Centripetal Force

Forces causing circular motion in horizontal and vertical circles

💫 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:

| Situation | Source of Centripetal Force | |-----------|----------------------------| | Ball on string | Tension in string | | Car on flat curve | Friction between tires and road | | Satellite orbiting Earth | Gravitational force | | Electron orbiting nucleus | Electric force | | Car on banked curve | Component of normal force (+ friction) | | Clothes in dryer | Normal 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=mv2rmgN = \frac{mv^2}{r} - mg

Minimum speed at top:

For the object to maintain contact (N0N \geq 0):

mvmin2rmg\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=Nmg=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: mgcosθmg\cos\theta
  • Fc=Nmgcosθ=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:

Nsinθ=mv2rN\sin\theta = \frac{mv^2}{r}

The vertical component balances weight:

Ncosθ=mgN\cos\theta = mg

Dividing these equations:

tanθ=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=mv2rmgN = \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)

tanθ=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=mv2rmgN_{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 vgrv \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}

tanθbanking=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

fsμsN=μsmgf_s \leq \mu_s N = \mu_s mg

For the car not to slip:

mv2rμsmg\frac{mv^2}{r} \leq \mu_s mg


Step 4: Solve for μs\mu_s

μsv2rg\mu_s \geq \frac{v^2}{rg}

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

μs225294\mu_s \geq \frac{225}{294}

μs0.77\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=mv2rmgN = \frac{mv^2}{r} - mg

N=m(v2rg)N = m\left(\frac{v^2}{r} - g\right)

N=60((12)289.8)N = 60\left(\frac{(12)^2}{8} - 9.8\right)

N=60(14489.8)N = 60\left(\frac{144}{8} - 9.8\right)

N=60(189.8)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=78.48.85 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): Nsinθ=mv2rN\sin\theta = \frac{mv^2}{r}

Vertical (equilibrium): Ncosθ=mgN\cos\theta = mg


Step 2: Divide equations to eliminate NN and mm

NsinθNcosθ=mv2/rmg\frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg}

tanθ=v2rg\tan\theta = \frac{v^2}{rg}


Step 3: Solve for vv

v=rgtanθv = \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}

v16.1 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): Nsinθ+fcosθ=mv2rN\sin\theta + f\cos\theta = \frac{mv^2}{r}

Vertical (equilibrium): Ncosθfsinθ=mgN\cos\theta - f\sin\theta = mg

Also: f=μsNf = \mu_s N


Step 5: Substitute and solve

From vertical equation: NcosθμsNsinθ=mgN\cos\theta - \mu_s N\sin\theta = mg N(cosθμssinθ)=mgN(\cos\theta - \mu_s\sin\theta) = mg N=mgcosθμssinθN = \frac{mg}{\cos\theta - \mu_s\sin\theta}

From horizontal equation: Nsinθ+μsNcosθ=mv2rN\sin\theta + \mu_s N\cos\theta = \frac{mv^2}{r} N(sinθ+μscosθ)=mv2rN(\sin\theta + \mu_s\cos\theta) = \frac{mv^2}{r}

Divide: sinθ+μscosθcosθμssinθ=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

sinθ+μscosθ=k(cosθμssinθ)\sin\theta + \mu_s\cos\theta = k(\cos\theta - \mu_s\sin\theta)

sinθ+μscosθ=kcosθkμssinθ\sin\theta + \mu_s\cos\theta = k\cos\theta - k\mu_s\sin\theta

μscosθ+kμssinθ=kcosθsinθ\mu_s\cos\theta + k\mu_s\sin\theta = k\cos\theta - \sin\theta

μs(cosθ+ksinθ)=kcosθsinθ\mu_s(\cos\theta + k\sin\theta) = k\cos\theta - \sin\theta

μs=kcosθsinθcosθ+ksinθ\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=0.6030.1740.985+0.106\mu_s = \frac{0.603 - 0.174}{0.985 + 0.106}

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

μs0.39\mu_s \approx 0.39


Answers:

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