Circular Motion and Gravitation (Newton’s Laws on Curved Paths)

Uniform Circular Motion

Uniform circular motion is motion along a circular path at constant speed. Even though the speed stays the same, the velocity is not constant—because velocity includes direction, and the direction continuously changes as you go around the circle. That single fact explains why circular motion is such an important application of Newton’s laws: changing velocity means there must be an acceleration, and acceleration requires a net force.

What makes circular motion “circular” (and why acceleration appears)

To see why a force is required, imagine a puck gliding on perfectly frictionless ice. If nothing pulls it inward, it travels in a straight line (Newton’s first law). A circular path is “trying” to curve inward at every instant, so something must keep redirecting the velocity vector.

In uniform circular motion:

• The speed is constant.
• The velocity vector is always tangent to the circle.
• The acceleration points toward the center of the circle, changing the direction of velocity.

This inward acceleration is called centripetal acceleration (centripetal means “center-seeking”). It is not a new type of force by itself; it is an acceleration that must be produced by the net inward force.

Kinematics relationships for a circle

If an object moves on a circle of radius $r$ with angular speed $\omega$:

$v = \omega r$

Here, $v$ is tangential speed (magnitude of velocity), and $\omega$ is in radians per second.

If the period (time for one full revolution) is $T$ and the frequency is $f$:

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

$\omega = 2\pi f$

These matter on AP Physics C because problems often give revolutions per second (frequency) or period and ask for acceleration or force.

Centripetal acceleration

The magnitude of the centripetal acceleration is

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

Using $v = \omega r$, you also get

$a_c = \omega^2 r$

Direction is everything: $a_c$ points toward the center at every instant. A common slip is treating $a_c$ as tangential; it is purely radial inward for uniform circular motion.

Newton’s 2nd law in the radial direction (the big idea)

Newton’s second law says

$\sum F = ma$

For uniform circular motion, it’s most powerful when you apply it along the radial (inward) direction, because the acceleration is entirely radial. If you choose “inward” as positive, then

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

This equation does not mean there is a “centripetal force” you add to the free-body diagram. Instead:

• Draw the real forces (tension, gravity, normal force, friction, thrust, etc.).
• Add their radial components.
• Set the net radial force equal to $m v^2/r$.

A helpful way to say it:

“Centripetal force” is the name we give to the net inward force required for circular motion.

Notation and component reference (radial vs tangential)

In circular motion problems, it’s common to resolve vectors into radial (toward/away from center) and tangential (along the direction of motion) components.

Real-world forces that can supply the needed inward net force

The inward net force can come from many sources:

• Tension (ball on a string)
• Static friction (car turning on a flat road)
• Normal force components (banked curve)
• Gravity (orbiting satellites, conical pendulum components)

The physics is the same: whatever combination of real forces points inward must add up to $m v^2/r$.

Worked example 1: Car rounding a flat curve (friction provides centripetal force)

A car of mass $m$ goes around an unbanked curve of radius $r$ at speed $v$. The only horizontal force available is static friction. What minimum coefficient of static friction $\mu_s$ is required?

Concept first: On a flat road, the normal force and weight cancel vertically, so the horizontal net force must come from friction alone. To keep the car in circular motion, friction must provide the required inward net force.

Forces:

• Weight $mg$ downward
• Normal $N$ upward
• Static friction $f_s$ inward (toward center)

Vertical equilibrium:

$N - mg = 0$

So $N = mg$.

Radial Newton’s 2nd law:

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

But $f_s \le \mu_s N = \mu_s mg$, so to avoid slipping,

$m\frac{v^2}{r} \le \mu_s mg$

Cancel $m$ and solve:

$\mu_s \ge \frac{v^2}{rg}$

Common misunderstanding: Students sometimes use kinetic friction or assume friction points opposite motion. In a steady turn at constant speed, friction points inward, perpendicular to the velocity.

Worked example 2: Conical pendulum (tension components)

A mass $m$ hangs from a string of length $L$ and moves in a horizontal circle at constant speed, making an angle $\theta$ with the vertical.

Concept first: The object has inward (horizontal) centripetal acceleration but no vertical acceleration. Tension must both support the weight and provide the inward net force.

Forces:

• Tension $T$ along string
• Weight $mg$ downward

Resolve tension:

• Vertical component: $T\cos\theta$ upward
• Horizontal (radial inward) component: $T\sin\theta$ inward

Vertical force balance:

$T\cos\theta - mg = 0$

Radial force requirement:

$T\sin\theta = m\frac{v^2}{r}$

Geometrically, the radius is

$r = L\sin\theta$

From here, you can solve for $T$, $v$, or $\omega$ depending on what’s given.

A frequent mistake is setting $T = m v^2/r$ directly. Only the horizontal component of tension contributes to centripetal acceleration.

Exam Focus

• Typical question patterns:
  • Given $r$ and $v$ (or $\omega$ and $r$), find the required net force and identify which real force supplies it.
  • Free-body diagram with tension/normal/friction and a radial equation $\sum F_r = m v^2/r$.
  • Banked curve or conical pendulum setups requiring component resolution.
• Common mistakes:
  • Adding a separate “centripetal force” to the free-body diagram instead of using the net radial force.
  • Mixing up directions: using tangential direction for $m v^2/r$, or drawing friction opposite velocity instead of inward.
  • Forgetting that only components of forces may point radially inward.

Non-Uniform Circular Motion

Non-uniform circular motion occurs when an object moves along a circular path but its speed changes with time. Now the velocity changes in two ways:

1. Its direction changes (like in uniform circular motion).
2. Its magnitude changes (because speed is changing).

That means the acceleration is no longer purely radial. Instead, it has both:

• a radial (centripetal) component due to direction change
• a tangential component due to speed change

Understanding this decomposition is essential in AP Physics C because it lets you apply Newton’s second law cleanly in two perpendicular directions.

Decomposing acceleration: radial and tangential components

For motion on a circle of radius $r$:

• Radial (inward) component:

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

• Tangential component (changes the speed):

$a_t = \frac{dv}{dt}$

The tangential component points:

• in the direction of motion if the object speeds up
• opposite the direction of motion if it slows down

The total acceleration magnitude is

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

and the direction is some angle between inward and tangential.

Angular acceleration connections

If you work with angular variables, define angular acceleration $\alpha$ as

$\alpha = \frac{d\omega}{dt}$

The tangential acceleration relates to $\alpha$ by

$a_t = r\alpha$

This is especially useful when a problem gives rotational information (like a motor driving a turntable) but asks for linear acceleration at radius $r$.

Newton’s 2nd law in two perpendicular directions

In non-uniform circular motion, it’s usually best to write Newton’s second law separately:

Radial direction:

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

Tangential direction:

$\sum F_t = m\frac{dv}{dt}$

These equations do different jobs:

• The radial equation tells you what net inward force is required to “bend” the path.
• The tangential equation tells you what net force is needed to change the speed.

A subtle but important point: even if the speed changes, the radial acceleration still uses $v^2/r$ with the _instantaneous_ speed $v$.

Vertical circular motion: why tension changes around the circle

A classic AP Physics C application is an object moving in a vertical circle (like a mass on a string or a roller coaster loop). Here, gravity contributes differently to the radial equation depending on where the object is.

Take a mass $m$ on a string moving in a vertical circle of radius $r$.

• At the top of the circle, “inward” points downward. Both tension $T$ (downward along the string) and weight $mg$ (downward) point inward, so:

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

• At the bottom, “inward” points upward. Tension points inward (upward), but weight points outward (downward), so:

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

This explains why the string tension is usually larger at the bottom than at the top for the same speed.

A common misconception is that “centripetal force” is always tension in a string problem. In vertical motion, the net inward force may be tension plus gravity (top) or tension minus gravity (bottom).

Worked example 1: Speeding up a mass on a string (radial and tangential forces)

A mass $m$ moves on a horizontal circle of radius $r$. At some instant its speed is $v$ and it is speeding up at rate $dv/dt$. The string tension provides the radial force, and a small tangential push provides the tangential force.

Concept first: Because the motion is in a horizontal plane, you can treat radial and tangential as perpendicular directions in that plane. Tension points radially inward; the push is tangential.

Radial equation (tension supplies centripetal requirement):

$T = m\frac{v^2}{r}$

Tangential equation (push changes speed):

$F_t = m\frac{dv}{dt}$

Key takeaway: The force that changes direction (radial) and the force that changes speed (tangential) are independent components. Students often try to combine them into one equation too early and lose the geometry.

Worked example 2: Minimum speed at the top of a vertical circle (string just stays taut)

A mass $m$ moves in a vertical circle of radius $r$ on a string. What minimum speed at the top keeps the string from going slack?

Concept first: A slack string means $T = 0$. At the top, inward is downward, and gravity already points inward. The minimum speed occurs when gravity alone provides exactly the required centripetal acceleration.

At the top:

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

Set $T = 0$ for the “just taut” condition:

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

Solve for $v$:

$v = \sqrt{gr}$

Common mistake: Using $T - mg$ at the top (sign error). Always decide what direction is “inward” at the point you’re analyzing.

How to avoid sign errors: choose directions deliberately

Non-uniform circular motion problems become much easier when you adopt a consistent method:

1. Choose radial inward as positive for $\sum F_r$.
2. Choose tangential positive in the direction of motion for $\sum F_t$.
3. Project each real force onto those directions at the instant of interest.

If you skip step 3, it’s easy to accidentally treat a force as fully radial when it actually has both radial and tangential components.

Exam Focus

• Typical question patterns:
  • Vertical circle questions asking for tension at top/bottom/side using $\sum F_r = m v^2/r$ with correct signs.
  • Problems where speed changes: use both $\sum F_r$ and $\sum F_t$ (two equations, two unknowns).
  • Given $\alpha$ or changing $\omega$, find tangential acceleration using $a_t = r\alpha$.
• Common mistakes:
  • Treating $a_t$ and $a_r$ as the same thing, or using $a = v^2/r$ even when speed is changing.
  • Using the wrong inward direction at the top vs bottom of a vertical circle, leading to incorrect tension equations.
  • Plugging $v = \omega r$ without consistent units (degrees per second instead of radians per second).

Newtonian Gravitation and Circular Orbits

Newton’s law of universal gravitation describes the attractive force between any two masses. In many AP Physics C problems, gravity supplies (all or part of) the centripetal force that keeps an object in orbit, making gravitation a direct application of circular motion plus Newton’s second law.

The gravitational force law (what it is and what it means)

For two point masses $m_1$ and $m_2$ separated by distance $r$, the magnitude of the gravitational force is

$F_g = \frac{G m_1 m_2}{r^2}$

• $G$ is the universal gravitational constant.
• The force acts along the line connecting the masses.
• The force is always attractive.

This inverse-square dependence is why gravity weakens quickly with distance, and why orbital motion depends strongly on altitude.

Gravitational field and the meaning of $g$

Near a planet of mass $M$, an object of mass $m$ experiences gravitational force

$F_g = \frac{G M m}{r^2}$

If you divide by $m$, you get the gravitational acceleration magnitude at distance $r$ from the planet’s center:

$g = \frac{G M}{r^2}$

This is a powerful idea: $g$ is not “always $9.8$.” It depends on $r$. (Near Earth’s surface, $r$ is approximately Earth’s radius, so $g$ is approximately constant in many problems.)

Circular orbits: gravity as the centripetal agent

For a satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$, the inward net force needed is $m v^2/r$. If gravity is the only significant force, then

$\frac{G M m}{r^2} = m\frac{v^2}{r}$

Cancel $m$ and solve for orbital speed:

$v = \sqrt{\frac{G M}{r}}$

This result shows:

• Higher orbits (larger $r$) have smaller orbital speed.
• Orbital speed depends on the central mass $M$, not on the satellite mass $m$.

Orbital period (a bridge to Kepler’s third law)

The period $T$ of a circular orbit is circumference divided by speed:

$T = \frac{2\pi r}{v}$

Substitute $v = \sqrt{GM/r}$:

$T = 2\pi\sqrt{\frac{r^3}{G M}}$

The proportionality $T^2 \propto r^3$ is the same relationship as Kepler’s third law for circular orbits, derived here directly from Newton’s laws.

Worked example: Orbital speed and period at a given radius

A satellite orbits Earth in a circular orbit of radius $r$ from Earth’s center. Using Earth’s mass $M_E$, find expressions for speed and period.

Concept first: Gravity supplies the centripetal force. Set gravitational force equal to the required net inward force.

Start with

$\frac{G M_E m}{r^2} = m\frac{v^2}{r}$

Solve:

$v = \sqrt{\frac{G M_E}{r}}$

Then

$T = \frac{2\pi r}{v}$

Substitute for $v$:

$T = 2\pi\sqrt{\frac{r^3}{G M_E}}$

Common mistake: Using $r$ as altitude above Earth’s surface rather than distance from Earth’s center. If altitude is $h$ and Earth’s radius is $R_E$, then the orbital radius is $r = R_E + h$.

Connecting back to circular motion skills

Orbit problems are circular motion problems in disguise:

• Identify the inward direction (toward the planet).
• Use $\sum F_r = m v^2/r$.
• Recognize that gravity is already radial, so it often drops in cleanly.

But be careful: in more complex scenarios (like a satellite with thrust, or an object inside a tunnel through Earth), gravity may not be the only force, or may not be constant.

Exam Focus

• Typical question patterns:
  • “Derive the orbital speed” or “derive the period” by equating gravity to centripetal requirement.
  • Compare two circular orbits (different radii) and relate speeds or periods using the formulas.
  • Compute $g$ at altitude using $g = GM/r^2$ and connect it to orbital motion.
• Common mistakes:
  • Confusing orbital radius $r$ with altitude, forgetting $r = R + h$.
  • Assuming $g$ is constant in orbital contexts where $r$ changes significantly.
  • Treating the satellite’s mass as affecting orbital speed (it cancels out in circular orbit equations).