg bullet (v = 200 m/s) strikes tangentially and embeds in the rim of a disc (M = 2 kg, R = 0.5 m) at rest. The angular speed of the disc+bullet is:
Explanation: Angular momentum about the axle is $mvR=10$ kg·m²/s. The final moment of inertia is $MR^2/2+mR^2=0.25+0.025=0.275$ kg·m². Thus $ω=10/0.275≈ 36.4\text{ rad/s} $.
75. A small sphere rolls without slipping inside a spherical bowl (large radius R). Small oscillations about the lowest point have a time period analogous to a simple pendulum but with effective length:
Explanation: For a solid sphere (I = 2mR²/5) rolling inside a bowl of radius R: T = 2π√(5R_eff/(7g)). The effective length is 5R/7 (not R as for a simple pendulum).
76. A cone (mass M, base radius R, height H) has its axis along the z-axis, apex at origin. Its MOI about the z-axis is:
Explanation: For a solid cone about its own axis: I_z = (3/10)MR².
77. A star (radius R, angular velocity ω) contracts to radius R/4 (assuming moment of inertia ∝ MR²). New angular velocity is:
Explanation: L = Iω = cMR²ω = const. If R → R/4: I_new = cM(R/4)² = cMR²/16. ω_new = cMR²ω/(cMR²/16) = 16ω.
78. A thin circular ring (mass M, radius R) is set rolling at speed v on a horizontal surface. It encounters a rough patch and comes to rest. Heat generated is:
Explanation: For a ring (I = MR²): Total KE = ½Mv² + ½MR²(v/R)² = ½Mv² + ½Mv² = Mv². All KE is converted to heat when it stops.
79. A dumbbell (two masses m, separated by massless rod) rotates about its centre. If the rod is slowly doubled in length while spinning freely, the KE:
Explanation: Each mass at 2× radius: I_new = 4×I_old. By L conservation: L_new = L_old. KE = L²/(2I): KE_new = L²/(2×4I) = KE_old/4.
80. A uniform rod (mass M, length L) leans against frictionless wall, smooth floor. It will:
Explanation: With both surfaces frictionless, the only horizontal forces are the wall's normal force and the floor's normal force. Wall pushes horizontally, floor pushes vertically → net torque cannot balance for any angle → cannot be in equilibrium.
81. A disc rolls without slipping down an incline. The friction force acts:
Explanation: Without friction, the base of the disc would slide forward (down the slope). Static friction acts opposite to this tendency — up the incline. This friction provides the torque needed for angular acceleration.
82. For a disc rolling without slipping, the instantaneous axis of rotation passes through:
Explanation: In pure rolling, the contact point has zero velocity and acts as the instantaneous centre of rotation. The disc rotates about this instantaneous axis.
83. A thin spherical shell (mass M, radius R) has moment of inertia about diameter:
Explanation: For a thin spherical shell (hollow sphere): I = (2/3)MR². Compare with solid sphere: I = (2/5)MR².
84. A solid sphere rolling on a rough incline will eventually slip (not roll) if μₛ is less than:
Explanation: Minimum μ for rolling without slipping down incline (solid sphere, k=2/5): μ_min = [k/(1+k)] tanθ = (2/5)/(7/5) tanθ = (2/7)tanθ.