Circular Motion: When the Path Curves Back

Kinematics of Circular Motion: The Angular Language

In straight-line kinematics you described the world with three quantities: position \(x\), velocity \(v = dx/dt\), and acceleration \(a = dv/dt\). Every equation of motion was built from these and the assumption of constant acceleration.

On a circle, that description becomes awkward. A particle moving on a circle of radius \(r\) returns to the same \((x, y)\) coordinates after every revolution, yet it has not “stopped”—it has been moving the entire time. The Cartesian coordinates \(x(t)\) and \(y(t)\) oscillate sinusoidally, which is a needlessly complicated way to describe what is, geometrically, the simplest possible curved path.

The natural coordinate for circular motion is not a distance along a line. It is an angle.

A single number—the angle \(\theta\) that the radius vector makes with a chosen reference direction—specifies the particle's position on the circle completely. As the particle moves, \(\theta\) changes monotonically (it does not oscillate), increasing steadily for counterclockwise motion and decreasing for clockwise. Every tool of one-dimensional kinematics—velocity, acceleration, equations of motion—carries over intact, with \(\theta\) playing the role that \(x\) played in Chapter 1.

Angular Displacement and the Radian

Fig 5.1 The radian. The arc \(AB\) has length \(s = r\theta\), so the angle in radians is \(\theta = s/r\) — a pure ratio of two lengths, hence dimensionless.

Consider a particle that moves along the arc of a circle of radius \(r\), sweeping through an angle \(\Delta\theta\) at the centre. The arc length \(s\) traced out is related to the angle by:

\[ \boxed{s = r\,\theta} \qquad (\theta \text{ in radians}) \]

This is not a derived result; it is the definition of the radian. One radian is the angle subtended at the centre of a circle when the arc length equals the radius. Equivalently: \[ \theta = \frac{s}{r} \]

Since the circumference of a full circle is \(2\pi r\), one complete revolution corresponds to \(\theta = 2\pi r / r = 2\pi\;\text{rad} = 360^\circ\). From this: \[ 1\;\text{rad} = \frac{180^\circ}{\pi} \approx 57.3^\circ \]

Why insist on radians? Because the relation \(s = r\theta\) is exact only when \(\theta\) is measured in radians. If \(\theta\) is in degrees, you must write \(s = r \times (\theta_{\text{deg}} \times \pi/180)\)—an ugly conversion factor that propagates into every subsequent formula. Radians are not merely a convention; they are the unit in which the geometry of circles has its simplest algebraic form.

Convention. Throughout this chapter, all angles are in radians unless explicitly stated otherwise. Every formula—\(v = r\omega\), \(a_t = r\alpha\), the angular equations of motion—assumes this. Substituting degrees into any of them is a guaranteed error.

Angular Velocity

Average Angular Velocity

If a particle sweeps through an angular displacement \(\Delta\theta\) in a time interval \(\Delta t\), its average angular velocity is: \[ \bar{\omega} = \frac{\Delta\theta}{\Delta t} \]

This tells you the rate of turning “on average” over the interval, just as \(\bar{v} = \Delta x / \Delta t\) gives the average linear speed over a displacement.

Instantaneous Angular Velocity

As always, the instantaneous quantity is the limit of the average as the interval shrinks to zero:

\[ \boxed{\omega = \frac{d\theta}{dt}} \]

The SI unit of \(\omega\) is \(\mathrm{rad/s}\). By the standard convention, \(\omega \gt 0\) means counterclockwise rotation (when viewed from the positive \(z\)-axis looking down on the \(xy\)-plane), and \(\omega \lt 0\) means clockwise.

Physical picture. Angular velocity measures how fast the radius vector sweeps out angle. A second hand on a clock completes \(2\pi\) radians every \(60\;\text{s}\), so \(\omega_{\text{second}} = 2\pi/60 \approx 0.105\;\text{rad/s}\). The hour hand completes \(2\pi\) radians every \(12\) hours: \(\omega_{\text{hour}} = 2\pi/(12 \times 3600) \approx 1.45 \times 10^{-4}\;\text{rad/s}\). Both hands undergo uniform circular motion; only the rate differs.

From Arc Length to Speed: Deriving \(v = r\omega\)

Fig 5.2 Tangential speed. A particle turning with angular speed \(\omega\) moves along the rim at \(v = r\omega\) — the velocity is always tangent to the circle, perpendicular to the radius.

Birth of the formula: Linear–Angular Bridge: \(v = r\omega\) and \(a_t = r\alpha\)

Part 1: Linear speed from angular velocity. Start from the arc-length relation : \[ s = r\,\theta \] Differentiate both sides with respect to time. The radius \(r\) is constant for a particle on a fixed circle, so: \[ \frac{ds}{dt} = r\,\frac{d\theta}{dt} \] The left-hand side is the linear speed \(v\); the right-hand side is \(r\omega\): \[ \boxed{v = r\omega} \] What this says: at any instant, the linear speed of a point on a circle equals the radius times the angular velocity. Points farther from the centre move faster even though they share the same \(\omega\)—on a spinning disc, the rim covers a larger circumference per revolution than a point near the hub. What this does not say: \(v = r\omega\) gives the magnitude of the velocity, not the vector. The velocity direction is always tangent to the circle and rotates continuously—this is the origin of centripetal acceleration (another section). Critical condition: \(\theta\) must be in radians. If \(\omega\) is given in degrees per second, convert first.

Part 2: Tangential acceleration from angular acceleration. Differentiate \(v = r\omega\) once more with respect to time: \[ \frac{dv}{dt} = r\,\frac{d\omega}{dt} \] The left-hand side, \(dv/dt\), is the tangential acceleration—the component along the direction of motion that changes the speed. The right-hand side is \(r\alpha\): \[ \boxed{a_t = r\alpha} \] What this gives: if \(\alpha = 0\) (uniform circular motion), then \(a_t = 0\) and the speed is constant. If \(\alpha \neq 0\), the speed changes (another section). What this does not give: \(a_t = r\alpha\) is the tangential component only. The radially inward centripetal component \(a_c = v^2/r\) is separate; the total acceleration is the vector sum of both.

Angular Acceleration

When a particle speeds up or slows down on its circular path, \(\omega\) changes with time. The rate of that change is the angular acceleration:

\[ \boxed{\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}} \]

The SI unit is \(\mathrm{rad/s^2}\). A positive \(\alpha\) (with the counterclockwise convention) means the particle is spinning faster in the positive sense; a negative \(\alpha\) means it is slowing down (or speeding up clockwise).

Just as with linear deceleration, the sign of \(\alpha\) alone does not tell you whether the particle is “speeding up” or “slowing down.” What matters is whether \(\alpha\) and \(\omega\) have the same sign (speeding up) or opposite signs (slowing down). This is the rotational analogue of the rule you learnt in Chapter 1: a car with \(v \gt 0\) and \(a \lt 0\) is decelerating. The derivation of \(a_t = r\alpha\) is given above in the derivationbox.

Period, Frequency, and Angular Velocity

Three quantities describe how fast a particle goes around a circle. They are all saying the same thing in different units.

Period (\(T\)): the time for one complete revolution. Unit: seconds.

Frequency (\(f\)): the number of revolutions per second. Unit: hertz (\(\mathrm{Hz} = \mathrm{s}^{-1}\)). By definition: \[ f = \frac{1}{T} \]

Angular velocity (\(\omega\)): the angle swept per second. In one period the particle sweeps \(2\pi\) radians, so:

\[ \boxed{\omega = \frac{2\pi}{T} = 2\pi f} \]

These conversions are used constantly. A problem may give the period; the formula \(a_c = \omega^2 r\) needs \(\omega\). Convert first, compute second.

Practical conversion. Many problems (especially in engineering contexts and NEET) give rotational speed in revolutions per minute (rpm). To convert: \[ \omega\;(\mathrm{rad/s}) = n\;(\mathrm{rpm}) \times \frac{2\pi}{60} \] A motor running at \(3000\;\text{rpm}\) has \(\omega = 3000 \times 2\pi/60 = 100\pi \approx 314\;\text{rad/s}\).

The Angular Equations of Motion

If \(\alpha\) is constant, the angular variables obey exactly the same equations of motion as their linear counterparts under constant linear acceleration. The correspondence is one-to-one:

Core intuition: Linear–Angular Dictionary

Linear (constant \(a\))	Angular (constant \(\alpha\))	Bridge
\(x\) (displacement)	\(\theta\) (angular displacement)	\(s = r\theta\)
\(v\) (velocity)	\(\omega\) (angular velocity)	\(v = r\omega\)
\(a\) (acceleration)	\(\alpha\) (angular acceleration)	\(a_t = r\alpha\)
\(v = v_0 + at\)	\(\omega = \omega_0 + \alpha t\)
[3pt] \(x = v_0 t + \tfrac{1}{2}at^2\)	\(\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2\)
[3pt] \(v^2 = v_0^2 + 2ax\)	\(\omega^2 = \omega_0^2 + 2\alpha\theta\)

You do not need to memorise a new set of equations. If you know the linear equations of motion from Chapter 1, replace \(x \to \theta\), \(v \to \omega\), \(a \to \alpha\), and every formula carries over unchanged. The angular equations are not analogies—they are the same mathematics with different symbols.

A word on units. In the angular equations, \(\theta\) must be in radians, \(\omega\) in \(\mathrm{rad/s}\), and \(\alpha\) in \(\mathrm{rad/s^2}\). Mixing units—for instance, using rpm for \(\omega\) in \(\omega^2 = \omega_0^2 + 2\alpha\theta\)—is one of the most common arithmetic errors in exam halls. Always convert before substituting.

A Trap Worth Naming

Misconception alert: Angular Velocity \(\neq\) Linear Velocity

Students frequently treat \(\omega\) and \(v\) as interchangeable. They are not. They differ in three fundamental ways.

Different dimensions. \(\omega\) has units \(\mathrm{rad/s}\); \(v\) has units \(\mathrm{m/s}\). They cannot be equal, added, or compared.
Different dependence on radius. On a spinning disc, every point shares the same \(\omega\). But points at different radii have different speeds: \(v = r\omega\). The rim of a bicycle wheel moves much faster than a spoke near the hub, even though both complete one revolution in the same time.
Constant \(\omega\) does not mean zero acceleration. If \(\omega\) is constant, then \(\alpha = 0\) and \(a_t = 0\)—the speed does not change. But the velocity direction changes continuously, producing a centripetal acceleration \(a_c = v^2/r = \omega^2 r\) directed radially inward. Uniform circular motion has zero tangential acceleration but nonzero centripetal acceleration. This is the single most important conceptual distinction in the chapter.

Exam trap: A JEE question gives a disc rotating at constant \(\omega\) and asks for the acceleration of a point on the rim. Students who confuse “constant \(\omega\)” with “zero acceleration” answer \(a = 0\). The correct answer is \(a = \omega^2 r\), directed toward the centre.

✓ Checkpoint · Circular Kinematics

A wheel of radius \(0.4\;\mathrm{m}\) completes \(5\) revolutions.
1. What is the angular displacement in radians?
2. What is the total arc length traced by a point on the rim?
(a) \(\Delta\theta = 5 \times 2\pi = 10\pi\;\mathrm{rad}\). (b) \(s = r\,\Delta\theta = 0.4 \times 10\pi = 4\pi \approx 12.6\;\mathrm{m}\).
A ceiling fan rotates at \(300\;\mathrm{rpm}\).
1. Convert this to \(\mathrm{rad/s}\).
2. If the blade is \(0.6\;\mathrm{m}\) long, find the linear speed of the tip.
3. Find the linear speed of a point \(0.2\;\mathrm{m}\) from the centre.
(a) \(\omega = 300 \times 2\pi/60 = 10\pi\;\mathrm{rad/s} \approx 31.4\;\mathrm{rad/s}\). (b) \(v_{\text{tip}} = r\omega = 0.6 \times 10\pi = 6\pi \approx 18.8\;\mathrm{m/s}\). (c) \(v = 0.2 \times 10\pi = 2\pi \approx 6.3\;\mathrm{m/s}\). Same \(\omega\), different \(v\)—this is the key distinction.
A turntable starts from rest and reaches \(\omega = 4\pi\;\mathrm{rad/s}\) in \(8\;\mathrm{s}\) with constant angular acceleration.
1. Find \(\alpha\).
2. How many revolutions does it complete in these \(8\;\mathrm{s}\)?
3. Find the angular velocity after \(4\;\mathrm{s}\).
(a) \(\alpha = (\omega - \omega_0)/t = (4\pi - 0)/8 = \pi/2\;\mathrm{rad/s^2}\). (b) \(\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = 0 + \tfrac{1}{2}(\pi/2)(64) = 16\pi\;\mathrm{rad}\). Number of revolutions \(= 16\pi / 2\pi = 8\). (c) \(\omega(4) = 0 + (\pi/2)(4) = 2\pi\;\mathrm{rad/s}\). This is half the final \(\omega\), as expected for constant \(\alpha\) starting from rest.
A grinding wheel rotating at \(\omega_0 = 40\;\mathrm{rad/s}\) is switched off and decelerates uniformly, coming to rest after \(200\;\mathrm{rad}\) of rotation.
1. Find the angular deceleration.
2. How long does it take to stop?
(a) \(\omega^2 = \omega_0^2 + 2\alpha\theta\). \(0 = 40^2 + 2\alpha(200) \Rightarrow \alpha = -1600/400 = -4\;\mathrm{rad/s^2}\). (b) \(\omega = \omega_0 + \alpha t \Rightarrow 0 = 40 - 4t \Rightarrow t = 10\;\mathrm{s}\).
Conceptual. Two points \(A\) and \(B\) lie on the same rotating disc. \(A\) is at radius \(r\); \(B\) is at radius \(2r\).
1. Compare their angular velocities.
2. Compare their linear speeds.
3. Compare their centripetal accelerations.
(a) Same \(\omega\) (rigid body: every point shares the same angular velocity). (b) \(v_B = 2r\omega = 2v_A\). \(B\) moves twice as fast. (c) \(a_{cB} = \omega^2(2r) = 2\omega^2 r = 2\,a_{cA}\). Centripetal acceleration is proportional to \(r\) at fixed \(\omega\).
True or False (justify in one sentence each):
1. If \(\omega\) is constant, the particle has zero acceleration.
2. Angular velocity has the same dimensions as frequency.
3. Doubling the radius while keeping \(\omega\) fixed doubles the centripetal acceleration.
(a) False. \(\omega\) constant \(\Rightarrow\) \(a_t = 0\), but \(a_c = \omega^2 r \neq 0\); the direction of velocity still changes. (b) True. Both have dimensions \(\mathrm{T}^{-1}\); they differ only by a factor of \(2\pi\) (which is dimensionless). (c) True. \(a_c = \omega^2 r\); at fixed \(\omega\), doubling \(r\) doubles \(a_c\).

Centripetal Acceleration: The Invisible Turn

The Question That Changes Everything

A particle moves at constant speed around a circle. The speedometer reads the same number at every instant. Is the particle accelerating?

If your instinct says “no,” you are in good company—and you are wrong.

Acceleration is not the rate of change of speed. It is the rate of change of velocity, and velocity is a vector. A vector can change in two ways: its magnitude can change (the object speeds up or slows down), or its direction can change (the object turns). On a straight road at constant speed, neither changes, and the acceleration is zero. On a circular path, the velocity vector is always tangent to the circle. As the particle moves, that tangent direction rotates continuously. A rotating vector has a nonzero time derivative, and a nonzero time derivative of velocity is, by definition, acceleration.

So a particle in uniform circular motion is permanently accelerating, even though it never changes speed. The natural follow-up is: how large is this acceleration, and which way does it point?

Why the Acceleration Must Point Inward

Before doing any calculation, we can pin down the direction by pure logic.

At every instant, the velocity \(\vec{v}\) is tangent to the circle. If the acceleration had a component along \(\vec{v}\) (forward or backward), it would change the speed—but we stipulated that the speed is constant. Therefore the acceleration has no tangential component. The only remaining possibility is that the acceleration is entirely perpendicular to \(\vec{v}\).

On a circle, the direction perpendicular to the tangent and lying in the plane of the circle is the radial direction. There are two radial options: inward (toward the centre) or outward (away from the centre). Which one?

Think of it this way. If no force—and therefore no acceleration—acted on the particle, it would fly off in a straight line tangent to the circle (Newton's First Law). The fact that it doesn't fly off, but instead curves inward to stay on the circle, means the acceleration must be directed inward, toward the centre.

This is why it is called centripetal acceleration, from the Latin centrum (centre) + petere (to seek). The acceleration seeks the centre.

Geometric Derivation: The Velocity Triangle

Fig 5.3 Why centripetal acceleration points inward. Between two nearby instants the velocity turns; the change \(\Delta\vec v\) points toward the centre, so \(\vec a = \Delta\vec v/\Delta t\) is centre-seeking.

Birth of the formula: Centripetal Acceleration from the Velocity Triangle

Consider a particle moving at constant speed \(v\) on a circle of radius \(r\). At time \(t\) it is at point \(P\); at time \(t + \Delta t\) it is at point \(Q\). The radius vector has swept through a small angle \(\Delta\theta\). The velocity at \(P\) is \(\vec{v}_P\), tangent to the circle at \(P\), with magnitude \(v\). The velocity at \(Q\) is \(\vec{v}_Q\), tangent to the circle at \(Q\), also with magnitude \(v\). Step 1: Construct \(\Delta\vec{v}\). The change in velocity is: \[ \Delta\vec{v} = \vec{v}_Q - \vec{v}_P \] Place \(\vec{v}_P\) and \(\vec{v}_Q\) tail-to-tail. Because each velocity vector is perpendicular to its respective radius, and the two radii make an angle \(\Delta\theta\) with each other, the angle between \(\vec{v}_P\) and \(\vec{v}_Q\) is also \(\Delta\theta\). Step 2: Find \(|\Delta\vec{v}|\). The vectors \(\vec{v}_P\) and \(\vec{v}_Q\) have equal magnitudes \(v\) and enclose an angle \(\Delta\theta\). They form an isosceles triangle whose base has length: \[ |\Delta\vec{v}| = 2v\sin\!\left(\frac{\Delta\theta}{2}\right) \] For small angles, \(\sin(\Delta\theta/2) \approx \Delta\theta/2\), so: \[ |\Delta\vec{v}| \approx v\,\Delta\theta \] This approximation becomes exact in the limit \(\Delta\theta \to 0\). Step 3: Divide by \(\Delta t\) and take the limit. \[ |\vec{a}| = \lim_{\Delta t \to 0} \frac{|\Delta\vec{v}|}{\Delta t} = v \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} = v\,\omega \] Substituting \(\omega = v/r\) from : \[ \boxed{a_c = \frac{v^2}{r}} \] Step 4: Direction. In the isosceles velocity triangle, the vector \(\Delta\vec{v}\) points from the tip of \(\vec{v}_P\) to the tip of \(\vec{v}_Q\). As \(\Delta\theta \to 0\), this direction becomes perpendicular to \(\vec{v}\) and points radially inward—toward the centre of the circle. Summary. A particle moving at constant speed \(v\) on a circle of radius \(r\) has an acceleration of magnitude \(v^2/r\), directed at every instant toward the centre.

A note on the “similar triangles” argument. Some textbooks derive \(a_c = v^2/r\) by observing that the position triangle \(OPQ\) (with sides \(r\), \(r\), and arc \(\approx v\,\Delta t\)) is similar to the velocity triangle (with sides \(v\), \(v\), and base \(|\Delta\vec{v}|\)). From this similarity: \[ \frac{|\Delta\vec{v}|}{v} = \frac{v\,\Delta t}{r} \quad\Longrightarrow\quad \frac{|\Delta\vec{v}|}{\Delta t} = \frac{v^2}{r} \] This is a slightly different route to the same result. Both are correct; use whichever is clearer to you.

Calculus Derivation: Differentiating the Position Vector

Birth of the formula: Centripetal Acceleration from Calculus

Place the centre of the circle at the origin. The position vector of a particle in uniform circular motion is: \[ \vec{r}(t) = r\cos(\omega t)\,\hat{x} + r\sin(\omega t)\,\hat{y} \] First derivative (velocity): \[ \vec{v} = \frac{d\vec{r}}{dt} = -r\omega\sin(\omega t)\,\hat{x} + r\omega\cos(\omega t)\,\hat{y} \] The speed is \(|\vec{v}| = r\omega = v\), constant as expected. Second derivative (acceleration): \[ \vec{a} = \frac{d\vec{v}}{dt} = -r\omega^2\cos(\omega t)\,\hat{x} - r\omega^2\sin(\omega t)\,\hat{y} \] This can be written compactly as: \[ \vec{a} = -\omega^2\,\vec{r} \] Two conclusions fall out immediately:

Magnitude: \(|\vec{a}| = \omega^2 r\). Using \(v = r\omega\), this is \(v^2/r\)—confirming .
Direction: \(\vec{a} = -\omega^2\,\vec{r}\) is antiparallel to \(\vec{r}\). Since \(\vec{r}\) points from the centre to the particle, \(-\vec{r}\) points from the particle to the centre. The acceleration is directed radially inward.

Verification: \(\vec{v}\cdot\vec{a}\): \[ \vec{v}\cdot\vec{a} = (-r\omega\sin\omega t)(-r\omega^2\cos\omega t) + (r\omega\cos\omega t)(-r\omega^2\sin\omega t) = 0 \] The acceleration is perpendicular to the velocity—it changes the direction of \(\vec{v}\) without altering its magnitude. This is the mathematical proof of what we argued qualitatively in another section.

Alternative Forms of \(a_c\)

Using \(v = r\omega\) and \(\omega = 2\pi/T = 2\pi f\), the centripetal acceleration can be written in several equivalent forms. Use whichever matches the quantities given in the problem:

\[ \boxed{a_c = \frac{v^2}{r} = \omega^2 r = v\omega = \frac{4\pi^2 r}{T^2} = 4\pi^2 f^2 r} \]

Each form highlights a different physical relationship:

Form	Use when you know…	Dependence
\(v^2/r\)	speed and radius	\(a_c \propto v^2\) at fixed \(r\)
\(\omega^2 r\)	angular velocity and radius	\(a_c \propto r\) at fixed \(\omega\)
\(v\omega\)	speed and angular velocity	—
\(4\pi^2 r/T^2\)	period and radius	\(a_c \propto 1/T^2\) at fixed \(r\)

The second column is the practical lesson: if a problem gives \(v\) and \(r\), use \(v^2/r\). If it gives \(\omega\) and \(r\), use \(\omega^2 r\). If it gives the period \(T\), use \(4\pi^2 r/T^2\). Choosing the right form avoids unnecessary intermediate calculations.

What Centripetal Acceleration Does—and Does Not Do

The centripetal acceleration is always perpendicular to the velocity. This perpendicularity has a precise consequence: centripetal acceleration changes the direction of the velocity vector without altering its magnitude.

Think of it as a steering wheel that is permanently turned but never touches the throttle or the brake. The car turns continuously but its speedometer does not budge.

Core intuition: Acceleration Decomposition on a Circle

In general, a particle on a circular path can have two components of acceleration:

Centripetal (radial): \(a_c = v^2/r\), directed toward the centre. Changes the direction of \(\vec{v}\).
Tangential: \(a_t = dv/dt = r\alpha\), directed along \(\vec{v}\) (forward if speeding up, backward if slowing down). Changes the magnitude of \(\vec{v}\).

The net acceleration is the vector sum: \[ |\vec{a}| = \sqrt{a_c^2 + a_t^2} \] In uniform circular motion, \(a_t = 0\) and \(\vec{a}\) is purely centripetal. In non-uniform circular motion (another section), both components are present.

Centripetal Force: Not a New Force

◈ Centripetal acceleration — the turning pull

live bench

Circle a particle at steady speed and watch the velocity stay tangent while the acceleration points dead centre. Dial speed and radius and the inward pull scales as a = v²/r — double the speed, quadruple the demand.

◈ loading the interactive bench…

↗ open in the Simulations page

Newton's second law applied in the radial direction gives:

\[ \sum F_{\text{radial}} = m\,a_c = \frac{mv^2}{r} \]

The left-hand side is the net inward component of all real forces acting on the body. The right-hand side is mass times centripetal acceleration. This is simply Newton's second law; it is not a new equation.

The term “centripetal force” refers to the value \(mv^2/r\)—the net radial force required to maintain circular motion. It is not a new fundamental force. It is a label for whatever combination of real forces produces the inward push.

Situation	What provides centripetal force	Radial equation
Satellite in orbit	Gravity	\(mg = mv^2/r\)
Ball on a string (horizontal)	Tension	\(T = mv^2/r\)
Car on a flat curve	Static friction	\(f_s = mv^2/r\)
Car on a banked road (no friction)	Component of normal force	\(N\sin\theta = mv^2/r\)
Electron around a nucleus	Coulomb force	\(kQq/r^2 = mv^2/r\)

In every case, you identify the real forces, resolve them radially, and set the inward component equal to \(mv^2/r\). You never add a separate “\(F_c\)” arrow to the free-body diagram.

Misconception alert: Three Traps in One Diagram

“Centripetal force is a separate force.” Wrong. “Centripetal force” is the net radial component of real forces—gravity, tension, friction, normal force, or some combination. Drawing an additional arrow labelled “\(F_c\)” on your FBD double-counts the inward force and produces incorrect equations. This is one of the most penalised errors on JEE and NEET papers. The rule: draw only forces you can name a physical source for. Then write \(\sum F_{\text{radial}} = mv^2/r\).
“Centrifugal force pushes the object outward.” Not in the inertial frame. In the ground (inertial) frame, there is no outward force on a body moving in a circle. What you feel as an outward push on a merry-go-round is your body's inertia resisting the inward acceleration imposed by the platform. In a rotating (non-inertial) reference frame, one introduces a fictitious “centrifugal force” \(m\omega^2 r\) directed outward to make Newton's laws work. It is a pseudo-force—not exerted by any physical object. If a problem says “work in the ground frame” (as most JEE/NEET problems implicitly do), centrifugal force does not exist. Never draw it on an inertial-frame FBD.
“The object ‘wants' to fly outward.” Objects do not want anything. A ball released from a circular path does not fly radially outward. It flies tangentially—in the direction of its instantaneous velocity at the moment of release (Newton's First Law). The confusion between “tangential departure” and “radial ejection” is common and costs marks on problems involving strings that snap or objects leaving a turntable.

✓ Checkpoint · Centripetal Acceleration

A car travels at \(20\;\mathrm{m/s}\) around a circular curve of radius \(100\;\mathrm{m}\).
1. Find the centripetal acceleration.
2. What provides this centripetal acceleration?
3. If the car doubles its speed, by what factor does the required centripetal force change?
(a) \(a_c = v^2/r = 400/100 = 4\;\mathrm{m/s^2}\). (b) Static friction between the tyres and the road. (c) \(F_c = mv^2/r \propto v^2\). Doubling \(v\) quadruples \(F_c\). This is why high-speed turns are dangerous on wet roads (reduced friction).
The Moon orbits the Earth at a mean distance of approximately \(3.84 \times 10^8\;\mathrm{m}\) with a period of \(27.3\) days. Calculate its centripetal acceleration toward the Earth.
\(\omega = 2\pi/T = 2\pi/(27.3 \times 86400) \approx 2.66 \times 10^{-6}\;\mathrm{rad/s}\). \(a_c = \omega^2 r = (2.66\times10^{-6})^2 \times 3.84\times10^8 \approx 2.72 \times 10^{-3}\;\mathrm{m/s^2}\). This is about \(g/3600\)—consistent with Newton's inverse-square law, since the Moon is roughly \(60\) Earth-radii away and \(1/60^2 \approx 1/3600\).
A stone of mass \(0.5\;\mathrm{kg}\) is tied to a string and whirled in a horizontal circle of radius \(1.2\;\mathrm{m}\) at \(3\;\mathrm{rev/s}\).
1. Find the centripetal acceleration of the stone.
2. Find the tension in the string.
(a) \(\omega = 2\pi \times 3 = 6\pi\;\mathrm{rad/s}\). \(a_c = \omega^2 r = 36\pi^2 \times 1.2 \approx 426\;\mathrm{m/s^2}\). (b) Tension provides centripetal force: \(T = ma_c = 0.5 \times 426 \approx 213\;\mathrm{N}\). This is a very large force—roughly the weight of a \(21\;\mathrm{kg}\) mass. Fast rotation generates surprisingly large centripetal forces.
An aircraft executes a horizontal circular turn of radius \(500\;\mathrm{m}\). The pilot should not experience more than \(4g\) of centripetal acceleration (\(g = 10\;\mathrm{m/s^2}\)). What is the maximum speed of the aircraft during this turn?
\(a_c \le 4g = 40\;\mathrm{m/s^2}\). \(v^2/r \le 40 \Rightarrow v \le \sqrt{40 \times 500} = \sqrt{20000} = 100\sqrt{2} \approx 141\;\mathrm{m/s} \approx 509\;\mathrm{km/h}\).
Conceptual. A ball moves in a horizontal circle on a frictionless table, held by a string attached to a peg at the centre. The string suddenly breaks.
1. Describe the path of the ball immediately after the string breaks.
2. A student says: “The ball flies radially outward because centrifugal force pushes it out.” Critique this statement.
(a) The ball moves in a straight line tangent to the circle at the point of release (Newton's First Law: no net force, no acceleration, straight-line motion at constant velocity). (b) The student is wrong on two counts. First, in the inertial frame there is no centrifugal force. Second, even if one mistakenly invoked such a force, the resulting motion would be radially outward—which is not what happens. The ball departs tangentially, not radially. The direction of departure is the direction of velocity at the instant of release, which is tangent to the circle.
True or False (justify in one sentence each):
1. Centripetal acceleration changes the speed of the particle.
2. If the radius of the circular path is halved while keeping speed constant, the centripetal acceleration is halved.
3. The centripetal force does work on the particle in uniform circular motion.
(a) False. Centripetal acceleration is perpendicular to velocity; it changes the direction, not the magnitude. (b) False. \(a_c = v^2/r\); halving \(r\) at constant \(v\) doubles \(a_c\). (c) False. The centripetal force is always perpendicular to the displacement (which is tangential); \(W = \vec{F}\cdot d\vec{s} = 0\) at every instant.
Two discs. Disc A has radius \(R\) and spins at \(\omega\). Disc B has radius \(2R\) and spins at \(\omega/2\). Compare the centripetal acceleration of a point on the rim of each disc.
Disc A: \(a_A = \omega^2 R\). Disc B: \(a_B = (\omega/2)^2 (2R) = \omega^2 R / 2\). So \(a_B = a_A/2\). Despite having twice the radius, Disc B has half the centripetal acceleration because \(a_c \propto \omega^2\)—angular velocity matters more than radius.
Satellites at different altitudes. Two satellites orbit the Earth in circular orbits. Satellite X is at radius \(r\); Satellite Y is at radius \(4r\). Gravity provides centripetal force: \(mg' = mv^2/r\), where \(g'\) decreases as \(1/r^2\).
1. Show that \(v \propto 1/\sqrt{r}\).
2. Find the ratio \(a_{cX}/a_{cY}\).
(a) \(mg' = mv^2/r\) and \(g' \propto 1/r^2\), so \(v^2/r \propto 1/r^2 \Rightarrow v^2 \propto 1/r \Rightarrow v \propto 1/\sqrt{r}\). (b) \(a_c = v^2/r \propto (1/r)/r = 1/r^2\). \(a_{cX}/a_{cY} = (4r)^2/r^2 = 16\). The closer satellite has \(16\) times the centripetal acceleration.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Stone on a Horizontal String

A stone of mass \(m = 0.4\;\mathrm{kg}\) is tied to a light, inextensible string of length \(L = 0.8\;\mathrm{m}\) and whirled in a horizontal circle on a smooth table at \(2\;\mathrm{rev/s}\). Find the tension in the string.

Solution. Sign convention: Radial direction positive toward the centre (inward along the string). FBD (horizontal plane): The only horizontal force on the stone is the tension \(T\), directed radially inward. (Gravity and the normal force act vertically and cancel; the table is smooth, so there is no friction.) NLM (radial): \[ T = \frac{mv^2}{r} \] We need \(v\). The radius of the circle is \(r = L = 0.8\;\mathrm{m}\), and \(\omega = 2\pi f = 2\pi \times 2 = 4\pi\;\mathrm{rad/s}\). So \(v = r\omega = 0.8 \times 4\pi = 3.2\pi\;\mathrm{m/s}\). Alternatively, use \(T = m\omega^2 r\) directly: \[ T = (0.4)(4\pi)^2(0.8) = (0.4)(16\pi^2)(0.8) = 5.12\pi^2 \approx \boxed{50.5\;\mathrm{N}} \] Check: The expression \(m\omega^2 r\) carries units \(\mathrm{kg}\cdot(\mathrm{rad/s})^2\cdot\mathrm{m}=\mathrm{kg\,m/s^2}=\mathrm{N}\), so the answer is a force. Magnitude: \(T\approx 50.5\;\mathrm{N}\) is about \(12.6\) times the weight \(mg = 4.0\;\mathrm{N}\), reasonable for rapid whirling—and a reminder that whirling tension far exceeds the static hanging load. Common error: Forgetting to convert frequency (\(\mathrm{rev/s}\)) to angular velocity (\(\mathrm{rad/s}\)) before substituting. \(f = 2\;\mathrm{rev/s}\) does not mean \(\omega = 2\). It means \(\omega = 4\pi\).

Example 2 · Satellite in Circular Orbit

A satellite of mass \(m\) orbits the Earth at height \(h\) above the surface. The Earth has mass \(M\) and radius \(R\). Derive an expression for the orbital speed.

Solution. Sign convention: Radial direction positive toward the Earth's centre. FBD: The only force on the satellite is gravitational attraction: \[ F_g = \frac{GMm}{(R+h)^2} \] directed toward the centre. There is no normal force, no tension, no friction. NLM (radial): \[ \frac{GMm}{(R+h)^2} = \frac{mv^2}{R+h} \] Cancel \(m\) (the satellite's mass drops out—all objects orbit at the same speed at the same altitude, regardless of mass): \[ \frac{GM}{(R+h)^2} = \frac{v^2}{R+h} \] \[ v^2 = \frac{GM}{R+h} \qquad\Longrightarrow\qquad \boxed{v = \sqrt{\frac{GM}{R+h}}} \] Key observations:

The orbital speed is independent of \(m\). A \(1\;\mathrm{kg}\) instrument and a \(10{,}000\;\mathrm{kg}\) space station orbit at the same speed at the same altitude. This is because gravity is proportional to \(m\), and so is the centripetal force requirement. The \(m\) cancels.
\(v\) decreases with increasing altitude: \(v \propto 1/\sqrt{R+h}\). Higher satellites move slower—a counterintuitive result. Near the surface (\(h \ll R\)): \(v \approx \sqrt{gR} \approx \sqrt{9.8 \times 6.4\times10^6} \approx 7.9\;\mathrm{km/s}\).
No fuel is being burned during a circular orbit. Gravity provides the centripetal force “for free.” The satellite is in permanent free fall—falling toward the Earth at every instant, but moving tangentially fast enough that the curved surface drops away at the same rate. This resolves Perplexing Question 1.

Check: Dimensionally \(\sqrt{GM/(R+h)}\) gives \(\sqrt{(\mathrm{m^3/s^2})/\mathrm{m}}=\mathrm{m/s}\), a speed. Limiting cases: as \(h\to\infty\), \(v\to0\) (distant orbits crawl); for \(h\ll R\), \(v\to\sqrt{gR}\approx 7.9\;\mathrm{km/s}\), the familiar low-orbit speed.

Example 3 · Angular Velocity, Linear Speed, Centripetal Acceleration

A particle moves in a circle of radius \(R = 1.5\;\mathrm{m}\), completing \(4\;\mathrm{rev/s}\). Find: (a) the angular velocity, (b) the linear speed, (c) the centripetal acceleration.

Solution. Sign convention: Radial direction positive toward the centre. (a) Convert frequency to angular velocity: \[ \omega = 2\pi f = 2\pi \times 4 = 8\pi \approx 25.1\;\mathrm{rad/s} \] (b) Linear speed using \(v = r\omega\): \[ v = 1.5 \times 8\pi = 12\pi \approx \boxed{37.7\;\mathrm{m/s}} \] (c) Centripetal acceleration. Both equivalent forms give the same answer: \[ a_c = \omega^2 r = (8\pi)^2 \times 1.5 = 64\pi^2 \times 1.5 = 96\pi^2 \approx \boxed{947\;\mathrm{m/s^2}} \] Cross-check: \(a_c = v^2/r = (12\pi)^2/1.5 = 144\pi^2/1.5 = 96\pi^2\) ✓ Check: The centripetal acceleration is about \(97g\) — reasonable for a particle spinning at 4 revolutions per second. Dimensional check: \(\omega^2 r = (\mathrm{rad/s})^2\cdot\mathrm{m} = \mathrm{m/s^2}\) ✓.

Example 4 · Normal Force at the Bottom of a Valley

A car of mass \(800\;\mathrm{kg}\) travels at \(v = 15\;\mathrm{m/s}\) through the lowest point of a road dip whose radius of curvature is \(R = 60\;\mathrm{m}\). Find the normal force on the car and compare it to the car's weight. (\(g = 10\;\mathrm{m/s^2}\).)

Solution. Sign convention: Radial direction positive upward (toward the centre of curvature, which is above the car at the lowest point). FBD:

Weight \(mg\) downward (away from centre).
Normal force \(N\) upward (toward centre).

NLM (radial): \[ N - mg = \frac{mv^2}{R} \] \[ N = m\!\left(g + \frac{v^2}{R}\right) = 800\!\left(10 + \frac{225}{60}\right) = 800 \times 13.75 = \boxed{11{,}000\;\mathrm{N}} \] The car's weight is \(mg = 8{,}000\;\mathrm{N}\), so the normal force is \(N/mg = 11000/8000 = 1.375\): the road pushes the car upward with \(1.375\) times its weight. Check: \(N \gt mg\) ✓ (passengers feel heavier at the bottom of a dip). Limiting case: as \(v \to 0\), \(N \to mg\) (stationary car on flat road). As \(R \to \infty\) (nearly straight road), \(N \to mg\) also ✓.

Example 5 · Ideal Speed on a Frictionless Banked Road

A highway circular curve of radius \(r = 200\;\mathrm{m}\) is banked at \(\theta = 30°\). Find the speed at which a car can navigate the curve without any friction. (\(g = 10\;\mathrm{m/s^2}\).)

Solution. Sign convention: Radial direction positive toward the centre of the curve (horizontal); vertical direction positive upward. FBD (friction-free): Only two forces act: weight \(mg\) downward and normal force \(N\) perpendicular to the banked surface. Resolving \(N\): \[\begin{aligned} \text{Vertical:} &\quad N\cos\theta = mg \quad\Longrightarrow\quad N = \frac{mg}{\cos\theta} \\ \text{Horizontal (centripetal):} &\quad N\sin\theta = \frac{mv^2}{r} \end{aligned}\] Dividing the second equation by the first: \[ \tan\theta = \frac{v^2}{rg} \qquad\Longrightarrow\qquad v = \sqrt{rg\tan\theta} = \sqrt{200\times10\times\tan30°} \] \[ v = \sqrt{200\times10\times\tfrac{1}{\sqrt{3}}} = \sqrt{\frac{2000}{\sqrt{3}}} \approx \boxed{34.0\;\mathrm{m/s}\;\approx122\;\mathrm{km/h}} \] Key observation: This speed is mass-independent: any car, regardless of mass, takes this banked curve friction-free at \(34.0\;\mathrm{m/s}\). Speed limits at banked curves are set from this formula, plus a friction margin. Check: \(\sqrt{rg\tan\theta}\) has units \(\sqrt{\mathrm{m\cdot m/s^2}} = \mathrm{m/s}\) ✓. As \(\theta\to0\), \(v\to0\): a flat road needs friction to turn ✓. As \(\theta\to90°\), \(v\to\infty\): a vertical wall cannot support circular motion at any finite speed ✓.

Perplexing Questions

Why Circular Motion Deserves Its Own Chapter