Circular motion

Centripetal acceleration points inward

Circular motion requires continuous turning of the velocity vector. The required acceleration points toward the center; ‘centripetal force’ names the net inward force, not an additional interaction.

Compare nearby velocity vectors

For speed v on a circle of radius r, the velocity direction turns through Δθ while the velocity change has magnitude Δv ≈ vΔθ. Arc length gives vΔt = rΔθ, so Δv/Δt ≈ v(vΔt/r)/Δt = v²/r. In vector form a_c = −(v²/r) r̂.

A 0.50 kg ball moves at 6 m/s on a 2 m string. Required inward force is m v²/r = 0.5(36)/2 = 9.0 N. If the string can supply only 7 N, the motion cannot remain circular at that speed.

Constant speed still has acceleration

Tangential acceleration is zero only when speed is constant. If the path radius grows without bound, v²/r tends to zero and the circular result approaches straight-line motion.

Work it: A 1200 kg car rounds a 50 m curve at 15 m/s. Compute the inward force and minimum static-friction coefficient on a level road; explain why doubling speed quadruples the requirement.