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

Problem 1

A car of mass $1200$ kg travels at $15$ m s$^{-1}$ around a circular bend of radius $50$ m. Calculate the centripetal force required.

Answer. $F_c = mv^2/r = 1200 \times 225/50 = 5400$ N.

If you get this wrong, revise: Centripetal Force

Problem 2

A conical pendulum has a string of length $1.5$ m and the bob moves in a horizontal circle of radius $0.90$ m with a period of $1.8$ s. Find the angle the string makes with the vertical and the Tension.

Answer. $\cos\alpha = \sqrt{1 - (0.9/1.5)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.80$. $\alpha = 36.9^\circ$.

$\omega = 2\pi/T = 2\pi/1.8 = 3.49$ rad s$^{-1}$. $v = \omega r = 3.49 \times 0.90 = 3.14$ m S$^{-1}$.

$T\sin\alpha = mv^2/r = m \times 3.14^2/0.90 = 10.96m$. $T = 10.96m/\sin 36.9° = 10.96m/0.60 = 18.3m$ N.

If you get this wrong, revise: Horizontal Circle: Conical Pendulum

Problem 3

A curve of radius $80$ m is banked at $15^\circ$. Calculate the optimum speed for the curve.

Answer. $v = \sqrt{rg\tan\theta} = \sqrt{80 \times 9.81 \times \tan 15°} = \sqrt{80 \times 9.81 \times 0.268} = \sqrt{210.3} = 14.5$ M s$^{-1}$.

If you get this wrong, revise: Banked Curves

Problem 4

A mass of $0.50$ kg on a string of length $1.0$ m is whirled in a vertical circle. What is the Minimum speed at the lowest point for the mass to complete the circle?

Answer. $v_{\mathrm{bottom}} = \sqrt{5gr} = \sqrt{5 \times 9.81 \times 1.0} = \sqrt{49.05} = 7.00$ m S$^{-1}$.

If you get this wrong, revise: Vertical Circles

Problem 5

A satellite orbits Earth at an altitude where the gravitational field strength is $4.5$ N kg$^{-1}$. If the orbit radius is $8.0 \times 10^6$ m, find the orbital speed and period.

Answer. $mg' = mv^2/r \implies v = \sqrt{g'r} = \sqrt{4.5 \times 8.0 \times 10^6} = \sqrt{3.6 \times 10^7} = 6000$ M s$^{-1}$.

$T = 2\pi r/v = 2\pi \times 8.0 \times 10^6/6000 = 8378$ s $\approx 2.3$ hours.

If you get this wrong, revise: Centripetal Force

Problem 6

A car of mass $1000$ kg rounds a level (unbanked) curve of radius $40$ m at $12$ m s$^{-1}$. What Minimum coefficient of static friction is required?

Answer. $F_c = mv^2/r = 1000 \times 144/40 = 3600$ N. This must be provided by friction: $\mu_s mg = \mu_s \times 9810 = 3600$. $\mu_s = 3600/9810 = 0.367$.

If you get this wrong, revise: Centripetal Force

Problem 7

A mass of $0.20$ kg is attached to a string of length $0.80$ m and whirled in a vertical circle. At the highest point, the speed is $3.0$ m s$^{-1}$. Find: (a) the tension at the highest point, (b) The speed at the lowest point, (c) the tension at the lowest point.

Answer. (a) $T + mg = mv^2/r$. $T = 0.20 \times 9.0/0.80 - 0.20 \times 9.81 = 2.25 - 1.962 = 0.288$ N.

(b) $v_b^2 = v_t^2 + 4gr = 9 + 4 \times 9.81 \times 0.80 = 9 + 31.4 = 40.4$. $v_b = 6.36$ m S$^{-1}$.

(c) $T - mg = mv_b^2/r$. $T = 0.20 \times 40.4/0.80 + 1.962 = 10.1 + 1.962 = 12.1$ N.

If you get this wrong, revise: Vertical Circles

Problem 8

Derive the centripetal acceleration formula $a = v^2/r$ using the geometric method (considering two Positions separated by a small angle).

Answer. Consider the velocity vector triangle: two velocity vectors of length $v$ separated by Angle $\delta\theta$. The change in velocity $\delta v$ is the chord of this arc. For small $\delta\theta$: $|\delta\mathbf{v}| = v\delta\theta$.

The time for this change is $\delta t = \delta s/v = r\delta\theta/v$.

$a = |\delta\mathbf{v}|/\delta t = v\delta\theta \cdot v/(r\delta\theta) = v^2/r$. The direction is towards the centre. $\square$

If you get this wrong, revise: Centripetal Acceleration, Derivation

Problem 9

A cyclist of total mass $80$ kg rides around a banked track of radius $25$ m at $8.0$ m s$^{-1}$. If the track is banked at $20^\circ$ Find the normal reaction and whether friction is needed (and in which Direction).

Answer. Optimum speed: $v_{\mathrm{opt}} = \sqrt{25 \times 9.81 \times \tan 20°} = \sqrt{25 \times 9.81 \times 0.364} = \sqrt{89.2} = 9.45$ M s$^{-1}$.

Since $8.0 < 9.45$ The cyclist is going too slowly, so friction must act up the slope to Prevent sliding down.

$N\cos 20° = mg + F\sin 20^\circ$ and $N\sin 20° - F\cos 20° = mv^2/r$.

This requires solving two simultaneous equations. $N\sin 20° - F\cos 20° = 80 \times 64/25 = 204.8$ and $N\cos 20° + F\sin 20° = 784.8$.

If you get this wrong, revise: Banked Curves

Problem 10

Explain why a particle released from circular motion moves tangentially, not radially outward.

Answer. By Newton’s first law, in the absence of a net force, a body continues in a straight Line at constant speed. When the centripetal force is removed (e.g., the string is cut), the only Velocity the particle has is tangential (perpendicular to the radius). There is no radial velocity Component — the particle was never moving radially outward during the circular motion; it was Accelerating radially inward. The apparent “outward fling” is an illusion of perspective from the Rotating frame.

If you get this wrong, revise: Intuition: Circular Motion is NOT Equilibrium

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 $g$.

Example: Bucket of Water in a Vertical Circle

A bucket of water of total mass $1.5$ kg is whirled in a vertical circle of radius $0.80$ m. Find: (a) the minimum speed at the top of the circle for the water to remain in the bucket, (b) the Tension in the arm (modelled as a string) at the top and bottom for this minimum-speed case, (c) The speed at the bottom.

Answer. (a) The water remains in the bucket if the bucket exerts a non-negative normal force on The water. This is equivalent to the string-remaining-taut condition: $v_{\min} = \sqrt{gr} = \sqrt{9.81 \times 0.80} = \sqrt{7.85} = 2.80$ m s$^{-1}$.

(b) At the top (minimum speed): $T = mv^2/r - mg = 1.5 \times 7.85/0.80 - 1.5 \times 9.81 = 14.7 - 14.7 = 0$ N.

This confirms the condition: at minimum speed, the tension is exactly zero — gravity alone provides the centripetal force.

At the bottom: $v_b = \sqrt{5gr} = \sqrt{5 \times 9.81 \times 0.80} = \sqrt{39.24} = 6.26$ m S$^{-1}$.

$T = mv_b^2/r + mg = 1.5 \times 39.24/0.80 + 14.7 = 73.6 + 14.7 = 88.3$ N.

(c) As calculated in (b): $v_b = 6.26$ m s$^{-1}$.

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:

Feature	Inertial frame (ground)	Rotating frame (car)
Real forces	Friction / normal from door (inward)	Friction / normal from door (inward)
Additional force	None	Centrifugal (fictitious, outward)
Passenger acceleration	Centripetal (inward)	Zero (apparent equilibrium)

Further Problems

Problem 11

A mass of $0.40$ kg on a string of length $1.2$ m is whirled in a vertical circle. At the highest Point, the tension in the string is $1.8$ N. Find: (a) the speed at the highest point, (b) the Speed at the lowest point, (c) the tension at the lowest point, (d) the difference $T_{\mathrm{bottom}} - T_{\mathrm{top}}$ and verify that it equals $6mg$.

Answer. (a) At the top: $T + mg = mv^2/r$.

$1.8 + 0.40 \times 9.81 = 0.40v_t^2/1.2$.

$1.8 + 3.924 = 0.333v_t^2$.

$v_t^2 = 5.724/0.333 = 17.18$ So $v_t = 4.14$ m s$^{-1}$.

(b) $v_b^2 = v_t^2 + 4gr = 17.18 + 4 \times 9.81 \times 1.2 = 17.18 + 47.09 = 64.27$.

$v_b = 8.02$ m s$^{-1}$.

(c) $T_b - mg = mv_b^2/r$.

$T_b = 0.40 \times 64.27/1.2 + 3.924 = 21.42 + 3.924 = 25.3$ N.

(d) $T_b - T_t = 25.3 - 1.8 = 23.5$ N. $6mg = 6 \times 0.40 \times 9.81 = 23.5$ N. $\checkmark$

If you get this wrong, revise: Vertical Circles — Detailed Analysis

Problem 12

A conical pendulum has a string of length $2.0$ m and the bob has mass $0.50$ kg. The bob moves in a Horizontal circle of radius $1.0$ m. Find: (a) the angle the string makes with the vertical, (b) the Tension, (c) the period, (d) the linear speed of the bob.

Answer. (a) $\sin\alpha = r/L = 1.0/2.0 = 0.50$ So $\alpha = 30.0^\circ$.

(b) $T = mg/\cos\alpha = 0.50 \times 9.81/\cos 30° = 4.905/0.866 = 5.66$ N.

(c) $T_{\mathrm{period}} = 2\pi\sqrt{L\cos\alpha/g} = 2\pi\sqrt{2.0 \times 0.866/9.81} = 2\pi\sqrt{0.1766} = 2.64$ S.

(d) $v = 2\pi r/T_{\mathrm{period}} = 2\pi \times 1.0/2.64 = 2.38$ m s$^{-1}$.

If you get this wrong, revise: Conical Pendulum — Extended Analysis

Problem 13

A car of mass $800$ kg travels over a humpback bridge of radius of curvature $20$ m. At what speed would the car lose contact with the road at the top of the bridge?

Answer. The car loses contact when the normal reaction is zero. At the top of the bridge:

$mg - N = mv^2/r$.

Setting $N = 0$: $mg = mv^2/r$ So $v = \sqrt{gr} = \sqrt{9.81 \times 20} = \sqrt{196.2} = 14.0$ m S$^{-1}$.

If you get this wrong, revise: Vertical Circles — Detailed Analysis

Problem 14

A bead slides without friction on a vertical circular wire of radius $0.50$ m. If it is given just enough speed at the lowest point to reach the highest point, what is its speed: (a) at the highest Point, (b) when the radius to the bead makes an angle of $60^\circ$ with the upward vertical?

Answer. “Just enough to reach the highest point” means the speed at the top is zero (for a Wire/rod, unlike a string, there is no minimum speed requirement since the wire can provide a normal Force in either direction).

(a) $v_{\mathrm{top}} = 0$.

By energy conservation: $\tfrac{1}{2}mv_b^2 = mg(2r) + 0$.

$v_b = \sqrt{4gr} = \sqrt{4 \times 9.81 \times 0.50} = \sqrt{19.62} = 4.43$ m s$^{-1}$.

(b) At angle $\theta = 60^\circ$ from the upward vertical, the height above the bottom is $r + r\cos 60° = r + r/2 = 3r/2 = 0.75$ m.

$\tfrac{1}{2}mv^2 = \tfrac{1}{2}mv_b^2 - mg \times 0.75$.

$v^2 = v_b^2 - 2g \times 0.75 = 19.62 - 14.72 = 4.90$.

$v = 2.21$ m s$^{-1}$.

If you get this wrong, revise: Vertical Circles — Detailed Analysis

Problem 15

Two identical conical pendulums have the same string length $L$ but different angles $\alpha_1 = 20^\circ$ and $\alpha_2 = 40^\circ$. Show that the pendulum with the larger angle has the shorter period, and find the ratio of their periods.

Answer. $T_1/T_2 = \sqrt\{\cos\alpha_1/\cos\alpha_2\} = \sqrt\{\cos 20°/\cos 40°\} = \sqrt\{0.9397/0.7660\} = \sqrt\{1.227\} = 1.108$.

Since $T_1/T_2 \gt 1$, $T_1 \gt T_2$: the pendulum at the larger angle ($40^\circ$) has the shorter Period, as expected.

The ratio is $T_1:T_2 = 1.108:1$ Or equivalently $T_2/T_1 = 0.902$.

If you get this wrong, revise: Conical Pendulum — Extended Analysis

Problem 16

A small sphere of mass $m$ is attached to a light rigid rod of length $r$ (rather than a string) and Whirled in a vertical circle. Explain why the rod can maintain circular motion even when the sphere is at rest at the top, and find the force exerted by the rod at the top and bottom when the sphere has the minimum possible speed at the top.

Answer. Unlike a string (which can only pull), a rod can both push and pull. At the top of the Circle, if the speed is zero, gravity provides more centripetal acceleration than needed ($g \gt v^2/r = 0$), so the rod must push outward (exert a compressive normal force) to reduce the Net inward force to zero.

At the top with $v = 0$: $mg - N = mv^2/r = 0$ So $N = mg$ (rod pushes outward with force $mg$).

At the bottom, by energy conservation: $v_b^2 = 0 + 4gr = 4gr$ So $v_b = 2\sqrt{gr}$.

$T - mg = mv_b^2/r = 4mg$ So $T = 5mg$ (rod pulls inward with force $5mg$).

If you get this wrong, revise: Vertical Circles — Detailed Analysis

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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.

flowchart TD
    A[06 Circular Motionx] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

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.

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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.