Circular Motion and Polar Coordinates

Uniform Circular Motion

Part 1 of 7 — Circular Motion

An object moving in a circle at constant speed is undergoing uniform circular motion. Despite constant speed, the object is accelerating because the direction of velocity is always changing.

Kinematics of Circular Motion

Quantity	Symbol	Formula
Period	$T$	Time for one revolution
Frequency	$f$	$f = 1/T$
Angular velocity	$\omega$	$\omega = 2\pi f = 2\pi/T$
Speed	$v$	$v = \omega r$
Arc length	$s$	$s = r\theta$

Position Vector

For circular motion of radius $r$ in the $xy$-plane:

$\vec{r}(t) = r\cos(\omega t)\,\hat{x} + r\sin(\omega t)\,\hat{y}$

Velocity Vector

$\vec{v}(t) = \frac{d\vec{r}}{dt} = -r\omega\sin(\omega t)\,\hat{x} + r\omega\cos(\omega t)\,\hat{y}$

The magnitude: $|\vec{v}| = r\omega = v$ (constant). The direction is tangent to the circle — perpendicular to $\vec{r}$ at every instant.

Centripetal Acceleration

Differentiating velocity:

$\vec{a}(t) = \frac{d\vec{v}}{dt} = -r\omega^2\cos(\omega t)\,\hat{x} - r\omega^2\sin(\omega t)\,\hat{y}$

$\vec{a}(t) = -\omega^2\vec{r}(t)$

This acceleration points radially inward (toward the center) and has magnitude:

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

Important: Centripetal Force is NOT a New Force

Centripetal acceleration is caused by a real force (or combination of forces) directed toward the center. Common sources:

Situation	Centripetal Force
Ball on a string	Tension
Car on a curve	Static friction
Planet orbiting	Gravity
Conical pendulum	Horizontal component of tension
Banked curve (no friction)	Normal force component

Newton's second law in the radial direction:

$\sum F_r = ma_c = m\frac{v^2}{r} = m\omega^2 r$

AP Tip: Never list "centripetal force" as a force on a free body diagram. It's a net force requirement, not a separate force.

Flat Curve: Car on a Turn

For a car of mass $m$ on a flat circular turn of radius $r$:

The static friction provides the centripetal force:

$f_s = m\frac{v^2}{r} \leq \mu_s mg$

Maximum speed before skidding:

$v_{\max} = \sqrt{\mu_s g r}$

Worked Example

A car takes a flat turn of radius 50 m. If $\mu_s = 0.8$ and $g = 10$ $m/s^{2}$:

$v_{\max} = \sqrt{0.8 \times 10 \times 50} = \sqrt{400} = 20 \text{ m/s} = 72 \text{ km/h}$

Conical Pendulum

A mass $m$ swings in a horizontal circle on a string of length $L$ at angle $\theta$ from the vertical:

Vertical equilibrium: $T\cos\theta = mg$

Radial (centripetal): $T\sin\theta = m\omega^2 r = m\omega^2 L\sin\theta$

Dividing: $\tan\theta = \omega^2 r / g = \omega^2 L\sin\theta / g$

$\omega = \sqrt{\frac{g}{L\cos\theta}}$

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

Notice: the period depends on the angle $\theta$ and string length $L$, but not on the mass.

Part 1 Summary

Concept	Formula
Speed	$v = \omega r = 2\pi r/T$
Centripetal acceleration	$a_c = v^2/r = \omega^2 r$
Centripetal force	$F_c = mv^2/r$
Max speed (flat curve)	$v_{\max} = \sqrt{\mu_s g r}$
Position vector	$\vec{r}(t) = r\cos(\omega t)\hat{x} + r\sin(\omega t)\hat{y}$

Next up: Part 2 — Centripetal Acceleration Derivation, proving $a_c = v^2/r$ rigorously using calculus.

Centripetal Acceleration Derivation

Part 2 of 7 — Circular Motion

On the AP Physics C exam, you may need to derive the centripetal acceleration formula from first principles. Let's build this up rigorously.

Method 1: Calculus Derivation

Start with the position vector in Cartesian unit vectors:

$\vec{r}(t) = r\cos(\omega t)\,\hat{x} + r\sin(\omega t)\,\hat{y}$

First derivative (velocity):

$\vec{v} = \frac{d\vec{r}}{dt} = -r\omega\sin(\omega t)\,\hat{x} + r\omega\cos(\omega t)\,\hat{y}$

Second derivative (acceleration):

$\vec{a} = \frac{d\vec{v}}{dt} = -r\omega^2\cos(\omega t)\,\hat{x} - r\omega^2\sin(\omega t)\,\hat{y}$

$\boxed{\vec{a} = -\omega^2\vec{r}}$

The magnitude: $|\vec{a}| = \omega^2 r$. Since $v = \omega r$:

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

The minus sign in $\vec{a} = -\omega^2\vec{r}$ shows the acceleration is anti-parallel to $\vec{r}$ — it points toward the center.

Method 2: Geometric/Limit Derivation

Consider an object moving at constant speed $v$ around a circle of radius $r$. In a small time $\Delta t$, the velocity vector rotates by angle $\Delta\theta = \omega\Delta t$.

Change in Velocity

The velocity vectors at times $t$ and $t + \Delta t$ both have magnitude $v$ but differ in direction by $\Delta\theta$.

Using the isoceles triangle formed by $\vec{v}(t)$, $\vec{v}(t+\Delta t)$, and $\Delta\vec{v}$:

$|\Delta\vec{v}| = 2v\sin\left(\frac{\Delta\theta}{2}\right)$

For small $\Delta\theta$: $\sin(\Delta\theta/2) \approx \Delta\theta/2$:

$|\Delta\vec{v}| \approx v\Delta\theta = v\omega\Delta t$

Average Acceleration

$|\vec{a}_{\text{avg}}| = \frac{|\Delta\vec{v}|}{\Delta t} = v\omega = \frac{v^2}{r}$

Taking the limit $\Delta t \to 0$ gives the instantaneous acceleration:

$a_c = \lim_{\Delta t \to 0} \frac{|\Delta\vec{v}|}{\Delta t} = v\omega = \frac{v^2}{r}$

Direction of $\Delta\vec{v}$

As $\Delta t \to 0$, the direction of $\Delta\vec{v}$ becomes perpendicular to $\vec{v}$, which means it points radially inward. This completes the derivation.

Method 3: Period and Frequency Form

Problems often give the period $T$ (time per revolution) or frequency $f = 1/T$ instead of the speed. Since $v = 2\pi r/T$:

$a_c = \frac{v^2}{r} = \frac{(2\pi r/T)^2}{r} = \frac{4\pi^2 r}{T^2} = 4\pi^2 r f^2$

Worked Example

A point on a wheel of radius $0.20$ m makes $5.0$ revolutions per second.

$a_c = 4\pi^2 (0.20)(5.0)^2 = 4\pi^2(5.0) \approx 197 \text{ m/s}^2$

Check with speed: $v = 2\pi(0.20)(5.0) = 6.28$ m/s, and $v^2/r = 39.5/0.20 \approx 197$ m/s$^2$ ✓

At a fixed rotation rate, $a_c \propto r$: points farther from the axle have larger centripetal acceleration.

Part 2 Summary

Three equivalent derivations of $a_c = v^2/r$:

Method	Key Step
Calculus	Differentiate $\vec{r}(t)$ twice
Geometric	Small-angle limit of $\Delta\vec{v}$ triangle
Period form	$a_c = 4\pi^2 r/T^2$

Key Results:

• $\vec{a} = -\omega^2 \vec{r}$ (always points toward center)
• $\vec{v} \perp \vec{a}$ in UCM (constant speed)
• $\vec{v} \cdot \vec{a} = 0 \implies |\vec{v}|$ is constant

Next up: Part 3 — Banked Curves, applying centripetal force analysis to tilted roadways.

Banked Curves

Part 3 of 7 — Circular Motion

Banked curves are inclined roadways designed so that the normal force provides some (or all) of the centripetal force needed for turning.

Frictionless Banked Curve

For a car on a banked curve (angle $\beta$) with no friction, the only forces are gravity and the normal force.

Vertical equilibrium: $N\cos\beta = mg$

Radial (horizontal, toward center): $N\sin\beta = \frac{mv^2}{r}$

Dividing these equations:

$\tan\beta = \frac{v^2}{rg}$

$\boxed{v_{\text{design}} = \sqrt{rg\tan\beta}}$

This is the design speed — the one speed at which friction is unnecessary.

Worked Example

A highway curve has $r = 200$ m and is banked at $\beta = 15°$.

$v_{\text{design}} = \sqrt{200(10)\tan 15°} = \sqrt{2000(0.268)} = \sqrt{536} = 23.2 \text{ m/s} \approx 83 \text{ km/h}$

Banked Curve with Friction

When the car's speed differs from the design speed, friction is needed.

Case 1: $v > v_{\text{design}}$ (Too Fast)

The car tends to slide up the bank. Friction acts down the incline (inward component helps centripetal force).

Vertical: $N\cos\beta - f\sin\beta = mg$

Radial: $N\sin\beta + f\cos\beta = mv^2/r$

With $f = \mu_s N$ at the maximum speed:

$v_{\max} = \sqrt{rg\cdot\frac{\tan\beta + \mu_s}{1 - \mu_s\tan\beta}}$

Case 2: $v < v_{\text{design}}$ (Too Slow)

The car tends to slide down the bank. Friction acts up the incline.

$v_{\min} = \sqrt{rg\cdot\frac{\tan\beta - \mu_s}{1 + \mu_s\tan\beta}}$

If $\mu_s > \tan\beta$, then $v_{\min} = 0$ (friction is strong enough to hold the car stationary on the bank).

Speed Range

$v_{\min} \leq v \leq v_{\max}$

This range widens with more friction and narrows to a single value ($v_{\text{design}}$) as $\mu_s \to 0$.

Detailed Derivation: $v_{\max}$

Let's carefully derive the maximum speed formula. Forces on the car when going too fast (friction points down the bank):

Decompose forces along vertical and horizontal (centripetal):

Normal force: magnitude $N$, tilted at angle $\beta$ from vertical. Friction: magnitude $f = \mu_s N$, along the bank surface (perpendicular to $N$), pointing down and inward.

Vertical equilibrium ($a_y = 0$): $N\cos\beta - \mu_s N\sin\beta = mg$ $N(\cos\beta - \mu_s\sin\beta) = mg$ $N = \frac{mg}{\cos\beta - \mu_s\sin\beta}$

Horizontal (centripetal, $a_r = v^2/r$): $N\sin\beta + \mu_s N\cos\beta = \frac{mv^2}{r}$ $N(\sin\beta + \mu_s\cos\beta) = \frac{mv^2}{r}$

Dividing: $\frac{v^2}{rg} = \frac{\sin\beta + \mu_s\cos\beta}{\cos\beta - \mu_s\sin\beta}$

Dividing numerator and denominator by $\cos\beta$:

$\frac{v^2}{rg} = \frac{\tan\beta + \mu_s}{1 - \mu_s\tan\beta}$

$\boxed{v_{\max} = \sqrt{rg\cdot\frac{\tan\beta + \mu_s}{1 - \mu_s\tan\beta}}}$

Note: The formula resembles the tangent addition formula: $\tan(\beta + \phi) = \frac{\tan\beta + \tan\phi}{1 - \tan\beta\tan\phi}$ where $\tan\phi = \mu_s$. So $v_{\max}^2 = rg\tan(\beta + \phi)$ where $\phi = \arctan(\mu_s)$.

Part 3 Summary

Concept	Formula
Design speed (no friction)	$v = \sqrt{rg\tan\beta}$
Max speed (with friction)	$v_{\max} = \sqrt{rg\frac{\tan\beta + \mu_s}{1 - \mu_s\tan\beta}}$
Min speed (with friction)	$v_{\min} = \sqrt{rg\frac{\tan\beta - \mu_s}{1 + \mu_s\tan\beta}}$
Compact form	$v^2 = rg\tan(\beta \pm \phi)$, $\phi = \arctan\mu_s$

Next up: Part 4 — Vertical Circles, where the centripetal force requirement changes with position.

Vertical Circles

Part 4 of 7 — Circular Motion

In vertical circular motion, gravity alternately aids and opposes the centripetal acceleration, making the analysis position-dependent. This is a favorite AP Physics C topic.

Forces at Key Positions

For a mass $m$ on a string (or track) of radius $r$, moving at speed $v$:

At the Top

Both gravity and tension point toward the center (downward): $mg + T_{\text{top}} = \frac{mv_{\text{top}}^2}{r}$

$T_{\text{top}} = \frac{mv_{\text{top}}^2}{r} - mg$

At the Bottom

Tension points toward center (up); gravity points away (down): $T_{\text{bot}} - mg = \frac{mv_{\text{bot}}^2}{r}$

$T_{\text{bot}} = \frac{mv_{\text{bot}}^2}{r} + mg$

At an Arbitrary Angle $\theta$ from Bottom

Measure $\theta$ from the bottom of the circle. Tension points toward the center, and the component of gravity along the string points away from the center with magnitude $mg\cos\theta$:

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

At $\theta = 0$ this reduces to the bottom equation, and at $\theta = 180°$ ($\cos\theta = -1$) it becomes the top equation.

Minimum Speed at the Top

At the top of a vertical circle, the minimum speed occurs when the tension (or normal force) drops to zero. At that point, gravity alone provides the centripetal force:

$mg = \frac{mv_{\min}^2}{r}$

$\boxed{v_{\min,\text{top}} = \sqrt{gr}}$

If the speed drops below this, the object leaves the circular path.

Minimum Speed at the Bottom (for Complete Loop)

Using energy conservation from bottom to top:

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

With $v_{\text{top}} = \sqrt{gr}$:

$\frac{1}{2}v_{\text{bot}}^2 = \frac{1}{2}gr + 2gr = \frac{5}{2}gr$

$\boxed{v_{\min,\text{bot}} = \sqrt{5gr}}$

Tension Difference (Top vs Bottom)

$T_{\text{bot}} - T_{\text{top}} = \frac{m(v_{\text{bot}}^2 - v_{\text{top}}^2)}{r} + 2mg$

Using energy conservation: $v_{\text{bot}}^2 - v_{\text{top}}^2 = 4gr$:

$T_{\text{bot}} - T_{\text{top}} = \frac{m(4gr)}{r} + 2mg = 4mg + 2mg = 6mg$

This result is independent of the speed — the tension difference is always $6mg$.

Vertical Circle on a Track (Normal Force)

For a ball rolling inside a circular loop track, the analysis is similar but the normal force replaces tension:

At the Top (Inside of Loop)

$N_{\text{top}} + mg = \frac{mv_{\text{top}}^2}{r}$

(both $N$ and $mg$ point toward center)

At the Bottom (Inside of Loop)

$N_{\text{bot}} - mg = \frac{mv_{\text{bot}}^2}{r}$

The minimum speed condition at the top is the same ($v_{\min} = \sqrt{gr}$), when $N_{\text{top}} = 0$.

Outside of a Hump

For a car going over a hill of radius $r$:

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

$N = mg - \frac{mv^2}{r} = m\left(g - \frac{v^2}{r}\right)$

The car leaves the road when $N = 0$:

$v_{\text{leave}} = \sqrt{gr}$

Worked Example: Roller Coaster

A roller coaster starts from rest at height $h$ above the bottom of a circular loop of radius $r$ (no friction). What is the normal force at the top of the loop?

By energy conservation from start to top of loop (height $= 2r$): $mgh = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)$ $v_{\text{top}}^2 = 2g(h - 2r)$

At the top: $N + mg = mv_{\text{top}}^2/r$ $N = \frac{m \cdot 2g(h-2r)}{r} - mg = mg\left(\frac{2h}{r} - 5\right)$

For $N \geq 0$: $h \geq 5r/2$. This means the starting height must be at least $2.5r$ above the bottom of the loop.

Part 4 Summary

Position	Centripetal Equation
Bottom	$T - mg = mv^2/r$
Top	$T + mg = mv^2/r$ (string)
Top	$N + mg = mv^2/r$ (track)
Hill crest	$mg - N = mv^2/r$

Key Result	Value
Min speed at top	$\sqrt{gr}$
Min speed at bottom (full loop)	$\sqrt{5gr}$
Tension difference (top vs bottom)	$6mg$

Next up: Part 5 — Non-Uniform Circular Motion, where both speed and direction change simultaneously.

Non-Uniform Circular Motion

Part 5 of 7 — Circular Motion

When an object moves in a circle with changing speed, it has both centripetal (radial) and tangential components of acceleration.

Decomposing Acceleration

The acceleration has two perpendicular components: one toward the center and one along the velocity.

Component	Direction	Formula	Cause
Centripetal $a_c$	Toward center	$v^2/r$	Changes direction
Tangential $a_t$	Along velocity	$dv/dt$	Changes speed

The total acceleration magnitude:

$|\vec{a}| = \sqrt{a_c^2 + a_t^2} = \sqrt{\left(\frac{v^2}{r}\right)^2 + \left(\frac{dv}{dt}\right)^2}$

Angular Quantities

Linear	Angular	Relation
$s$ (arc length)	$\theta$ (angle)	$s = r\theta$
$v$ (speed)	$\omega$ (angular velocity)	$v = r\omega$
$a_t$ (tangential)	$\alpha$ (angular acceleration)	$a_t = r\alpha$

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

Components from Angular Quantities

For a point at fixed radius $r$ on a rotating object, both components come straight from the angular velocity $\omega$ and angular acceleration $\alpha$:

$a_c = r\omega^2 \quad \text{(toward the center)}, \qquad a_t = r\alpha \quad \text{(along the velocity)}$

The net acceleration is the vector sum of these two perpendicular components.

Example: Constant Angular Acceleration

A disk starts from rest and accelerates at $\alpha = 2$ $rad/s^{2}$. At $t = 3$ s, a point at $r = 0.5$ m from the center has:

$\omega = \alpha t = 6 \text{ rad/s}$ $v = r\omega = 3 \text{ m/s}$

$a_c = r\omega^2 = 0.5(36) = 18 \text{ m/s}^2$ $a_t = r\alpha = 0.5(2) = 1 \text{ m/s}^2$

$|\vec{a}| = \sqrt{18^2 + 1^2} = \sqrt{325} = 18.03 \text{ m/s}^2$

Angle of net acceleration from radial direction: $\phi = \arctan\left(\frac{a_t}{a_c}\right) = \arctan\left(\frac{1}{18}\right) \approx 3.2°$

The acceleration is almost purely centripetal because $\omega$ has built up substantially.

Energy and Work in Non-Uniform Circular Motion

The work-energy theorem applies component-by-component:

• Centripetal force does no work (perpendicular to displacement).
• Tangential force does work: $W_t = \int F_t \, ds = \int F_t \, r\,d\theta$

$\frac{dK}{dt} = \vec{F} \cdot \vec{v} = F_t v$

Since centripetal force $\perp$ velocity, only the tangential component changes the kinetic energy.

Power in Circular Motion

$P = F_t v = F_t r\omega = \tau\omega$

where $\tau = F_t r$ is the torque.

Worked Example: Vertical Circle with Gravity

A ball on a string swings in a vertical circle. At angle $\theta$ from the bottom:

Tangential acceleration (due to gravity): $a_t = -g\sin\theta$

(negative because gravity decelerates the ball as it rises)

Speed as a function of angle (using energy conservation): $\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 - mgr(1 - \cos\theta)$ $v^2 = v_0^2 - 2gr(1 - \cos\theta)$

Centripetal acceleration at angle $\theta$: $a_c = \frac{v^2}{r} = \frac{v_0^2}{r} - 2g(1 - \cos\theta)$

Tension at angle $\theta$: $T = \frac{mv^2}{r} + mg\cos\theta = \frac{mv_0^2}{r} - 2mg(1-\cos\theta) + mg\cos\theta$ $T = \frac{mv_0^2}{r} - mg(2 - 3\cos\theta)$

Part 5 Summary

Concept	Formula
Tangential acceleration	$a_t = r\alpha = dv/dt$
Centripetal acceleration	$a_c = v^2/r = r\omega^2$
Total acceleration	$\vert a\vert = \sqrt{a_c^2 + a_t^2}$
Only $F_t$ does work	$P = F_t v = \tau\omega$
Vertical circle: $v^2(\theta)$	$v_0^2 - 2gr(1-\cos\theta)$

Next up: Part 6 — Problem-Solving Workshop with challenging circular motion problems.

Problem-Solving Workshop

Part 6 of 7 — Circular Motion

Strategy for Circular Motion Problems

Step	Action
1	Identify the circular path and radius $r$
2	Draw FBD at the specified point
3	Establish radial direction (toward center) and tangential direction
4	Apply $\sum F_r = mv^2/r$ (radial)
5	Apply $\sum F_t = ma_t$ (tangential, if non-uniform)
6	Use energy conservation to relate speeds at different positions

Problem 2: Conical Pendulum with Calculus

A mass $m$ hangs from a string of length $L$ and swings in a horizontal circle, making angle $\theta$ with the vertical. Show that the period is $T = 2\pi\sqrt{L\cos\theta/g}$ and find $dT/d\theta$.

Solution

From Part 1: $\omega = \sqrt{g/(L\cos\theta)}$, so:

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

Differentiating with respect to $\theta$:

$\frac{dT}{d\theta} = 2\pi \cdot \frac{1}{2}\left(\frac{L\cos\theta}{g}\right)^{-1/2} \cdot \frac{-L\sin\theta}{g}$

$\frac{dT}{d\theta} = -\pi\sin\theta\sqrt{\frac{L}{g\cos\theta}} = -\frac{\pi L \sin\theta}{\sqrt{gL\cos\theta}}$

Since $dT/d\theta < 0$ for $0 < \theta < 90°$: as the angle increases, the period decreases. A wider swing means a faster revolution (shorter period).

For small angles: $T \approx 2\pi\sqrt{L/g}$ (simple pendulum period) — this is the maximum period.

Problem 3: Bead on a Rotating Hoop

A bead slides frictionlessly on a circular hoop of radius $R$ that rotates about a vertical axis with angular speed $\omega$. Find the equilibrium angle $\theta$ from the bottom of the hoop.

Setup

The bead is at angle $\theta$ from the bottom of the hoop. Its distance from the rotation axis is $r = R\sin\theta$.

Forces on the bead:

• Normal force $N$ (perpendicular to the hoop, along the radius)
• Gravity $mg$ (downward)

Radial direction (toward rotation axis): $N\sin\theta = m\omega^2 R\sin\theta$

Vertical direction: $N\cos\theta = mg$

From the radial equation (for $\theta \neq 0$): $N = m\omega^2 R$

Substituting into vertical: $m\omega^2 R \cos\theta = mg$

$\cos\theta = \frac{g}{\omega^2 R}$

Analysis

• If $\omega^2 R > g$ (fast rotation): $\cos\theta < 1$, so $\theta > 0$. The bead rises.
• If $\omega^2 R < g$ (slow rotation): $\cos\theta > 1$ is impossible — the bead stays at the bottom ($\theta = 0$).
• Critical angular speed: $\omega_c = \sqrt{g/R}$

Above $\omega_c$ the bead rides up the hoop, and the faster the hoop spins, the closer $\theta$ gets to $90°$ (it never quite reaches it).

Workshop Summary

Key Problem Types

Type	Strategy
Flat curve	Friction = centripetal, $v_{\max} = \sqrt{\mu_s gr}$
Banked curve	Resolve $N$ and $f$ into radial and vertical
Vertical circle	Energy for speeds, N2L for forces at each position
Conical pendulum	Vertical = $mg$, horizontal = centripetal
Rotating hoop	$\cos\theta = g/(\omega^2 R)$; rises only if $\omega > \sqrt{g/R}$

Next up: Part 7 — Review & Applications, bringing it all together.

Review & Applications

Part 7 of 7 — Circular Motion

Complete Topic Reference

Concept	Formula	Part
Speed	$v = r\omega = 2\pi r/T$	1
Centripetal acceleration	$a_c = v^2/r = \omega^2 r$	1, 2
Position vector	$\vec{r} = r\cos(\omega t)\hat{x} + r\sin(\omega t)\hat{y}$	2
acceleration derivation	$\vec{a} = -\omega^2\vec{r}$	2
Design speed (banked)	$v = \sqrt{rg\tan\beta}$	3
Min speed at top of loop	$v_{\min} = \sqrt{gr}$	4
Min speed at bottom of loop	$v_{\min} = \sqrt{5gr}$	4
Tangential acceleration	$a_t = r\alpha = dv/dt$	5
Total acceleration	$\vert\vec{a}\vert = \sqrt{a_c^2 + a_t^2}$	5

Application: Centrifuge

A centrifuge of radius $R = 0.15$ m spins at $n = 3000$ RPM.

Angular velocity: $\omega = 2\pi \times \frac{3000}{60} = 100\pi \approx 314 \text{ rad/s}$

Centripetal acceleration: $a_c = \omega^2 R = (100\pi)^2(0.15) = 14\,804 \text{ m/s}^2$

In units of $g$: $\frac{a_c}{g} = \frac{14\,804}{9.8} \approx 1\,511\,g$

Application: Car on a Banked Curve

An engineer designs a highway exit ramp with $r = 200$ m for a design speed of 60 km/h (16.7 m/s).

Required banking angle: $\tan\beta = \frac{v^2}{rg} = \frac{278}{2000} = 0.139$ $\beta = \arctan(0.139) \approx 7.9°$

With $\mu_s = 0.7$, maximum safe speed: $v_{\max} = \sqrt{rg\frac{\tan\beta + \mu_s}{1 - \mu_s\tan\beta}} = \sqrt{2000\frac{0.139 + 0.7}{1 - 0.098}}$ $= \sqrt{2000 \times 0.931} = \sqrt{1862} \approx 43.1 \text{ m/s} \approx 155 \text{ km/h}$

Application: Kepler's Third Law (Preview)

For a planet of mass $m$ orbiting a star of mass $M$ in a circular orbit:

Gravitational force = centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$

$v = \sqrt{\frac{GM}{r}}$

Period: $T = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^3}{GM}}$

$\boxed{T^2 = \frac{4\pi^2}{GM}r^3}$

This is Kepler's Third Law: $T^2 \propto r^3$.

Geostationary Orbit

For a satellite orbiting with $T = 24$ hours:

$r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$

$r = \left(\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{4\pi^2}\right)^{1/3} \approx 42\,200 \text{ km}$

This is about 35,800 km above Earth's surface.

🎉 Topic Complete: Circular Motion

You've mastered the full AP Physics C treatment of circular motion:

Part	Topic	Status
1	Uniform circular motion	✅
2	Centripetal acceleration derivation	✅
3	Banked curves	✅
4	Vertical circles	✅
5	Non-uniform circular motion	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Key Takeaway: Circular motion problems require (1) identifying the center and radius, (2) applying Newton's second law in the radial direction with $\sum F_r = mv^2/r$, and (3) using energy conservation for speed relationships. The calculus-level content (derivations, angular kinematics, energy methods) is what distinguishes AP Physics C from AP Physics 1.