Circular Motion Notes | Leaving Cert Applied Maths

Kinematics, Angular Velocity, and Vector Calculus Derivation

A particle travelling along a circular path of radius $r$ has an instantaneous velocity vector directed along the tangent to the circle. Even if its speed $v$ remains constant, the continuous change in direction means the particle accelerates. This inward acceleration always points towards the centre of the circle and is termed the centripetal acceleration, $a_c$ .

A particle on a circle has tangential velocity perpendicular to its radius and acceleration directed towards the centre.

Angular velocity, $\omega = \frac{d\theta}{dt}$ , measures the rate at which the radius vector sweeps out an angle in radians per second (rad s $^{-1}$ ). Since an arc length $s = r\theta$ , differentiating with respect to time yields the relationship between linear speed and angular velocity:

v = r\omega

The periodic time $T_p$ taken to complete one full revolution of $2\pi$ radians is:

T_p = \frac{2\pi}{\omega} = \frac{2\pi r}{v}

If rotation is specified as $n$ revolutions per minute (rev/min), convert to radians per second using $\omega = \frac{2\pi n}{60}$ .

Vector Calculus Derivation of Centripetal Acceleration

We can prove the formula for centripetal acceleration using elementary vector calculus. Set the origin $(0,0)$ at the centre of the circle. At any time $t$ , the position vector $\vec{r}(t)$ of a particle moving at constant angular speed $\omega$ is:

\vec{r} = r\cos(\omega t)\vec{i} + r\sin(\omega t)\vec{j}

Differentiating with respect to time gives the velocity vector $\vec{v}$ :

\vec{v} = \frac{d\vec{r}}{dt} = -r\omega\sin(\omega t)\vec{i} + r\omega\cos(\omega t)\vec{j}

The magnitude of velocity is $|\vec{v}| = \sqrt{(-r\omega\sin(\omega t))^2 + (r\omega\cos(\omega t))^2} = r\omega = v$ .

Differentiating a second time yields the acceleration vector $\vec{a}$ :

\vec{a} = \frac{d\vec{v}}{dt} = -r\omega^2\cos(\omega t)\vec{i} - r\omega^2\sin(\omega t)\vec{j} = -\omega^2(r\cos(\omega t)\vec{i} + r\sin(\omega t)\vec{j}) = -\omega^2\vec{r}

The negative sign confirms that the acceleration vector points in the exact opposite direction to $\vec{r}$ , directed straight inward towards the centre. Its magnitude is:

a_c = \omega^2 r = \frac{v^2}{r}

When the Speed is Not Constant

The derivation above assumes a constant angular speed $\omega$ . In a vertical circle the speed changes as the particle rises and falls, so the acceleration has two parts: a component $v^2/r$ towards the centre (because the direction of motion is turning) and a tangential component $dv/dt$ along the tangent (because the speed is changing). In examinations we resolve forces along the radius, so the tangential component does not appear and $mv^2/r$ remains the exact radial equation. The changing speed is handled separately using conservation of energy.

Dynamics of Horizontal Circles, Conical Pendulums, and Bowls

Centripetal force is not an independent physical force; it is simply the net inward radial force provided by real forces such as tension, friction, gravity, or normal reaction. Applying Newton's second law along the inward radial line gives:

\sum F_{\text{inward}} = \frac{mv^2}{r} = mr\omega^2

In horizontal circular motion, the particle remains at a constant vertical height. Because there is no vertical acceleration, forces in the vertical direction balance in static equilibrium, while horizontal components provide the inward acceleration:

\sum F_{\text{vertical}} = 0 \qquad \text{and} \qquad \sum F_{\text{horizontal}} = mr\omega^2 = \frac{mv^2}{r}

Drawing and Labelling the Diagram

In SEC examinations, method marks are awarded for clear free-body diagrams:

Draw the string or contact surface with tension $T$ or normal reaction $R$ clearly directed.
Draw the weight $mg$ acting vertically downwards from the particle.
Mark the radius $r$ horizontally from the particle to the vertical axis of rotation.
State your angle explicitly. If $\theta$ is the angle between the string and the vertical, the components are $T\cos\theta = mg$ and $T\sin\theta = mr\omega^2$ . If $\theta$ is measured from the horizontal, the trigonometric ratios reverse ( $T\sin\theta = mg$ and $T\cos\theta = mr\omega^2$ ). Never draw an extra arrow labelled 'centripetal force'.

The Conical Pendulum

For a bob of mass $m$ on a string of length $L$ making angle $\theta$ with the vertical:

Radius: $r = L\sin\theta$
Vertically: $T\cos\theta = mg$
Radially: $T\sin\theta = mr\omega^2 = \frac{mv^2}{r}$

Dividing the radial equation by the vertical equation eliminates both tension $T$ and mass $m$ :

\tan\theta = \frac{v^2}{rg} = \frac{r\omega^2}{g}

If the vertical depth of the circle below the suspension point is $h = L\cos\theta$ , then substituting $r = L\sin\theta$ gives:

\omega^2 = \frac{g}{h}

A suspended bob follows a horizontal circle. Its string length L, orbit radius r and vertical depth h form a right triangle; tension points towards the pivot and weight points down.

Inside a Smooth Hemispherical Bowl or Cone

A particle moving in a horizontal circle on the smooth inside surface of a hemispherical bowl of radius $a$ is supported by the normal reaction $R$ , which points along the radius of the bowl towards its centre $O$ . If the radius from $O$ to the particle makes an angle $\theta$ with the downward vertical:

Radius of the circular path: $r = a\sin\theta$
Vertically: $R\cos\theta = mg$
Radially: $R\sin\theta = mr\omega^2$

Dividing gives $\tan\theta = \frac{r\omega^2}{g}$ . Since the depth of the circle below the centre $O$ is $h = a\cos\theta$ , we obtain the identical shortcut:

\omega^2 = \frac{g}{h}

For any horizontal circle held by a single contact force pointing towards a fixed axis point, $\omega^2 = g/h$ , where $h$ is the vertical depth below that point. The required angular speed does not depend on the mass of the particle.

Rotating Rides with Suspension Offsets

In amusement park rides such as a chairoplane, chains are hung from the rim of a rotating disc of radius $d$ rather than from a central point. If a chain of length $L$ hangs at angle $\theta$ to the vertical, the horizontal displacement from the pivot is $L\sin\theta$ . Adding the disc offset gives the total orbit radius:

r = d + L\sin\theta

A rotating ride has a pivot offset d from its axis and a chain projecting a further L sin θ horizontally, giving total radius r.

The resolved equations remain $T\cos\theta = mg$ and $T\sin\theta = mr\omega^2$ , leading to:

\tan\theta = \frac{(d + L\sin\theta)\omega^2}{g} \implies \omega = \sqrt{\frac{g\tan\theta}{d + L\sin\theta}}

Banked Tracks and Lateral Friction Limits

When a vehicle rounds a bend of radius $r$ , centripetal acceleration must be supplied horizontally.

Unbanked Level Bend

On a flat horizontal road, the normal reaction is $R = mg$ . The only horizontal force available is sideways static friction $F_f$ directed towards the centre of the turn:

F_f = \frac{mv^2}{r}

Because static friction has an upper limit $F_{\text{max}} = \mu R = \mu mg$ , where $\mu$ is the coefficient of friction, the maximum safe cornering speed is:

v_{\text{max}} = \sqrt{\mu gr}

Smooth Banked Track

To eliminate reliance on tyre friction, tracks and roads are banked at an angle $\theta$ to the horizontal. With zero friction, the normal reaction $R$ tilts inward:

R\cos\theta = mg \qquad \text{and} \qquad R\sin\theta = \frac{mv^2}{r}

Dividing the radial equation by the vertical equation gives the speed at which no friction is required (often referred to as the design speed):

\tan\theta = \frac{v^2}{rg} \implies v = \sqrt{rg\tan\theta}

Rough Banked Track with Friction

When a vehicle travels faster than the no-friction speed, it tends to slide up the banking, so limiting friction $\mu R$ acts down the slope:

\sum F_{\text{vertical}} = 0 \implies R\cos\theta - \mu R\sin\theta = mg

\sum F_{\text{radial}} = \frac{mv^2}{r} \implies R\sin\theta + \mu R\cos\theta = \frac{mv^2}{r}

Dividing the equations eliminates $R$ and $m$ , giving the maximum speed before slipping outward:

v_{\text{max}}^2 = rg\left(\frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}\right)

When travelling below the no-friction speed, the vehicle tends to slip down the incline, so friction acts up the slope. Changing the sign of $\mu$ yields the minimum speed:

v_{\text{min}}^2 = rg\left(\frac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta}\right)

If $\mu \ge \tan\theta$ , the numerator in the minimum speed formula becomes zero or negative. This means friction alone can hold the car on the slope even at rest, so no minimum speed exists.

Two bank cross-sections show weight downwards and reaction perpendicular to the road. Friction acts down the bank above design speed and up the bank below design speed.

Hooke's Law in Circular Motion

Hooke's law states that the tension in an elastic spring or string is proportional to its extension: $T = k(l - l_0)$ , where $l_0$ is the natural length and $k$ is the elastic stiffness in N m $^{-1}$ .

When a particle of mass $m$ attached to an elastic string rotates on a smooth horizontal table, the extended string forms the radius of the circle: $r = l_0 + x$ . The tension supplies the centripetal force:

k(r - l_0) = mr\omega^2

Rearranging to isolate the orbit radius $r$ :

kr - kl_0 = mr\omega^2 \implies r(k - m\omega^2) = kl_0 \implies r = \frac{kl_0}{k - m\omega^2}

The stiffness $k$ must be greater than $m\omega^2$ , otherwise the string cannot hold the particle in a circle at that rate of rotation. For example, if $m = 0.5\text{ kg}$ , $l_0 = 0.4\text{ m}$ , $k = 50\text{ N m}^{-1}$ , and $\omega = 6\text{ rad s}^{-1}$ , then $m\omega^2 = 0.5(36) = 18\text{ N m}^{-1} < 50$ , yielding $r = \frac{50(0.4)}{50 - 18} = \frac{20}{32} = 0.625\text{ m}$ .

Dynamics of Vertical Circular Motion

In a vertical circle, gravity acts in the plane of motion, meaning the particle's speed changes continuously with height. The constant-acceleration formulae ( $v = u + at$ , $v^2 = u^2 + 2as$ ) cannot be used here, because neither the size nor the direction of the acceleration is constant.

The Four-Step Solution Method

State your datum level and angle convention clearly on your sketch.
Apply conservation of mechanical energy from the launch point to the point of interest to express $v^2$ in terms of the angle.
Formulate Newton's second law along the inward radius to express tension $T$ or normal reaction $R$ .
Apply the required physical condition ( $T = 0$ , $R = 0$ , or $v = 0$ ) to solve for the unknown angle or speed.

Writing the Radial Equation: One Rule, Four Geometries

Add up all force components pointing towards the centre, subtract those pointing away from the centre, and set the resultant equal to $mv^2/r$ :

Situation	Radial equation (inward)	Contact or slack condition
Particle on a string, at angle $\theta$ to the downward vertical	$T - mg\cos\theta = \frac{mv^2}{r}$	String slacks when $T = 0$
Inside a loop or bowl, at angle $\alpha$ to the upward vertical (above centre)	$R + mg\cos\alpha = \frac{mv^2}{r}$	Leaves track when $R = 0$
Inside a concave track or valley, at angle $\theta$ to the downward vertical (below centre)	$R - mg\cos\theta = \frac{mv^2}{r}$	$R$ is greatest at bottom; cannot leave track here
Outside a convex surface (sphere, hill crest), at angle $\alpha$ to upward vertical	$mg\cos\alpha - R = \frac{mv^2}{a}$	Leaves surface when $R = 0$ ( $v^2 = ga\cos\alpha$ )

At the crest of a humpback bridge of radius $r$ ( $\alpha = 0$ ), $R = m(g - v^2/r)$ , so wheels leave the road once $v \ge \sqrt{gr}$ . At the base of a valley, $R = m(g + v^2/r)$ , so an occupant feels heavier than normal.

General Tension Formula for a String

Taking the lowest point as the reference datum ( $h = 0$ ), the height at an angle $\theta$ from the downward vertical is $h = r(1 - \cos\theta)$ :

\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgr(1 - \cos\theta) \implies v^2 = u^2 - 2gr(1 - \cos\theta)

A particle below and to the right of the centre has height r(1 − cos θ) above the bottom. Tension points inward and weight vertically down.

Substituting this $v^2$ into the radial equation $T = mg\cos\theta + \frac{mv^2}{r}$ :

T = \frac{mu^2}{r} - 2mg + 3mg\cos\theta

Evaluating tension at the bottom ( $\theta = 0^\circ$ , $\cos\theta = 1$ ) and top ( $\theta = 180^\circ$ , $\cos\theta = -1$ ):

$T_{\text{bottom}} = \frac{mu^2}{r} + mg$
$T_{\text{top}} = \frac{mu^2}{r} - 5mg$

Subtracting gives the standard relationship:

$ $T_{\text{bottom}} - T_{\text{top}} = 6mg$ $

Projectile Motion After the String Goes Slack

When $T = 0$ , the particle leaves the circular path. Its velocity at that instant is tangential (perpendicular to the radius). If the radius makes an angle $\theta$ with the upward vertical, the tangent makes the same angle $\theta$ with the horizontal. The velocity components at departure are:

Horizontal component: $v_x = v\cos\theta$
Vertical component (upward): $v_y = v\sin\theta$

From that point onward, the particle moves as a free projectile under gravity. Its maximum additional height above the slack point is found using $v_y^2 = u_y^2 - 2gh_{\text{extra}}$ :

h_{\text{extra}} = \frac{(v\sin\theta)^2}{2g}

At an upper-left departure point, velocity is tangent to the circle and directed upwards and rightwards. A parabolic path rises to an apex above the slack point.

Critical Boundary Conditions: Loops, Rods, and Dimensional Checks

Depending on the launch speed $u$ at the lowest point, a particle on a string exhibits three distinct behaviours:

Launch speed $u$	Motion regime	What actually happens
$u \le \sqrt{2gr}$	Oscillation	Particle comes to rest ( $v = 0$ ) at or below the horizontal equator. The string stays taut ( $T \ge 0$ ) throughout; in the limiting case $u = \sqrt{2gr}$ the particle comes to rest level with the centre, where $T = 0$ momentarily and the particle swings back.
$\sqrt{2gr} < u < \sqrt{5gr}$	Slack / departs circle	Tension vanishes ( $T = 0$ ) above the equator while speed $v > 0$ . The particle enters parabolic projectile motion.
$u \ge \sqrt{5gr}$	Complete circle	The particle maintains tension right over the top ( $T_{\text{top}} \ge 0$ ) and completes full revolutions.

Deriving the Full Loop Conditions

Flexible String or Inside a Smooth Loop: At the summit, both tension (or normal reaction) and gravity act downwards:

T_{\text{top}} + mg = \frac{mv_{\text{top}}^2}{r} \implies \text{for } T_{\text{top}} \ge 0, \; v_{\text{top}}^2 \ge gr

By conservation of energy, $u^2 = v_{\text{top}}^2 + 4gr \ge gr + 4gr = 5gr$ , giving $u \ge \sqrt{5gr}$ .

Light Rigid Rod or Threaded Wire Ring: A rigid rod can support compression, and a bead threaded on a wire cannot leave the path. The particle keeps going as long as $v^2 > 0$ all the way round. Since $v_{\text{top}}^2 = u^2 - 4gr$ , the particle just reaches the top when $u = 2\sqrt{gr}$ (arriving with zero speed) and completes the circle for $u > 2\sqrt{gr}$ .

Dimensional Analysis and Model Verification

You are expected to be able to check whether an answer is sensible by looking at its dimensions in terms of mass $[\text{M}]$ , length $[\text{L}]$ , and time $[\text{T}]$ :

$[a_c] = [v^2/r] = (\text{LT}^{-1})^2 / \text{L} = \text{LT}^{-2}$ (acceleration)
$[F_c] = [mv^2/r] = \text{M} \times \text{LT}^{-2} = \text{MLT}^{-2}$ (force in newtons)
$[\sqrt{5gr}] = (\text{LT}^{-2} \cdot \text{L})^{1/2} = \text{LT}^{-1}$ (speed in m s $^{-1}$ )

In energy equations, check each term separately: every term must carry dimensions of energy, $[\text{ML}^2\text{T}^{-2}]$ . Terms like $\frac{1}{2}mv^2$ , $mgh$ , and $mg \cdot r(1 - \cos\theta)$ all balance. Cancelling $m$ across every term is mathematically correct; an error occurs only if one term lacks a factor present in the others.

Key terms

Centripetal Acceleration: The inward radial acceleration of a particle moving in a circular path, directed towards the centre of curvature, with magnitude v²/r or ω²r.
Angular Velocity: The rate of change of angular displacement with respect to time, measured in radians per second (rad s⁻¹), linked to linear speed by v = rω.
Centripetal Force: The resultant net inward force acting towards the centre of a circular path, equal to mv²/r or mrω²; supplied by real forces like tension, friction, or gravity.
Conical Pendulum: A particle suspended by a light string from a fixed point that sweeps out a horizontal circle while the string describes the surface of a cone.
Banked Track: A curved road or track inclined at an angle to the horizontal, allowing the horizontal component of the normal reaction to provide centripetal acceleration.
Coefficient of Friction: The dimensionless ratio μ between the maximum limiting static frictional force and the normal reaction force between two contacting surfaces.
Conservation of Mechanical Energy: The principle that the sum of kinetic energy and gravitational potential energy remains constant in a conservative system with no work done against dissipative forces.
Slack Condition: The physical state in vertical circular motion where string tension drops to zero (T = 0), causing the particle to leave the circular path and enter projectile motion.
Hooke's Law: The physical principle stating that the tension in an elastic spring or string is directly proportional to its extension beyond natural length: T = k(l - l₀).
Dimensional Analysis: The algebraic verification of physical equations by comparing the fundamental base units of mass [M], length [L], and time [T] across all terms.

Check yourself

A 0.2 kg ball on a string of length 0.8 m swings in a horizontal circle 0.5 m below the fixed suspension point. Find its angular velocity ω and the tension in the string. (Take g = 9.8 m s⁻²)
Using the shortcut ω² = g/h: ω² = 9.8 / 0.5 = 19.6, so ω = √19.6 ≈ 4.43 rad s⁻¹. For tension, cos θ = h/L = 0.5 / 0.8 = 5/8. Resolving vertically: T cos θ = mg gives T(5/8) = (0.2)(9.8) = 1.96 N, so T = 1.96 × (8/5) = 3.14 N.
A particle moves in a horizontal circle 0.45 m below the centre of a smooth hemispherical bowl. Find its angular velocity.
For a smooth bowl, the normal reaction points towards the centre, giving ω² = g/h. With h = 0.45 m: ω² = 9.8 / 0.45 ≈ 21.78, so ω ≈ 4.67 rad s⁻¹.
What is the difference between the tension at the bottom and the tension at the top for any particle completing a full vertical circle on a string?
T_bottom - T_top = 6mg. This difference is constant and independent of the initial projection speed.
A car negotiates an unbanked circular bend of radius 80 m on a flat road where μ = 0.5. Calculate the maximum safe speed without skidding. (Take g = 9.8 m s⁻²)
v_max = √(μgr) = √(0.5 × 9.8 × 80) = √392 ≈ 19.8 m s⁻¹.
Use dimensional analysis to test whether the expression v = √(5gr) is dimensionally consistent.
[√(5gr)] = ([g][r])^(1/2) = (LT⁻² · L)^(1/2) = (L²T⁻²)^(1/2) = LT⁻¹, which matches the dimensions of velocity [v].

Circular Motion