🌍Centripetal Force
What is Centripetal Force?
Centripetal force is the net inward force required to keep an object moving in a circle. By Newton's Second Law:
F_c = ma_c = mv²/r
Important: centripetal force is NOT a new type of force! It's just a name for whatever force (or combination of forces) is pointing toward the center of the circle.
Examples of what provides centripetal force: - Ball on a string: tension in the string - Car on a flat curve: friction from the road - Earth orbiting the Sun: gravitational force - Roller coaster at the top of a loop: gravity + normal force
Centripetal Force Explorer
Explore all four scenarios. Switch scenarios, adjust speed and radius, then hit "Cut the String" to see what really happens when the centripetal force is removed — spoiler: the ball goes tangent, not outward!
Think About It
You're at the top of a roller coaster loop. You feel lighter than normal. What forces act on you? Which direction is the centripetal direction?
✏️ Worked Example: Car on Flat Curve
Problem: A 1000 kg car takes a flat curve of radius 50 m at 20 m/s. What minimum friction force is needed to keep it on the road?
✏️ Worked Example: Vertical Circle
Problem: A ball on a 2 m string swings in a vertical circle. At the top, it has speed 5 m/s. Mass = 0.5 kg. Find the tension at the top.
📐 Key Equations
Centripetal force
F_c = (mv²)/(r)Σ F_inward = (mv²)/(r)⚠️ Common Mistakes
Misconception: Treating "centripetal force" as a separate force to add to a free-body diagram.
✓ Correct thinking: Centripetal force is not a new force — it is the name for the net inward force produced by real forces (tension, friction, gravity, normal force, etc.).
Why: Drawing a "centripetal force arrow" in addition to the real forces double-counts and leads to wrong equations. Always identify which real force provides the centripetal force.
Misconception: Believing there is an outward "centrifugal force" pushing objects away from the center.
✓ Correct thinking: In an inertial reference frame there is no centrifugal force. Objects tend to move in a straight line; it only feels like an outward push because you are being accelerated inward.
Why: The "centrifugal force" is a fictitious (pseudo) force that appears only in the rotating reference frame. AP Physics 1 uses inertial frames, so it does not exist here.
Misconception: Forgetting to account for gravity when analyzing vertical circles (e.g., loop-the-loop).
✓ Correct thinking: At every point in a vertical circle, gravity acts downward and must be included in the net force equation. At the top, both tension and gravity point inward; at the bottom, tension points inward and gravity points outward.
Why: Omitting gravity in vertical-circle problems gives wrong tension values and fails to find the minimum speed condition correctly.
📝 Practice Problems
Try these problems. Check your answer when ready.
A 0.3 kg ball on a 0.8 m string moves in a horizontal circle at 4 m/s. What is the tension in the string?
F_c = (mv²)/(r)A 1200 kg car travels around a flat curve of radius 80 m at 15 m/s. What friction force does the road exert on the car?
f = (mv²)/(r)At the bottom of a circular dip (r = 30 m), a 60 kg rider moves at 12 m/s. What is the normal force on the rider? (g = 10 m/s²)
N - mg = (mv²)/(r)A roller coaster car (mass 500 kg) is at the top of a loop of radius 12 m. The speed at the top is 14 m/s. Find the normal force on the car from the track. (g = 10 m/s²)
N + mg = (mv²)/(r)What is the minimum speed a roller coaster car must have at the top of a loop of radius 10 m to maintain contact with the track? (g = 10 m/s²)
mg = (mv²)/(r) (N=0)A satellite of mass 800 kg orbits at a height where the gravitational pull is 7200 N. The orbital radius is 7×10⁶ m. Find the orbital speed.
F_g = (mv²)/(r)A conical pendulum: a 0.5 kg ball on a 1.2 m string makes an angle of 30° with the vertical as it moves in a horizontal circle. Find the tension in the string and the ball's speed. (g = 10 m/s²)
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