Speed is constant, yet velocity changes every instant — because direction keeps changing. This is the beautiful paradox of uniform circular motion. From NCERT Chapter 4 (Exploration edition) Class 9 Science. Aligned with CBSE syllabus 2026-27.
Q. What do a satellite orbiting Earth, a child on a merry-go-round, and an athlete running on a circular track have in common?
All three are in circular motion — they move along a path that is (approximately) circular. The distinguishing feature of circular motion is that the object keeps returning to regions of the space it has already passed through, tracing out a closed curved path.
Circular motion is the motion of an object along a circular path (or curved path that approximates a circle). The distance from the centre of the circle to the object (the radius $R$) remains constant throughout.
Examples:
Q. What is fundamentally different about circular motion compared to straight-line motion?
In linear (straight-line) motion, an object moves along a single direction — it can speed up, slow down, or reverse, but it always moves along the same line. The direction of motion does not change (or changes only by 180° on reversal).
In circular motion, the direction of motion changes continuously, even if the speed stays constant. At every new point on the circle, the object is moving in a new direction — the direction of velocity rotates through a full 360° in every complete revolution.
Velocity is a vector — it has both magnitude (speed) and direction. For an object on a circular path, the direction it is heading (the direction of velocity) must be along the circle at each point — otherwise the object would leave the circular path. As the object moves around the circle, this direction rotates. Even if the speed (magnitude) is unchanged, the velocity vector changes because its direction changes. A changing velocity means acceleration exists — even in uniform circular motion.
Q. How do we calculate distance and displacement for circular paths?
Consider a child on a merry-go-round of radius $R$ who moves from point A along the arc to point C (through point B on the way), covering a curved arc ABC.
The distance is always measured along the path; the displacement is always the straight-line shortcut from start to finish.
This is the same principle as the Sarang swimming pool example (C02): you can have non-zero average speed alongside zero average velocity, whenever the object returns to its starting position.
After half a revolution (semicircle), the object is on the opposite side of the circle from where it started. The displacement is the diameter of the circle = $2R$, directed from the start point straight across to the end point. The distance covered is the semicircle = $\pi R$. Note that $\pi R > 2R$ (since $\pi > 2$), so distance is greater than displacement — as always.
Uniform circular motion is the motion of an object along a circular path with constant speed. The magnitude of velocity (speed) does not change, but the direction of velocity changes continuously as the object moves around the circle.
The word "uniform" here refers to uniform speed — not uniform velocity (which would require both magnitude AND direction to be constant).
Q. Is uniform circular motion possible in real life?
Approximately, yes. The Moon's orbit around Earth is close to uniform circular motion (it is actually elliptical, but close enough for many calculations). An electric motor rotating a wheel at constant RPM produces near-UCM. A ball on a string swung in a horizontal circle at constant speed also approximates UCM.
Q. How do we calculate the speed in uniform circular motion?
In uniform circular motion, the object covers the circumference $2\pi R$ in exactly one time period $T$. Since speed is constant, average speed equals instantaneous speed at every moment:
Where:
Derivation: Speed = distance ÷ time. In one full revolution, distance = $2\pi R$ (circumference) and time = $T$. Therefore $v = 2\pi R / T$.
The formula $v = 2\pi R/T$ gives speed (scalar), not velocity (vector). The average velocity over one complete revolution is always zero (because displacement = 0). The speed $v$ is the magnitude of the instantaneous velocity at any point on the circle.
Q. At any point on the circular path, in which direction is the object moving?
The direction of velocity is always along the tangent to the circle at that point — not along the radius and not towards the centre. This can be understood intuitively and demonstrated by Activity 4.5.
Place a small marble inside a circular ring (like a bangle or a circular plastic tube). Spin the ring so the marble goes around in a circle inside it. Now suddenly lift one side of the ring to remove a section — the marble escapes through the gap.
Observation: When the ring is removed, the marble does not fly outward (radially) — it moves in a straight line in the direction it was moving at the moment of release. That direction is the tangent to the circle at the escape point.
Conclusion: The instantaneous velocity of an object in circular motion is always directed along the tangent to the circle at that point.
NCERT explains the tangential direction through a beautiful progression:
This progression makes the concept of tangential velocity intuitive — you build it up from straight-line motion in polygons until the circle appears naturally.
Q. The speed is constant in UCM — so why do we say the object is accelerating?
This is one of the most important conceptual questions in Chapter 4, and it challenges a common misunderstanding.
Acceleration is defined as the rate of change of velocity — not the rate of change of speed. Velocity is a vector with both magnitude (speed) and direction.
In UCM:
Conclusion: UCM is an example of accelerated motion — even though the speed is constant. The acceleration is directed towards the centre of the circle (centripetal acceleration), and it causes the object to keep turning towards the centre rather than flying off in a straight line.
In everyday life, when we say a vehicle is "accelerating," we mean it is speeding up. But in physics, acceleration means any change in velocity — including a change in direction alone.
UCM is a perfect example of this distinction: speed does not change, but direction does — so there is acceleration in the physics sense. A car going around a roundabout at constant speed is accelerating physically, even though the speedometer shows no change.
The acceleration in UCM is called centripetal acceleration (from Latin: "centre-seeking"). Its magnitude is $a_c = v^2/R$ (directed towards the centre), and the force causing it is the centripetal force. You will study these in detail in Class 11 physics. For Class 9, it is sufficient to know that the acceleration exists and is directed towards the centre.
Uniform circular motion — perfectly constant speed on a perfectly circular path — is an idealisation that rarely exists in nature with exact precision. In real situations:
Despite these limitations, the UCM model is incredibly useful as a first approximation for circular and orbital motion. It gives correct-order-of-magnitude answers for satellite speeds, centripetal forces, and orbital periods.
| Parameter | Uniform (Linear) Motion | Uniform Circular Motion |
|---|---|---|
| Path | Straight line | Circle |
| Speed | Constant | Constant |
| Velocity direction | Constant (same direction always) | Continuously changing (tangential) |
| Velocity (vector) | Constant | Changing (direction changes) |
| Acceleration | Zero | Non-zero (centripetal, towards centre) |
A child sits on a merry-go-round of radius 2 m. The merry-go-round completes one full revolution in 8 s. Calculate the speed of the child.
Given: $R = 2$ m, $T = 8$ s
Using Eq. 4.5:
$$v = \frac{2\pi R}{T} = \frac{2 \times 3.14 \times 2}{8} = \frac{12.57}{8} \approx \mathbf{1.57 \text{ m s}^{-1}}$$The child moves at approximately 1.57 m s⁻¹, always tangentially to the circular platform.
A satellite orbits Earth at a speed of 8 km s⁻¹ in a circular orbit of radius 6,400 km (approximately at Earth's surface). Find its time period $T$.
Given: $v = 8$ km s⁻¹ = 8000 m s⁻¹, $R = 6400$ km = 6,400,000 m
From Eq. 4.5: $v = 2\pi R / T \Rightarrow T = 2\pi R / v$
$$T = \frac{2 \times 3.14 \times 6{,}400{,}000}{8000} = \frac{40{,}192{,}000}{8000} \approx \mathbf{5024 \text{ s} \approx 84 \text{ min}}$$The satellite completes one orbit in approximately 84 minutes — this is consistent with the actual orbital period of the International Space Station!
An athlete runs along a circular track of radius 100 m at a constant speed of 5 m s⁻¹. (a) Find the time period. (b) What is the displacement after one complete lap? (c) What is the average velocity for one full lap?
(a) Time period:
$$T = \frac{2\pi R}{v} = \frac{2 \times 3.14 \times 100}{5} = \frac{628}{5} = \mathbf{125.7 \text{ s} \approx 126 \text{ s}}$$(b) Displacement after one lap: The athlete returns to the starting point. Displacement = 0 m.
(c) Average velocity for one full lap:
$$\vec{v}_{av} = \frac{\text{displacement}}{T} = \frac{0}{125.7} = \mathbf{0 \text{ m s}^{-1}}$$Note: The average speed is 5 m s⁻¹ (non-zero), but average velocity is zero. Same pattern as the Sarang swimming pool example.
Q1. An object moves along a circular track at constant speed. Is this motion uniform or non-uniform? Is it accelerated? Justify your answer.
Speed is constant → this is uniform motion (equal distances in equal time). However, it is accelerated motion — because the direction of velocity changes continuously as the object goes around the circle. A changing velocity (even if only in direction) means acceleration is non-zero. The acceleration is centripetal (directed towards the centre).
Q2. In uniform circular motion, what is the direction of velocity at the topmost point of the circle? What about at the rightmost point?
At the topmost point: the tangent to the circle is horizontal. If the object is moving anti-clockwise, velocity points to the left; if clockwise, to the right.
At the rightmost point: the tangent is vertical. If moving anti-clockwise, velocity points upward; if clockwise, downward.
At every point, velocity is perpendicular to the radius at that point (i.e., tangential).
Q3. Why does average velocity equal zero for a complete revolution but average speed does not?
After one complete revolution, the object is back at the starting point, so net displacement = 0. Average velocity = displacement/time = 0/T = 0.
But the object did cover the full circumference $2\pi R$ as distance. Average speed = distance/time = $2\pi R / T \neq 0$ (as long as $R > 0$).
This is a fundamental difference between speed (uses distance) and velocity (uses displacement).
N1. A fan blade tip moves in a circle of radius 0.4 m. The fan makes 3 revolutions per second. Find the speed of the tip.
$T = 1/3$ s (3 revolutions per second), $R = 0.4$ m
$$v = \frac{2\pi R}{T} = \frac{2 \times 3.14 \times 0.4}{1/3} = 2 \times 3.14 \times 0.4 \times 3 \approx \mathbf{7.5 \text{ m s}^{-1}}$$N2. The Moon completes one orbit around Earth (radius ≈ 3.84 × 10⁸ m) in 27.3 days. Calculate the Moon's orbital speed in m s⁻¹.
$T = 27.3 \times 24 \times 3600 = 2,358,720$ s; $R = 3.84 \times 10^8$ m
$$v = \frac{2\pi R}{T} = \frac{2 \times 3.14 \times 3.84 \times 10^8}{2{,}358{,}720} \approx \frac{2.411 \times 10^9}{2.359 \times 10^6} \approx \mathbf{1022 \text{ m s}^{-1} \approx 1.02 \text{ km s}^{-1}}$$N3. An object moves in a circular path of radius 7 m with a speed of 11 m s⁻¹. Find the time taken to complete 5 revolutions.
Time for 1 revolution: $T = 2\pi R/v = (2 \times 22/7 \times 7)/11 = 44/11 = 4$ s
Time for 5 revolutions = $5 \times 4 = \mathbf{20 \text{ s}}$
N4. A stone tied to a string of length 1.5 m is whirled in a horizontal circle and completes 10 revolutions in 5 s. Find the speed of the stone.
$T = 5/10 = 0.5$ s (time per revolution), $R = 1.5$ m
$$v = \frac{2\pi R}{T} = \frac{2 \times 3.14 \times 1.5}{0.5} = \frac{9.42}{0.5} \approx \mathbf{18.8 \text{ m s}^{-1}}$$