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Centripetal Force

Forces causing circular motion in horizontal and vertical circles

Written and reviewed by the Study Mondo Education TeamLast updated
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💫 Centripetal Force

What is Centripetal Force?

Centripetal force is NOT a new type of force! It's the net force that causes an object to move in a circular path.

Newton's Second Law for Circular Motion

Fc=mac=mv2rF_c = ma_c = \frac{mv^2}{r}

where:

  • FcF_c = centripetal force (net force toward center)
  • mm = mass (kg)
  • vv = speed (m/s)
  • rr = 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:

SituationSource of Centripetal Force
Ball on stringTension in string
Car on flat curveFriction between tires and road
Satellite orbiting EarthGravitational force
Electron orbiting nucleusElectric force
Car on banked curveComponent of normal force (+ friction)
Clothes in dryerNormal force from drum wall

Horizontal Circular Motion

Example: Object on String (Horizontal Circle)

For a ball swung in a horizontal circle:

T=mv2rT = \frac{mv^2}{r}

where TT is the tension in the string.

Free Body Diagram:

  • Tension TT points toward center (provides all centripetal force)
  • Weight mgmg 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:

fs=mv2rf_s = \frac{mv^2}{r}

The static friction provides centripetal force.

Maximum safe speed:

fs,max=μsN=μsmgf_{s,max} = \mu_s N = \mu_s mg

mvmax2r=μsmg\frac{mv_{max}^2}{r} = \mu_s mg

vmax=μsgrv_{max} = \sqrt{\mu_s gr}

Important: Tighter curves (smaller rr) 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)
  • Fc=N+mg=mv2rF_c = N + mg = \frac{mv^2}{r}
  • N=mv2r−mgN = \frac{mv^2}{r} - mg

Minimum speed at top:

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

mvmin2r≥mg\frac{mv_{min}^2}{r} \geq mg

vmin=grv_{min} = \sqrt{gr}

If v<grv < \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)
  • Fc=N−mg=mv2rF_c = N - mg = \frac{mv^2}{r}
  • N=mv2r+mgN = \frac{mv^2}{r} + mg

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

General Point

At angle θ\theta from bottom:

  • Must resolve weight into components
  • Component toward center: mgcos⁡θmg\cos\theta
  • Fc=N−mgcos⁡θ=mv2rF_c = N - mg\cos\theta = \frac{mv^2}{r}

Banked Curves

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

Without Friction (Ideal Banking)

The horizontal component of normal force provides centripetal force:

Nsin⁡θ=mv2rN\sin\theta = \frac{mv^2}{r}

The vertical component balances weight:

Ncos⁡θ=mgN\cos\theta = mg

Dividing these equations:

tan⁡θ=v2rg\tan\theta = \frac{v^2}{rg}

This gives the ideal banking angle for speed vv.

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: ∑Fc=mv2r\sum F_c = \frac{mv^2}{r}
  5. Apply Newton's 2nd Law perpendicular to centripetal direction (often: ∑F=0\sum F = 0)
  6. Solve for the unknown

Common Scenarios and Formulas

1. Horizontal Circle with String

T=mv2rT = \frac{mv^2}{r}

2. Car on Flat Curve

vmax=μsgrv_{max} = \sqrt{\mu_s gr}

3. Top of Vertical Circle

N=mv2r−mgN = \frac{mv^2}{r} - mg vmin=grv_{min} = \sqrt{gr}

4. Bottom of Vertical Circle

N=mv2r+mgN = \frac{mv^2}{r} + mg

5. Banked Curve (No Friction)

tan⁡θ=v2rg\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: Nbottom=mv2r+mgN_{bottom} = \frac{mv^2}{r} + mg vs Ntop=mv2r−mgN_{top} = \frac{mv^2}{r} - mg

Mistake 4: Minimum Speed

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


Real-World Applications

Loop-the-Loop Roller Coasters

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

Centrifuges

  • Large vv and small rr create huge centripetal acceleration
  • Separate substances by density
  • Can create accelerations of thousands of gg'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

Fc=mv2r=mω2rF_c = \frac{mv^2}{r} = m\omega^2 r

vmax,flat curve=μsgrv_{max,\text{flat curve}} = \sqrt{\mu_s gr}

vmin,top of loop=grv_{min,\text{top of loop}} = \sqrt{gr}

tan⁡θbanking=v2rg\tan\theta_{\text{banking}} = \frac{v^2}{rg}

📚 Practice Problems

1Problem 1easy

❓ Question:

A 1200 kg car travels at 15 m/s around a curve with a radius of 30 m on a flat road. What is the minimum coefficient of static friction needed to prevent the car from slipping?

💡 Show Solution

Given Information:

  • Mass: m=1200m = 1200 kg
  • Speed: v=15v = 15 m/s
  • Radius: r=30r = 30 m
  • Flat road (no banking)

Find: Minimum coefficient of static friction μs\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=mgN = mg (vertical equilibrium on flat road)
  • Friction: fsf_s points toward center (provides FcF_c)

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

fs=Fcf_s = F_c

fs=mv2rf_s = \frac{mv^2}{r}


Step 3: Use friction relationship

fs≤μsN=μsmgf_s \leq \mu_s N = \mu_s mg

For the car not to slip:

mv2r≤μsmg\frac{mv^2}{r} \leq \mu_s mg


Step 4: Solve for μs\mu_s

μs≥v2rg\mu_s \geq \frac{v^2}{rg}

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

μs≥225294\mu_s \geq \frac{225}{294}

μs≥0.77\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!

2Problem 2medium

❓ Question:

A 60 kg student rides a roller coaster through a vertical loop with radius 8 m. At the top of the loop, the student is moving at 12 m/s. What is the normal force exerted by the seat on the student at this point?

💡 Show Solution

Given Information:

  • Mass: m=60m = 60 kg
  • Radius: r=8r = 8 m
  • Speed at top: v=12v = 12 m/s
  • Location: top of vertical loop

Find: Normal force NN at the top


Analysis:

At the top of the loop, both weight and normal force point downward (toward center). Their sum provides the centripetal force.


Step 1: Draw FBD at top of loop

Forces on student:

  • Weight: W=mg=60×9.8=588W = mg = 60 \times 9.8 = 588 N (down, toward center)
  • Normal force: NN (down, toward center)

Step 2: Apply Newton's 2nd Law toward center (downward)

∑Fc=N+mg=mv2r\sum F_c = N + mg = \frac{mv^2}{r}


Step 3: Solve for NN

N=mv2r−mgN = \frac{mv^2}{r} - mg

N=m(v2r−g)N = m\left(\frac{v^2}{r} - g\right)

N=60((12)28−9.8)N = 60\left(\frac{(12)^2}{8} - 9.8\right)

N=60(1448−9.8)N = 60\left(\frac{144}{8} - 9.8\right)

N=60(18−9.8)N = 60(18 - 9.8)

N=60(8.2)N = 60(8.2)

N=492 NN = 492 \text{ N}


Check minimum speed:

vmin=gr=9.8×8=78.4≈8.85 m/sv_{min} = \sqrt{gr} = \sqrt{9.8 \times 8} = \sqrt{78.4} \approx 8.85 \text{ m/s}

Since 12>8.8512 > 8.85 m/s, the student maintains contact ✓


Answer: The normal force is 492 N (downward on the student, or upward on the seat).

Interpretation: The student feels lighter than normal since N<mgN < mg (588 N). The apparent weight is about 84% of the actual weight.

3Problem 3hard

❓ Question:

A highway curve with radius 150 m is banked at an angle of 10°. (a) What is the "design speed" for which no friction is needed? (b) If a car travels at 30 m/s on this curve, what minimum coefficient of friction is required?

💡 Show Solution

Given Information:

  • Radius: r=150r = 150 m
  • Banking angle: θ=10°\theta = 10°
  • For part (b): speed v=30v = 30 m/s

(a) Find design speed (no friction needed)


Analysis:

At the design speed, the horizontal component of the normal force alone provides the centripetal force.


Step 1: Set up equations

Horizontal (toward center): Nsin⁡θ=mv2rN\sin\theta = \frac{mv^2}{r}

Vertical (equilibrium): Ncos⁡θ=mgN\cos\theta = mg


Step 2: Divide equations to eliminate NN and mm

Nsin⁡θNcos⁡θ=mv2/rmg\frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg}

tan⁡θ=v2rg\tan\theta = \frac{v^2}{rg}


Step 3: Solve for vv

v=rgtan⁡θv = \sqrt{rg\tan\theta}

v=150×9.8×tan⁡(10°)v = \sqrt{150 \times 9.8 \times \tan(10°)}

v=1470×0.1763v = \sqrt{1470 \times 0.1763}

v=259.2v = \sqrt{259.2}

v≈16.1 m/sv \approx 16.1 \text{ m/s}


(b) Find minimum μs\mu_s for v=30v = 30 m/s


Analysis:

At 30 m/s, the car is going faster than the design speed, so friction must help provide additional centripetal force. Friction points down the slope (toward center and down).


Step 4: Forces with friction

Let ff be friction force down the slope.

Horizontal (toward center): Nsin⁡θ+fcos⁡θ=mv2rN\sin\theta + f\cos\theta = \frac{mv^2}{r}

Vertical (equilibrium): Ncos⁡θ−fsin⁡θ=mgN\cos\theta - f\sin\theta = mg

Also: f=μsNf = \mu_s N


Step 5: Substitute and solve

From vertical equation: Ncos⁡θ−μsNsin⁡θ=mgN\cos\theta - \mu_s N\sin\theta = mg N(cos⁡θ−μssin⁡θ)=mgN(\cos\theta - \mu_s\sin\theta) = mg N=mgcos⁡θ−μssin⁡θN = \frac{mg}{\cos\theta - \mu_s\sin\theta}

From horizontal equation: Nsin⁡θ+μsNcos⁡θ=mv2rN\sin\theta + \mu_s N\cos\theta = \frac{mv^2}{r} N(sin⁡θ+μscos⁡θ)=mv2rN(\sin\theta + \mu_s\cos\theta) = \frac{mv^2}{r}

Divide: sin⁡θ+μscos⁡θcos⁡θ−μssin⁡θ=v2rg\frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta} = \frac{v^2}{rg}

Let k=v2rg=9001470=0.612k = \frac{v^2}{rg} = \frac{900}{1470} = 0.612

sin⁡θ+μscos⁡θ=k(cos⁡θ−μssin⁡θ)\sin\theta + \mu_s\cos\theta = k(\cos\theta - \mu_s\sin\theta)

sin⁡θ+μscos⁡θ=kcos⁡θ−kμssin⁡θ\sin\theta + \mu_s\cos\theta = k\cos\theta - k\mu_s\sin\theta

μscos⁡θ+kμssin⁡θ=kcos⁡θ−sin⁡θ\mu_s\cos\theta + k\mu_s\sin\theta = k\cos\theta - \sin\theta

μs(cos⁡θ+ksin⁡θ)=kcos⁡θ−sin⁡θ\mu_s(\cos\theta + k\sin\theta) = k\cos\theta - \sin\theta

μs=kcos⁡θ−sin⁡θcos⁡θ+ksin⁡θ\mu_s = \frac{k\cos\theta - \sin\theta}{\cos\theta + k\sin\theta}

With θ=10°\theta = 10°, cos⁡(10°)=0.985\cos(10°) = 0.985, sin⁡(10°)=0.174\sin(10°) = 0.174:

μs=0.612(0.985)−0.1740.985+0.612(0.174)\mu_s = \frac{0.612(0.985) - 0.174}{0.985 + 0.612(0.174)}

μs=0.603−0.1740.985+0.106\mu_s = \frac{0.603 - 0.174}{0.985 + 0.106}

μs=0.4291.091\mu_s = \frac{0.429}{1.091}

μs≈0.39\mu_s \approx 0.39


Answers:

  • (a) Design speed: 16.1 m/s (about 36 mph)
  • (b) Minimum coefficient of friction: 0.39
Explain using:

📌 Related Topics in Circular Motion & Gravitation

❓ Frequently Asked Questions

What is Centripetal Force?▾
Forces causing circular motion in horizontal and vertical circles
How can I study Centripetal Force effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Centripetal Force?▾
Centripetal Force is part of the AP Physics 1 course on Study Mondo, specifically in the Circular Motion & Gravitation section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Centripetal Force?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.