AP Physics 1 Unit 2 Notes: Circular Motion and the Radial Force Model

Describing Circular Motion (What moves in a circle and how you describe it)

Circular motion is motion where an object’s path curves around a center point. In AP Physics 1, you focus most on uniform circular motion (UCM): motion in a circle at constant speed. Even though the speed is constant, the motion is still accelerated—because the velocity is changing direction continuously.

This is one of the most important “mind shifts” in mechanics: acceleration is not only about speeding up or slowing down. Any change in velocity—change in magnitude or direction—requires acceleration.

To describe circular motion, you need a few geometric quantities:

• Radius r: distance from the center of the circle to the object.
• Period T: time for one full revolution.
• Frequency f: number of revolutions per second.
• Speed v: how fast the object moves along the circular path (tangent to the circle).

A key “translation” skill is converting between “one trip around the circle” language and linear distance. One revolution covers the circumference:

\text{distance per revolution} = 2\pi r

So if it takes time T to go that distance, the speed is:

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

Also, frequency and period are reciprocals:

f = \frac{1}{T}

Notation connections you should be fluent with

AP Physics 1 sometimes uses angular ideas informally (especially period/frequency), even if you do not do full calculus-based rotational dynamics yet.

Even if \omega isn’t emphasized in every classroom, it’s a convenient connector: if something goes around faster, both \omega and v increase.

Example 1: Speed from radius and period

A cart moves in a circle of radius r = 2.0\ \text{m} and completes one revolution every T = 1.6\ \text{s}. Find its speed.

1. Use circumference and period:

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

2. Substitute:

v = \frac{2\pi(2.0)}{1.6}

3. Compute:

v \approx 7.85\ \text{m/s}

What to notice: The speed depends on both how big the circle is (bigger r means farther per lap) and how quickly laps happen (smaller T means faster).

Exam Focus

• Typical question patterns:
  • Given r and T (or f), find v.
  • Given speed and radius, find period or frequency.
  • Interpret what “constant speed” does (and does not) imply about acceleration.
• Common mistakes:
  • Treating constant speed as zero acceleration (forgetting direction change).
  • Using diameter instead of radius in 2\pi r.
  • Mixing up T and f (remember: period is time per cycle).

Centripetal Acceleration (Why you accelerate even at constant speed)

In uniform circular motion, the velocity vector is always tangent to the circle, but it continually rotates. The acceleration that changes the direction of velocity points toward the center of the circle.

Centripetal acceleration is the inward acceleration required to keep an object moving in a circular path. “Centripetal” means “center-seeking.”

What it is

For an object moving at speed v in a circle of radius r, the magnitude of centripetal acceleration is:

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

• a_c is centripetal acceleration (m/s²)
• v is tangential speed (m/s)
• r is radius (m)

Why it matters

This relationship is the bridge between motion (kinematics-style quantities like v and r) and forces (Newton’s laws). Once you know the required inward acceleration, Newton’s Second Law tells you the required net inward force.

How it works (conceptually)

Imagine taking a snapshot of the velocity vector at two nearby points on the circle. Each vector has the same length (same speed) but a different direction. The change in velocity points inward. That inward change per unit time is what acceleration measures.

A very common misconception is: “The acceleration points along the motion because the object is moving forward.” In UCM, the acceleration is perpendicular to the velocity at every instant.

Alternative forms using period

If you know the period instead of speed, substitute v = \frac{2\pi r}{T} into a_c = \frac{v^2}{r}:

a_c = \frac{4\pi^2 r}{T^2}

This is useful when the problem describes “revolutions per second” or “time for one lap.”

Example 2: Centripetal acceleration from speed

A ball on a string moves at v = 6.0\ \text{m/s} in a circle of radius r = 1.5\ \text{m}. Find a_c.

1. Use centripetal acceleration:

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

2. Substitute:

a_c = \frac{(6.0)^2}{1.5}

3. Compute:

a_c = 24\ \text{m/s}^2

What to notice: Because v is squared, doubling speed quadruples the required centripetal acceleration (and therefore the required inward net force).

Exam Focus

• Typical question patterns:
  • Compute a_c from v and r (or from T and r).
  • Compare scenarios (what happens to a_c if v doubles or r triples?).
  • Direction questions: identify inward acceleration and tangential velocity directions on a diagram.
• Common mistakes:
  • Forgetting that a_c points toward the center (not along the path).
  • Treating a_c as a separate “extra” acceleration you add in a random direction rather than a component of net acceleration.
  • Algebra slips with the squared speed in \frac{v^2}{r}.

Centripetal Force and Newton’s Second Law (The “radial force model”)

The most important idea in AP Physics 1 circular motion is that there is no special new force called “centripetal force.”

Centripetal force is a name for the net force directed toward the center that produces centripetal acceleration.

What it is

Newton’s Second Law in the radial (inward) direction is:

\sum F_{\text{radial}} = m\frac{v^2}{r}

• \sum F_{\text{radial}} is the net inward force (N)
• m is mass (kg)
• v is speed (m/s)
• r is radius (m)

Why it matters

This is how you solve almost every AP circular-motion force problem:

1. Choose a direction as “radially inward” (toward the center).
2. Draw a free-body diagram (real forces only: tension, weight, normal, friction, etc.).
3. Add up the inward components of those real forces.
4. Set that net inward force equal to m\frac{v^2}{r}.

The circle is the path; the inward net force is what bends the path.

How it works (step-by-step problem strategy)

When you see “moves in a circle” in a forces unit, your brain should immediately do two things:

• Identify the center of the circular path, and draw an arrow from the object to the center (that’s inward).
• Separate radial vs tangential directions.

In uniform circular motion, the tangential acceleration is zero, so the net force tangent to the path is zero. But the net radial force is generally not zero.

A subtle but crucial point: the object might have several forces acting, and some can point outward. That’s fine. What matters is the sum in the radial direction.

Example 3: Required inward net force

A 0.50\ \text{kg} object moves at 4.0\ \text{m/s} in a circle of radius 2.0\ \text{m}. What net inward force is required?

1. Use radial Newton’s Second Law:

F_{\text{net, inward}} = m\frac{v^2}{r}

2. Substitute:

F_{\text{net, inward}} = 0.50\frac{(4.0)^2}{2.0}

3. Compute:

F_{\text{net, inward}} = 4.0\ \text{N}

What to notice: This is not telling you which force is 4.0 N. It’s telling you the net inward force must be 4.0 N. In a real setup, that inward net might come from tension, friction, a component of the normal force, or a combination.

Exam Focus

• Typical question patterns:
  • “What force provides the centripetal force?” (Answer: identify which real force(s) point inward.)
  • Use a free-body diagram and \sum F_{\text{radial}} = m\frac{v^2}{r} to solve for tension, normal force, or friction.
  • Concept questions about what happens if the inward net force is removed (the object moves tangent in a straight line).
• Common mistakes:
  • Drawing “centripetal force” as an extra force on the free-body diagram.
  • Setting one force equal to m\frac{v^2}{r} without checking if multiple forces contribute inward/outward.
  • Confusing “inward” with “up” (inward depends on the circle’s center, not gravity).

Tension, Normal Force, and Apparent Weight in Circular Motion (Common force providers)

In many AP problems, circular motion happens because a familiar contact force or tension continually redirects the object.

Horizontal circle on a string (tension provides inward net force)

If an object swings in a horizontal circle (like a ball on a string above your head), the string’s tension usually has an inward component that supplies the required centripetal acceleration.

If the motion is a perfectly horizontal circle at constant height, then forces must also balance vertically (no vertical acceleration). This often leads to two equations:

• Vertical equilibrium: net vertical force equals zero.
• Radial inward: net inward force equals m\frac{v^2}{r}.

A common case is the conical pendulum, where the string makes an angle with the vertical and the bob moves in a horizontal circle.

Conical pendulum relationships

Let tension be T and the string angle from vertical be \theta.

Vertical balance:

T\cos\theta = mg

Radial inward:

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

This pair shows a deep idea: the same tension force both holds the object up (vertical component) and curves its path (horizontal inward component).

Example 4: Conical pendulum speed

A 0.20\ \text{kg} bob on a string moves in a horizontal circle of radius 0.60\ \text{m}. The string makes an angle \theta = 30^\circ from vertical. Find the speed v.

1. Start with the two component equations:

T\cos\theta = mg

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

2. Divide the radial equation by the vertical equation to eliminate T and m:

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

3. Solve for v:

v = \sqrt{rg\tan\theta}

4. Substitute values (using g = 9.8\ \text{m/s}^2):

v = \sqrt{(0.60)(9.8)\tan 30^\circ}

5. Compute:

v \approx 1.84\ \text{m/s}

What to notice: The mass canceled. For this ideal conical pendulum, the speed depends on geometry and gravity, not on the bob’s mass.

Vertical circle (apparent weight changes)

When an object moves in a vertical circle (like a roller coaster loop or a bucket swung in a circle), gravity is sometimes inward, sometimes outward, depending on where you are on the circle.

This leads to “apparent weight” effects. If you’re in a roller-coaster seat, the normal force from the seat is what you feel as your weight. In circular motion, that normal force can change dramatically with position and speed.

Bottom of a vertical circle

At the bottom, inward is upward (toward the center). Forces on a rider typically include normal force N upward and weight mg downward.

Radial (inward) equation:

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

So:

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

This shows why you feel “heavier” at the bottom when moving fast.

Top of a vertical circle

At the top, inward is downward. Both weight and the normal force (if you are still in contact) can point inward.

Radial (inward) equation (inward is downward):

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

So:

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

If N becomes zero, you are just barely maintaining contact (the seat is not pushing on you). That “just-contact” condition is common in AP questions.

Example 5: Minimum speed at the top to maintain contact

A car goes over the top of a circular hill of radius r = 15\ \text{m}. What minimum speed keeps the car in contact with the road at the top?

1. At the top, inward is downward. Forces: weight mg downward, normal N upward (which is outward, opposite inward). For just contact, set N = 0.

2. Radial inward equation at the top with N = 0:

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

3. Solve for v:

v = \sqrt{rg}

4. Substitute:

v = \sqrt{(15)(9.8)}

5. Compute:

v \approx 12\ \text{m/s}

What to notice: Again, mass cancels. The minimum speed depends on radius and gravity.

Exam Focus

• Typical question patterns:
  • Conical pendulum: use components of tension to find speed, angle, or tension.
  • Vertical circle: find normal force at top/bottom, or find minimum speed for contact.
  • “Apparent weight” phrasing: interpret apparent weight as the normal force.
• Common mistakes:
  • Using mg = m\frac{v^2}{r} at the bottom (that condition applies at the top for just-contact, not generally at the bottom).
  • Getting inward direction wrong: “inward” changes around the circle.
  • Forgetting that normal force can be zero (contact lost) but cannot be negative in a simple contact model.

Friction and Banked Curves (Circular motion without strings)

Objects often move in circles because the ground (or a track) pushes on them. Two common ways a surface can provide inward net force are:

• Static friction on a flat curve
• A component of the normal force on a banked curve

Flat curve: friction supplies the centripetal force

On a level (unbanked) curve, the only horizontal force available is typically static friction. If a car of mass m goes around a flat curve of radius r at speed v, the required inward net force is:

F_{\text{net, inward}} = m\frac{v^2}{r}

If friction is the only inward force, then:

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

But static friction has a maximum:

f_s \le \mu_s N

On level ground, N = mg, so the “no-slip” condition becomes:

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

Mass cancels, leaving a maximum safe speed:

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

This is a powerful result: on a flat curve, it’s the friction coefficient, radius, and gravity that set the speed limit, not the car’s mass.

Example 6: Maximum speed on a flat curve

A car rounds a flat curve of radius r = 50\ \text{m}. The coefficient of static friction is \mu_s = 0.40. Find v_{\text{max}}.

1. Use:

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

2. Substitute:

v_{\text{max}} = \sqrt{(0.40)(9.8)(50)}

3. Compute:

v_{\text{max}} = 14\ \text{m/s}

What to notice: If you double the radius, the maximum speed increases by a factor of \sqrt{2}, not by 2.

Banked curve: normal force contributes inward

On a banked curve, the road is tilted. Then the normal force is not straight up; it is perpendicular to the road surface and has a horizontal component that can point toward the center of the circular path.

Conceptually, banking is useful because it can reduce reliance on friction. At a particular “design speed,” the curve can be negotiated with no friction needed (idealized).

If friction is neglected, the bank angle \theta and speed relate (one common form) by:

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

This resembles the conical pendulum result because the geometry of components is similar: you’re using components of a force to supply both vertical balance and horizontal inward acceleration.

Exam Focus

• Typical question patterns:
  • Flat curve: find v_{\text{max}} or \mu_s using the radial force model.
  • Identify which way static friction points (it points in whatever direction prevents slipping).
  • Banked curve (often qualitative or algebraic): relate angle, speed, and radius.
• Common mistakes:
  • Using kinetic friction automatically; most “no slipping” turning problems require static friction.
  • Assuming friction always points opposite the direction of motion; in turning, friction often points sideways (toward the center).
  • Forgetting to check the limit condition f_s \le \mu_s N (many students set f_s = \mu_s N even when not at the threshold of slipping).

Non-Uniform Circular Motion: Radial vs Tangential Acceleration (When speed changes too)

Not all circular motion is uniform. If the object’s speed changes while moving along a circular path, you have non-uniform circular motion.

The key idea: acceleration can have two perpendicular components

The net acceleration can be thought of as the sum of:

• Radial (centripetal) acceleration: points inward, changes direction of velocity.
• Tangential acceleration: points along the tangent, changes the speed.

The radial component is still:

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

Tangential acceleration is related to how quickly the speed changes (similar to straight-line motion along the path). If you know speed changes from v_i to v_f over time \Delta t, a common algebra-based approximation is:

a_t = \frac{v_f - v_i}{\Delta t}

The total acceleration magnitude (since these components are perpendicular) is:

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

Why it matters

AP problems sometimes include wording like “speeding up as it turns” or show a curved path with changing spacing between dots. If speed increases, you still need inward force to curve the motion, but you also need a tangential net force to increase speed.

A frequent misconception is to treat centripetal acceleration as the “whole” acceleration even when the object speeds up. In that case, centripetal is only one component.

Example 7: Total acceleration with both components

A particle moves in a circle of radius r = 10\ \text{m}. At an instant its speed is v = 8.0\ \text{m/s} and it is speeding up at a_t = 2.0\ \text{m/s}^2. Find the magnitude of the total acceleration.

1. Compute centripetal acceleration:

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

a_c = \frac{(8.0)^2}{10} = 6.4\ \text{m/s}^2

2. Combine perpendicular components:

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

a = \sqrt{(6.4)^2 + (2.0)^2}

3. Compute:

a \approx 6.7\ \text{m/s}^2

What to notice: Even a modest tangential acceleration changes the total only a little when centripetal is already large.

Exam Focus

• Typical question patterns:
  • Identify direction of acceleration vectors on curved paths (inward component always exists if path is curved).
  • Compute a_c and combine with a given a_t.
  • Explain which forces cause speed change vs direction change.
• Common mistakes:
  • Setting tangential net force to zero even when speed is changing.
  • Adding a_c and a_t as scalars instead of perpendicular components.
  • Confusing “tangent” direction with “toward the center” (they are perpendicular).

Building the Free-Body Diagram Correctly for Circular Motion (The skill AP tests)

Circular motion problems in Unit 2 are ultimately Newton’s Laws problems. The physics is rarely “memorize a special equation” and more “choose directions wisely and apply Newton’s Second Law.”

A reliable setup procedure

1. Draw the circle and mark the center. You need this to know what “radially inward” means at the object’s location.
2. Draw a free-body diagram with only real forces:
   • Weight mg (always downward)
   • Normal force N (perpendicular to surface)
   • Tension T (along the string)
   • Friction f (along the surface, direction based on impending motion)
3. Choose axes that match the physics:
   • One axis radially inward
   • One axis tangential
4. Write Newton’s Second Law by component:
   • Radial: \sum F_{\text{radial}} = m\frac{v^2}{r} (inward positive is a common choice)
   • Tangential: \sum F_{\text{tangent}} = ma_t

Example 8: Tension at the bottom of a vertical circle

A 0.30\ \text{kg} mass is attached to a string and swings in a vertical circle of radius r = 0.80\ \text{m}. At the bottom, its speed is v = 5.0\ \text{m/s}. Find the string tension at the bottom.

1. At the bottom, inward is upward.
2. Forces: tension T upward (inward), weight mg downward (outward relative to inward).
3. Apply radial Newton’s Second Law:

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

4. Solve for tension:

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

5. Substitute:

T = (0.30)(9.8) + (0.30)\frac{(5.0)^2}{0.80}

6. Compute:

T = 2.94 + 9.38 = 12.32\ \text{N}

What to notice: Tension is larger than weight because it must both support the mass against gravity and provide the additional inward net force for circular motion.

Exam Focus

• Typical question patterns:
  • “Draw and label a free-body diagram” at a specified point on a circular path.
  • Solve for an unknown force (tension/normal/friction) using the radial equation.
  • Conceptual explanations: what happens when a string breaks (object travels tangent).
• Common mistakes:
  • Putting m\frac{v^2}{r} as a force on the free-body diagram (it is not a force).
  • Choosing axes as horizontal/vertical when radial/tangential would be simpler, then getting components wrong.
  • Forgetting that gravity is constant downward even though “inward” changes around the circle.