Unit 2 · Topic 2.10

2.10 Circular Motion

An object moving in a circle accelerates toward the center even at constant speed, with $a_c = \frac{v^2}{r}$ . Real forces such as gravity, tension, normal force and friction must provide that inward net force. If the speed changes, there's also a tangential acceleration, and for circular orbits gravity alone supplies the centripetal force, leading to Kepler's third law.

Key terms

centripetal acceleration
tangential acceleration
period
frequency
banked curve
Kepler's third law

Centripetal acceleration

When an object moves in a circle of radius r at speed v, its velocity keeps changing direction, so it's accelerating. That acceleration points toward the center of the circle and has size:

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

"Centripetal" just means "center-seeking". The time for one revolution is the period T, and the frequency is f = 1/T, revolutions per second. For uniform circular motion, $v = \frac{2\pi r}{T}$ .

What provides the centripetal force

Newton's second law toward the center is $\sum F_{in} = \frac{mv^2}{r}$ . The left side is built from real forces on your FBD; $\frac{mv^2}{r}$ is not an extra force.

A car on a flat curve: static friction toward the center. Maximum speed $v = \sqrt{\mu_sgr}$ .
A car on a banked curve with no friction: the horizontal component of the normal force. $\tan\theta = \frac{v^2}{rg}$ . If the bank also has friction, components of both static friction and the normal force point toward the center, so cars can go a range of speeds without sliding.
A conical pendulum (a mass swinging in a horizontal circle on a string): the horizontal component of tension, while the vertical component balances the weight.
A ball on a string in a vertical circle: tension plus the component of gravity toward the center, which changes around the loop.
A satellite in a circular orbit: gravity alone.

Vertical circles and changing speed

In a vertical circle, gravity speeds the object up on the way down and slows it on the way up, so the speed changes. At the top, both tension and gravity point toward the center: $T + mg = \frac{mv^2}{r}$ . At the bottom, tension points in and gravity out: $T - mg = \frac{mv^2}{r}$ , so the tension is greatest at the bottom.

At the top, the string goes slack if $\frac{mv^2}{r} < mg$ . The minimum speed to stay in the circle there is $v = \sqrt{gr}$ .

Whenever the speed changes, there's also a tangential acceleration, $a_t = \frac{dv}{dt}$ , along the direction of motion. The total acceleration is the vector sum of the two perpendicular parts: $a = \sqrt{a_c^2 + a_t^2}$ .

Circular orbits and Kepler's third law

For a satellite of mass m in a circular orbit of radius r around a body of mass M, gravity is the centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r} \quad\Rightarrow\quad v = \sqrt{\frac{GM}{r}}$

The satellite's mass cancels. Substituting $v = \frac{2\pi r}{T}$ gives Kepler's third law, $T^2 = \frac{4\pi^2}{GM}r^3$ : the square of the period is proportional to the cube of the orbit's radius. Farther satellites move more slowly and take longer to go around. You'll use this third law. Elliptical orbits come back in 6.6, where conservation of energy and angular momentum do the work, so you don't need Kepler's first and second laws by name.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Maximum speed on a flat curve
A car takes a flat curve of radius 50 m. The coefficient of static friction between tires and road is 0.80. Find the maximum speed without skidding. Use g = 9.8 m/s².
Show the solutionHide the solution
1. Step 1: Static friction provides the centripetal force (the tires aren't sliding sideways). The normal force is mg on a flat road.
2. Step 2: At the maximum speed, friction is at its limit: $\mu_smg = \frac{mv^2}{r}$ .
3. Step 3: The mass cancels: $v = \sqrt{\mu_sgr} = \sqrt{(0.80)(9.8)(50)} \approx 19.8$ m/s.
4. Step 4: For comparison, a frictionless curve banked for 20 m/s on an 80 m radius needs $\tan\theta = \frac{20^2}{(80)(9.8)}$ , about 27°.
Answer: About 19.8 m/s (around 71 km/h).
Example 2Calculator allowed
Tension at the top of a vertical circle (classic trap)
A 0.50 kg ball on a 0.80 m string swings in a vertical circle. At the top it moves at 4.0 m/s. Find the tension there, and the minimum speed at the top for the string to stay taut. Use g = 9.8 m/s².
Show the solutionHide the solution
1. Step 1: At the top, tension and gravity both point down, toward the center: $T + mg = \frac{mv^2}{r}$ .
2. Step 2: $T = \frac{(0.50)(4.0)^2}{0.80} - (0.50)(9.8) = 10 - 4.9 = 5.1$ N.
3. Step 3: Minimum speed: set T = 0, so $mg = \frac{mv^2}{r}$ and $v = \sqrt{gr} = \sqrt{(9.8)(0.80)} = 2.8$ m/s.
4. Step 4: The trap is drawing a separate "centripetal force" arrow on the FBD and adding it to tension and weight. Only tension and weight act; their sum is the centripetal force.
Answer: T = 5.1 N; minimum speed 2.8 m/s.
Example 3Calculator allowed
Period of a high orbit
A satellite orbits Earth in a circle of radius $4.22 \times 10^7$ m. Earth's mass is $5.97 \times 10^{24}$ kg. Find the orbital period and speed.
Show the solutionHide the solution
1. Step 1: Kepler's third law: $T = 2\pi\sqrt{\frac{r^3}{GM}}$ .
2. Step 2: $GM = (6.67 \times 10^{-11})(5.97 \times 10^{24}) \approx 3.98 \times 10^{14}$ m³/s².
3. Step 3: $T = 2\pi\sqrt{\frac{(4.22 \times 10^7)^3}{3.98 \times 10^{14}}} \approx 8.63 \times 10^4$ s, which is about 24.0 hours. This is the geosynchronous orbit used by many communications satellites: one placed above the equator keeps pace with Earth's rotation and stays over the same spot.
4. Step 4: Speed: $v = \sqrt{\frac{GM}{r}} \approx 3.07 \times 10^3$ m/s, about 3.1 km/s.
Answer: About $8.6 \times 10^4$ s (24 hours), at about 3.1 km/s.

Common mistakes

Drawing "centripetal force" or $\frac{mv^2}{r}$ as its own arrow on a free-body diagram.
Saying an object moving in a circle at constant speed has zero acceleration. Its direction changes, so it accelerates toward the center.
Thinking an object released from circular motion flies straight outward. It moves off along the tangent, in a straight line.
Expecting a satellite's mass to change its orbital speed or period. It cancels.

On the exam

Expect derivations of a maximum or minimum speed (flat curve, banked curve, top of a loop) in terms of r, g and μ or θ. Write the inward net force from the FBD and set it equal to $\frac{mv^2}{r}$ .
Lab questions sometimes ask you to graph $T^2$ against $r^3$ (or $v^2$ against r) to get a straight line, then use the slope to find a mass or g.

Connected topics

Videos

Topic 2.10 - Circular Motion
Lessons With LondotWatch on YouTube (opens in a new tab)
Centripetal force | Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Centripetal Acceleration Introduction
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C Mechanics - Unit 2 - Lesson 15C - Banked Roads
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
High School Physics - Vertical Circular Motion
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
Kepler's Third Law Derivation
Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 2.10 Circular Motion. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A 2.0 kg ball moves in a horizontal circle of radius 2.0 m at a constant 4.0 m/s. What is the net force on the ball?

Question 2 of 4Calculator allowed

A curve of radius 80 m is to be banked so that a car moving at 20 m/s needs no friction to stay on the road. What bank angle above the horizontal is needed? Use g = 10 m/s².

Question 3 of 4Calculator allowed

A roller coaster car goes around the inside of a vertical loop of radius 10 m. What is the slowest speed it can have at the top and still keep contact with the track? Use g = 10 m/s².

Question 4 of 4Calculator allowed

Satellite B orbits a planet in a circle with 4 times the radius of satellite A's circular orbit. How does B's orbital period compare with A's?

0 of 4 answered