💪 Centripetal Force: $F_c = mv^2/r$

Part 1 of 7 — Centripetal Force

"Centripetal force" is NOT a new type of force. It's a label for whichever real force (or combination of forces) causes an object to move in a circle. Understanding this distinction is one of the most important concepts in AP Physics 1.

In this lesson you will learn:

• The formula $F_c = mv^2/r$
• Why centripetal force is not a new force
• How to apply Newton's Second Law to circular motion
• The connection between $F = ma$ and $a_c = v^2/r$

Newton's Second Law for Circular Motion

Newton's Second Law says $\vec{F}_{net} = m\vec{a}$.

For circular motion, the acceleration is centripetal: $a_c = v^2/r$, directed toward the center.

Therefore, the net force toward the center must be:

$F_c = ma_c = \frac{mv^2}{r}$

Equivalent Forms

$F_c = \frac{mv^2}{r} = m\omega^2 r = \frac{4\pi^2 mr}{T^2}$

THE KEY INSIGHT

Centripetal force is not a new force! It is the net radial force — the sum of all real forces pointing toward (or away from) the center.

Common AP Mistake

❌ Drawing "centripetal force" as a separate force on a free body diagram ✅ Identifying which real force(s) provide the centripetal acceleration

Centripetal Force Concepts 🎯

Centripetal Force Calculations 🧮

Use $g = 10$ m/s².

1. A 0.5 kg ball moves in a circle of radius 2 m at 4 m/s. What centripetal force is required (in N)?

2. A 1500 kg car rounds a curve of radius 100 m at 25 m/s. What centripetal force is needed (in N)?

3. A 60 kg person on a merry-go-round sits 3 m from the center. The ride makes one revolution every 6 s. What centripetal force acts on the person (in N, round to nearest whole number)?

Force Identification 🔍

Exit Quiz — Centripetal Force ✅

🔍 Identifying the Real Force

Part 2 of 7 — Centripetal Force

The heart of circular motion problems is identifying which real force provides the centripetal acceleration. In this lesson, you'll learn to analyze different scenarios and recognize the source of centripetal force.

In this lesson you will learn:

• How to identify centripetal force in various scenarios
• Tension, gravity, friction, and normal force as centripetal forces
• How to set up Newton's 2nd Law equations for circular motion
• When multiple forces contribute to centripetal force

Force Analysis for Circular Motion

The Recipe

1. Draw a free body diagram with only real forces
2. Choose a coordinate system: radial (toward center = positive) and tangential
3. Apply Newton's 2nd Law in the radial direction:

$\sum F_{\text{radial}} = \frac{mv^2}{r}$

Key Scenarios

Ball on a String (Horizontal Circle)

• Tension provides centripetal force
• $T = mv^2/r$

Car on a Flat Curve

• Static friction provides centripetal force
• $f_s = mv^2/r$
• Maximum speed before skidding: $\mu_s mg = mv^2/r \Rightarrow v_{max} = \sqrt{\mu_s g r}$

Satellite in Orbit

• Gravity provides centripetal force
• $\frac{GMm}{r^2} = \frac{mv^2}{r}$

Object on a Rotating Platform

• Static friction provides centripetal force
• $f_s = mv^2/r$ (inward toward center)
• Object slides off when $f_s > \mu_s N$

Multiple Forces Acting Together

Sometimes two or more forces combine to provide centripetal force:

• Vertical circle at the top: gravity + normal force both point toward center
• Conical pendulum: horizontal component of tension provides centripetal force

Identify the Force 🎯

Force Analysis Calculations 🧮

Use $g = 10$ m/s².

1. A 1200 kg car rounds a flat curve of radius 80 m. If $\mu_s = 0.6$, what is the maximum speed before the car skids (in m/s, round to 3 significant figures)?

2. A 0.2 kg ball on a 0.5 m string moves in a horizontal circle. If the tension in the string is 10 N, what is the ball's speed (in m/s)?

3. A coin on a turntable sits 0.15 m from the center. If $\mu_s = 0.4$, what is the maximum rotation frequency before the coin slides (in Hz, round to 3 significant figures)?

Scenario Analysis 🔍

Exit Quiz — Identifying Forces ✅

🚗 Horizontal Circles

Part 3 of 7 — Centripetal Force

Horizontal circles are the most common circular motion scenarios on the AP exam. We'll analyze cars on curves, balls on strings, and objects on turntables.

In this lesson you will learn:

• Cars on flat curves — friction as centripetal force
• Ball on a horizontal string — tension as centripetal force
• Conical pendulums — component analysis
• Maximum speed problems

Car on a Flat Curve

Setup

A car moves at speed $v$ around a flat (unbanked) curve of radius $r$.

Free Body Diagram

• Weight: $mg$ (down)
• Normal force: $N = mg$ (up, since flat road)
• Static friction: $f_s$ (toward center — this is the centripetal force!)

Newton's 2nd Law

Vertical: $N - mg = 0 \Rightarrow N = mg$

Radial: $f_s = \frac{mv^2}{r}$

Maximum Speed

The car skids when friction reaches its maximum: $f_{s,max} = \mu_s N = \mu_s mg$

$\mu_s mg = \frac{mv^2_{max}}{r}$

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

Key Insights

• $v_{max}$ doesn't depend on mass! A truck and a sports car have the same $v_{max}$ (same $\mu_s$, same curve)
• Wet roads reduce $\mu_s$ → lower $v_{max}$
• Tighter curves (smaller $r$) → lower $v_{max}$
• On ice ($\mu_s \approx 0.1$), $v_{max}$ drops dramatically

Cars on Curves 🎯

The Conical Pendulum

A ball on a string swings in a horizontal circle, with the string making angle $\theta$ with the vertical.

Free Body Diagram

• Weight: $mg$ (down)
• Tension: $T$ along the string (up and toward center)

Component Analysis

Vertical: $T\cos\theta = mg \Rightarrow T = \frac{mg}{\cos\theta}$

Radial: $T\sin\theta = \frac{mv^2}{r}$

where $r = L\sin\theta$ (the radius of the circle, $L$ = string length).

Solving for Speed

Dividing the radial equation by the vertical equation:

$\tan\theta = \frac{v^2}{rg} = \frac{v^2}{gL\sin\theta}$

$v = \sqrt{gL\sin\theta\tan\theta}$

Solving for Period

$T_{period} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$

Key Insight

The period depends on $\cos\theta$ and $L$, but NOT on the mass. Faster spinning → larger $\theta$ → shorter period.

Horizontal Circle Problems 🧮

Use $g = 10$ m/s².

1. A car rounds a flat curve of radius 40 m. If $\mu_s = 0.5$, what is the maximum speed (in m/s, round to 3 significant figures)?

2. A conical pendulum has string length $L = 0.8$ m and makes angle $\theta = 30°$ with the vertical. What is the period of revolution (in s, round to 3 significant figures)?

3. A 0.3 kg ball on a 1.2 m string swings as a conical pendulum at $\theta = 45°$. What is the tension in the string (in N, round to 3 significant figures)?

Horizontal Circle Concepts 🔍

Exit Quiz — Horizontal Circles ✅

🎡 Vertical Circles

Part 4 of 7 — Centripetal Force

Vertical circles are among the most challenging (and most tested!) topics on the AP Physics 1 exam. The key is that gravity's role changes at different positions — sometimes adding to centripetal force, sometimes opposing it.

In this lesson you will learn:

• Force analysis at the top and bottom of a vertical circle
• Minimum speed at the top to maintain circular motion
• Normal force vs. weight at different positions
• Tension in a string at various points

Forces at Top and Bottom

At the BOTTOM of the Circle

Forces on the object:

• Weight $mg$ → downward (away from center)
• Normal force $N$ or Tension $T$ → upward (toward center)

Newton's 2nd Law (toward center = positive):

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

Key result: $N > mg$ — you feel heavier at the bottom! This is why you feel pressed into your seat on a roller coaster valley.

At the TOP of the Circle

Forces on the object:

• Weight $mg$ → downward (toward center ✓)
• Normal force $N$ or Tension $T$ → downward (toward center ✓) for inside of loop

Newton's 2nd Law (toward center = positive):

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

Key result: $N < mg$ — you feel lighter at the top!

Summary Table

Minimum Speed at the Top

At the top of a vertical circle, $N = mv^2/r - mg$.

The minimum speed occurs when $N = 0$ (the object barely maintains contact):

$0 = \frac{mv^2_{min}}{r} - mg$

$v_{min} = \sqrt{gr}$

What This Means

• If $v > \sqrt{gr}$: object stays on the circular path, $N > 0$
• If $v = \sqrt{gr}$: object barely maintains contact, $N = 0$
• If $v < \sqrt{gr}$: object can't maintain circular motion at that radius — it "falls" inside the circle

For a Ball on a String

At the top, tension plays the role of $N$:

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

Wait — note the sign! At the top of a vertical circle with a string, both tension and gravity point toward the center (down):

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

Minimum speed (when $T = 0$): $v_{min} = \sqrt{gr}$ — same result!

Vertical Circle Problems 🎯

Vertical Circle Calculations 🧮

Use $g = 10$ m/s².

1. A 0.5 kg ball on a 1 m string is swung in a vertical circle at 5 m/s at the bottom. What is the string tension at the bottom (in N)?

2. What is the tension in the same string at the top, if the speed at the top is 3 m/s (in N)?

3. A roller coaster loop has radius 15 m. What minimum height must the car start from (released from rest) to safely complete the loop? Assume frictionless. Hint: use energy conservation. (in m)

Round all answers to 3 significant figures.

Vertical Circle Concepts 🔍

Exit Quiz — Vertical Circles ✅

🏗️ Banked Curves

Part 5 of 7 — Centripetal Force

On a banked curve, the road is tilted at an angle so that a component of the normal force provides centripetal force — even without friction! This is a classic AP Physics 1 topic.

In this lesson you will learn:

• Why we bank curves
• The ideal banking angle (no friction needed)
• Banked curves with friction
• Engineering applications

The Ideal Banking Angle

Setup

A road is banked at angle $\theta$. We want to find the angle where a car can turn without any friction.

Free Body Diagram

• Weight: $mg$ (down)
• Normal force: $N$ (perpendicular to road surface — tilted inward)

Component Analysis

Vertical: $N\cos\theta = mg$ → $N = mg/\cos\theta$

Radial (horizontal, toward center): $N\sin\theta = mv^2/r$

Solving for the Banking Angle

Dividing radial by vertical:

$\frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg}$

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

$\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)$

Key Insights

• The ideal angle depends on speed and radius, not mass
• Each speed has its own ideal angle — there's only one "design speed"
• At the design speed: no friction needed, safe even on ice
• Below design speed: car tends to slide down the bank
• Above design speed: car tends to slide up the bank

Ideal Banking Angle 🎯

Banked Curves with Friction

Going Faster Than Design Speed

If $v > v_{design}$, the car tends to slide up the bank. Friction acts down the bank (and inward).

Both $N\sin\theta$ and $f\cos\theta$ point toward the center:

$N\sin\theta + f\cos\theta = \frac{mv^2}{r}$

$N\cos\theta + f\sin\theta = mg + \text{...wait}$

Actually, let's be careful:

Radial: $N\sin\theta + f_s\cos\theta = mv^2/r$

Vertical: $N\cos\theta - f_s\sin\theta = mg$

Going Slower Than Design Speed

If $v < v_{design}$, the car tends to slide down the bank. Friction acts up the bank.

Radial: $N\sin\theta - f_s\cos\theta = mv^2/r$

Vertical: $N\cos\theta + f_s\sin\theta = mg$

Maximum Speed on a Banked Curve

Set $f_s = \mu_s N$ (friction at maximum) in the "going fast" equations:

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

Minimum Speed on a Banked Curve

Set $f_s = \mu_s N$ in the "going slow" equations:

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

Banking Angle Calculations 🧮

Use $g = 10$ m/s².

1. A highway curve has radius 200 m. What banking angle is needed for a design speed of 30 m/s (in degrees, round to 3 significant figures)?

2. A track is banked at $30°$ with radius 100 m. What is the design speed — the speed requiring no friction (in m/s, round to 3 significant figures)?

3. A curve is banked at $20°$ with radius 150 m. What is the maximum speed if $\mu_s = 0.3$ (in m/s, round to nearest whole number)? Use $v_{max} = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}$.

Banking Concepts 🔍

Exit Quiz — Banked Curves ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — Centripetal Force

Time to bring together everything you've learned about centripetal force — horizontal circles, vertical circles, banked curves, and force identification. These problems combine multiple concepts.

In this lesson you will:

• Solve multi-step centripetal force problems
• Combine energy conservation with circular motion
• Tackle compound scenarios (ramps leading to loops, etc.)
• Practice AP-level free response strategies

Problem-Solving Framework

Step 1: Identify the Circular Path

• What is the radius?
• Is it horizontal or vertical?
• Where are you analyzing (top, bottom, side)?

Step 2: Draw the Free Body Diagram

• Draw ONLY real forces
• NEVER draw "centripetal force" as a separate arrow
• Identify which forces have radial components

Step 3: Apply Newton's 2nd Law (Radial Direction)

$\sum F_{\text{toward center}} = \frac{mv^2}{r}$

Step 4: Use Energy Conservation if Needed

For problems involving height changes:

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

Common Combined Problems

1. Ramp → loop: Use energy to find speed at any point, then use $F = mv^2/r$
2. Spring → circle: Spring PE converts to KE, then centripetal force analysis
3. Swinging on a rope: Pendulum energy → tension analysis at various angles

Worked Example: Ramp to Loop

A block starts from rest at height $h$ and slides down a frictionless ramp into a circular loop of radius $r = 5$ m.

Find the minimum $h$ for the block to complete the loop.

Step 1: Minimum speed at the top of the loop

$v_{top,min} = \sqrt{gr} = \sqrt{10 \times 5} = \sqrt{50}$

Step 2: Energy conservation (ground to top of loop)

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

$h = \frac{v_{top}^2}{2g} + 2r = \frac{gr}{2g} + 2r = \frac{r}{2} + 2r = \frac{5r}{2}$

$h = \frac{5(5)}{2} = 12.5 \text{ m}$

Step 3: Normal force at the bottom at this minimum condition

Speed at bottom: $\frac{1}{2}mv_{bot}^2 = mgh = mg(5r/2)$

$v_{bot}^2 = 5gr = 250$

$N - mg = mv_{bot}^2/r = m(5g) = 5mg$

$N = 6mg$ — the rider feels 6g at the bottom!

Multi-Step Problems 🧮

Use $g = 10$ m/s².

1. A 2 kg block slides from rest down a frictionless ramp of height 8 m and enters a circular loop of radius 2 m. What is the block's speed at the top of the loop (in m/s, round to 3 significant figures)?

2. In problem 1, what is the normal force on the block at the top of the loop (in N)?

3. A spring ($k = 500$ N/m) compressed 0.6 m launches a 0.5 kg ball into a vertical loop of radius 1 m (loop bottom is at spring level). What is the ball's speed at the top of the loop (in m/s, round to 3 significant figures)?

Applied Centripetal Force Problems 🎯

Challenge Problems 🧮

Use $g = 10$ m/s².

1. A 1500 kg car travels at 20 m/s over a circular dip in the road (radius 80 m). What is the normal force at the bottom of the dip (in N)?

2. A conical pendulum has string length 2 m and the ball moves at 4 m/s in a horizontal circle. What is the radius of the circle (in m, round to 3 significant figures)?

3. A banked frictionless curve has radius 100 m and banking angle $15°$. What is the design speed (in m/s, round to 3 significant figures)?

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Centripetal Force

This final lesson puts all centripetal force concepts together for AP exam preparation. We'll review common question types, practice FRQ strategies, and address the trickiest conceptual questions.

In this lesson you will:

• Tackle AP-style multiple choice questions
• Practice FRQ structure and scoring
• Review all centripetal force scenarios
• Master the most common exam traps

Centripetal Force Toolkit

Core Equation

$F_c = \frac{mv^2}{r} = m\omega^2 r = \frac{4\pi^2 mr}{T^2}$

Scenario Quick Reference

FRQ Key Phrases

• "The net force toward the center provides centripetal acceleration"
• "By Newton's 2nd Law in the radial direction..."
• "The centripetal force is provided by [tension/friction/gravity/normal force]"

AP-Style Multiple Choice 🎯

FRQ Practice Structure

Typical AP FRQ: "Loop the Loop"

A small block of mass $m$ starts from rest at height $h$ on a frictionless ramp and enters a circular loop of radius $R$.

(a) Derive an expression for the block's speed at the top of the loop in terms of $m$, $h$, $R$, and $g$.

Energy conservation: $mgh = \frac{1}{2}mv_{top}^2 + mg(2R)$

$v_{top} = \sqrt{2g(h - 2R)}$

(b) Derive an expression for the normal force on the block at the top of the loop.

At the top: $mg + N = mv_{top}^2/R$

$N = mv_{top}^2/R - mg = m[2g(h - 2R)]/R - mg = mg(2h/R - 4 - 1) = mg(2h/R - 5)$

(c) Find the minimum height $h_{min}$ for the block to complete the loop.

Set $N = 0$: $0 = mg(2h_{min}/R - 5)$

$h_{min} = 5R/2$

(d) How would the answer to (c) change if the block had twice the mass?

It wouldn't! Mass cancels from all equations. The minimum height is independent of mass: $h_{min} = 5R/2$ regardless of $m$.

AP-Style Calculations 🧮

Use $g = 10$ m/s².

1. A car ($m = 1000$ kg) rounds a flat curve ($r = 50$ m, $\mu_s = 0.6$). What is the maximum speed (in m/s, round to 3 significant figures)?

2. A block starts from height $h = 3R$ on a frictionless ramp and enters a loop of radius $R = 4$ m. What is the normal force at the top of the loop, expressed as a multiple of $mg$? (Just give the number, e.g., "2" for $2mg$)

3. A 0.2 kg ball on a 0.8 m string moves at 6 m/s at the bottom of a vertical circle. What is the tension at the bottom (in N)?

Common AP Traps 🔍

Final Exit Quiz — Centripetal Force ✅