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=ma
📚 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?
Forces causing circular motion in horizontal and vertical circles
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c
=
rmv2
where:
Fc = 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:
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=rmv2
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)
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=rmv2
N=rmv2−mg
Minimum speed at top:
For the object to maintain contact (N≥0):
rmvmin2≥mg
vmin=gr
If v<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=rmv2
N=rmv2+mg
Normal force is largest at the bottom (you feel "heavier").
General Point
At angle θ from bottom:
Must resolve weight into components
Component toward center: mgcosθ
Fc=N−mgcosθ=rmv2
Banked Curves
A banked curve is tilted at angle θ to help cars turn without relying solely on friction.
Without Friction (Ideal Banking)
The horizontal component of normal force provides centripetal force:
Nsinθ=rmv2
The vertical component balances weight:
Ncosθ=mg
Dividing these equations:
tanθ=rgv2
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
Draw a free body diagram
Identify the center of the circular path
Choose a coordinate system with one axis toward the center
Apply Newton's 2nd Law in the centripetal direction: ∑Fc=rmv2
Apply Newton's 2nd Law perpendicular to centripetal direction (often: ∑F=0)
Solve for the unknown
Common Scenarios and Formulas
1. Horizontal Circle with String
T=rmv2
2. Car on Flat Curve
vmax=μsgr
3. Top of Vertical Circle
N=rmv2−mgvmin=gr
4. Bottom of Vertical Circle
N=rmv2+mg
5. Banked Curve (No Friction)
tanθ=rgv2
⚠️ 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=rmv2+mg vs Ntop=rmv
Mistake 4: Minimum Speed
❌ Wrong: Minimum speed is zero
✅ Right: At the top of a vertical circle, vmin=gr to maintain circular motion
Real-World Applications
Loop-the-Loop Roller Coasters
Must have v≥gr at top to maintain contact
Designed with extra speed for safety
Riders feel "weightless" if v=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
Fc=rmv2=mω2r
vmax,flat curve=μsgr
vmin,top of loop=gr
tanθbanking=rgv2
v=15
Radius: r=30 m
Flat road (no banking)
Find: Minimum coefficient of static friction μs
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: fs points toward center (provides Fc)
Step 2: Apply Newton's 2nd Law in centripetal direction
fs=Fc
fs=rmv2
Step 3: Use friction relationship
fs≤μsN=μsmg
For the car not to slip:
rmv2≤μsmg
Step 4: Solve for μs
μs≥rgv2
μs≥(30)(9.8)(15)2
μs≥294225
μs≥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=60 kg
Radius: r=8 m
Speed at top: v=12 m/s
Location: top of vertical loop
Find: Normal force N 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=588 N (down, toward center)
Normal force: N (down, toward center)
Step 2: Apply Newton's 2nd Law toward center (downward)
∑Fc=N+mg=r
Step 3: Solve for N
N=rmv2−mg
N=m(rv2−g)
N=60(8(12)2−9.8
N=60(8144−9.8)
N=60(18−9.8)
N=60(8.2)
N=492 N
Check minimum speed:
vmin=gr
Since 12>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<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=150 m
Banking angle: θ=10°
For part (b): speed v=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=rmv2
Vertical (equilibrium):
Ncosθ=mg
Step 2: Divide equations to eliminate N and m
NcosθNsinθ=
tanθ=rgv2
Step 3: Solve for v
v=rgtanθ
v=150×9.8×tan(10°)
v=1470×0.1763
v=259.2
v≈16.1 m/s
(b) Find minimum μs for v=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 f be friction force down the slope.
Horizontal (toward center):
Nsinθ+fcosθ=rmv
Vertical (equilibrium):
Ncosθ−fsinθ=mg
Also: f=μsN
Step 5: Substitute and solve
From vertical equation:
Ncosθ−μsNsinθ=mg
From horizontal equation:
Nsinθ+μsNcosθ=r
Divide:
cosθ−μssinθ
Let k=rgv2=
sinθ+μscosθ=k(cosθ−μ
sinθ+μscosθ=kcosθ−k
μscosθ+kμssinθ=
μs(cosθ+ksinθ)=kcosθ−
μs=cosθ+ksinθ
With θ=10°, cos(10°)=0.985, sin(10°)=0.174:
μs=0.985+0.612(0.174)0.612(0.985)−0.174
μs=0.985+0.1060.603−0.174
μs=1.0910.429
μs≈0.39
Answers:
(a) Design speed: 16.1 m/s (about 36 mph)
(b) Minimum coefficient of friction: 0.39
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.