Circular Motion

Circular Motion

Info: Board Coverage AQA Paper 1 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

Explore the simulation above to develop intuition for this topic.

1. Angular Quantities

Definition. The angular displacement $\theta$ is the angle swept out by a radius vector. It Is measured in radians (rad).

Definition. The angular velocity $\omega$ is the rate of change of angular displacement:

$\boxed{\omega = \frac{d\theta}{dt}}$

The SI unit is rad s$^{-1}$. The relationship with linear velocity is:

$v = \omega r$

Period and frequency. For uniform circular motion:

$\omega = \frac{2\pi}{T} = 2\pi f$

Where $T$ is the period (time for one revolution) and $f$ is the frequency.

Example. A CD rotates at $480$ rpm. Converting: $\omega = 480 \times 2\pi/60 = 16\pi \approx 50.3$ rad s$^{-1}$. A point $0.06$ m from the centre Has linear speed $v = \omega r = 50.3 \times 0.06 = 3.02$ m s$^{-1}$.

2. Centripetal Acceleration — Derivation

We prove that a body moving in a circle of radius $r$ at constant speed $v$ has an acceleration of Magnitude $a = v^2/r = \omega^2 r$ directed towards the centre.

Method 1: Calculus

The position vector of a particle moving in a circle in the $xy$-plane is:

$\mathbf{r}(t) = r\cos(\omega t)\,\mathbf{i} + r\sin(\omega t)\,\mathbf{j}$

Differentiating to find velocity:

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

Note: $|\mathbf{v}| = r\omega\sqrt{\sin^2(\omega t) + \cos^2(\omega t)} = r\omega = v$. $\checkmark$

Differentiating again to find acceleration:

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

$\boxed{\mathbf{a} = -\omega^2 r\,\hat{\mathbf{r}}}$

The acceleration has magnitude $|\mathbf{a}| = \omega^2 r = \frac{v^2}{r}$ and is directed radially inward (towards the centre). The negative sign indicates this inward direction. $\square$

Key insight from the calculus approach. The acceleration vector $\mathbf{a}(t) = -\omega^2 \mathbf{r}(t)$ is always antiparallel to the position vector. This means It always points towards the centre, regardless of where the particle is on the circle. Even though The speed is constant, the direction of velocity changes continuously, requiring acceleration.

Method 2: Geometry

Consider two positions of the particle separated by a small angle $\delta\theta$. The change in Velocity $\delta\mathbf{v}$ is directed towards the centre. From the isosceles triangle formed:

$\frac{|\delta\mathbf{v}|}{v} = \frac{|\delta\mathbf{s}|}{r} \implies |\delta\mathbf{v}| = \frac{v}{r}|\delta\mathbf{s}|$

Dividing by $\delta t$ and taking the limit:

$a = \frac{|\delta\mathbf{v}|}{\delta t} = \frac{v}{r}\frac{|\delta\mathbf{s}|}{\delta t} = \frac{v^2}{r}$

3. Centripetal Force

By Newton’s second law, the net force producing centripetal acceleration is:

$\boxed{F_c = \frac{mv^2}{r} = m\omega^2 r}$

Intuition: Circular Motion is NOT Equilibrium

A particle in uniform circular motion is accelerating (towards the centre) even though its speed is Constant. The velocity vector is changing direction. There is always a net inward force. If you Cut the string, the particle does not fly radially outward — it moves tangentially (Newton’s First law).

4. Applications

Horizontal Circle: Conical Pendulum

A mass $m$ on a string of length $L$ moves in a horizontal circle of radius $r = L\sin\alpha$Where $\alpha$ is the angle the string makes with the vertical.

Vertical: $T\cos\alpha = mg$ … (i)

Horizontal (centripetal): $T\sin\alpha = \frac{mv^2}{r}$ … (ii)

Dividing (ii) by (i):

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

From (i), the tension is:

$\boxed{T = \frac{mg}{\cos\alpha} = \frac{mgL}{\sqrt{L^2 - r^2}}}$

The period of the conical pendulum can be expressed by substituting $r = L\sin\alpha$ and $v = \omega r$ into the result for $\tan\alpha$:

$\tan\alpha = \frac{\omega^2 L\sin\alpha}{g} \implies \omega^2 = \frac{g}{L\cos\alpha}$

$\boxed{T = 2\pi\sqrt{\frac{L\cos\alpha}{g}}}$

Example. A conical pendulum with $L = 1.0$ m and $\alpha = 30^\circ$ has period $T = 2\pi\sqrt{1.0 \times \cos 30°/9.81} = 2\pi\sqrt{0.0883} = 1.86$ s.

Banked Curves

A road is banked at angle $\theta$ so that a vehicle travelling at speed $v$ can negotiate the curve Without friction.

Resolving vertically: $N\cos\theta = mg$ … (i)

Resolving horizontally: $N\sin\theta = \frac{mv^2}{r}$ … (ii)

Dividing (ii) by (i):

$\boxed{\tan\theta = \frac{v^2}{rg}}$

$\boxed{v_{\mathrm{optimum}} = \sqrt{rg\tan\theta}}$

Intuition. At the optimum speed, the horizontal component of the normal reaction provides Exactly the centripetal force. If the vehicle goes faster, friction acts down the slope; if slower, Friction acts up the slope.

Proof of Maximum Speed on a Banked Curve with Friction

When friction is present, the maximum safe speed is found by resolving forces with friction acting down the slope (opposing the tendency to slide up).

Resolving perpendicular to the slope: $N = mg\cos\theta + \frac{mv^2}{r}\sin\theta$

Resolving parallel to the slope (friction acts down the slope for maximum speed):

$N\sin\theta + \mu N\cos\theta = \frac{mv^2}{r}$

Combining:

$\frac{mv^2}{r} = N(\sin\theta + \mu\cos\theta) = (mg\cos\theta + \frac{mv^2}{r}\sin\theta)(\sin\theta + \mu\cos\theta)$

This gives:

$\frac{v^2}{r}\left[1 - \sin\theta(\sin\theta + \mu\cos\theta)\right] = g\cos\theta(\sin\theta + \mu\cos\theta)$

$\boxed{v_{\max} = \sqrt{\frac{rg(\sin\theta + \mu\cos\theta)}{\cos\theta - \mu\sin\theta}} = \sqrt{\frac{rg(\tan\theta + \mu)}{1 - \mu\tan\theta}}}$

For the minimum speed, friction acts up the slope, giving:

$\boxed{v_{\min} = \sqrt{\frac{rg(\tan\theta - \mu)}{1 + \mu\tan\theta}}}$

Real-world example. Motorways are banked at about $2$—$3^\circ$ for drainage and slight curve Assistance. At higher angles, velodromes use banking up to $45^\circ$ so cyclists can maintain speed Through tight turns. The normal reaction alone provides the centripetal force at the design speed.

Vertical Circles

Consider a mass $m$ on a string of length $r$ moving in a vertical circle.

At any angle, the forces on the mass are tension $T$ (towards centre) and weight $mg$ (vertically Down).

At the top of the circle (both $T$ and $mg$ act towards the centre):

$T + mg = \frac{mv^2}{r}$

Condition for the string to remain taut: $T \geq 0$So:

$\frac{mv^2}{r} \geq mg \implies v \geq \sqrt{gr}$

At the bottom of the circle (centripetal direction is upward):

$T - mg = \frac{mv^2}{r}$

Energy conservation relates speeds at top and bottom:

$\frac{1}{2}mv_{\mathrm{bottom}}^2 = \frac{1}{2}mv_{\mathrm{top}}^2 + mg(2r)$

$v_{\mathrm{bottom}}^2 = v_{\mathrm{top}}^2 + 4gr$

For the minimum speed at the top ($v_{\mathrm{top}} = \sqrt{gr}$):

$v_{\mathrm{bottom}}^2 = gr + 4gr = 5gr \implies v_{\mathrm{bottom}} = \sqrt{5gr}$

Problem Set

5. Vertical Circles — Detailed Analysis

The basic equations were introduced in Section 4. Here we examine vertical circular motion in Greater detail, including general positions on the circle and a full worked example.

General Position on the Circle

Consider a mass $m$ attached to a string of length $r$Moving in a vertical circle. At a general Position where the string makes angle $\theta$ with the downward vertical, the forces are:

• Tension $T$ along the string (towards the centre)
• Weight $mg$ vertically downward

Resolving along the radial direction (towards the centre):

$T - mg\cos\theta = \frac{mv^2}{r}$

$\boxed{T = \frac{mv^2}{r} + mg\cos\theta}$

At the top ($\theta = 180^\circ$So $\cos\theta = -1$):

$\boxed{T_{\mathrm{top}} = \frac{mv^2}{r} - mg}$

Both $T$ and $mg$ point towards the centre. The string remains taut if $T_{\mathrm{top}} \geq 0$ Giving the minimum speed at the top:

$\boxed{v_{\min} = \sqrt{gr}}$

At the bottom ($\theta = 0^\circ$So $\cos\theta = 1$):

$\boxed{T_{\mathrm{bottom}} = \frac{mv^2}{r} + mg}$

The tension must overcome gravity and provide the centripetal force, so $T_{\mathrm{bottom}}$ is Always greater than $T_{\mathrm{top}}$ for the same speed.

Energy Conservation Between Top and Bottom

$\tfrac{1}{2}mv_b^2 = \tfrac{1}{2}mv_t^2 + mg(2r)$

$\boxed{v_b^2 = v_t^2 + 4gr}$

For the minimum case ($v_t = \sqrt{gr}$):

$v_b^2 = gr + 4gr = 5gr \implies \boxed{v_b = \sqrt{5gr}}$

Difference in Tension: Top vs Bottom

$T_{\mathrm{bottom}} - T_{\mathrm{top}} = \left(\frac{mv_b^2}{r} + mg\right) - \left(\frac{mv_t^2}{r} - mg\right) = \frac{m(v_b^2 - v_t^2)}{r} + 2mg$

Using $v_b^2 - v_t^2 = 4gr$:

$T_{\mathrm{bottom}} - T_{\mathrm{top}} = \frac{m \cdot 4gr}{r} + 2mg = 4mg + 2mg = 6mg$

$\boxed{T_{\mathrm{bottom}} - T_{\mathrm{top}} = 6mg}$

This result is independent of the radius and speed — only on the mass and $g$.

Info: Board Coverage AQA Paper 1 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

6. Conical Pendulum — Extended Analysis

Section 4 derived the basic results for the conical pendulum. Here we present a complete derivation And discuss additional features.

Full Derivation

A mass $m$ is suspended from a string of length $L$ and set into motion so that it traces a Horizontal circle of radius $r$. The string makes a constant angle $\alpha$ with the vertical.

Geometry. The horizontal circle radius and string length are related by:

$\boxed{r = L\sin\alpha}$

Forces. Two forces act on the mass: tension $T$ along the string (at angle $\alpha$ to the Vertical) and weight $mg$ downward.

Resolving vertically (no vertical acceleration):

$T\cos\alpha = mg \implies \boxed{T = \frac{mg}{\cos\alpha}}$

Resolving horizontally (centripetal acceleration towards centre):

$T\sin\alpha = \frac{mv^2}{r} = m\omega^2 r$

Substituting $T = mg/\cos\alpha$ and $r = L\sin\alpha$Then simplifying for $\sin\alpha \neq 0$:

$\frac{mg}{\cos\alpha}\sin\alpha = m\omega^2 L\sin\alpha \implies \omega^2 = \frac{g}{L\cos\alpha}$

The period is therefore:

$\boxed{T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L\cos\alpha}{g}}}$

Key features of the period formula:

The period depends on $L\cos\alpha$ (the vertical height of the bob below the pivot), not on the Mass. For a given vertical height $h = L\cos\alpha$All conical pendulums have the same period, Regardless of the angle or string length. A larger angle $\alpha$ (wider circle) gives a shorter Period (faster motion). As $\alpha \to 0$The period approaches $T = 2\pi\sqrt{L/g}$Which is the Period of a simple pendulum for small oscillations.

7. Centrifugal Force — Rotating Reference Frames

Definition. The centrifugal force is a fictitious (pseudo) force that appears to act on a Body in a rotating reference frame, directed radially outward from the axis of rotation.

Intuition. Passengers in a car going around a bend feel “thrown outward”. This is not because an Outward force acts on them. In the inertial (ground) frame, the passengers tend to move in a Straight line (Newton’s first law) while the car turns underneath them. The door or seatbelt must Exert an inward force to change the passengers’ direction — this is the centripetal force.

In the rotating frame of the car, the passengers appear stationary. To make Newton’s second law work In this non-inertial frame, we must introduce a fictitious outward force:

$\boxed{F_{\mathrm{centrifugal}} = m\omega^2 r = \frac{mv^2}{r}}$

This has the same magnitude as the centripetal force but points in the opposite direction.

Summary of the two descriptions:

Further Problems

Common Pitfalls

1. Using the wrong equation from the data sheet. Take time to read the full equation, including conditions and variable definitions.

2. Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.

3. Forgetting to include units in final answers, especially when working with derived units like $\text{N}\,\text{kg}^{-1}\,\text{m}^2$.

4. Confusing scalar and vector quantities. Always check whether direction matters for the quantity in question.

Common Mistakes

1. Confusing centripetal force with centrifugal force. Centripetal force is the REAL force acting towards the centre (gravity, tension, friction, normal reaction). Centrifugal force is a FICTITIOUS force that only appears in a rotating frame of reference. In an inertial frame (which exam questions use), there is no centrifugal force — only centripetal force.

2. Adding centripetal force to the force diagram. Centripetal force is not a separate force — it is the RESULTANT of the existing forces. In a conical pendulum, the centripetal force is the horizontal component of tension, not tension plus a separate “centripetal force.” Never draw centripetal force as an extra arrow on a free-body diagram.

3. Forgetting that speed is constant but velocity is not. In uniform circular motion, the speed (magnitude of velocity) is constant, but the direction of velocity continuously changes. This means there IS acceleration (centripetal acceleration a = v²/r), even though the speed does not change.

4. Using the wrong formula for centripetal acceleration. Centripetal acceleration can be written as a = v²/r OR a = ω²r OR a = 4π²r/T². Use whichever form matches the variables given in the question. Converting between them incorrectly is a major source of errors.

5. Assuming the string remains taut at all speeds. For a mass on a string in a vertical circle, the string only remains taut if the speed at the top is at least √(gr). Below this speed, the string goes slack and the mass follows a parabolic trajectory, not a circular one.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

• Dynamics — Centripetal force is the net inward force derived from Newton’s second law applied to curved paths.
• Momentum — Collisions in circular motion problems can be analysed using conservation of momentum and vector components.
• Gravitational Fields — Satellite orbits are circular motion problems where gravity provides the centripetal force.
• Work, Energy and Power — Centripetal force does no work because it is always perpendicular to the displacement.

Intuition

Think of circular motion as a constant tug-of-war between inertia and constraint. Your object wants to fly off in a straight line (inertia), but something keeps pulling it inward — a string, friction, gravity, or a normal reaction. This inward pull is the centripetal force. It doesn’t add energy (it’s always perpendicular to motion), but it continuously changes the direction of velocity.

The key mental model: imagine swinging a ball on a string. At every instant, the ball’s velocity is tangential — perpendicular to the string. The string pulls it inward, curving the path into a circle. If the string breaks, the ball doesn’t fly radially outward; it flies off tangentially in whatever direction it was moving at that instant. This is why the “centrifugal force” feeling in a car cornering is actually your body’s inertia — you want to go straight, but the car turns underneath you.