where:

• $F_c$ = centripetal force (net force toward center)
• $m$ = mass (kg)
• $v$ = speed (m/s)
• $r$ = radius (m)

💡 Key Point: "Centripetal force" is just a name for whatever force (or combination of forces) points toward the center and causes circular motion.

Sources of Centripetal Force

Different situations provide centripetal force from different sources:

Horizontal Circular Motion

Example: Object on String (Horizontal Circle)

For a ball swung in a horizontal circle:

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

where $T$ is the tension in the string.

Free Body Diagram:

• Tension $T$ points toward center (provides all centripetal force)
• Weight $mg$ points down
• If truly horizontal, tension must also support weight (actually moves in slight cone shape)

Example: Car on Flat Curve

For a car turning on a flat road:

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

The static friction provides centripetal force.

Maximum safe speed:

$f_{s,max} = \mu_s N = \mu_s mg$

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

$v_{max} = \sqrt{\mu_s gr}$

Important: Tighter curves (smaller $r$) require lower speeds!

Vertical Circular Motion

Vertical circles are more complex because gravity acts differently at different points.

Top of Circle

At the highest point:

• Both weight and normal force point toward center (downward)
• $F_c = N + mg = \frac{mv^2}{r}$
• $N = \frac{mv^2}{r} - mg$

Minimum speed at top:

For the object to maintain contact ($N \geq 0$):

$\frac{mv_{min}^2}{r} \geq mg$

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

If $v < \sqrt{gr}$, the object falls away from the circle!

Bottom of Circle

At the lowest point:

• Normal force points up (toward center)
• Weight points down (away from center)
• $F_c = N - mg = \frac{mv^2}{r}$
• $N = \frac{mv^2}{r} + mg$

Normal force is largest at the bottom (you feel "heavier").

General Point

At angle $\theta$ from bottom:

• Must resolve weight into components
• Component toward center: $mg\cos\theta$
• $F_c = N - mg\cos\theta = \frac{mv^2}{r}$

Banked Curves

A banked curve is tilted at angle $\theta$ to help cars turn without relying solely on friction.

Without Friction (Ideal Banking)

The horizontal component of normal force provides centripetal force:

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

The vertical component balances weight:

$N\cos\theta = mg$

Dividing these equations:

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

This gives the ideal banking angle for speed $v$.

With Friction

If friction is present:

• Can handle range of speeds
• Friction helps at high speeds, opposes at low speeds
• Banking reduces friction needed

Problem-Solving Strategy

1. Draw a free body diagram
2. Identify the center of the circular path
3. Choose a coordinate system with one axis toward the center
4. Apply Newton's 2nd Law in the centripetal direction: $\sum F_c = \frac{mv^2}{r}$
5. Apply Newton's 2nd Law perpendicular to centripetal direction (often: $\sum F = 0$)
6. Solve for the unknown

Common Scenarios and Formulas

1. Horizontal Circle with String

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

2. Car on Flat Curve

$v_{max} = \sqrt{\mu_s gr}$

3. Top of Vertical Circle

$N = \frac{mv^2}{r} - mg$ $v_{min} = \sqrt{gr}$

4. Bottom of Vertical Circle

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

5. Banked Curve (No Friction)

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

⚠️ Common Mistakes

Mistake 1: Centripetal Force as a Separate Force

❌ Wrong: Drawing "centripetal force" as an additional force on FBD ✅ Right: Centripetal force is the NET force toward center from real forces (tension, friction, gravity, normal, etc.)

Mistake 2: Direction of Friction on Curves

❌ Wrong: Friction always opposes motion (points backward) ✅ Right: On a curve, friction points toward the center (perpendicular to velocity) to provide centripetal force

Mistake 3: Tension in Vertical Circles

❌ Wrong: Tension is the same at top and bottom ✅ Right: Tension is much larger at bottom than at top: $N_{bottom} = \frac{mv^2}{r} + mg$ vs $N_{top} = \frac{mv^2}{r} - mg$

Mistake 4: Minimum Speed

❌ Wrong: Minimum speed is zero ✅ Right: At the top of a vertical circle, $v_{min} = \sqrt{gr}$ to maintain circular motion

Real-World Applications

Loop-the-Loop Roller Coasters

• Must have $v \geq \sqrt{gr}$ at top to maintain contact
• Designed with extra speed for safety
• Riders feel "weightless" if $v = \sqrt{gr}$ (normal force = 0)

Centrifuges

• Large $v$ and small $r$ create huge centripetal acceleration
• Separate substances by density
• Can create accelerations of thousands of $g$'s

Banked Highways

• Interstate highway curves are banked for typical speed limits
• Reduces wear on tires and reliance on friction
• Can navigate safely even on ice (at design speed)

📝 Key Formulas Summary

$F_c = \frac{mv^2}{r} = m\omega^2 r$

$v_{max,\text{flat curve}} = \sqrt{\mu_s gr}$

$v_{min,\text{top of loop}} = \sqrt{gr}$

$\tan\theta_{\text{banking}} = \frac{v^2}{rg}$

Find: Minimum coefficient of static friction $\mu_s$

Analysis:

The static friction force provides the centripetal force needed for circular motion.

Step 1: Draw FBD and identify forces

• Normal force: $N = mg$ (vertical equilibrium on flat road)
• Friction: $f_s$ points toward center (provides $F_c$)

Step 2: Apply Newton's 2nd Law in centripetal direction

$f_s = F_c$

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

Step 3: Use friction relationship

$f_s \leq \mu_s N = \mu_s mg$

For the car not to slip:

$\frac{mv^2}{r} \leq \mu_s mg$

Step 4: Solve for $\mu_s$

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

$\mu_s \geq \frac{(15)^2}{(30)(9.8)}$

$\mu_s \geq \frac{225}{294}$

$\mu_s \geq 0.77$

Answer: The minimum coefficient of static friction is 0.77.

Note: This is relatively high - dry concrete on rubber is about 1.0, but wet roads are much lower (~0.4), which is why cars slip on curves when wet!