Centripetal force in circular motion

Identify the circle’s centre and radius $r>0$. Velocity points along the tangent while radial acceleration points towards the centre. For a classical object of constant mass, the required radial resultant is $F_z=\frac{mv^2}{r}$. Do not draw a new interaction in addition to real forces: “centripetal” means inward and describes their radial sum. This radial balance is Newton’s second law for the inward component.[1]

For uniform circular motion, the entire net force is radial. If speed changes along the same circle, a tangential resultant also exists. The expression $\frac{mv^2}{r}$ still gives the radial component but no longer the full force magnitude.[2]

Compare two nearby instants of uniform circular motion. The velocity vectors have equal lengths but different directions. Move their tails together and connect their tips: that connecting vector is the velocity change. The velocity vectors enclose the same angle as the radius vectors, so their triangles are similar.[2]

Higher speed turns the direction of motion more rapidly; it also makes the velocity vector being redirected longer. These two factors explain why doubling speed on the same circle requires four times the radial acceleration. Newton’s second law then multiplies this kinematic quantity by mass.[2][1]

A circle with constant angular velocity can be parameterised directly. Put the origin at its centre. One derivative gives tangential velocity; two derivatives give an inward vector. This exposes the direction change without inventing an additional force law.[2]

For nonuniform motion, replace ωt by θ(t). Speed is $v=r|\dot\theta|$, while radial acceleration points inward with $a_r=r\dot\theta^2=\frac{v^2}{r}$. Taking the positive tangent towards increasing θ gives the signed tangential component $a_\theta=r\ddot\theta$. This equals $\frac{\mathrm dv}{\mathrm dt}$ only for positive angular velocity; for negative angular velocity their signs are opposite. Total acceleration magnitude remains $a=\sqrt{a_r^2+a_\theta^2}$.

90 revolutions per minute corresponds to f = 1.5 s⁻¹ and ω = 3π rad/s. Include the factor 60 before squaring. v is tangential speed, not angular displacement. kg·(m/s)²/m gives kg·m/s² = N. Angular velocity instead describes angular change per time; for uniform circular motion the period connects it to frequency.

At fixed v, a larger r reduces required force. At fixed ω, force increases with r because $v=\omega r$ increases too. These statements hold different quantities constant and are consistent. Doubling v at unchanged r and m requires four times the radial force.

On a level unbanked road, static friction can provide horizontal radial force. In the simple model $N=mg$, $\frac{mv^2}{r}\le\mu_smg$. Static friction is not always μs mg; that is its maximum available magnitude. Without other horizontal forces, the limiting speed is $v_{\mathrm{max}}=\sqrt{\mu_sgr}$.[1]

In a vertical circle, add all radial contributions. At the bottom of a tethered body’s circle, tension S obeys $S-mg=\frac{mv^2}{r}$. At the top, $S+mg=\frac{mv^2}{r}$. Tension is therefore not everywhere equal to required radial force. A string can pull but cannot push.

Force diagrams need real forces at their application points. Do not add an extra $\frac{mv^2}{r}$ arrow to tension and weight already drawn: $\frac{mv^2}{r}$ is the result of the radial balance, not another term. In an idealised circular orbit around a spherically symmetric mass, gravity can supply the radial resultant. The gravitational law determines the available force; the circular-motion balance determines the corresponding orbital speed.[3]

At the top of a vertical tethered circle, the centre lies below the object. Both weight and tension therefore point inward. Choose inward as positive here. Their sum supplies the required radial force; subtract the weight contribution to find tension.[1]

If the rearrangement demanded negative tension, a taut-string model would be impossible: a string cannot push. The limiting case $S=0$ at the top gives $v=\sqrt{gr}$. That condition concerns this point only; reaching it through the entire circle also requires checking earlier motion, energy and contact conditions.[1]

In an inertial frame, the real radial resultant accelerates the body inward. In a frame rotating at constant angular velocity, the same body can be stationary. Writing its motion equation in the usual form requires an outward centrifugal inertial force. This is not an additional interaction with another body.[1]

Centrifugal inertial force is not the reaction to string tension. The Newton-third-law partner acts on the string, not additionally on the orbiting mass. Motion relative to a rotating frame may require other inertial forces; centrifugal force alone is not generally sufficient.

If guidance suddenly disappears and no forces remain, motion continues straight with instantaneous tangential velocity. It does not suddenly become radially outward. If guidance is insufficient while other forces remain, determine the new path from those remaining forces. On the constrained circular path, radial force is perpendicular to instantaneous velocity. Its own power contribution is therefore zero; a tangential force can still change speed.

The equation is classical and assumes specified circular geometry and constant mass. Relativistic particle trajectories cannot simply use F = ma with unchanged m. For an arbitrary smooth curve, local curvature radius can replace circle radius and may vary along the path.

Circular motion changes velocity direction and requires radial acceleration. Balance all real forces inward without adding another force type. Changing speed adds a tangential component. Check whether tension, contact or static friction can actually supply the required force, and use the balance appropriate to the inertial or rotating frame.