Uniform Circular Motion

What "uniform" really means in circular motion

NCERT §3.10 opens with a single sentence that hides three subtleties: "When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion." Notice what is constant — the speed, the magnitude of velocity — and what is not — the velocity vector itself. The velocity at every point is tangential to the circle, so as the particle moves from $P$ to $P'$ the tangent rotates with it. A rotating vector of constant length is still a changing vector. By the very definition of acceleration as $\mathrm{d}\vec{v}/\mathrm{d}t$, uniform circular motion is accelerated motion.

NCERT's Points to Ponder #4 spells it out: the kinematic equations for uniform acceleration do not apply to uniform circular motion, because here acceleration has constant magnitude but a continuously changing direction. Uniform circular motion cannot be reduced to the constant-acceleration toolkit you used for projectile motion or motion with constant acceleration. It needs its own machinery — the angular description.

Angular displacement and angular speed

As the particle moves on a circle of radius $R$, the radius vector $\vec{r}$ from the centre $C$ to the particle sweeps through an angle. NCERT calls this the angular distance $\Delta\theta$, measured in radians by convention. The arc length swept is $\Delta s = R\,\Delta\theta$ — this is the definition of the radian. One full revolution corresponds to an angular distance of $2\pi$ rad.

The angular speed $\omega$ is the rate of change of angular distance with time:

$$\omega = \frac{\Delta\theta}{\Delta t} \quad\Longrightarrow\quad \omega = \frac{\mathrm{d}\theta}{\mathrm{d}t}\quad(\text{NCERT eq. 3.44})$$

Its SI unit is radians per second (rad s$^{-1}$). Because radians are dimensionless, the dimensions of $\omega$ reduce to $[T^{-1}]$ — the same as frequency. Yet $\omega$ and the cyclic frequency $f$ differ numerically by a factor of $2\pi$. If a wheel makes $n$ revolutions per second, $\omega = 2\pi n$ rad s$^{-1}$.

The bridge formula v = omega r

The single most useful identity in this chapter connects the linear and angular descriptions. NCERT derives it from the arc-length identity. In time $\Delta t$ the particle traces an arc $\Delta s = R\,\Delta\theta$ on the circle, so its linear speed is

$$v = \frac{\Delta s}{\Delta t} = R\,\frac{\Delta\theta}{\Delta t} = R\,\omega \quad(\text{NCERT eq. 3.45})$$

Two readings are useful. First, for a fixed $\omega$, a particle farther from the axis moves faster — outer riders on a merry-go-round travel faster than inner ones. Second, $\omega$ is a property of the whole rotating body (every particle on a rigid wheel shares the same $\omega$), while $v$ varies with $R$.

Centripetal acceleration — the geometric derivation

NCERT figure 3.18 carries the proof in three panels. Let the particle be at $P$ at time $t$ with velocity $\vec{v}$ and at $P'$ at time $t + \Delta t$ with velocity $\vec{v}\,'$. Both speeds equal $v$, but the directions are different — each is tangential to the circle at the respective point. Construct the velocity-change vector $\Delta\vec{v} = \vec{v}\,' - \vec{v}$ using the triangle rule.

Because $\vec{v}\perp\vec{r}$ and $\vec{v}\,'\perp\vec{r}\,'$, the angle between $\vec{v}$ and $\vec{v}\,'$ equals the angle $\Delta\theta$ between the position vectors. The triangle $CPP'$ (position vectors and chord) and the triangle formed by $\vec{v}$, $\vec{v}\,'$ and $\Delta\vec{v}$ are similar isosceles triangles — same vertex angle, same equal-side ratio. The proportionality of sides gives

$$\frac{|\Delta\vec{v}|}{v} = \frac{|\Delta\vec{r}|}{R} \quad\Longrightarrow\quad |\Delta\vec{v}| = \frac{v}{R}\,|\Delta\vec{r}|.$$

For small $\Delta t$ the chord $|\Delta\vec{r}|$ approaches the arc $v\,\Delta t$. Dividing both sides by $\Delta t$ and taking the limit,

$$a_c = \lim_{\Delta t\to 0}\frac{|\Delta\vec{v}|}{\Delta t} = \frac{v}{R}\cdot\lim_{\Delta t\to 0}\frac{|\Delta\vec{r}|}{\Delta t} = \frac{v}{R}\cdot v = \frac{v^2}{R}\quad(\text{NCERT eq. 3.43})$$

The direction argument is equally important. As $\Delta t \to 0$, $\Delta\vec{v}$ becomes perpendicular to $\vec{v}$ itself. Since $\vec{v}$ is tangential, the perpendicular direction radially-inward is the only option, and that direction points to the centre. Hence the name centripetal — Newton's coinage, from the Greek for "centre-seeking". Substituting $v = \omega R$ gives the equivalent form

$$a_c = \frac{v^2}{R} = \omega^2 R = 4\pi^2 f^2 R\quad(\text{NCERT eq. 3.46, 3.48}).$$

Period, frequency and the constants of the motion

Three time-scales describe the same rotation. The time period $T$ is the time for one revolution. During $T$ the particle covers an arc of $2\pi R$ at speed $v$, so

$$T = \frac{2\pi R}{v} = \frac{2\pi}{\omega}.$$

The frequency $f$ (NCERT uses $\nu$) is the number of revolutions per second, $f = 1/T = \omega/(2\pi)$. Its SI unit is the hertz (Hz). Combining these,

Vector form and omega as an axial vector

NCERT writes the centripetal acceleration in vector form as $\vec{a}_c = -\omega^2\,\vec{r}$, where $\vec{r}$ points from the centre outward and the minus sign encodes the inward direction. NIOS uses the equivalent $\vec{a}_c = -(v^2/r)\,\hat{r}$.

Angular velocity itself can be promoted to a vector. By the right-hand rule, $\vec{\omega}$ points along the axis of rotation — curl the fingers along the rotation, thumb gives $\vec{\omega}$. The linear velocity then satisfies the cross-product identity $\vec{v} = \vec{\omega}\times\vec{r}$. For anticlockwise motion in the $xy$-plane, $\vec{\omega}$ points along $+\hat{z}$.

Centripetal force is a role, not a separate force

Multiplying the centripetal acceleration by the particle's mass gives, via Newton's second law, the centripetal force:

$$F_c = m\,a_c = \frac{m v^2}{R} = m\omega^2 R.$$

NIOS §4.4 is emphatic on a point NCERT only hints at: centripetal force is not a new fundamental interaction. It is the net inward radial force that must already be present for circular motion. The physical agent varies: gravity for the moon's orbit, string tension for a whirled stone, friction for a car on a level curve, the horizontal component of the normal reaction on a banked road, electrostatic attraction in the Bohr atom. The full development belongs to the Laws of Motion chapter.

Uniform vs non-uniform circular motion

When the speed itself changes — a roller-coaster cart, a stone on a string in a vertical loop — the motion is called non-uniform circular motion. The acceleration then has two perpendicular components: the centripetal $a_c = v^2/R$ pointing to the centre, and a tangential $a_t = \mathrm{d}v/\mathrm{d}t$ along the velocity. The net acceleration has magnitude $|\vec{a}| = \sqrt{a_c^2 + a_t^2}$. For uniform circular motion $a_t = 0$; the vertical-loop section below trespasses briefly into the non-uniform case.

Banking of roads

When a car of mass $m$ rounds a curve of radius $r$ at speed $v$ on a level road, the centripetal force comes from friction between the tyres and the road. Friction is unreliable — water, oil or worn rubber drops the coefficient drastically. Engineers therefore bank the curve, raising the outer edge of the road through an angle $\theta$ so that the horizontal component of the normal reaction itself supplies the inward force, even on a frictionless surface.

NIOS §4.4.1 carries the free-body analysis. Two forces act on the car when friction is ignored: the weight $mg$ vertically down and the normal reaction $F_N$ perpendicular to the road surface. Resolving $F_N$ horizontally and vertically:

$$F_N\sin\theta = \frac{m v^2}{r},\qquad F_N\cos\theta = m g.$$

Dividing the first by the second eliminates $F_N$ and $m$ together:

$$\boxed{\;\tan\theta = \frac{v^2}{r g}\;}\quad(\text{NIOS eq. 4.21})$$

Two consequences NEET examiners love. First, the banking angle is independent of the mass of the vehicle — the same road is safe for a scooter and a heavy truck. Second, $\theta$ scales with $v^2$ and inversely with $r$, so high-speed expressways and sharp curves both demand steeper banks. In reality friction shares the load, so the engineered angle is gentler than this frictionless ideal.

Conical pendulum — the same equation in disguise

A small mass on the end of a string of length $L$ traces a horizontal circle while the string sweeps out a cone of half-angle $\theta$. Two forces act on the bob: the string tension $T$ along the string, and gravity $mg$ downward. The vertical component of $T$ balances the weight; the horizontal component supplies the centripetal force for the circle of radius $r = L\sin\theta$.

$$T\cos\theta = m g,\qquad T\sin\theta = \frac{m v^2}{r}\quad\Longrightarrow\quad \tan\theta = \frac{v^2}{r g}.$$

The conical pendulum and the banked road obey the same equation. Same algebra, same answer — only the physical agent supplying the inward force changes (tension here, normal reaction there). NEET problems sometimes swap one for the other to test whether you actually understood the geometry.

Vertical-circle minima

NIOS §4.3.2 covers the motion of a body of mass $m$ tied to a string and whirled in a vertical circle of radius $r$. At the lowest point $P$ the tension $T_1$ and weight $mg$ act in opposite directions; at the topmost point $Q$ they act in the same direction (both downward). Writing the radial equation at each point:

$$\text{At }Q:\quad T_2 + mg = \frac{m v_{\text{top}}^2}{r};\qquad \text{At }P:\quad T_1 - mg = \frac{m v_{\text{bot}}^2}{r}.$$

For the body to just complete the loop, the string must remain taut everywhere; the critical point is the top, where the tension can drop to zero. Setting $T_2 = 0$ gives

$$v_{\text{top, min}} = \sqrt{g r}.$$

Energy conservation between bottom and top, with a vertical drop of $2r$, then yields

$$v_{\text{bot, min}}^2 = v_{\text{top, min}}^2 + 4 g r = g r + 4 g r = 5 g r\quad\Longrightarrow\quad v_{\text{bot, min}} = \sqrt{5 g r}.$$

The two thresholds $\sqrt{g r}$ and $\sqrt{5 g r}$ are NEET catechism — memorise them. The same result governs a bucket of water swung in a vertical loop: at the top, gravity itself supplies the inward pull, water clings to the bucket, nothing spills.

Worked examples