Derivation

Uniform Circular Motion and Centripetal Acceleration

D-421 Home PU-101 Threads force · symmetry Depends on Galilean Invariance of Newton's Laws

Statement

For a particle of mass \(m\) moving in a circle of radius \(R\) with constant angular speed \(\omega\), the acceleration is directed radially inward with magnitude \(a_c = \omega^2 R = v^2/R\), where \(v = \omega R\) is the constant speed; Newton's second law then requires a net inward force \(F_c = m v^2/R = m \omega^2 R\), and the period of revolution is \(T = 2\pi/\omega = 2\pi R/v\).

Why it matters

Uniform circular motion is the simplest non-trivial trajectory in which the speed is constant yet the velocity is not. It isolates the purely geometric fact that a changing direction of velocity requires an acceleration, and it gives the single formula \(a = v^2/R\) that appears in every curved-path problem from a car on a roundabout to a satellite in orbit.

The result is also the template for the more general non-uniform case, where a tangential component \(a_t = dv/dt\) is added to the radial component. Mastering the uniform case first lets one read off the radial part of any circular motion instantly and focus on the dynamics that change the speed.

Assumptions

The radius \(R\) is constant.If the path is not a circle (an ellipse, a spiral, a general curve), the instantaneous radius of curvature \(\rho\) replaces \(R\) and the radial acceleration is \(v^2/\rho\), which varies along the path; the simple constant-\(R\) formula no longer applies at a glance.

The angular speed \(\omega\) is constant (uniform motion).If \(\omega\) varies, a tangential acceleration \(a_t = R\,\dot\omega = dv/dt\) appears in addition to the radial term, and the total acceleration is no longer purely centripetal; the magnitude is \(\sqrt{(v^2/R)^2 + (dv/dt)^2}\), not \(v^2/R\).

The particle is treated as a point mass.For an extended rigid body in circular motion, different points trace circles of different radii and the body may also rotate about its own centre of mass; one must then add the rotational kinetic energy and torque about the centre of mass separately.

The analysis is in an inertial frame.In a frame co-rotating with the particle, the particle appears at rest and a fictitious “centrifugal force” \(m\omega^2 R\) appears to balance the real inward force. This is not a real force; the centripetal-acceleration statement \(a = v^2/R\) is valid only in an inertial frame where Newton's second law holds in its standard form.

Speeds are non-relativistic, \(v \ll c\).At relativistic speeds the relation between force and acceleration is no longer \(\vec F = m\vec a\) with a constant \(m\); the transverse (centripetal) force required for circular motion grows as \(\gamma m v^2/R\), and the simple \(F = mv^2/R\) underestimates the required force by a factor of \(\gamma = 1/\sqrt{1-v^2/c^2}\).

Derivation

\[ \vec r(t) = R\cos(\omega t)\,\hat{\mathbf x} + R\sin(\omega t)\,\hat{\mathbf y} \]

Parametrise the circular path of radius \(R\) by the angle \(\theta = \omega t\); the position vector has constant magnitude \(R\). A

\[ \vec v(t) = \frac{d\vec r}{dt} = -R\omega\sin(\omega t)\,\hat{\mathbf x} + R\omega\cos(\omega t)\,\hat{\mathbf y} \]

Differentiate each component with respect to time; the chain rule gives a factor \(\omega\) from the argument \(\omega t\). A

\[ v = |\vec v| = R\omega\sqrt{\sin^2(\omega t)+\cos^2(\omega t)} = R\omega \]

The speed is the magnitude of \(\vec v\); the trigonometric identity \(\sin^2\theta+\cos^2\theta=1\) makes it constant. A

\[ \vec a(t) = \frac{d\vec v}{dt} = -R\omega^2\cos(\omega t)\,\hat{\mathbf x} - R\omega^2\sin(\omega t)\,\hat{\mathbf y} \]

Differentiate \(\vec v\) once more; the second derivative of the trigonometric functions brings down another factor \(\omega\) and flips the sign. A

\[ \vec a(t) = -\omega^2\bigl[R\cos(\omega t)\,\hat{\mathbf x} + R\sin(\omega t)\,\hat{\mathbf y}\bigr] = -\omega^2\,\vec r(t) \]

Factor out \(-\omega^2\); the remaining vector in brackets is exactly \(\vec r(t)\). The acceleration is antiparallel to the position vector, i.e. directed toward the centre. B

\[ a = |\vec a| = \omega^2 R = \frac{v^2}{R} \]

Take the magnitude of Step 5; substitute \(v = R\omega\) from Step 3 to obtain the two equivalent forms. A

\[ \vec v\cdot\vec a = \bigl(-R\omega\sin\omega t\bigr)\bigl(-R\omega^2\cos\omega t\bigr) + \bigl(R\omega\cos\omega t\bigr)\bigl(-R\omega^2\sin\omega t\bigr) = 0 \]

The dot product vanishes: acceleration is perpendicular to velocity. Hence \(\frac{d}{dt}\bigl(\tfrac12 v^2\bigr) = \vec v\cdot\vec a = 0\), confirming that the speed is constant. C

\[ T = \frac{2\pi}{\omega} = \frac{2\pi R}{v} \]

The period is the time for \(\omega t\) to advance by \(2\pi\); substituting \(\omega = v/R\) gives the second form. A

\[ F_c = m a_c = \frac{m v^2}{R} = m\omega^2 R \]

Newton's second law: the net force must supply the centripetal acceleration. This is the force required for the circular motion, not a new kind of force. B

\[ \vec a_{\text{total}} = \frac{dv}{dt}\,\hat{\mathbf t} + \frac{v^2}{R}\,\hat{\mathbf r}_{\text{in}} \]

For non-uniform circular motion the acceleration splits into a tangential part (changing speed) and the radial part (changing direction); the uniform case has \(dv/dt=0\). B

Result

\[ a_c = \frac{v^2}{R} = \omega^2 R, \qquad F_c = \frac{m v^2}{R} = m\omega^2 R, \qquad T = \frac{2\pi R}{v} \]

Reading. A particle moving at constant speed \(v\) around a circle of radius \(R\) accelerates toward the centre with magnitude \(v^2/R\). The net force on the particle must point inward with magnitude \(mv^2/R\); whatever physical interaction supplies that force (tension, friction, gravity, normal reaction) is doing the job of “centripetal force.” The period is the time for one full revolution.

Units check. \([v^2/R] = \mathrm{(m^2/s^2)/m = m/s^2}\), matching acceleration. \([mv^2/R] = \mathrm{kg\,(m^2/s^2)/m = kg\,m/s^2 = N}\), matching force. \([2\pi R/v] = \mathrm{m/(m/s) = s}\), matching time. All consistent.

Limiting cases

Very large radius, fixed speed. As \(R\to\infty\) at fixed \(v\), \(a_c = v^2/R\to 0\): the path becomes a straight line and no centripetal force is needed, recovering uniform linear motion.
Very small radius, fixed speed. As \(R\to 0\) at fixed \(v\), \(a_c = v^2/R\to\infty\): an arbitrarily large inward force is required, which is why tight turns at high speed demand enormous friction or tension.
Zero speed. If \(v=0\) the particle is at rest at a point on the circle; \(a_c=0\) and \(F_c=0\), consistent with no motion.
Non-uniform circular motion. When \(v\) changes, the total acceleration is \(\vec a = (dv/dt)\,\hat{\mathbf t} + (v^2/R)\,\hat{\mathbf r}_{\text{in}}\); the radial part still equals \(v^2/R\) at each instant but the tangential part adds a component along the path.
Relativistic limit. As \(v\to c\), the required centripetal force in an inertial frame is \(F = \gamma m v^2/R\) with \(\gamma\to\infty\); the non-relativistic formula \(mv^2/R\) underestimates the force by the Lorentz factor.

Breaks when

Path is not circular. For an ellipse, a spiral, or any general curve the radius of curvature \(\rho\) varies with position; the radial acceleration is \(v^2/\rho\) at each point, not \(v^2/R\) with a single \(R\). One must compute \(\rho\) from the curvature of the specific path.
Non-inertial (rotating) frame. In a frame co-rotating with the particle, the particle is at rest and a fictitious centrifugal force \(m\omega^2 R\) appears to balance the real inward force. The statement \(a = v^2/R\) in the inertial sense does not hold in that frame; one must include the fictitious forces explicitly.
Relativistic speeds. When \(v\) is a significant fraction of \(c\), the transverse force needed for circular motion is \(\gamma m v^2/R\), not \(mv^2/R\). The simple formula fails by a factor of \(\gamma\), which can be large for particles in accelerators.
Extended or deformable body. For a rigid body rotating about a fixed axis, different mass elements have different radii and speeds; the net centripetal force on the body is the integral of \(dm\,v^2/R\) over the body, and internal stresses (tension in a rod, shear in a shaft) must supply the force for each element. The point-mass formula applies only to the centre of mass if the body is not rotating about its own axis.

Failure modes

“Centrifugal force” in the inertial frame. In an inertial frame there is no outward force on the particle. The only real force is inward (tension, friction, gravity, normal reaction). The “centrifugal force” is a fictitious force that appears only in a co-rotating (non-inertial) frame. Writing \(F_{\text{centrifugal}} = mv^2/R\) as a real force in the inertial frame and then setting it equal to the centripetal force is a double-counting error.
Confusing centripetal force with a new type of force. “Centripetal force” is not a fundamental interaction; it is the name for the net inward force required. The actual force may be tension, friction, gravity, or the normal reaction. Students sometimes draw a separate “centripetal force” arrow on a free-body diagram in addition to the real forces, which is incorrect.
Using \(a = v^2/R\) when the speed is not constant. If the particle is speeding up or slowing down, the total acceleration has a tangential component \(a_t = dv/dt\) in addition to the radial component \(v^2/R\). Using only \(v^2/R\) for the total acceleration underestimates the net force when \(dv/dt \neq 0\).
Mixing up \(v\) and \(\omega\). The two forms \(a = v^2/R\) and \(a = \omega^2 R\) are equivalent only when \(v = \omega R\). If one substitutes a linear speed into the \(\omega^2 R\) formula (or vice versa) without the correct conversion, the result is wrong by a factor of \(R\) or \(1/R\).

Discussion

The derivation is almost tautological once the parametrisation \(\vec r = R(\cos\omega t,\sin\omega t)\) is written down: two differentiations give \(\vec a = -\omega^2\vec r\), and the rest is algebra. The physical content is that a vector of constant magnitude whose direction rotates at rate \(\omega\) has a derivative of magnitude \(\omega\) times the original magnitude, perpendicular to it. This is the same reason the velocity is perpendicular to the position vector in circular motion: the derivative of a rotating vector is always perpendicular to the vector itself.

The formula \(a_c = v^2/R\) has a simple geometric interpretation. In a small time \(dt\), the particle moves an arc length \(v\,dt\) and its velocity vector rotates by an angle \(d\theta = \omega\,dt = (v/R)\,dt\). The change in velocity is \(|d\vec v| = v\,d\theta = v^2\,dt/R\), so \(a = |d\vec v|/dt = v^2/R\). This “velocity-vector rotation” argument avoids calculus and makes clear why the acceleration is proportional to \(v^2\) and inversely proportional to \(R\): faster speed means the velocity vector rotates through a larger angle in the same time, and a smaller radius means the same arc length corresponds to a larger angle.

The connection to Newton's second law is straightforward but conceptually important. The “centripetal force” is not a new force; it is the name for whatever net inward force the real interactions provide. In a conical pendulum it is the horizontal component of the string tension; in a car on a flat curve it is static friction; in a satellite orbit it is gravity. The formula \(F_c = mv^2/R\) tells you how much inward force is required for a given \(v\) and \(R\); the physics of the specific situation tells you whether that force is actually available.

At a deeper level, the circular orbit is the simplest solution of the Kepler problem: an inverse-square attraction (gravity or electrostatics) with exactly the right energy for its angular momentum. In Lagrangian mechanics the circular orbit is a critical point of the effective potential \(V_{\text{eff}}(r) = -GMm/r + L^2/(2mr^2)\), and the condition \(dV_{\text{eff}}/dr = 0\) reproduces \(mv^2/r = GMm/r^2\). The stability of the orbit (small radial perturbations oscillate) is the harmonic-oscillator limit of the Kepler problem, and the frequency of those oscillations is the epicyclic frequency. Thus the simple formula \(a = v^2/R\) is the zeroth-order term in a hierarchy that includes precession, rosette orbits, and the full Keplerian dynamics.

Common misconceptions. The most persistent is the belief that a “centrifugal force” pushes the particle outward in the inertial frame. It does not. In the inertial frame the only force is inward, and the particle's inertia (its tendency to move in a straight line) is what makes it “want” to fly off tangentially if the inward force is removed. The centrifugal force is a bookkeeping device in a rotating frame, not a physical interaction. A second misconception is to think that the centripetal force does work on the particle. It does not: the force is always perpendicular to the velocity, so \(\vec F\cdot d\vec r = 0\) and the kinetic energy is constant. A third is to confuse the centripetal acceleration with the total acceleration in non-uniform circular motion; when the speed changes, there is also a tangential acceleration, and the total acceleration is the vector sum of the two components.

Worked examples

Example 1. A 1200 kg car negotiates a flat (unbanked) circular curve of radius \(R = 80\ \mathrm{m}\). The coefficient of static friction between the tyres and the road is \(\mu_s = 0.65\). Find the maximum speed at which the car can negotiate the curve without skidding. (\(g = 9.8\ \mathrm{m/s^2}\).)

\[ F_c = \frac{m v^2}{R} \]

The centripetal force required for circular motion of radius \(R\) at speed \(v\). A

\[ F_{\text{friction}} \le \mu_s N = \mu_s m g \]

On a flat road the normal force is \(N = mg\); the maximum static friction force is \(\mu_s mg\). This friction force supplies the centripetal force. A

\[ \frac{m v_{\max}^2}{R} = \mu_s m g \]

At the maximum speed the required centripetal force equals the maximum available friction; mass cancels. B

\[ v_{\max} = \sqrt{\mu_s g R} \]

Solve for \(v_{\max}\). A

\[ v_{\max} = \sqrt{(0.65)(9.8\ \mathrm{m/s^2})(80\ \mathrm{m})} = \sqrt{509.6\ \mathrm{m^2/s^2}} \]

Insert the numerical values. A

\[ v_{\max} = \sqrt{509.6} \approx 22.6\ \mathrm{m/s} \]

Evaluate the square root. A

\[ v_{\max} \approx 22.6\ \mathrm{m/s} \approx 81\ \mathrm{km/h} \]

Reading. The car can safely negotiate the curve at up to about 22.6 m/s (81 km/h). Above this speed the required centripetal force \(mv^2/R\) exceeds the maximum static friction \(\mu_s mg\), and the car skids outward.

Units check. \([\mu_s g R] = \mathrm{(m/s^2)(m) = m^2/s^2}\), whose square root is m/s. Correct.

Example 2. A 0.50 kg puck on a smooth horizontal table is attached to a string through a hole at the centre and moves in a horizontal circle of radius \(R = 0.80\ \mathrm{m}\) at a constant speed \(v = 3.0\ \mathrm{m/s}\). Find (a) the tension in the string, (b) the period of revolution, and (c) the centripetal acceleration. (\(g = 9.8\ \mathrm{m/s^2}\).)

\[ T = \frac{m v^2}{R} \]

The string tension is the only horizontal force, so it supplies the centripetal force; gravity is balanced by the table’s normal force. A

\[ T = \frac{(0.50\ \mathrm{kg})(3.0\ \mathrm{m/s})^2}{0.80\ \mathrm{m}} = \frac{0.50 \times 9.0}{0.80}\ \mathrm{N} \]

Insert the numerical values. A

\[ T = \frac{4.5}{0.80}\ \mathrm{N} = 5.6\ \mathrm{N} \]

Evaluate. A

\[ T_{\text{period}} = \frac{2\pi R}{v} = \frac{2\pi(0.80\ \mathrm{m})}{3.0\ \mathrm{m/s}} \]

The period is the circumference divided by the speed. A

\[ T_{\text{period}} = \frac{5.03}{3.0}\ \mathrm{s} \approx 1.68\ \mathrm{s} \]

Evaluate: \(2\pi(0.80) = 5.027\ \mathrm{m}\); \(5.027/3.0 = 1.676\ \mathrm{s}\). A

\[ a_c = \frac{v^2}{R} = \frac{(3.0\ \mathrm{m/s})^2}{0.80\ \mathrm{m}} = \frac{9.0}{0.80}\ \mathrm{m/s^2} = 11.3\ \mathrm{m/s^2} \]

The centripetal acceleration. A

\[ T = 5.6\ \mathrm{N}, \qquad T_{\text{period}} \approx 1.7\ \mathrm{s}, \qquad a_c \approx 11\ \mathrm{m/s^2} \]

Reading. The string tension is 5.6 N, directed inward along the string. The mass completes one revolution every 1.7 s. The centripetal acceleration is 11.3 m/s\(^2\), about 1.15 times \(g\), directed toward the centre of the circle.

Units check. \([mv^2/R] = \mathrm{kg\,(m^2/s^2)/m = kg\,m/s^2 = N}\). \([2\pi R/v] = \mathrm{m/(m/s) = s}\). \([v^2/R] = \mathrm{(m^2/s^2)/m = m/s^2}\). All consistent.

Problems

A stone of mass 0.25 kg slides on smooth ice, tied to a post by a string, in a horizontal circle of radius 1.0 m at a constant speed of 4.0 m/s. Find the tension in the string and the period of revolution.
Solution
Tension: \(T = mv^2/R = (0.25)(4.0)^2/1.0 = 0.25 \times 16.0 = 4.0\ \mathrm{N}\). Period: \(T_{\text{period}} = 2\pi R/v = 2\pi(1.0)/4.0 = 6.283/4.0 = 1.57\ \mathrm{s}\).
A 1500 kg car travels around a flat circular curve of radius 50 m. If the coefficient of static friction is 0.70, find the maximum safe speed. (\(g = 9.8\ \mathrm{m/s^2}\).)
Solution
\(v_{\max} = \sqrt{\mu_s g R} = \sqrt{(0.70)(9.8)(50)} = \sqrt{343} = 18.5\ \mathrm{m/s}\). In km/h: \(18.5 \times 3.6 = 66.7\ \mathrm{km/h}\).
A satellite orbits the Earth in a circular orbit of radius \(r = 7000\ \mathrm{km}\). Find its orbital speed and period. (Use \(G = 6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}\), \(M_{\oplus} = 5.97\times10^{24}\ \mathrm{kg}\).)
Solution
Orbital speed: \(v = \sqrt{GM/r} = \sqrt{(6.67\times10^{-11})(5.97\times10^{24})/(7.0\times10^6)} = \sqrt{3.982\times10^{14}/7.0\times10^6} = \sqrt{5.69\times10^7} = 7.54\times10^3\ \mathrm{m/s} = 7.54\ \mathrm{km/s}\). Period: \(T = 2\pi r/v = 2\pi(7.0\times10^6)/(7.54\times10^3) = 4.398\times10^7/7.54\times10^3 = 5833\ \mathrm{s} \approx 97.2\ \mathrm{min}\).
A 2.0 kg mass is attached to a string and whirled in a vertical circle of radius 0.60 m. At the top of the circle the speed is 3.0 m/s. Find the tension in the string at the top and at the bottom of the circle. (\(g = 9.8\ \mathrm{m/s^2}\).)
Solution
At the top, both tension and gravity point downward (toward the centre): \(T_{\text{top}} + mg = mv^2/R\), so \(T_{\text{top}} = mv^2/R - mg = (2.0)(9.0)/0.60 - (2.0)(9.8) = 30.0 - 19.6 = 10.4\ \mathrm{N}\). The speed is not constant in a vertical circle; energy conservation from top to bottom (a drop of \(2R\)) gives \(v_b^2 = v_t^2 + 4gR = 9.0 + 4(9.8)(0.60) = 32.5\ \mathrm{m^2/s^2}\). At the bottom, tension points up (toward the centre) and gravity down: \(T_{\text{bottom}} = mv_b^2/R + mg = (2.0)(32.5)/0.60 + 19.6 = 108.4 + 19.6 = 128\ \mathrm{N}\). The difference, \(T_{\text{bottom}} - T_{\text{top}} = 6mg\), is independent of the speed.
A 1000 kg car enters a banked curve of radius \(R = 100\ \mathrm{m}\) at a speed of 20 m/s. The curve is banked at an angle \(\theta\) such that no friction is required. Find \(\theta\). (\(g = 9.8\ \mathrm{m/s^2}\).)
Solution
For a frictionless banked curve, the horizontal component of the normal force supplies the centripetal force: \(N\sin\theta = mv^2/R\), and the vertical component balances gravity: \(N\cos\theta = mg\). Dividing: \(\tan\theta = v^2/(Rg) = (20)^2/(100 \times 9.8) = 400/980 = 0.408\). Thus \(\theta = \arctan(0.408) = 22.2^\circ\).