Exploring Mechanics / 6. Circular Motion
Moving in a circle always requires a net force pointed at the center — everything in this section is a variation on that one fact.
Make a prediction first: you whirl a ball on a string in a horizontal circle at constant speed. Its speed never changes — so is there any net force acting on it at all?
ac = v²/r · Fc = mv²/r (always pointed toward the center)
Yes — there is a net force, and it's the entire point of this section. Velocity is a vector; even at perfectly constant speed, a circling object's velocity direction is constantly changing, which is itself an acceleration. That acceleration points directly at the circle's center (never forward or backward along the path), so the net force causing it — tension, friction, gravity, whatever's actually supplying it — always points at the center too.
Velocity is tangent to the circle at every instant. A short time later, the velocity vector has the same length but a slightly different direction — subtracting the two velocity vectors to get Δv gives a vector that, in the limit of a very short time interval, points exactly toward the center, perpendicular to the velocity itself. That's the geometric origin of "centripetal" ("center-seeking") acceleration, and working through the geometry carefully gives the exact formula ac = v²/r.
Watch the velocity arrow (always tangent, always changing direction) and the tension arrow (always pointing straight at the center). Then cut the string and see exactly what "centripetal force" was preventing.
The instant tension disappears, the net force on the ball drops to zero — by Newton's First Law it then travels in a straight line at constant velocity, tangent to the circle at the exact point of release. It does not fly straight outward (radially), which is the most common misconception this demonstration corrects.
Switch viewpoints on the exact same spinning platform. From outside, looking down (an inertial frame), everything is explained by one real, center-pointing force. From inside the spinning platform itself, that same physics only makes sense if you invent an outward "centrifugal" force — which is why it's called a fictitious force: it's an artifact of watching from a rotating frame, not a force any object is actually exerting.
Newton's laws in their simple F = ma form only work directly in inertial (non-accelerating) frames; a rotating frame is accelerating (it's constantly changing direction), so to keep using F = ma-style bookkeeping inside it, you have to add fictitious terms — the centrifugal force here — that don't correspond to any actual push or pull. This connects directly back to the intro to rotating platforms in the Toolkit → Reference Frames tab.
Swing a bob in a steady horizontal circle on a string tilted away from vertical — a conical pendulum is uniform circular motion with the centripetal force supplied by a component of the string's own tension.
The same orbiting ball, watched from the ground and from inside the spinning platform at the same instant — one panel needs a single real force, the other needs two that cancel.
Make a prediction first: a race track curve is banked (tilted inward) at just the right angle for a specific speed. At exactly that speed, how much friction does a car need from its tires to make the turn?
Ideal banking angle (no friction needed): tanθ = v²/(rg)
On a banked curve, the road's normal force itself tilts inward. Pick the bank angle so that the horizontal component of that tilted normal force alone supplies exactly the needed centripetal force — and the answer to the prediction is zero friction required at that one specific "design speed."
Set a curve's radius and design speed and watch the required bank angle update — steeper for faster curves, shallower for gentler ones.
Real curves are driven at a range of speeds, not just the exact design speed — friction fills the gap, up to its limit. Too slow and you'd slide down the bank; too fast and you'd slide up and off it.
Decomposing both the normal force and (up-slope or down-slope) friction into horizontal and vertical components and solving Newton's Second Law in each direction gives v² = rg(tanθ ± μ) / (1 ∓ μtanθ) — the "+"/"−" (top sign) case is the fastest safe speed, where friction acts down-slope to keep the car from sliding up and off; the "−"/"+" case is the slowest, where friction acts up-slope to keep the car from sliding down. Setting μ = 0 in either formula collapses both back to the single ideal-angle result from Widget 1.
Push speed above the safe range and the car visibly drifts up and off the bank; drop below it and the car slides down — the top-down view and the cross-section update together in real time.
Make a prediction first: a roller-coaster car goes around a vertical loop at constant track speed near the top. Is the normal force pushing on the rider from the track the same all the way around the loop, or does it change?
N = mv²/r + mg cosφ, where φ is the angle measured from the bottom of the loop
N changes continuously around the loop — largest at the bottom (where gravity and the centripetal requirement both point the same way, "up"), smallest at the top (where gravity now helps supply the centripetal force instead of fighting it). This tab develops the height-and-loop-radius relationship that Energy → Roller Coaster Physics already used, by tracking apparent weight around the whole circuit instead of just the top point.
Set a bottom speed and watch the normal-force readout swing as the marker travels around the loop. If it ever needs to go negative, the car has lost contact with the track before reaching that point.
At angle φ from the bottom, height above the bottom is h(φ) = r(1−cosφ), so by energy conservation v(φ)² = vbottom² − 2gr(1−cosφ). Newton's Second Law along the radial direction then gives N = mv(φ)²/r + mgcosφ. Requiring N ≥ 0 everywhere around the loop turns out to be most restrictive exactly at the top (φ = 180°), which is why checking only the top point (as the Roller Coaster Physics tab did) is enough to guarantee the whole loop is safe.
Swing a bucket of water in a fast vertical circle and the water stays in, even upside down at the top. It's the exact same N ≥ 0 condition — the "bucket" here is just gravity and the water's own inertia playing the role of the track.
Make a prediction first: the International Space Station orbits only about 400 km above Earth's surface — roughly a 10% increase in distance from Earth's center. Is gravity up there dramatically weaker than at the surface, or does something else entirely explain why astronauts float?
F = Gm₁m₂/r²
Drag the distance slider and watch how sharply the force drops — doubling the distance cuts the force to a quarter, not a half.
Real astronomical masses span such an enormous range (a kilogram-scale probe next to a 5.97×10²⁴ kg planet) that a normal linear slider would be useless — nearly the entire slider would represent "basically zero" on a planetary scale. Sliding the exponent instead (10 raised to whatever the slider says) is a standard trick for exploring quantities across many orders of magnitude at once.
Gravity at the ISS's altitude is barely weaker than at the surface — compute it yourself below. Astronauts float for a completely different reason: they (and their station) are in continuous free fall, moving forward fast enough that the ground curves away from them exactly as fast as they fall toward it. That's an orbit, and it's developed fully in the next tab, Orbits.
Make a prediction first: a satellite is launched sideways, exactly fast enough to circle its planet. Launch a second, identical satellite from the same height but a little slower — does it just fall to a lower circular orbit, or does something else happen?
Circular speed: vcirc = √(GM/r) · Escape speed: vesc = √(2GM/r) = vcirc√2
Launch at exactly circular speed and you get a circle. Anything slower or faster (up to escape speed) gives an ellipse — not a smaller or bigger circle — with the launch point as one extreme of the ellipse. That answers the prediction: the slower satellite doesn't settle into a new circular orbit at all.
Every possible gravitational orbit is a conic section, sorted purely by total mechanical energy: bound orbits (energy < 0) are ellipses, with the circle as the special case of zero eccentricity; the borderline (energy = 0, exactly escape speed) is a parabola; unbound orbits (energy > 0, faster than escape speed) are hyperbolas that never return. This is the same "conservation of energy decides the outcome" idea from Section 3, just applied to gravity specifically instead of springs or ramps.
Escape velocity is always exactly √2 times the circular orbital velocity at the same distance — compare real worlds below.
Slide the orbital radius and watch T²/r³ stay exactly constant — that ratio is Kepler's Third Law, and it depends only on the central mass, never on the orbiting object's own mass or speed.
For a circular orbit, the centripetal requirement mv²/r = GMm/r² combined with v = 2πr/T gives, after simplifying, T² = (4π²/GM)r³ — a constant (4π²/GM) times r³, exactly Kepler's Third Law. Using years and astronomical units around the Sun specifically, that constant works out to almost exactly 1, which is why T²/r³ ≈ 1 above regardless of r.