Circular Motion

This topic explores fundamental concepts that shape our understanding of the world.

Circular motion in further mathematics extends the basic treatment to include banked tracks, conical Pendulums, vertical circles with energy methods, and problems where the circular path constraints Determine unknown forces.

Definition. The angular displacement $\theta$ is the angle swept by a radius vector, Measured in radians.

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$^{-1}$.

Definition. The angular acceleration $\alpha$ is the rate of change of angular velocity:

$\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$

For a particle moving in a circle of radius $r$:

$\boxed{v = \omega r}$

$\boxed{a_{\mathrm{tangential}} = \alpha r}$

For uniform circular motion:

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

The position vector of a particle in a circle of radius $r$ in the $xy$-plane:

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

Velocity (first derivative):

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

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

Acceleration (second derivative):

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

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

Magnitude: $a = \omega^2 r = \dfrac{v^2}{r}$Directed radially inward. $\blacksquare$

By Newton”s second law:

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

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 equilibrium: $T\cos\alpha = mg$ … (i)

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

Dividing (ii) by (i):

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

Since $r = L\sin\alpha$ and $\omega = \dfrac{v}{r}$:

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

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

A vehicle on a banked track of angle $\theta$ and radius $r$.

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

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

Dividing (ii) by (i):

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

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

At the optimum speed, no friction is needed. If $v > v_{\mathrm{opt}}$Friction acts down the Slope. If $v < v_{\mathrm{opt}}$Friction acts up the slope.

A particle moves in a horizontal circle on the smooth inner surface of a sphere of radius $r$ at Height $h$ below the centre.

The radius of the circle is $a = \sqrt{r^2 - h^2}$.

Normal reaction $R$ acts radially outward. Resolving vertically: $R\cos\phi = mg$ where $\sin\phi = a/r$.

Horizontal: $R\sin\phi = \dfrac{mv^2}{a}$.

$\frac{v^2}{a} = \frac{R\sin\phi}{m} = \frac{g\sin\phi}{\cos\phi} = g\tan\phi = \frac{ga}{h}$

Consider a mass $m$ on a string of length $r$ moving in a vertical circle. At angle $\theta$ from The downward vertical:

Along the string (towards centre): $T - mg\cos\theta = \dfrac{mv^2}{r}$

At the top ($\theta = 180^\circ$): both $T$ and $mg$ act towards centre:

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

At the bottom ($\theta = 0^\circ$): centripetal direction is upward:

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

For the string to remain taut at the top: $T \geq 0$.

$T + mg = \frac{mv^2}{r} \implies T = \frac{mv^2}{r} - mg \geq 0$

$\frac{mv^2}{r} \geq mg \implies v^2 \geq gr \implies \boxed{v_{\min} = \sqrt{gr}}$

At this minimum speed, $T = 0$ — the weight alone provides the centripetal force. $\blacksquare$

Using conservation of mechanical energy between two points on a vertical circle:

$\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2$

Where $h$ is the height above a reference level.

Between top and bottom (height difference $2r$):

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

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

For minimum complete circle ($v_{\mathrm{top}} = \sqrt{gr}$):

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

Using energy conservation, the speed at angle $\theta$ from the bottom is:

$\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 - mgr(1-\cos\theta)$

Where $v_0$ is the speed at the bottom.

$v^2 = v_0^2 - 2gr(1-\cos\theta)$

Tension at angle $\theta$ (measuring from the bottom):

$T = \frac{mv^2}{r} + mg\cos\theta = \frac{m}{r}[v_0^2 - 2gr(1-\cos\theta)] + mg\cos\theta$

$T = \frac{mv_0^2}{r} - 2mg + 2mg\cos\theta + mg\cos\theta = \frac{mv_0^2}{r} - 2mg + 3mg\cos\theta$

A particle slides on the smooth outer surface of a sphere of radius $r$. It leaves the surface when The normal reaction $R = 0$.

At angle $\theta$ from the top: $R + mg\cos\theta = \dfrac{mv^2}{r}$.

Energy: $\dfrac{1}{2}mv^2 = mgr(1-\cos\theta)$ (from rest at the top).

When $R = 0$: $mg\cos\theta = \dfrac{mv^2}{r} = \dfrac{2mg(1-\cos\theta)}{r}$.

$g\cos\theta = 2g(1-\cos\theta) \implies \cos\theta = 2 - 2\cos\theta \implies 3\cos\theta = 2$.

$\boxed{\theta = \arccos\!\left(\frac{2}{3}\right) \approx 48.2°}$

When the speed varies, there is both centripetal and tangential acceleration:

$a_c = \frac{v^2}{r} \quad \mathrm{(radially inward)}, \qquad a_t = \frac{dv}{dt} \quad \mathrm{(tangential)}$

The total acceleration has magnitude:

$a = \sqrt{a_c^2 + a_t^2}$

The resultant force has a radial component providing $a_c$ and a tangential component providing $a_t$.

Consider a particle of mass $m$ on a string of length $r$ moving in a vertical circle. Let $v_0$ be The speed at the lowest point (the reference level for energy).

At angle $\theta$ measured from the upward vertical (so the top is $\theta = 0$ and the bottom Is $\theta = \pi$), the height above the lowest point is:

$h = r + r\cos\theta = r(1 + \cos\theta)$

By conservation of energy:

$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgr(1 + \cos\theta)$

$\boxed{v^2 = v_0^2 - 2gr(1 + \cos\theta)}$

At angle $\theta$ from the upward vertical, the radial direction (towards the centre) has component Of weight $mg\cos\theta$ pointing towards the centre:

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

Substituting the speed:

$T = \frac{m}{r}[v_0^2 - 2gr(1 + \cos\theta)] - mg\cos\theta = \frac{mv_0^2}{r} - 2mg - 2mg\cos\theta - mg\cos\theta$

$\boxed{T = \frac{mv_0^2}{r} - 2mg - 3mg\cos\theta}$

At the top ($\theta = 0$, $\cos\theta = 1$): $T = \dfrac{mv_0^2}{r} - 5mg$.

At the bottom ($\theta = \pi$, $\cos\theta = -1$): $T = \dfrac{mv_0^2}{r} + mg$.

At the midpoint ($\theta = \pi/2$, $\cos\theta = 0$): $T = \dfrac{mv_0^2}{r} - 2mg$.

For a complete circle, we need $T \geq 0$ everywhere. The minimum tension occurs at the top ($\theta = 0$):

$\frac{mv_0^2}{r} - 5mg \geq 0 \implies v_0^2 \geq 5gr \implies \boxed{v_0 = \sqrt{5gr}}$

At this speed, $T_{\mathrm{top}} = 0$ and the weight alone provides the centripetal acceleration at The top.

The speed at the top is: $v_{\mathrm{top}}^2 = v_0^2 - 2gr(2) = 5gr - 4gr = gr$Confirming $v_{\mathrm{top}} = \sqrt{gr}$.

A car of mass $m$ travels on a banked curve of radius $r$ and angle $\theta$With coefficient of Friction $\mu$.

When the car travels faster than the optimum speed, friction acts down the slope to provide Additional centripetal force.

Resolving perpendicular to the surface: $N = mg\cos\theta + \dfrac{mv^2}{r}\sin\theta$ … (i)

Resolving along the surface (towards centre): $F + mg\sin\theta = \dfrac{mv^2}{r}\cos\theta$ … (ii)

At limiting friction: $F = \mu N$.

Substituting (i) into $F = \mu N$ and then into (ii):

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

$\mu g\cos\theta + \frac{\mu v^2\sin\theta}{r} + g\sin\theta = \frac{v^2\cos\theta}{r}$

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

$\boxed{v_{\max}^2 = \frac{rg(\sin\theta + \mu\cos\theta)}{\cos\theta - \mu\sin\theta}}$

Similarly, when travelling slower than the optimum speed, friction acts up the slope:

$\boxed{v_{\min}^2 = \frac{rg(\sin\theta - \mu\cos\theta)}{\cos\theta + \mu\sin\theta}}$

Note: $v_{\min}$ only exists if $\sin\theta > \mu\cos\theta$I.e., $\tan\theta > \mu$. If the bank Angle is too shallow, the car can come to rest without sliding down.

Example. A curve of radius $80\,\mathrm{m}$ is banked at $30^\circ$ with $\mu = 0.3$. Find the Maximum and minimum safe speeds for a car on this curve.

$v_{\max}^2 = \dfrac{80 \times 9.8(\sin 30° + 0.3\cos 30°)}{\cos 30° - 0.3\sin 30°} = \dfrac{784(0.5 + 0.2598)}{0.8660 - 0.15} = \dfrac{784 \times 0.7598}{0.7160}$

$= \dfrac{595.7}{0.7160} \approx 831.9 \implies v_{\max} \approx 28.8\,\mathrm{m s}^{-1}$.

$v_{\min}^2 = \dfrac{80 \times 9.8(\sin 30° - 0.3\cos 30°)}{\cos 30° + 0.3\sin 30°} = \dfrac{784(0.5 - 0.2598)}{0.8660 + 0.15} = \dfrac{784 \times 0.2402}{1.016}$

$= \dfrac{188.3}{1.016} \approx 185.3 \implies v_{\min} \approx 13.6\,\mathrm{m s}^{-1}$.

The optimum speed (no friction needed) is $v_{\mathrm{opt}} = \sqrt{80 \times 9.8 \times \tan 30°} = \sqrt{452.6} \approx 21.3\,\mathrm{m s}^{-1}$ Which lies between $v_{\min}$ and $v_{\max}$ as expected.

A mass $m$ is attached to a string of length $L$ and moves in a horizontal circle of radius $r$ at Constant speed $v$. The string makes a constant angle $\alpha$ with the vertical.

Forces on the mass:

• Tension $T$ along the string, directed towards the pivot
• Weight $mg$ vertically downward

Since the mass moves in a horizontal circle, there is no vertical acceleration:

$T\cos\alpha = mg \implies T = \frac{mg}{\cos\alpha}$ … (i)

The horizontal component of tension provides the centripetal force:

$T\sin\alpha = \frac{mv^2}{r}$ … (ii)

Since $r = L\sin\alpha$ and $v = \omega r = \omega L\sin\alpha$:

From (i) and (ii): $\dfrac{mg\sin\alpha}{\cos\alpha} = \dfrac{m\omega^2 L^2\sin^2\alpha}{L\sin\alpha}$

$g\tan\alpha = \omega^2 L\sin\alpha$

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

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

$\boxed{T_{\mathrm{period}} = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L\cos\alpha}{g}}}$

Key observations:

• The period depends only on $L$, $\alpha$And $g$ — it is independent of mass
• As $\alpha \to 0$The period approaches $2\pi\sqrt{L/g}$ (simple pendulum for small angles)
• As $\alpha \to 90^\circ$The period $\to 0$ (impractical: requires infinite speed)
• A larger angle $\alpha$ means a faster rotation (shorter period)

Example. A motorcyclist rides around a vertical circular track of radius $8\,\mathrm{m}$. The Motorcycle and rider have combined mass $200\,\mathrm{kg}$. Find the minimum speed at the lowest Point to complete the loop, the normal reaction at the top and bottom at this minimum speed, and the Speed and reaction at a point $90^\circ$ from the bottom.

Minimum speed at the bottom: $v_{\min} = \sqrt{5gr} = \sqrt{5 \times 9.8 \times 8} = \sqrt{392} \approx 19.8\,\mathrm{m s}^{-1}$.

At the top (minimum speed): $v_{\mathrm{top}} = \sqrt{gr} = \sqrt{78.4} \approx 8.85\,\mathrm{m s}^{-1}$.

$R + mg = \dfrac{mv_{\mathrm{top}}^2}{r} \implies R = \dfrac{200 \times 78.4}{8} - 200 \times 9.8 = 1960 - 1960 = 0\,\mathrm{N}$.

At minimum speed, the motorcycle is just in contact with the track at the top.

At the bottom:

$v_{\mathrm{bottom}} = \sqrt{5gr} \approx 19.8\,\mathrm{m s}^{-1}$.

$R - mg = \dfrac{mv_{\mathrm{bottom}}^2}{r} \implies R = \dfrac{200 \times 392}{8} + 1960 = 9800 + 1960 = 11760\,\mathrm{N}$.

Note: $R = 6mg$ at the bottom when $v = \sqrt{5gr}$.

At $90^\circ$ from the bottom (measuring from the downward vertical, so $\theta = 90^\circ$):

Height above bottom: $r(1 - \cos 90°) = r = 8\,\mathrm{m}$.

$v^2 = v_0^2 - 2gr = 392 - 156.8 = 235.2 \implies v \approx 15.3\,\mathrm{m s}^{-1}$.

At this point, the weight acts radially (towards the centre) and the reaction acts radially outward:

$mg - R = \dfrac{mv^2}{r} \implies R = mg - \dfrac{200 \times 235.2}{8} = 1960 - 5880$.

$R = -3920\,\mathrm{N}$.

The negative sign confirms the track pushes inward (downward at this point) to maintain the Circular path. This is expected: at this speed, the motorcyclist would fly off without the track Pushing them inward.

Centripetal force is not a force in its own right. It is the resultant of the real forces (tension, normal reaction, weight, friction) directed towards the centre of the circle. The most Common mistake is to draw “centripetal force” as an additional arrow on a free body diagram Alongside tension, weight, etc. This double-counts and produces incorrect equations.

Correct approach: draw only the physical forces, then apply Newton’s second law towards the centre: “sum of force components towards centre $= mv^2/r$”.

At different points on a vertical circle, the normal reaction can point in different directions:

• At the bottom: reaction points upward (away from centre)
• At the top: reaction points downward (towards centre)
• At the sides: reaction points horizontally (towards or away from centre depending on the speed)

The sign of $R$ in your equations should emerge from the physics. If you get a negative $R$It means the contact force acts in the opposite direction to what you assumed.

• String: can only pull (tension $\geq 0$). Minimum speed at top is $\sqrt{gr}$Minimum at bottom is $\sqrt{5gr}$.
• Rod: can push and pull. Particle can reach the top with $v = 0$Minimum at bottom is $\sqrt{4gr} = 2\sqrt{gr}$.
• Smooth wire: like a rod in that it can provide a reaction in either direction.
• Rough surface: friction can provide tangential force, making the problem significantly more complex.

$\omega = v/r$ is only valid when $v$ is the tangential speed and $r$ is the radius of the circular Path. In a conical pendulum, the radius of the circle is $L\sin\alpha$Not $L$.

Problem. A conical pendulum consists of a particle of mass $0.5\,\mathrm{kg}$ on a string of Length $1\,\mathrm{m}$ making angle $\theta$ with the vertical. It rotates with angular speed $4\,\mathrm{rad\,s^{-1}}$. Find $\theta$.

Solution. Resolving vertically: $T\cos\theta = mg = 4.9\,\mathrm{N}$.

Resolving horizontally: $T\sin\theta = m\omega^2 r = 0.5 \times 16 \times \sin\theta = 8\sin\theta$.

Dividing: $\tan\theta = \dfrac{8\sin\theta}{4.9/\cos\theta}$. Actually:

$\frac{T\sin\theta}{T\cos\theta} = \tan\theta = \frac{m\omega^2 L\sin\theta}{mg} = \frac{\omega^2 L \sin\theta}{g}$

$\cos\theta = \frac{g}{\omega^2 L} = \frac{9.8}{16} = 0.6125$

$\boxed{\theta \approx 52.2°}$

Problem. A car of mass $m$ travels around a banked curve of radius $r$ at banking angle $\alpha$ With friction coefficient $\mu$. Find the maximum safe speed.

Solution. At maximum speed, friction acts down the slope. Resolving perpendicular to surface: $N = mg\cos\alpha + \dfrac{mv^2}{r}\sin\alpha$.

Resolving along surface: $\dfrac{mv^2}{r}\cos\alpha = mg\sin\alpha + \mu N$.

$\frac{mv^2}{r}\cos\alpha = mg\sin\alpha + \mu mg\cos\alpha + \frac{\mu mv^2}{r}\sin\alpha$

$v^2\!\left(\frac{\cos\alpha}{r} - \frac{\mu\sin\alpha}{r}\right) = g(\sin\alpha + \mu\cos\alpha)$

$\boxed{v_{\max}^2 = \frac{rg(\sin\alpha + \mu\cos\alpha)}{\cos\alpha - \mu\sin\alpha}}$

Problem. A particle of mass $m$ is attached to a light rod of length $l$ and rotates in a Vertical circle. Find the minimum angular speed for complete circles.

Solution. At the top: $T + mg = m\omega^2 l$. For a rod, $T \geq 0$ (can push). Minimum: $T = 0$:

$mg = m\omega_{\min}^2 l \implies \boxed{\omega_{\min} = \sqrt{\frac{g}{l}}}$

Problem. A particle of mass $0.2\,\mathrm{kg}$ on a light inextensible string of length $0.5\,\mathrm{m}$ is projected horizontally from the lowest point. Find the minimum speed for Complete circles.

Solution. At the top: $v_{\text{top}}^2 \geq gr = 4.9$.

Energy conservation: $\dfrac{1}{2}mu^2 = \dfrac{1}{2}mv_{\text{top}}^2 + mg(2r)$.

$u^2 = gr + 4gr = 5gr = 5 \times 9.8 \times 0.5 = 24.5$

$\boxed{u_{\min} = \sqrt{24.5} \approx 4.95\,\mathrm{m\,s^{-1}}}$

Problem. A particle of mass $m$ on a string of length $r$ moves in a vertical circle. At angle $\theta$ from the downward vertical, find the tension.

Solution. Resolving toward the centre:

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

At the bottom ($\theta = 0$): $T = \dfrac{mv^2}{r} + mg$ (maximum). At the top ($\theta = \pi$): $T = \dfrac{mv^2}{r} - mg$ (minimum).

Problem. A disc rotates with angular acceleration $\alpha = -0.2\omega\,\mathrm{rad\,s^{-2}}$. If $\omega = 10\,\mathrm{rad\,s^{-1}}$ at $t = 0$Find $\omega(t)$.

Solution. $\dfrac{d\omega}{dt} = -0.2\omega$. Separable:

$\int \frac{d\omega}{\omega} = \int -0.2\,dt \implies \ln\omega = -0.2t + \ln 10$

$\boxed{\omega = 10e^{-0.2t}}$

A particle of mass $m$ moves in a horizontal circle of radius $r$ on the inside of a smooth Hemispherical bowl of radius $R$. Find the speed $v$ in terms of R$$rAnd $g$.

A bead of mass $m$ slides on a smooth circular wire of radius $a$ in a vertical plane. It is Projected from the lowest point with speed $\sqrt{6ga}$. Find the speed and reaction at $60°$ above The lowest point.

Prove that for a particle on a string in vertical circular motion, the string goes slack when $v^2 < gr$ at the highest point.