2.9 Circular Motion

Circular motion happens when an object moves along a curved path, and centripetal acceleration ($a_c = v^2/r$) always points toward the center to keep it turning. In AP Physics 1, the key move is identifying which real forces supply the inward net force.

AP Physics Circular Motion

In AP Physics 1, circular motion means an object has an inward, center-seeking component of acceleration called centripetal acceleration. Its magnitude is $a_c = v^2/r$, and the inward net force is supplied by real forces like gravity, tension, friction, or the normal force.

The big exam idea is that "centripetal force" is not a separate force to draw. Start with a free-body diagram, choose inward as the radial direction, and set the net inward force equal to $m v^2/r$.

Why This Matters for the AP Physics 1 Exam

Circular motion ties together almost everything from Unit 2: free-body diagrams, Newton's second law, gravity, friction, and tension. The key move is recognizing that "centripetal force" is not a new force. It is the net force from real forces (gravity, normal force, tension, friction) pointing toward the center.

This topic rewards you for translating between words, diagrams, and equations. You might describe why a car needs friction to turn, then derive the speed it can handle, then connect that math back to your explanation. That kind of reasoning across representations shows up in the qualitative-quantitative translation free-response question, which can pull content from any unit. Getting comfortable explaining circular motion conceptually, not just plugging into $a_c = v^2/r$, sets you up for both multiple-choice and free-response success.

Key Takeaways

• Centripetal acceleration points toward the center and equals $a_c = v^2/r$. It changes direction, not speed.
• "Centripetal force" is the net force directed toward the center, supplied by real forces like gravity, normal force, tension, or friction.
• Tangential acceleration changes an object's speed and points along the path. The net acceleration is the vector sum of centripetal and tangential parts.
• Period and frequency are reciprocals: $T = 1/f$. For constant-speed circular motion, $T = 2\pi r / v$.
• At the top of a vertical loop with minimum speed, gravity alone supplies the centripetal force, giving $v = \sqrt{gr}$.
• For a circular orbit, gravity provides the centripetal acceleration, leading to Kepler's third law: $T^2 = \frac{4\pi^2}{GM}R^3$.

Centripetal Acceleration

Centripetal acceleration keeps an object moving in a circle by constantly changing the direction of its velocity toward the center. Without a net force toward the center, the object would travel in a straight line, following Newton's first law.

• It is directed toward the center of the circular path and is a component of the object's total acceleration.
• Its magnitude is the tangential speed squared divided by the radius: $a_c = \frac{v^2}{r}$
• It always points perpendicular to the object's instantaneous velocity.
• It can come from a single force, several forces, or components of forces acting on the object.

A useful warning: there is no separate "centripetal force." When you draw a free-body diagram, you only draw the actual forces present (gravity, normal, tension, friction). The net of those forces, pointing toward the center, is what causes centripetal acceleration.

Forces that supply centripetal acceleration

• Top of a vertical loop: An object needs a minimum speed to stay on the circular path. At that minimum speed, gravity is the only force providing the centripetal acceleration, giving $v = \sqrt{gr}$
• Banked curves: Components of the normal force and static friction can point toward the center. For this course, you only solve banked curves quantitatively in the ideal case where no friction is needed; there, the horizontal component of the normal force alone provides the centripetal acceleration. If friction is required, you only describe it qualitatively.
• Conical pendulum: A horizontal component of the string tension supplies the centripetal acceleration while the vertical component balances gravity.

Tangential Acceleration and Net Acceleration

Tangential acceleration is the rate at which an object's speed changes, and it points tangent to the circular path along the direction of motion. When an object speeds up or slows down as it goes around a circle, tangential acceleration is present. This produces nonuniform circular motion.

The net acceleration of an object moving in a circle is the vector sum of the centripetal and tangential components:

• Centripetal part: perpendicular to velocity, changes direction only.
• Tangential part: parallel to velocity, changes speed only.
• Both present: the net acceleration points at an angle between the center and the tangent line, depending on the relative sizes of the two parts.

If a car speeds up while rounding a curve, it has both centripetal acceleration (turning) and tangential acceleration (speeding up) at the same time.

Period and Frequency

Uniform circular motion happens when an object moves at constant speed around a circle. You can describe it using period and frequency.

• Period ($T$) is the time to complete one full revolution, measured in seconds.
• Frequency ($f$) is the number of revolutions per unit time, measured in hertz (Hz).

They are reciprocals of each other: $T = \frac{1}{f}$

For an object moving at constant speed in a circle, the period relates to radius and speed: $T = \frac{2\pi r}{v}$

This shows that a larger circle takes more time per revolution at the same speed, while a faster object completes each revolution in less time.

Kepler's Third Law for Circular Orbits

For a satellite in a circular orbit, gravity is the only force directed toward the center, so gravity provides the entire centripetal acceleration. Because of this, the orbital period depends on the orbit radius and the mass of the central body:

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

Where:

• $T$ is the orbital period
• $R$ is the orbital radius (measured from the center of the central body)
• $M$ is the mass of the central body
• $G$ is the universal gravitational constant

This applies to planets orbiting the Sun, moons orbiting planets, and satellites orbiting Earth. Objects farther out take longer to orbit, since $T^2$ is proportional to $R^3$. For this course, you are not expected to know Kepler's first or second laws.

How to Use This on the AP Physics 1 Exam

Problem Solving

• Identify the radius and speed first, then apply $a_c = v^2/r$. Watch units; convert diameters to radii and km to m.
• For orbits, remember $R$ is measured from the center of the central body, so add the body's radius to the altitude.
• Use $T = 2\pi r / v$ to connect speed and timing, and $T = 1/f$ to switch between period and frequency.

Free Response

• Always start with a free-body diagram showing only real forces. Then write Newton's second law with the net force toward the center set equal to $m v^2 / r$.
• When asked to explain, say which specific force or force component points toward the center. Avoid labeling an arrow "centripetal force" on a diagram.
• For a qualitative-quantitative translation style question, first explain in words why the object turns, then derive the equation, then connect your math back to your explanation.

Common Trap

• Setting up a vertical loop: at the top, both gravity and the normal force can point downward (toward the center). The minimum-speed case is when the normal force drops to zero, leaving gravity alone.

Common Misconceptions

• "Centripetal force is its own force." It is not. It is the name for the net force pointing toward the center, supplied by gravity, tension, normal force, or friction.
• "Centripetal acceleration speeds the object up." It only changes direction. Speed changes come from tangential acceleration.
• "There is an outward force pushing you out of a turn." The feeling of being pushed outward is inertia. The actual net force points inward.
• "At the top of a loop, something pushes the object up." Gravity and the normal force both point downward there. The object stays on the path because that downward net force matches the required $m v^2 / r$.
• "In Kepler's third law, R is the altitude above the surface." $R$ is the distance from the center of the central body, so you must add the body's radius to the altitude.
• "Heavier satellites orbit slower." The orbital period depends on $R$ and the central body's mass $M$, not on the satellite's own mass.

Practice Problem 1: Centripetal Acceleration

Solution

Use $a_c = \frac{v^2}{r}$:

$a_c = \frac{(15 \text{ m/s})^2}{100 \text{ m}} = \frac{225 \text{ m}^2/\text{s}^2}{100 \text{ m}} = 2.25 \text{ m/s}^2$

The centripetal acceleration is $2.25 \text{ m/s}^2$ toward the center of the circle.

Practice Problem 2: Period and Frequency

Solution

a) The period is the time for one full revolution.

Period ($T$) = 60 seconds

b) Frequency is the reciprocal of the period:

$f = \frac{1}{T} = \frac{1}{60 \text{ s}} = 0.0167 \text{ Hz}$

c) Find the radius, then the linear speed.

Radius ($r$) = diameter/2 = 40 m / 2 = 20 m

$v = \frac{2\pi r}{T} = \frac{2\pi \times 20 \text{ m}}{60 \text{ s}} = \frac{40\pi \text{ m}}{60 \text{ s}} \approx 2.09 \text{ m/s}$

The Ferris wheel has a period of 60 seconds, a frequency of 0.0167 Hz, and passengers move at about 2.09 m/s.

Practice Problem 3: Kepler's Third Law

Solution

First find the orbital radius, measured from Earth's center:

Orbital radius ($R$) = Earth's radius + satellite altitude $R$ = 6370 km + 2000 km = 8370 km = 8.37 × 10⁶ m

Apply Kepler's third law: $T^2 = \frac{4\pi^2}{GM}R^3$

Substitute the values: $T^2 = \frac{4\pi^2(8.37 \times 10^6\,\text{m})^3}{(6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2)(5.97 \times 10^{24}\,\text{kg})}$

Compute the denominator: $(6.67 \times 10^{-11})(5.97 \times 10^{24}) = 3.98 \times 10^{14}$

Compute the orbital radius cubed: $(8.37 \times 10^6)^3 = 5.86 \times 10^{20}\,\text{m}^3$

Then: $T^2 = \frac{4\pi^2(5.86 \times 10^{20})}{3.98 \times 10^{14}} \approx 5.80 \times 10^7\,\text{s}^2$

Take the square root: $T = \sqrt{5.80 \times 10^7}\,\text{s} \approx 7.62 \times 10^3\,\text{s} \approx 7619\,\text{s}$

Convert to minutes: $T = 7619\,\text{s} \div 60 \approx 127.0\,\text{minutes}$

The satellite completes one orbit in about 127 minutes, or roughly 2 hours and 7 minutes.

Related AP Physics 1 Guides

• 2.1 Systems and Center of Mass
• 2.2 Forces and Free-Body Diagrams
• 2.3 Newton's Third Law
• 2.4 Newton's First Law
• 2.5 Newton's Second Law
• 2.6 Gravitational Force