Uniform circular motion
The idea
Moving in a circle at steady speed still means accelerating, because the direction of the velocity is constantly changing. That acceleration points toward the center of the circle — centripetal acceleration — with magnitude a = v²/r. By Newton's second law some real force must supply it: Fc = mv²/r. Centripetal force is never a new kind of force; it is a job description filled by tension, gravity, friction, or a normal force, depending on the situation.
The mental model: at every instant the object 'wants' to fly off along the tangent (Newton's first law), and the inward force continuously bends its path into the circle. Cut the force — let go of the string — and the object departs along the tangent line, not outward along the radius. Notice the strong dependence on speed: doubling v quadruples the required inward force.
The stubborn misconception is the outward 'centrifugal force' supposedly flinging riders to the outside of a turn. In an inertial frame no such force exists: your body tries to continue straight while the car door pushes you inward into the curve. What feels like an outward pull is your own inertia being overruled by an inward push.
Worked example
A 900 kg car rounds a flat, unbanked curve of radius 50 m at a steady 15 m/s. What friction force must the road supply, and what minimum coefficient of static friction between tires and road makes the turn possible?
- Find the required centripetal acceleration: a = v²/r = 15²/50 = 225/50 = 4.5 m/s², directed toward the center of the curve.
- Apply the second law toward the center: the only horizontal force available on a flat curve is friction from the road, so f = ma = 900 × 4.5 = 4050 N inward.
- This is static friction, not kinetic — the tire patches are not sliding on the pavement — and static friction can supply at most μₛN.
- On a flat road the normal force balances gravity: N = mg = 900 × 9.8 = 8820 N.
- Set the required friction equal to the maximum available to find the threshold: μₛ = 4050/8820 ≈ 0.46. Any grippier surface works; anything slicker and the car drifts outward along a wider arc.
- Sanity-check with the speed dependence: at 30 m/s the required force quadruples to 16200 N, needing μₛ ≈ 1.8 — beyond ordinary tires, which is why speed, not radius, is usually what betrays drivers on curves.
Answer. The road must supply 4050 N of friction toward the center, requiring a coefficient of static friction of at least about 0.46.
Check your understanding
- Why does a passenger feel pressed against the door in a tight turn if no outward force is actually acting on them?
- How does banking a curve reduce the friction needed, and which force component takes over the centripetal job?
- What happens to the required centripetal force if you double the speed but also double the radius of the turn?
- Why must the friction in this problem be static rather than kinetic, even though the car is clearly moving?
Build the foundations first
Uniform circular motion builds on these concepts. If any feel shaky, start there.