Prerequisites
Before studying this topic, make sure you understand:
- Circular Motion - Angular velocity, centripetal acceleration
- Newton’s Second Law - F = ma
- Friction - Coefficient of friction
The Hook: Why Don’t You Fall Out of a Roller Coaster Loop?
In Fast X (2023), cars perform insane loop-the-loop stunts. When the car is upside down at the top of the loop, why doesn’t the driver fall out? What keeps the car on the track?
The answer: Centripetal force — the net force toward the center that keeps objects moving in circles!
The Big Questions:
- What force makes objects move in circles?
- Why do roads bank on curves?
- How fast can you go on a circular path without slipping?
The Core Concept: Centripetal Force
What is Centripetal Force?
Centripetal force is the net force directed toward the center of the circular path, required for circular motion.
$$\boxed{F_c = \frac{mv^2}{r} = m\omega^2 r = mv\omega}$$Where:
- $F_c$ = Centripetal force (toward center)
- $m$ = mass
- $v$ = linear speed
- $r$ = radius of circular path
- $\omega$ = angular velocity
Direction: Always toward the center (radially inward)
Centripetal force is NOT a separate force of nature. It’s the net result of existing forces (tension, friction, gravity, normal force) that happens to point toward the center.
Examples:
- Planet orbiting Sun: Centripetal force = Gravitational force
- Car on curve: Centripetal force = Friction force
- Stone on string: Centripetal force = Tension in string
Interactive Demo: Visualize Circular Motion Forces
See how forces combine to create centripetal acceleration.
Centripetal Acceleration
$$\boxed{a_c = \frac{v^2}{r} = \omega^2 r}$$Direction: Toward center (same as centripetal force)
From Newton’s Second Law:
$$F_c = ma_c = m \cdot \frac{v^2}{r}$$Common Circular Motion Scenarios
1. Stone Tied to String (Horizontal Circle)
Setup: Stone of mass $m$ moves in horizontal circle of radius $r$ with speed $v$.
Free Body Diagram:
- Tension $T$ (toward center, horizontal)
- Weight $mg$ (downward)
For horizontal circular motion:
Only the horizontal component provides centripetal force.
If string is horizontal (ideal case):
$$T = \frac{mv^2}{r}$$Breaking point: String breaks if $T > T_{\text{max}}$
Maximum safe speed:
$$v_{\text{max}} = \sqrt{\frac{rT_{\text{max}}}{m}}$$2. Conical Pendulum
Setup: Bob of mass $m$ attached to string of length $L$ moves in horizontal circle while string makes angle $\theta$ with vertical.
Free Body Diagram:
- Tension $T$ (along string)
- Weight $mg$ (downward)
Components:
- Horizontal: $T\sin\theta$ (toward center)
- Vertical: $T\cos\theta$ (upward)
Vertical equilibrium:
$$T\cos\theta = mg \quad \text{...(1)}$$Horizontal (centripetal force):
$$T\sin\theta = \frac{mv^2}{r} \quad \text{...(2)}$$where $r = L\sin\theta$ (radius of horizontal circle)
Dividing (2) by (1):
$$\tan\theta = \frac{v^2}{rg} = \frac{v^2}{Lg\sin\theta}$$ $$v^2 = Lg\sin\theta\tan\theta$$Alternative form:
$$\cos\theta = \frac{g}{L\omega^2}$$where $\omega$ is angular velocity.
Period of revolution:
$$T_{\text{period}} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$$“Conical pendulum COS is Constant”
$$\cos\theta = \frac{g}{L\omega^2}$$Larger $\omega$ (faster spin) → smaller $\cos\theta$ → larger $\theta$ (more horizontal)
Vehicle on Circular Road
Level Road (No Banking)
Car moving on flat circular road of radius $r$ at speed $v$.
Centripetal force provided by: Friction between tires and road
$$f = \frac{mv^2}{r}$$Maximum friction available:
$$f_{\text{max}} = \mu_s N = \mu_s mg$$For no skidding:
$$\frac{mv^2}{r} \leq \mu_s mg$$ $$v \leq \sqrt{\mu_s rg}$$Maximum safe speed:
$$\boxed{v_{\text{max}} = \sqrt{\mu_s rg}}$$In Fast X, when cars take sharp turns at high speed on flat roads, they risk skidding. The maximum speed depends on:
- $\mu$ (tire grip) — better tires, higher speed
- $r$ (curve radius) — sharper turn (smaller $r$), lower safe speed
- $g$ (gravity) — can’t change this on Earth!
This is why racetracks have gentle curves (large $r$) for high-speed sections.
Key insight: Maximum speed is independent of mass!
- Heavy car and light car have same maximum safe speed
- Why? Both friction and required centripetal force scale with mass
Banked Road (Without Friction)
Road is tilted at angle $\theta$ to the horizontal.
Free Body Diagram:
- Normal force $N$ (perpendicular to road surface)
- Weight $mg$ (downward)
Components:
- Horizontal (toward center): $N\sin\theta$
- Vertical (upward): $N\cos\theta$
Vertical equilibrium:
$$N\cos\theta = mg \quad \text{...(1)}$$Horizontal (centripetal force):
$$N\sin\theta = \frac{mv^2}{r} \quad \text{...(2)}$$Dividing (2) by (1):
$$\tan\theta = \frac{v^2}{rg}$$Optimum speed (no friction needed):
$$\boxed{v_{\text{opt}} = \sqrt{rg\tan\theta}}$$At this speed, vehicle can navigate the curve without any friction — normal force alone provides centripetal force!
Banking angle for given speed:
$$\boxed{\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)}$$On a banked road:
- Component of normal force points toward center
- Reduces dependence on friction
- Safer at high speeds
- Reduces tire wear
Highways and racetracks use banking on curves!
Banked Road (With Friction)
Most general case: Banked road + friction available.
Two scenarios:
Scenario A: Maximum Speed (Friction Up the Slope)
Car trying to go too fast tends to slip up the slope. Friction acts down the slope.
Horizontal (toward center):
$$N\sin\theta + f\cos\theta = \frac{mv^2}{r}$$Vertical:
$$N\cos\theta = mg + f\sin\theta$$Using $f = \mu N$ and solving:
$$\boxed{v_{\text{max}} = \sqrt{rg \cdot \frac{\mu + \tan\theta}{1 - \mu\tan\theta}}}$$Scenario B: Minimum Speed (Friction Down the Slope)
Car going too slow tends to slip down the slope. Friction acts up the slope.
$$\boxed{v_{\text{min}} = \sqrt{rg \cdot \frac{\tan\theta - \mu}{1 + \mu\tan\theta}}}$$Range of safe speeds:
$$v_{\text{min}} \leq v \leq v_{\text{max}}$$Wrong: On banked road, friction always points toward center.
Correct: Direction of friction depends on speed:
- Too fast → friction acts down the slope (away from center component)
- Too slow → friction acts up the slope (toward center component)
- Just right ($v = v_{\text{opt}}$) → no friction needed!
Vertical Circular Motion
General Setup
Object moves in a vertical circle of radius $r$.
Key difference from horizontal: Weight acts throughout, changing the required force at different points.
At any point, centripetal force:
$$F_c = \frac{mv^2}{r}$$But this centripetal force is provided by different combinations of tension/normal and weight at different positions.
Case 1: Stone Tied to String (Vertical Circle)
At the Lowest Point (Bottom)
Forces:
- Tension $T_L$ (upward)
- Weight $mg$ (downward)
Net force toward center (upward):
$$T_L - mg = \frac{mv_L^2}{r}$$ $$\boxed{T_L = mg + \frac{mv_L^2}{r}}$$Tension is maximum at the bottom.
At the Highest Point (Top)
Forces:
- Tension $T_H$ (downward)
- Weight $mg$ (downward)
Net force toward center (downward):
$$T_H + mg = \frac{mv_H^2}{r}$$ $$\boxed{T_H = \frac{mv_H^2}{r} - mg}$$Tension is minimum at the top.
Minimum Speed at Top (Critical Condition)
For the stone to complete the circle, tension must be $\geq 0$ at the top.
At limiting case: $T_H = 0$
$$0 = \frac{mv_H^2}{r} - mg$$ $$v_H = \sqrt{gr}$$Minimum speed at top:
$$\boxed{v_{H,\text{min}} = \sqrt{gr}}$$Using energy conservation to find minimum speed at bottom:
$$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2r)$$ $$v_L^2 = v_H^2 + 4gr = gr + 4gr = 5gr$$Minimum speed at bottom:
$$\boxed{v_{L,\text{min}} = \sqrt{5gr}}$$“Top 1, Bottom 5”
- Minimum speed at top: $v_H = \sqrt{gr}$ (coefficient = 1)
- Minimum speed at bottom: $v_L = \sqrt{5gr}$ (coefficient = 5)
Easy to remember: 1 and 5!
Case 2: Object Inside a Smooth Sphere
Object moves inside a smooth hemispherical bowl.
At angle $\theta$ from vertical:
Radial (toward center):
$$N - mg\cos\theta = \frac{mv^2}{r}$$At the highest point the object can reach: $N = 0$ (loses contact)
$$mg\cos\theta = \frac{mv^2}{r}$$Using energy conservation from bottom to height $h = r - r\cos\theta = r(1-\cos\theta)$:
$$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgr(1-\cos\theta)$$where $u$ is initial speed at bottom.
Solving for the angle where object leaves contact gives conditions based on initial speed.
Case 3: Motor Cycle in Vertical Circle (Well of Death)
Motorcycle moves on the inside of a vertical cylindrical wall.
At horizontal level:
Centripetal force (horizontal, toward center):
$$N = \frac{mv^2}{r}$$Vertical equilibrium:
$$f = mg$$Using $f = \mu N$:
$$\mu N = mg$$ $$\mu \cdot \frac{mv^2}{r} = mg$$Minimum speed:
$$\boxed{v_{\text{min}} = \sqrt{\frac{rg}{\mu}}}$$In stunt shows (and movies!), motorcycles ride along the inside wall of a vertical cylinder. They need:
- High speed (to generate large normal force $N$)
- Good friction (so that $\mu N \geq mg$)
If speed drops below minimum, friction can’t support weight → rider falls!
Death-Defying Stunts: Loop-the-Loop
Car on Vertical Circular Track
At the top of the loop:
Forces:
- Normal force $N$ (downward, from track)
- Weight $mg$ (downward)
Centripetal force:
$$N + mg = \frac{mv^2}{r}$$Minimum speed (N = 0):
$$mg = \frac{mv_{\text{min}}^2}{r}$$ $$\boxed{v_{\text{min}} = \sqrt{gr}}$$Same as stone on string!
But: If car goes slower than this, it loses contact with track and falls.
At the bottom:
Energy conservation from bottom to top:
$$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2r)$$ $$v_L^2 = gr + 4gr = 5gr$$ $$\boxed{v_{L,\text{min}} = \sqrt{5gr}}$$In Fast X, for a car to successfully complete a vertical loop of radius 10 m:
Minimum speed at top: $v_H = \sqrt{10 \times 10} = 10$ m/s Minimum speed at bottom: $v_L = \sqrt{5 \times 10 \times 10} = 10\sqrt{5} \approx 22.4$ m/s
If the car enters slower than 22.4 m/s, it won’t make it over the top!
Comparison: Horizontal vs Vertical Circular Motion
| Aspect | Horizontal Circle | Vertical Circle |
|---|---|---|
| Speed | Constant | Varies (fastest at bottom, slowest at top) |
| Force variation | Constant centripetal force | Changes with position |
| Energy | Only KE (constant) | KE + PE (total constant) |
| Tension/Normal | Constant | Maximum at bottom, minimum at top |
| Critical speed | None (any speed works) | Minimum speed needed at top |
Memory Tricks & Patterns
Mnemonic for Centripetal Force
“Faster Cars Race More”
- Force = $\frac{mv^2}{r}$
Or: F = m × v²/r
Mnemonic for Banking
“Tan Takes Victory”
$$\tan\theta = \frac{v^2}{rg}$$Pattern Recognition for Vertical Circles
| Position | Tension/Normal | Formula |
|---|---|---|
| Bottom | Maximum | $T = mg + \frac{mv^2}{r}$ |
| Top | Minimum | $T = \frac{mv^2}{r} - mg$ |
| Side | Intermediate | $T = \frac{mv^2}{r}$ |
Quick Formulas Table
| Scenario | Formula |
|---|---|
| Level road max speed | $v_{\text{max}} = \sqrt{\mu rg}$ |
| Banked road (no friction) | $v = \sqrt{rg\tan\theta}$ |
| Vertical circle min speed (top) | $v_H = \sqrt{gr}$ |
| Vertical circle min speed (bottom) | $v_L = \sqrt{5gr}$ |
| Conical pendulum | $\cos\theta = \frac{g}{L\omega^2}$ |
Common Mistakes to Avoid
Wrong: Centripetal force is a new type of force, like friction or tension.
Correct: Centripetal force is the net force toward the center. It’s provided by existing forces:
- Tension
- Friction
- Normal force
- Gravity
- Or combination of these
Wrong: Objects in circular motion experience an outward centrifugal force.
Correct: In an inertial frame, there’s NO centrifugal force. Objects move in circles because of inward centripetal force.
Centrifugal force is a pseudo force that appears only in the rotating (non-inertial) frame.
JEE rule: Unless problem specifically asks for non-inertial frame, work in inertial frame (no centrifugal force!).
Wrong: Object maintains constant speed in vertical circular motion.
Correct: Speed varies! Energy is conserved, so:
- Faster at bottom (lower PE, higher KE)
- Slower at top (higher PE, lower KE)
Wrong: On a banked road, friction always provides centripetal force.
Correct: Friction direction depends on speed:
- At optimum speed: friction = 0
- Too fast: friction opposes (component away from center)
- Too slow: friction assists (component toward center)
Practice Problems
Level 1: Foundation (NCERT)
A stone of mass 0.5 kg is tied to a string and whirled in a horizontal circle of radius 2 m with speed 4 m/s. Find the tension in the string.
Solution:
For horizontal circular motion, tension provides centripetal force:
$$T = \frac{mv^2}{r} = \frac{0.5 \times 4^2}{2} = \frac{0.5 \times 16}{2} = 4 \text{ N}$$Answer: 4 N
A car moves on a circular road of radius 50 m. If coefficient of friction is 0.4, find maximum safe speed. (g = 10 m/s²)
Solution:
$$v_{\text{max}} = \sqrt{\mu rg} = \sqrt{0.4 \times 50 \times 10} = \sqrt{200} = 10\sqrt{2} \approx 14.14 \text{ m/s}$$Answer: 14.14 m/s (or 10√2 m/s)
A road of radius 100 m is banked for traffic moving at 20 m/s. Find the banking angle. (g = 10 m/s²)
Solution:
$$\tan\theta = \frac{v^2}{rg} = \frac{20^2}{100 \times 10} = \frac{400}{1000} = 0.4$$ $$\theta = \tan^{-1}(0.4) \approx 21.8°$$Answer: 21.8° (or tan⁻¹(0.4))
Level 2: JEE Main
A conical pendulum has string length 1 m and makes angle 60° with vertical. Find: (a) Speed of the bob (b) Period of revolution (g = 10 m/s²)
Solution:
(a) Speed:
$$\cos\theta = \frac{g}{L\omega^2}$$ $$\omega^2 = \frac{g}{L\cos\theta} = \frac{10}{1 \times \cos 60°} = \frac{10}{0.5} = 20$$ $$\omega = \sqrt{20} = 2\sqrt{5} \text{ rad/s}$$Radius of circle: $r = L\sin\theta = 1 \times \sin 60° = \frac{\sqrt{3}}{2}$ m
$$v = r\omega = \frac{\sqrt{3}}{2} \times 2\sqrt{5} = \sqrt{15} \approx 3.87 \text{ m/s}$$(b) Period:
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{2\sqrt{5}} = \frac{\pi}{\sqrt{5}} \approx 1.405 \text{ s}$$Answers: (a) √15 m/s ≈ 3.87 m/s (b) π/√5 s ≈ 1.41 s
A stone tied to a string of length 0.8 m is whirled in a vertical circle. Find the minimum speed at the highest point for the stone to complete the circle. (g = 10 m/s²)
Solution:
At the highest point, for minimum speed, tension = 0.
$$v_{H,\text{min}} = \sqrt{gr} = \sqrt{10 \times 0.8} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \text{ m/s}$$Answer: 2√2 m/s ≈ 2.83 m/s
A 2 kg stone whirls in a vertical circle of radius 1 m. At the lowest point, its speed is 8 m/s. Find the tension in the string at this point. (g = 10 m/s²)
Solution:
At the lowest point:
$$T_L = mg + \frac{mv^2}{r} = 2 \times 10 + \frac{2 \times 8^2}{1}$$ $$= 20 + \frac{2 \times 64}{1} = 20 + 128 = 148 \text{ N}$$Answer: 148 N
Level 3: JEE Advanced
A stone of mass 0.5 kg is attached to a string of length 2 m and whirled in a vertical circle. Find: (a) Minimum speed at the bottom to complete the circle (b) Tension at the bottom when moving with this minimum speed (c) Tension at the top at this speed (g = 10 m/s²)
Solution:
(a) Minimum speed at bottom:
$$v_{L,\text{min}} = \sqrt{5gr} = \sqrt{5 \times 10 \times 2} = \sqrt{100} = 10 \text{ m/s}$$(b) Tension at bottom (with $v_L = 10$ m/s):
$$T_L = mg + \frac{mv_L^2}{r} = 0.5 \times 10 + \frac{0.5 \times 10^2}{2}$$ $$= 5 + \frac{50}{2} = 5 + 25 = 30 \text{ N}$$(c) Speed at top (using energy conservation):
$$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2r)$$ $$v_H^2 = v_L^2 - 4gr = 100 - 4 \times 10 \times 2 = 100 - 80 = 20$$ $$v_H = \sqrt{20} = 2\sqrt{5} \text{ m/s}$$Tension at top:
$$T_H = \frac{mv_H^2}{r} - mg = \frac{0.5 \times 20}{2} - 5 = 5 - 5 = 0 \text{ N}$$(As expected for minimum speed condition!)
Answers: (a) 10 m/s (b) 30 N (c) 0 N
A circular road of radius 40 m is banked at 15°. Coefficient of friction is 0.2. Find: (a) Optimum speed (no friction needed) (b) Maximum safe speed (g = 10 m/s²)
Solution:
(a) Optimum speed (no friction):
$$v_{\text{opt}} = \sqrt{rg\tan\theta} = \sqrt{40 \times 10 \times \tan 15°}$$ $$\tan 15° \approx 0.268$$ $$v_{\text{opt}} = \sqrt{400 \times 0.268} = \sqrt{107.2} \approx 10.35 \text{ m/s}$$(b) Maximum speed (with friction):
$$v_{\text{max}} = \sqrt{rg \cdot \frac{\mu + \tan\theta}{1 - \mu\tan\theta}}$$ $$= \sqrt{40 \times 10 \times \frac{0.2 + 0.268}{1 - 0.2 \times 0.268}}$$ $$= \sqrt{400 \times \frac{0.468}{1 - 0.0536}}$$ $$= \sqrt{400 \times \frac{0.468}{0.9464}}$$ $$= \sqrt{400 \times 0.4945} = \sqrt{197.8} \approx 14.06 \text{ m/s}$$Answers: (a) 10.35 m/s (b) 14.06 m/s
A motorcyclist rides on the vertical wall of a cylindrical well of radius 5 m. If coefficient of friction is 0.5, find the minimum speed required. (g = 10 m/s²)
Solution:
For vertical wall at horizontal level:
Centripetal force (horizontal):
$$N = \frac{mv^2}{r}$$Vertical balance:
$$f = mg$$Using $f = \mu N$:
$$\mu \cdot \frac{mv^2}{r} = mg$$ $$v^2 = \frac{rg}{\mu} = \frac{5 \times 10}{0.5} = 100$$ $$v = 10 \text{ m/s}$$Answer: 10 m/s
Quick Revision Box
| Scenario | Formula | Key Point |
|---|---|---|
| Centripetal force | $F_c = \frac{mv^2}{r}$ | Net force toward center |
| Level road max speed | $v = \sqrt{\mu rg}$ | Friction provides $F_c$ |
| Banking (no friction) | $v = \sqrt{rg\tan\theta}$ | Normal provides $F_c$ |
| Vertical circle (top min) | $v_H = \sqrt{gr}$ | Tension = 0 at limit |
| Vertical circle (bottom min) | $v_L = \sqrt{5gr}$ | From energy conservation |
| Tension at bottom | $T_L = mg + \frac{mv^2}{r}$ | Maximum tension |
| Tension at top | $T_H = \frac{mv^2}{r} - mg$ | Minimum tension |
| Conical pendulum | $\cos\theta = \frac{g}{L\omega^2}$ | String angle relation |
When to Use These Concepts
For horizontal circular motion:
- Identify what provides centripetal force (friction, tension, normal component)
- Set $F_c = \frac{mv^2}{r}$
- Solve for unknown
For banked roads:
- Check if friction is mentioned
- No friction: Use $\tan\theta = \frac{v^2}{rg}$
- With friction: Use max/min speed formulas
For vertical circular motion:
- Use energy conservation to relate speeds at different points
- At each point, apply $F_c = \frac{mv^2}{r}$ with correct force directions
- For minimum speed: Set normal/tension = 0 at highest point
Key question: What provides the centripetal force?
- Level road → Friction
- Banked road → Normal (or normal + friction)
- String → Tension
- Inside sphere → Normal
- Gravity → Planets, satellites
JEE Exam Strategy
Weightage
- JEE Main: 3-4 questions per year (high yield!)
- JEE Advanced: 2-3 problems, often combined with energy (4-5 marks each)
Common Question Types
- Maximum speed on level road (easy, 2 marks)
- Banking angle calculation (easy-moderate, 2-3 marks)
- Vertical circle minimum speed (moderate, 3 marks)
- Tension at different points (moderate, 3-4 marks)
- Conical pendulum (moderate, 3 marks)
- Combined banking + friction (advanced, 4-5 marks)
- Well of death / rotor (moderate-advanced, 4 marks)
Time-Saving Tricks
Trick 1: Memorize vertical circle minimum speeds:
- Top: $\sqrt{gr}$ (coefficient = 1)
- Bottom: $\sqrt{5gr}$ (coefficient = 5)
Trick 2: Level road maximum speed: $v = \sqrt{\mu rg}$ (direct formula!)
Trick 3: Banking with no friction: $\tan\theta = \frac{v^2}{rg}$ (super quick!)
Trick 4: At any point in circle, always draw FBD first — shows which forces contribute to centripetal force
Trick 5: For conical pendulum, remember: $\cos\theta = \frac{g}{L\omega^2}$ (unusual but direct!)
Teacher’s Summary
- Centripetal force is NOT a new force — it’s the net force toward center from existing forces
- Always directed toward center — perpendicular to velocity, changes direction but not speed (in horizontal circles)
- $F_c = \frac{mv^2}{r}$ — fundamental formula for all circular motion
- Horizontal circles: Speed constant, force constant
- Vertical circles: Speed varies (fastest at bottom, slowest at top), need minimum speed at top
- Banking reduces friction dependence — optimum speed needs zero friction
- Minimum speeds for vertical circle: $\sqrt{gr}$ at top, $\sqrt{5gr}$ at bottom (remember 1 and 5!)
- No centrifugal force in inertial frames — pseudo force only in rotating frames
“Centripetal force doesn’t pull objects outward — it pulls them INWARD to keep them in circular paths!”
Related Topics
Within Laws of Motion
- Newton’s Second Law — F = ma applied to circular motion
- Friction — Provides centripetal force on level roads
- Newton’s First Law — Why objects “want” to go straight (inertia)
Connected Chapters
- Circular Motion — Kinematics (description without forces)
- Work-Energy-Power — Energy in vertical circular motion
- Gravitation — Planetary circular motion
- Rotational Motion — Angular momentum in circular paths
Math Connections
- Trigonometry — Resolving forces on banked roads
- Vectors — Centripetal acceleration as vector