Forces in Circular Motion

Newton's second law adds just one requirement to what the kinematics has already settled: the real forces must add up to m times that acceleration. Nothing about the drawing changes, so build the free-body diagram exactly as in any other problem, one arrow per interaction: the Earth's pull, a surface's normal force and friction, a string's tension. Only then ask which forces, or which components of them, point at the centre. A 0.050 kg puck circling on frictionless ice at 2.0 m s⁻¹, tied to a 0.40 m cord, has a weight and a normal force that are vertical and cancel; the cord is the only horizontal force, so T = 0.050 × 2.0²/0.40 = 0.50 N. For a satellite the inward force is gravity, for a car on a flat road it is friction. Centripetal force is the name of that inward total, never an extra arrow.

Choose axes by where the acceleration points, not by which way the surface tilts. A car on a banked curve travels in a horizontal circle, so the centre is level with the car and the radial axis is horizontal; the second axis is vertical, with no acceleration along it. That is the opposite of the inclined-plane choice, where the acceleration runs along the slope. Resolve the normal force: N sin θ points at the centre and N cos θ points up. With no friction, N sin θ = mv²/r and N cos θ = mg. Dividing removes N and m and leaves tan θ = v²/(rg), which fixes the design speed of the bank. A 120 m radius curve meant for 20 m s⁻¹ needs tan θ = 20²/(120 × 9.81) = 0.340, so θ = 18.8°. The vertical equation also gives N = mg/cos θ = 1.06mg: on a banked turn the road pushes harder than the car weighs, because only the vertical part of N holds the car up.

At any other speed the normal force alone supplies the wrong inward force, and friction along the road makes up the difference. Keep the horizontal and vertical axes and take f positive down the bank: N sin θ + f cos θ = mv²/r and N cos θ − f sin θ = mg. Solving the pair gives f = m((v²/r) cos θ − g sin θ) and N = m(g cos θ + (v²/r) sin θ). The sign of f carries the physics. Above the design speed f is positive: the car tends to slide up and out, so friction points down the bank. Below it f is negative and friction points up the bank, stopping a slow car from sliding in toward the centre. Static friction manages either only while |f| ≤ μₛN. A car at 18 m s⁻¹ on a 60 m radius curve banked at 10°, whose design speed is 10.2 m s⁻¹, needs f = 0.368mg down the bank with N = 1.08mg, so its tyres must provide μₛ ≥ 0.34.

A bob on a cord of length L, circling horizontally with the cord at angle β to the vertical, is a frictionless banked curve with tension in place of the normal force. The radius is r = L sin β, and the equations are T sin β = mω²r and T cos β = mg. Substituting r cancels sin β from the first, so T = mω²L whatever the angle. The second then gives cos β = g/(ω²L), a prediction you can test. Spin a 0.60 m pendulum at 1.0 rev s⁻¹, so ω = 2π rad s⁻¹, and cos β = 9.81/(39.48 × 0.60) = 0.414: the cord sits at 65.5° to the vertical and pulls with 2.41mg. Slow down and the cone narrows. Below ω = √(g/L) = 4.04 rad s⁻¹, or 0.64 rev s⁻¹, the equation asks for cos β > 1, which no angle gives, and the bob simply hangs straight down. As the cone opens, the period 2π√(L cos β/g) shortens.

In a vertical circle gravity has a component along the path, so the speed cannot stay constant. Measure φ from the lowest point. A string's tension is radial and has no component along the path, so the tangential equation is m dv/dt = −mg sin φ. The chain rule turns dv/dt into (dv/dφ)(v/r), and integrating from the bottom, where the speed is v₀, gives v² = v₀² − 2gr(1 − cos φ): the ball loses 4gr of v² on the way to the top. The radial equation then holds at every instant with the local speed. Toward the centre the string pulls in and gravity contributes −mg cos φ, so T = mv²/r + mg cos φ = mv₀²/r − 2mg + 3mg cos φ. The tension falls by exactly 6mg from bottom to top, whatever the launch speed. At the top T + mg = mv²/r, and T ≥ 0 demands vₜₒₚ ≥ √(gr): for a 1.2 m string that is 3.43 m s⁻¹ at the top, or v₀ ≥ √(5gr) = 7.67 m s⁻¹ at the bottom.

Solving the radial equation assumes the object stays on the circle, and every contact force carries a condition that can break that assumption. A surface can push but not pull, a string can pull but not push, and static friction cannot exceed μₛN. When the algebra returns a negative N, a negative T or a friction demand above μₛN, the object is not following that circle. A car cresting a hump of radius 40 m has mg − N = mv²/r, so N = m(g − v²/r) reaches zero at v = √(gr) = 19.8 m s⁻¹, about 71 km/h; any faster and the car leaves the road. A car too fast for a flat bend slides outward onto a wider arc, and a cart slower than √(gr) at the top of a loop drops off the track. The particle model has blind spots too: a tall vehicle can tip before its tyres slip, which takes torques to predict, and the outward lurch a passenger feels is the body carrying on straight until the door pushes it round; treating that as a force needs the accelerating frame of a later topic.