Original practice sets for Circular Motion 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. Angular velocity ω is defined as:
Explanation: ω = dθ/dt. The SI unit is rad/s.
2. Relation between angular velocity ω and time period T:
Explanation: One full revolution = 2π radians in time T: ω = 2π/T. Also ω = 2πf where f is the frequency.
3. Relation between linear speed v and angular speed ω for a point at radius r:
Explanation: v = rω. Points farther from the axis have greater linear speed for the same angular speed.
4. A particle moves in a circle of radius 2 m at speed 4 m/s. Its centripetal acceleration is:
Explanation: a_c = v²/r = 16/2 = 8 m/s².
5. The centripetal force on a body in circular motion is always directed:
Explanation: Centripetal means 'centre-seeking'. The net centripetal force is always directed radially inward toward the centre.
6. Centripetal force is:
Explanation: Centripetal force is NOT a new type of force — it's just the label for the net radial inward force required for circular motion, whatever force provides it.
7. Angular acceleration α is defined as:
Explanation: Angular acceleration α = dω/dt = d²θ/dt². Its SI unit is rad/s².
8. Centripetal force on a mass m moving at speed v in circle of radius r is:
Explanation: F_c = mv²/r = mω²r. This inward force maintains circular motion.
9. For ideal banking (no friction) of a road with bank angle θ, the speed at which a vehicle can navigate a curve of radius r is:
Explanation: N sinθ = mv²/r and N cosθ = mg → tanθ = v²/(rg) → v = √(rg tanθ).
10. A conical pendulum with string length L, half-cone angle θ, has time period:
Explanation: Vertical: T cosθ = mg → T = mg/cosθ. Horizontal: T sinθ = mω²r = mω²L sinθ. So ω² = g/(L cosθ), T = 2π/ω = 2π√(L cosθ/g).
11. Minimum speed at the top of a vertical circle (radius r) for a string to remain taut is:
Explanation: At the top, for minimum speed: T = 0. Then mg = mv²/r → v_min = √(gr).
12. If the minimum speed at the top of a vertical circle (radius r) is √(gr), the corresponding minimum speed at the bottom is:
Explanation: Energy conservation from top to bottom: ½mv_b² = ½mv_t² + mg(2r). With v_t = √(gr): v_b² = gr + 4gr = 5gr → v_b = √(5gr).
13. A car turns in a horizontal circle on a flat road. The centripetal force is provided by:
Explanation: On a flat (unbanked) road, the tyres don't slide — static friction provides the centripetal force directed toward the centre of the turn.
14. A person in a car going over a hill (radius r) at speed v. The apparent weight at the top is:
Explanation: At the top of the hill: mg − N = mv²/r → N = m(g−v²/r). Apparent weight decreases at high speed.
15. Due to Earth's rotation, the effective value of g at the equator is:
Explanation: At the equator, part of gravity provides centripetal acceleration for Earth's rotation: g_eff = g − ω²R. This reduces effective g at the equator. At poles, there's no rotation effect, so g_eff = g (maximum).
16. Work done by centripetal force in one complete revolution:
Explanation: Centripetal force is perpendicular to velocity (and displacement direction) at every point → W = Fs cos90° = 0 over any arc.
17. On a banked road (θ = 30°, r = 100 m, μ = 0.3, g = 10). Maximum safe speed is approximately:
Explanation: v_max = √(rg(tanθ+μ)/(1−μtanθ)) = √(100×10×(0.577+0.3)/(1−0.3×0.577)) = √(1000×0.877/0.827) ≈ √(1061) ≈ 32.6 m/s ≈ 33.8 m/s.
18. A wheel rotates at 600 rpm. A point on the rim at radius 0.5 m has centripetal acceleration:
Explanation: ω = 600×2π/60 = 20π rad/s. a_c = ω²r = (20π)²×0.5 = 400π²×0.5 = 200π² m/s².
19. A ball (mass m) moves in a vertical circle (radius r). Tension at the bottom when moving at speed v_b is:
Explanation: At the bottom: T − mg = mv_b²/r (centripetal direction is upward). T = mg + mv_b²/r.
20. At the top of a vertical circle, tension T when speed is v_t:
Explanation: At the top: T + mg = mv_t²/r (both T and mg point toward centre). T = mv_t²/r − mg.
21. On a banked road (θ = 30°, r = 100 m, μ = 0.3, g = 10). Minimum safe speed is approximately:
Explanation: v_min = √(rg(tanθ−μ)/(1+μtanθ)) = √(1000×(0.577−0.3)/(1+0.3×0.577)) = √(1000×0.277/1.173) ≈ √236 ≈ 15.4 m/s.
22. A car (mass 1500 kg) takes a flat horizontal turn (r = 50 m) at 20 m/s. Required centripetal force:
Explanation: F_c = mv²/r = 1500×400/50 = 12000 N.
23. A ball on a string (length 0.5 m) is whirled in a vertical circle. Minimum speed at the bottom to complete the circle is: (g = 10)
Explanation: v_bottom_min = √(5gr) = √(5×10×0.5) = √25 = 5 m/s. Wait: √(5gr) = √(5×10×0.5) = √25 = 5 m/s. So answer is 5 m/s.
24. A conical pendulum (string length L = 0.5 m) rotates at 4 rad/s. The cone half-angle θ satisfies:
Explanation: From conical pendulum: cosθ = g/(ω²L) = 10/(16×0.5) = 10/8 = 1.25. Since cosθ > 1 is impossible, the string cannot make a conical pendulum at ω = 4 rad/s with L = 0.5 m — it would require ω > √(g/L) = √20 ≈ 4.47 rad/s.
25. For a ball in a vertical circle on a string, the string goes slack (T = 0) when the speed at that height h (from bottom) satisfies:
Explanation: T = 0 when component of gravity provides all centripetal force: mg cosθ = mv²/r. Combined with energy conservation from bottom, this determines the exact angle where string goes slack (not always the top).
26. In a centrifuge rotating at ω rad/s, a particle at radius r from the axis experiences a centrifugal pseudo force (in rotating frame) of:
Explanation: In the rotating frame, the centrifugal pseudo force = mω²r directed radially outward.
27. For a ball on a rod (vs. string) in a vertical circle, the minimum speed at the top for maintaining circular motion is:
Explanation: A string can only pull (tension ≥ 0); minimum at top = √(gr). A rigid rod can push or pull; minimum speed at top = 0 (rod pushes outward when needed).
28. A particle moves in a circle of radius r at speed v. Its angular momentum about the centre has magnitude:
Explanation: L = r × p = r × mv = mvr (for circular motion where r ⊥ v).
29. On a roller coaster at the top of a loop (radius r), a person feels weightless when:
Explanation: Weightlessness means N = 0. At the top: mg + N = mv²/r. For N = 0: v = √(gr).
30. If a particle in circular motion is speeding up, it has:
Explanation: Centripetal acceleration (a_c = v²/r) is always present for circular motion. A change in speed gives tangential acceleration (a_t = dv/dt). Both are present when speed changes.
31. At the equator, if Earth's rotation were fast enough that centripetal acceleration = g, objects would:
Explanation: Effective weight = m(g − ω²R). If ω²R = g, effective weight = 0 → objects float (as in orbit).
32. A ball moving in a vertical circle (radius r) has speed v₀ at the bottom. At angle θ from the bottom, the speed is:
Explanation: Height gained = r − r cosθ = r(1−cosθ). Energy conservation: ½mv² = ½mv₀² − mg×r(1−cosθ) → v = √(v₀²−2gr(1−cosθ)).
33. A car (mass m) on a flat circular road (radius r) has μₛ = 0.6. Maximum safe speed is: (g = 10)
Explanation: Maximum centripetal force = friction = μmg. mv²/r = μmg → v_max = √(μrg) = √(0.6×r×10) = √(6r).
34. For non-uniform circular motion, the total acceleration vector points:
Explanation: Net acceleration = centripetal (inward) + tangential (along tangent). The resultant points between the two, not exactly toward centre nor along tangent.
35. For a satellite in circular orbit at radius r, the orbital speed is:
Explanation: Gravitational force = centripetal force: GMm/r² = mv²/r → v = √(GM/r).
36. A coin on a rotating turntable slides outward when friction is insufficient. In the ground frame, this is because:
Explanation: In the ground frame, there is no centrifugal force. The coin moves outward because there is not enough centripetal (inward) force (friction) to maintain circular motion.
37. A ball on a rigid rod (length r) in a vertical circle. At the top, the rod exerts a thrust (push) when speed v_top satisfies:
Explanation: At top: mg − F_rod = mv²/r. F_rod = mg − mv²/r. Rod pushes (thrust, positive outward) when F_rod
38. A planet orbits at radius r with period T. Another planet at 4r has period:
Explanation: By Kepler's 3rd law: T² ∝ r³. (T₂/T₁)² = (4r/r)³ = 64 → T₂/T₁ = 8 → T₂ = 8T.
39. A car (mass m) on the inside of a circular loop (radius R). At the top, normal force is N. Speed v satisfies:
Explanation: At the top of the loop (inside): both mg and N point toward the centre (downward). mg + N = mv²/R.
40. Two particles move in the same circle at angular speeds ω₁ and ω₂. The time for one to lap the other (relative angular speed) depends on:
Explanation: The relative angular speed is |ω₁ − ω₂|. The time for one to lap the other (gain one full revolution) is T = 2π/|ω₁ − ω₂|.
41. A ball (mass m) moves in a vertical circle (radius r). Speed at bottom = √(7gr). Tension at bottom T_b and at top T_t. T_b − T_t =
Explanation: T_b = m(v_b²/r + g) = m(7g+g) = 8mg. v_t² = v_b² − 4gr = 7gr−4gr = 3gr. T_t = m(v_t²/r − g) = m(3g−g) = 2mg. T_b − T_t = 8mg − 2mg = 6mg. (Note: T_b−T_t is always 6mg for vertical circles!)
42. A car (mass 1200 kg) travels at 20 m/s over a bridge (radius 50 m). Normal force on car at the top of the bridge: (g = 10)
Explanation: mg − N = mv²/r → N = m(g − v²/r) = 1200(10 − 400/50) = 1200(10−8) = 2400 N.
43. Conical pendulum (mass m, string L, angle θ). String tension is:
Explanation: Vertical equilibrium: T cosθ = mg → T = mg/cosθ.
44. A vehicle goes around a banked curve (θ = 37°, r = 40 m, g = 10). The ideal speed (no friction needed) is:
Explanation: v = √(rg tanθ) = √(40×10×tan37°) = √(400×0.75) = √300 ≈ 17.3 m/s.
45. A particle moves in a circle of radius 3 m. At a certain instant, v = 6 m/s and a_tangential = 4 m/s². The magnitude of total acceleration is:
Explanation: a_c = v²/r = 36/3 = 12 m/s². a_t = 4 m/s². a_total = √(12² + 4²) = √(144+16) = √160 ≈ 12.65 m/s².
46. A ball (0.2 kg) is swung in a horizontal circle (r = 1 m). String breaks at tension 25 N. Maximum angular velocity before breaking: (g = 10)
Explanation: String tension provides centripetal force: T = mω²r (if horizontal, ignoring gravity) → ω = √(T/mr) = √(25/0.2) = √125 ≈ 11.18 rad/s. If conical angle is involved, ω_max = √(T−mg cosθ)/(mL sinθ)...
47. A ball enters a circular loop (radius R = 2 m) at the bottom. Minimum speed at the bottom to complete the loop: (g = 10)
Explanation: v_min at bottom = √(5gR) = √(5×10×2) = √100 = 10 m/s.
48. A particle in circular motion doubles its speed. The centripetal acceleration becomes:
Explanation: a_c = v²/r. If v → 2v: a_c → (2v)²/r = 4v²/r = 4×a_c.
49. A ball on a string (radius r) swings in a vertical circle with speed v₀ at the bottom. It leaves the circular path (string goes slack) at angle θ from the vertical where cosθ =:
Explanation: String slack: mg cosθ = mv²/r. Energy: v² = v₀² − 2gr(1−cosθ). Substituting: rg cosθ = v₀²−2gr+2gr cosθ → cosθ(rg−2rg) = v₀²−2gr → cosθ = (v₀²−2gr)/(3gr) but only valid when 0 ≤ cosθ ≤ 1, i.e. top half.
50. In a rotating frame (angular velocity ω), a stationary object (in the lab frame) appears to move with centrifugal force mω²r and Coriolis force:
Explanation: Coriolis force = 2m(v' × ω) where v' is the velocity in the rotating frame. For a stationary object in lab frame moving with −ωr velocity in rotating frame, F_Cor = 2m(v'×ω).
51. A ball rolls in a vertical circular track (smooth) of radius r. At the top, the normal force from the track is N_top = mv²_top/r − mg. If the ball just completes the loop, N_top = 0 and:
Explanation: At the top of the track: gravity alone provides centripetal force → mg = mv²/r → v_top = √(gr). Same as for the string case.
52. A satellite orbits at height h above Earth (radius R). Its orbital speed is v = √(gR²/(R+h)). The orbital period is:
Explanation: Circumference of orbit = 2π(R+h). T = distance/speed = 2π(R+h)/v.
53. A particle in circular motion (speed v, radius r) rotates through 90°. The magnitude of change in velocity is:
Explanation: Initial velocity v (say, east). After 90°, velocity v (north). |Δv| = √(v²+v²) = v√2.
54. An electron (m = 9.1×10⁻³¹ kg) moves in a circle of radius 0.53×10⁻¹⁰ m at speed 2.2×10⁶ m/s. Centripetal force:
Explanation: F = mv²/r = 9.1×10⁻³¹ × (2.2×10⁶)² / (0.53×10⁻¹⁰) = 9.1×10⁻³¹ × 4.84×10¹² / 0.53×10⁻¹⁰ ≈ 8.3×10⁻⁸ N ≈ 8.2×10⁻⁸ N.
55. A ball is swung in a vertical circle overhead (as in a sling). At the lowest point, the string is taut. If the ball barely completes the circle (v_top = √gr), the speed at the lowest point is:
Explanation: Energy conservation from bottom to top: ½mv_b² = ½mv_t² + mg(2r). v_t = √(gr). v_b² = gr + 4gr = 5gr → v_b = √(5gr).
56. A car (mass 800 kg) moves in a horizontal circle (r = 40 m) at 20 m/s on a flat road (μ = 0.6, g = 10). Is the motion possible?
Explanation: Centripetal force needed = mv²/r = 800×400/40 = 8000 N. Max friction = μmg = 0.6×800×10 = 4800 N. 8000 > 4800 → friction insufficient → car CANNOT make the turn at that speed. Answer: No.
57. An astronaut in a space station orbiting Earth at height h feels weightless because:
Explanation: In orbit, both the astronaut and station accelerate toward Earth at the same rate (centripetal acceleration = g at that height). There is no relative acceleration between them → apparent weightlessness. Gravity is very much present.
58. In a rotor (cylindrical drum rotating at ω), a person stands against the wall (radius R). The person doesn't fall when floor is removed if:
Explanation: Normal force N = mω²R. Friction f = μN = μmω²R ≥ mg. Condition: μω²R ≥ g.
59. A ball on a string of length 2 m just completes a vertical circle. Speed at the top is: (g = 10)
Explanation: v_top_min = √(gr) = √(10×2) = √20 ≈ 4.47 m/s.
60. A car goes through the lowest point of a valley (radius r = 50 m) at 20 m/s. Apparent weight of a 60 kg person is: (g = 10)
Explanation: At bottom of valley: N − mg = mv²/r → N = m(g + v²/r) = 60(10+400/50) = 60(10+8) = 1080 N.
61. A car (r = 100 m) enters a curve at 20 m/s and accelerates at 2 m/s². Centripetal acceleration at entry:
Explanation: a_c = v²/r = 400/100 = 4 m/s² (tangential acceleration is 2 m/s² — these are perpendicular).
62. A satellite orbits at radius 2R from Earth's centre (R = Earth's radius). If g at surface = 10 m/s², orbital speed is:
Explanation: g at 2R = g×R²/(2R)² = g/4 = 2.5 m/s². v = √(g'×2R) = √(2.5×2R) = √(5R) m/s (where R in metres).
63. A ball on a string swings in a vertical circle. The string goes slack at a certain point. After going slack, the ball follows:
Explanation: Once the string goes slack (T = 0), the only force on the ball is gravity → projectile motion → parabolic path.
64. On a banked road, if speed > ideal speed, which force prevents outward skidding?
Explanation: When v > v_ideal, the required centripetal force exceeds what banking provides. The extra is provided by friction acting up the slope (toward the inner edge of the curve, i.e., with inward horizontal component).
65. A satellite is moved to a higher orbit. Its orbital speed:
Explanation: v = √(GM/r). Higher r → smaller v. Satellites move slower in higher orbits.
66. For any object moving in a vertical circle (radius r, mass m), the difference in tension at the bottom and top is:
Explanation: T_b = mv_b²/r + mg. T_t = mv_t²/r − mg. v_b² = v_t² + 4gr. T_b − T_t = m(v_b²−v_t²)/r + 2mg = m×4g + 2mg = 6mg. This is always 6mg regardless of speed.
67. In the inertial (ground) frame, there is:
Explanation: In the inertial frame, the real radially inward force (gravity, tension, friction, etc.) is called centripetal. Centrifugal force is a pseudo force that exists only in the rotating (non-inertial) frame.
68. A bead on a smooth horizontal rotating rod at initial radius r₀ with zero velocity (relative to rod). As the rod rotates at ω, the bead slides outward. The equation of motion in the rotating frame is:
Explanation: In the rotating frame: bead experiences centrifugal force mω²r (outward) and Coriolis force −2mω × v' (where v' is velocity in rotating frame). Both affect the bead's motion in the rotating frame.
69. A particle moves in a circle with tangential acceleration a_t. The angle at which the resultant acceleration makes with the radius (centripetal direction) is:
Explanation: a_c is radial (inward), a_t is tangential. The angle from the radial direction is φ where tanφ = a_t/a_c.
70. A ball moves in a vertical circle on a track with friction (μ). Work done by friction per revolution:
Explanation: For a ball on a circular track, the normal force varies with position. Work by friction per revolution = −∮μN ds, which requires integration. For a flat horizontal circle, W_friction = −μmg×2πr. For a vertical circle, the calculation is more complex.
71. A proton (mass m, charge q) moves in a circle of radius r in a magnetic field B. Its speed is:
Explanation: Magnetic force = centripetal force: qvB = mv²/r → v = qBr/m.
72. Deriving Kepler's third law from circular orbit: if v = √(GM/r) and T = 2πr/v, then T ∝:
Explanation: T = 2πr/v = 2πr/√(GM/r) = 2πr×√(r/GM) = 2π√(r³/GM). So T ∝ r^(3/2), confirming Kepler's third law.
73. For a ball in a vertical circle (string, radius r, speed v₀ at bottom), the tension as function of angle θ from the bottom is:
Explanation: At angle θ: height = r(1−cosθ). v² = v₀²−2gr(1−cosθ). Centripetal: T − mg cosθ' = mv²/r (where θ' is angle from vertical at that point). Full derivation gives T = mv₀²/r − mg(3−2cosθ) where θ is measured from the bottom.
74. A ball enters a circular loop (radius R) at the bottom with speed v₀. Friction coefficient μ. The speed at the top (after friction through half loop, treating friction as reducing energy proportionally):
Explanation: Energy loss to friction along the half-circle arc (length πR): W_friction = μmg×πR. Gravity PE gain = mg×2R. Energy equation: ½mv_top² = ½mv₀² − mg×2R − μmg×πR. v_top = √(v₀²−4gR−2πμgR). Approximating: √(v₀²−(4+πμ)gR×2/... careful with factor), the approximate answer is √(v₀²−(4+2πμ)gR).
75. What minimum speed at the bottom of a vertical loop (radius R) is needed for a block on the inside of the loop to stay on the track at the top?
Explanation: At the top, minimum condition: N = 0, mg = mv_top²/R → v_top = √(gR). Energy: ½mv_bottom² = ½mv_top² + mg×2R → v_bottom_min = √(v_top²+4gR) = √(gR+4gR) = √(5gR).
76. Earth's orbital speed around the Sun (~1.5×10¹¹ m, T = 3.15×10⁷ s) is approximately:
Explanation: v = 2πr/T = 2π×1.5×10¹¹/(3.15×10⁷) ≈ 9.42×10¹¹/(3.15×10⁷) ≈ 2.99×10⁴ m/s ≈ 29.9 km/s.
77. A particle in circular motion (r = 2 m) starts from rest with angular acceleration α = 3 rad/s². After 3 seconds, the magnitude of total linear acceleration is:
Explanation: After t = 3 s: ω = αt = 9 rad/s. a_c = ω²r = 81×2 = 162 m/s². a_t = αr = 3×2 = 6 m/s². a_total = √(162²+6²) = √(26244+36) = √26280 ≈ √26280. The exact value is √(1296+36) = √1332 if using different r... For r=2: √(162²+6²) = √(26280) = 6√730 ≈ 162 m/s².
78. A coin is placed at the edge (r = 0.5 m) of a turntable that gradually increases from 0 to ω. The coin starts to slide when:
Explanation: Coin slides when centripetal force needed > maximum static friction: mω²r > μmg → ω > √(μg/r).
79. At the equator, what angular velocity would make g_eff = 0 (objects become weightless)? (g = 9.8 m/s², R_Earth = 6.4×10⁶ m)
Explanation: g_eff = 0: g = ω²R → ω = √(g/R) = √(9.8/6.4×10⁶) = √(1.53×10⁻⁶) ≈ 1.24×10⁻³ rad/s. Compare with Earth's actual ω = 7.27×10⁻⁵ rad/s — Earth would need to spin ~17× faster!