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NDA Physics · Kinematics and Motion

Circular Motion

In uniform circular motion the speed is constant but the velocity changes continuously because its direction keeps turning; this change is a centripetal acceleration of magnitude v²/r directed toward the centre.

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Why this matters

Three PYQs, but they punch above their weight because they expose the chapter's deepest idea: constant speed is NOT constant velocity. The bank tests that a body going round a circle at steady speed is still accelerating (toward the centre), that a gentler curve (larger radius) means a smaller acceleration, and — in a HARD item — the average acceleration over half a circle. Hold v²/r and the velocity-is-a-vector idea and these are quick marks.

Concept 1 of 3

Uniform circular motion — constant speed, changing velocity

Intuition

Drive a car round a roundabout at a steady 30 km/h: the speedometer never moves, but the car is constantly changing direction. Since velocity includes direction, the velocity is changing every instant — and a changing velocity means there is an acceleration, even though the speed is fixed.

Definition

In uniform circular motion the speed is constant but the velocity changes continuously because its direction is always changing (the velocity is tangent to the circle). A changing velocity means a non-zero acceleration; that acceleration points toward the centre and is called centripetal acceleration.

Worked example

A car moves uniformly along a circular track at constant speed. What, if anything, is changing about its motion?
  1. Speed is constant (uniform motion), so the magnitude of velocity is fixed.
  2. But the direction of motion changes continuously around the circle.
  3. Velocity is a vector, so a change in direction is a change in velocity.
Answer:The velocity changes (in direction), due to the continuous change in direction of motion.
Practice this conceptself-check · 4 quick reps

Try it yourself

A satellite orbits Earth in a circle at constant speed. Is it accelerating? Justify in one line.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    In uniform circular motion, is the speed constant?
  2. 2.
    In uniform circular motion, is the velocity constant?
  3. 3.
    Which way does the centripetal acceleration point?
  4. 4.
    Is a body in uniform circular motion accelerating?

From the bank · past-year question

Example 1Kinematics and MotionEASY
A uniform motion of a car along a circular path experiences

[Q118 · Sep · 2021]

Constant speed is not constant velocity

The single most common error in circular motion: treating 'uniform speed' as 'no acceleration'. The direction of velocity changes every instant, so the velocity is changing and there IS an acceleration — directed toward the centre.

Concept 2 of 3

Centripetal acceleration — v²/r toward the centre

Intuition

The inward acceleration that keeps a body on its circular path has magnitude v²/r. For the same speed, a sharper curve (smaller radius) needs a bigger inward acceleration; a gentle curve (large radius) needs only a small one. That is why fast bends on a road are built with large radii.

Definition

The centripetal acceleration of a body moving at speed vv on a circle of radius rr has magnitude ac=v2ra_c = \dfrac{v^2}{r}, directed toward the centre. At fixed speed it is inversely proportional to the radius: a gentle curve (large rr) gives a smaller acceleration than a sharp curve (small rr).

Centripetal acceleration

ac=v2ra_c = \dfrac{v^2}{r}
  • a_ccentripetal acceleration (toward centre)
  • vspeed
  • rradius of the circular path

Worked example

A car takes a bend of radius 50 m at 10 m/s. Find its centripetal acceleration.
  1. Use ac=v2ra_c = \dfrac{v^2}{r}.
  2. ac=10250=10050=2a_c = \dfrac{10^2}{50} = \dfrac{100}{50} = 2 m/s2^2, directed toward the centre of the bend.
Answer:2 m/s² (toward the centre).
Practice this conceptself-check · 4 quick reps

Try it yourself

At the same speed, a car takes first a sharp curve (small radius) and then a gentle curve (large radius). On which curve is the centripetal acceleration larger?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    v = 6 m/s, r = 9 m. Centripetal acceleration?
  2. 2.
    At fixed speed, doubling the radius does what to a_c?
  3. 3.
    Direction of centripetal acceleration?
  4. 4.
    Gentle (large-radius) curve vs sharp (small-radius), same speed — which has smaller a_c?

From the bank · past-year question

Example 2Kinematics and MotionMODERATE
An object moves in a circular path with a constant speed. Which one of the following statements is correct?

[Q69 · Sep · 2017]

Centripetal acceleration does not change the speed

The centripetal acceleration is perpendicular to the velocity (it points inward, velocity is tangent), so it changes only the DIRECTION of motion, never the speed. An option claiming 'centripetal acceleration causes the object to slow down' is false.

Concept 3 of 3

Average acceleration over part of a circle

Intuition

Centripetal acceleration v²/r is the INSTANTANEOUS value. The AVERAGE acceleration over a stretch of the path is different — it uses the actual change in the velocity vector divided by the time. Over half a circle the velocity simply reverses, so the change in velocity is 2v.

Definition

Average acceleration over an interval is aˉ=∣Δv⃗∣Δt\bar{a} = \dfrac{|\Delta \vec{v}|}{\Delta t}, using the vector change in velocity. Over half a circle the velocity reverses direction, so ∣Δv⃗∣=2v|\Delta \vec{v}| = 2v; the time is the arc length over the speed, Δt=πrv\Delta t = \dfrac{\pi r}{v}. Hence the average acceleration over half a circle is 2vπr/v=2v2πr\dfrac{2v}{\pi r / v} = \dfrac{2v^2}{\pi r}.

Average acceleration over half a circle

aˉ=∣Δv⃗∣Δt=2vπr/v=2v2πr\bar{a} = \dfrac{|\Delta \vec{v}|}{\Delta t} = \dfrac{2v}{\pi r / v} = \dfrac{2v^2}{\pi r}
  • \Delta \vec{v}vector change in velocity (= 2v over half a circle)
  • \Delta ttime for the half circle (= πr/v)
  • vconstant speed

Worked example

A particle moves at constant speed v on a circle of radius R. Find its average acceleration over the time it covers half the circle.
  1. Over half a circle the velocity reverses, so ∣Δv⃗∣=2v|\Delta \vec{v}| = 2v.
  2. Time for the half circle: Δt=half circumferencev=πRv\Delta t = \dfrac{\text{half circumference}}{v} = \dfrac{\pi R}{v}.
  3. Average acceleration: aˉ=2vπR/v=2v2πR\bar{a} = \dfrac{2v}{\pi R / v} = \dfrac{2v^2}{\pi R}.
Answer:2v2πR\dfrac{2v^2}{\pi R}.
Practice this conceptself-check · 3 quick reps

Try it yourself

Over one COMPLETE revolution at constant speed, what is the average acceleration?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Over half a circle, the change in velocity magnitude is?
  2. 2.
    Average acceleration over one full revolution?
  3. 3.
    Time to cover half a circle of radius R at speed v?

From the bank · past-year question

Example 3Kinematics and MotionHARD
A particle is moving in a circle of radius R with a constant speed v. Its average acceleration over the time when it moves over half the circle is :

[Q55 · Apr · 2023]

Average acceleration is not the instantaneous v²/r

v²/r is the instantaneous centripetal acceleration. The AVERAGE over half a circle uses |Δv|/Δt = 2v ÷ (πr/v) = 2v²/πr — a different expression. And over a full revolution the average acceleration is zero (Δv = 0), even though the instantaneous value is never zero.

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