An object moving in a circle of radius r with constant speed v has acceleration of magnitude a_c = v² / r directed toward the centre of the circle (centripetal). In terms of angular speed ω = v/r, a_c = ω² r. The acceleration changes direction continuously even though the speed is constant.
-- NCERT Class 11 Physics, Ch. 3, p. 42Dynamics UCM
8 MCQs9 revision cards9-step worked example
Source: NCERT Laws of MotionOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Dynamics UCM, explained for NEET
The highest-yield trap in dynamics of uniform circular motion is treating centripetal force as a separate, additional force on the free-body diagram. It is not. Centripetal force is the label for the net inward radial component of whatever real forces act on the body — tension, friction, normal reaction, gravity, or some combination. Drawing a separate "centripetal force" arrow alongside real forces double-counts the inward force and wrecks the entire calculation.
In uniform circular motion, an object moves at constant speed v along a circle of radius r. Because the direction of velocity changes continuously, the object accelerates. This acceleration is centripetal — directed toward the centre — with magnitude a_c = v²/r (NCERT Class 11 Physics, Chapter 3, page 42). By Newton's second law, a net inward force F_c = mv²/r must exist to produce this acceleration (NCERT Class 11 Physics, Chapter 4, page 63).
The procedure is always the same: draw only real forces on the free-body diagram, resolve them along the radial direction, and set the net inward component equal to mv²/r. On a level road, static friction alone provides centripetal force, giving v_max = √(μ_s·g·r). On a banked road, the normal force's horizontal component contributes, leading to the banked-road formula.
A common confusion in NEET: the question asks which quantity is constant in UCM. Speed and kinetic energy are constant. Velocity is not — its direction changes every instant. Acceleration magnitude (v²/r) is constant, but acceleration direction keeps rotating inward. NEET distractors exploit the speed-versus-velocity conflation.
Watch out: if you ever draw "centripetal force" as a separate arrow on an FBD, you have invented a force that does not exist. Identify the real provider — then set it equal to mv²/r.
Can you answer these Dynamics UCM MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In uniform circular motion, which of the following quantities remains constant?
Show answer and why every option is right or wrong
Answer: A. Speed is constant in UCM, so KE = ½mv² is constant. Velocity, acceleration, and centripetal force all change direction continuously, so none of them is constant as a vector (NCERT Class 11 Physics, Chapter 3, page 42).
Why B is wrong: B is wrong because centripetal acceleration has constant magnitude v²/r but its direction rotates inward at every point — it is not a constant vector.
Why C is wrong: C is wrong because velocity is a vector; in UCM the direction changes continuously even though speed is constant (trap: conflating speed with velocity).
Why D is wrong: D is wrong because centripetal force = mv²/r directed inward; the direction rotates with the object, so the force vector is not constant.
MCQ 2Direct ApplicationPractice
A car takes an unbanked circular turn of radius 20 m on a level road. The coefficient of static friction between the tyres and the road is 0.50. Taking g = 10 m/s², the maximum speed at which the car can take the turn without skidding is:
Show answer and why every option is right or wrong
Answer: B. B is correct. On a level road the only horizontal force is static friction, so it must supply the whole centripetal force: mv²/r ≤ μₛmg. The mass cancels, giving v_max = √(μₛrg) = √(0.50 × 20 × 10) = √100 = 10 m/s.
Why A is wrong: A is wrong because 100 m/s is μₛrg without the square root. That product is v², in m²/s².
Why C is wrong: C is wrong because 14 m/s is √(rg), which leaves out the coefficient of friction — it is the answer only if μₛ were 1.
Why D is wrong: D is wrong because 20 m/s is √(rg/μₛ): dividing by the coefficient of friction instead of multiplying. Less friction must mean a LOWER safe speed, not a higher one.
MCQ 3Direct ApplicationPractice
A car moves along a level circular road of radius 50 m. If the coefficient of static friction between the tyres and the road is 0.40 and g = 10 m/s², the maximum safe speed of the car is:
Show answer and why every option is right or wrong
Answer: C. On a level road, v_max = √(μ_s·g·r) = √(0.40 × 10 × 50) = √200 ≈ 14 m/s (NCERT Class 11 Physics, Chapter 4, page 62).
Why A is wrong: A is wrong because 10 m/s comes from √(μ_s·g·r/2) = √100, halving 200 before taking the root; the friction limit μ_s·mg = mv²/r gives v² = μ_s·g·r = 200 with no factor ½.
Why B is wrong: B is wrong because 20 m/s comes from √(2μ_s·g·r) = √400, carrying a factor 2 over from v² = 2gh; the friction limit has no 2.
Why D is wrong: D is wrong because this uses μ_s·g·r = 0.40 × 10 × 50 = 200 without taking the square root — the formula requires v_max = √(μ_s·g·r), and the answer should be in m/s, not m²/s².
MCQ 4Easy RecallPractice
A student draws the free-body diagram of a ball on a string in horizontal circular motion. The student draws three arrows: tension (inward), weight (downward), and a separate "centripetal force" arrow (inward). What is wrong with this diagram?
Show answer and why every option is right or wrong
Answer: B. Centripetal force is not a new fundamental force. It is the net inward radial component of real forces (here, the horizontal component of tension). Drawing it as a separate arrow double-counts the inward force (NCERT Class 11 Physics, Chapter 4, page 63).
Why A is wrong: A is wrong because weight is a real force that acts on the ball regardless of the plane of motion — gravity does not vanish for horizontal circular motion.
Why C is wrong: C is wrong because tension in the string physically pulls the ball inward toward the centre; it does not point outward. 'Centrifugal force' is a pseudo-force in rotating frames, not a real force on the FBD in an inertial frame.
Why D is wrong: D is wrong because 'centripetal force' is not a separate real force — drawing it alongside tension creates a phantom force that does not exist (trap: treating centripetal force as an additional force).
MCQ 5Easy RecallPractice
In uniform circular motion at constant speed v, the centripetal acceleration has:
Show answer and why every option is right or wrong
Answer: C. The magnitude of centripetal acceleration is v²/r, which is constant when v and r are constant. However, the direction always points toward the centre, which rotates as the object moves around the circle — so the direction changes continuously (NCERT Class 11 Physics, Chapter 3, page 42).
Why A is wrong: A is wrong because although the magnitude v²/r is constant, the direction of centripetal acceleration rotates continuously inward — it is not fixed in space (trap: confusing constant magnitude with constant vector).
Why B is wrong: B is wrong because the magnitude v²/r does NOT change in uniform circular motion — speed v and radius r are both constant.
Why D is wrong: D is wrong because while the direction does change, the magnitude v²/r remains constant throughout uniform circular motion.
MCQ 6Direct ApplicationPractice
A body of mass 2.0 kg moves in a circle of radius 0.50 m. The centripetal force on it is 16 N. What is the speed of the body?
Show answer and why every option is right or wrong
Answer: D. F_c = mv²/r → v² = F_c·r/m = 16 × 0.50 / 2.0 = 4.0 → v = 2.0 m/s (NCERT Class 11 Physics, Chapter 4, page 63).
Why A is wrong: A is wrong because 1.0 m/s would give a centripetal force of mv²/r = 2.0 × 1.0/0.50 = 4.0 N, a quarter of the stated 16 N; speed enters squared, so halving v quarters the force.
Why B is wrong: B is wrong because 8.0 is F_c/m = 16/2.0, the centripetal acceleration in m/s², not a speed; v follows from a = v²/r, giving v² = 8.0 × 0.50 = 4.0.
Why C is wrong: C is wrong because 4.0 is v², not v: F_c·r/m = 16 × 0.50/2.0 = 4.0 m²/s², and the square root must still be taken.
MCQ 7Concept TrapPractice
A car negotiates a frictionless banked curve. The banking angle is θ. Which force provides the centripetal acceleration?
Show answer and why every option is right or wrong
Answer: D. On a frictionless banked road, the only contact force is the normal reaction (perpendicular to the road surface). Its horizontal component (N sin θ) points toward the centre of the curve and provides the centripetal force. Weight acts vertically downward and is balanced by N cos θ (NCERT Class 11 Physics, Chapter 4, page 63).
Why A is wrong: A is wrong because the road is explicitly frictionless — no friction force exists to provide any centripetal acceleration.
Why B is wrong: B is wrong because weight acts vertically downward, not toward the centre of the circular path. It cannot directly provide the horizontal centripetal force.
Why C is wrong: C is wrong because the vertical component of the normal reaction (N cos θ) balances the weight mg — it acts vertically upward, not toward the centre of the curve.
MCQ 8CalculationPractice
A 0.50 kg ball moves in a horizontal circle of radius 0.90 m, completing one full revolution every T = 2.0 s. Using v = 2πr/T and taking π² ≈ 10, what is the magnitude of the centripetal force acting on the ball?
Show answer and why every option is right or wrong
Answer: A. NCERT gives v = 2πr/T for uniform circular motion, so centripetal acceleration a_c = v²/r = (2πr/T)²/r = 4π²r/T² (combining NCERT Class 11 Physics, Chapter 3, page 42, equations for v and a_c). Substituting T² = (2.0)² = 4.0, r = 0.90 m, and π² ≈ 10: a_c = 4 × 10 × 0.90 / 4.0 = 9.0 m/s². By Newton's second law (NCERT Class 11 Physics, Chapter 4, pages 54–55), F = m·a_c = 0.50 × 9.0 = 4.5 N.
Why B is wrong: B (2.25 N) drops BOTH the factor of 4 in a_c = 4π²r/T² and the squaring of T, effectively computing π²mr/T instead of 4π²mr/T² — two errors that only partly cancel, understating the correct 4.5 N.
Why C is wrong: C (9.0 N) uses T instead of T² in the denominator, i.e. computes 4π²mr/T rather than 4π²mr/T² — forgetting that the period enters the centripetal acceleration formula squared, which exactly doubles the correct force.
Why D is wrong: D (1.125 N) drops the factor of 4 in a_c = 4π²r/T², computing π²mr/T² instead — this understates the correct force by exactly a factor of 4.
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Dynamics UCM: quick recall before you leave
How do you solve a Dynamics UCM question? A worked example
Pattern: UCM properties at an instant (anchored to PYQ 2024 Q3 Q17)
- 1
Given
• Mass m = 0.20 kg• Radius r = 0.80 m• Frequency f = 5.0 rev/s (2 sig figs)
- 2
Required
(a) Centripetal acceleration a_c
(b) Tension T in the string - 3
Concept
In UCM, centripetal acceleration = ω²r, directed toward the centre. The string tension is the real force providing centripetal force: T = ma_c = mω²r.
- 4
Formula
• ω = 2πf• a_c = ω²r• T = ma_c = mω²r
- 5
Substitution
• ω = 2π × 5.0 = 10π rad/s• a_c = (10π)² × 0.80 = 100π² × 0.80• T = 0.20 × a_c
- 6
Calculation
• ω = 31.4 rad/s (3 sig figs for intermediate)• a_c = 100 × 9.870 × 0.80 = 789.6 m/s² → 7.9 × 10² m/s² (2 sig figs, limited by f = 5.0)• T = 0.20 × 789.6 = 157.9 N → 1.6 × 10² N (2 sig figs)
Note on exact values: π is an exact mathematical constant and the integer 2 in ω = 2πf is exact — neither limits the significant-figure count. The sig-fig count is governed by the measured values (m, r, f), all given to 2 significant figures. - 7
Final answer
(a) a_c ≈ 7.9 × 10² m/s²
(b) T ≈ 1.6 × 10² N - 8
Common trap
A student might add a separate "centripetal force" on top of the string tension. That would double the inward force and yield T = mv²/r − F_centripetal, a meaningless equation. The tension IS the centripetal force here — no additional force exists.
- 9
Similar NEET-style question
A ball of mass 0.10 kg is attached to a string of length 0.50 m and whirled in a horizontal circle at 4.0 rev/s. Find the tension in the string. (Answer: T = mω²r = 0.10 × (8π)² × 0.50 ≈ 32 N.)
---
What to remember before solving Dynamics UCM questions
For a body of mass m moving in a circle of radius r at uniform speed v, the centripetal force required (directed toward the centre) is F_c = m v² / r = m ω² r. This is the net INWARD radial force; it is provided by some real force (tension, friction, gravity, etc.).
-- NCERT Class 11 Physics, Ch. 4, p. 63Which Dynamics UCM formulas do you need for NEET?
2 formulas — click to collapse
Centripetal acceleration in uniform circular motion
An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.
| Symbol | Quantity | SI Unit |
|---|---|---|
| a_c | Centripetal acceleration | m/s^2 |
| v | Tangential speed | m/s |
| r | Radius of circle | m |
| omega | Angular speed | rad/s |
Valid when
- Speed v is constant (uniform circular motion)
- r and the centre are well-defined (instantaneous radius of curvature for general curved motion)
Do NOT use when
- Non-uniform circular motion (then there is also a tangential acceleration component)
Centripetal force in uniform circular motion
The net inward force required to keep a body of mass m moving in a circle of radius r at speed v is m*v^2/r. This 'centripetal' force is NOT a new fundamental force — it is whichever real force (tension, friction, gravity, etc.) provides the inward acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_c | Centripetal force | N |
| m | Mass of the body | kg |
| v | Tangential speed | m/s |
| r | Radius of the circle | m |
| omega | Angular speed | rad/s |
Valid when
- Speed is uniform (a_t = 0, only radial acceleration matters)
- Identify the real force that provides F_c (tension, friction, normal component, etc.)
Where do students lose marks on Dynamics UCM?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Root cause: concept gap
Correction
Centripetal force is NOT a new fundamental force. It is the NET inward radial component of the real forces (tension, friction, normal, gravity, etc.). On a free-body diagram, draw only the real forces; their net inward component must equal m*v^2/r.
Wrong option pattern
Distractor sums tension + 'centripetal force' as separate inward forces.
More in Laws of Motion: 15 exam traps and mistakes · 7 formulas · 9 question patterns from its other lessons.
Dynamics UCM questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Dynamics UCM?
1 recurring pattern from past papers — click to collapse
Particle in uniform circular motion at constant speed; identify which of speed / velocity / acceleration / kinetic energy is constant. Speed and KE constant; velocity (vector) and acceleration (centripetal) are NOT constant — direction changes.
InterpretationEasy
Common distractors
treats velocity as scalar
Conflates speed (scalar) with velocity (vector)
Sources
NCERT refs: Class 11 Physics Chapter 3, p.42 | Class 11 Physics Chapter 4, p.63
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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