LocusBlog

Centripetal acceleration

A stone whirled on a string at steady speed is accelerating every instant, and the centripetal acceleration has magnitude

$$a_c = \frac{v^2}{r}$$

pointed at the centre of the circle. $v$ is the speed and $r$ is the radius. Nothing in that formula refers to mass, so it is a statement about the geometry of going round a corner, not about what is doing the pushing.

The word to fight over is “accelerating”. Speed is constant and velocity is not, because velocity is a vector and its direction turns continuously. Acceleration is the rate of change of the velocity vector, so a turning velocity of fixed length is an acceleration of fixed magnitude. Constant speed buys you nothing here.

Where the square comes from

Take two instants separated by $\Delta t$, in which the object turns through an angle $\Delta\theta$. The two position vectors $\vec r_1$ and $\vec r_2$ both have length $r$ and are separated by $\Delta\theta$, so they and their difference $\Delta\vec r$ form an isosceles triangle with apex angle $\Delta\theta$.

Now the velocity. In uniform circular motion the velocity is tangent to the circle, so it is perpendicular to the radius at every instant. When the radius turns by $\Delta\theta$, the velocity turns by the same $\Delta\theta$. Both velocity vectors have length $v$, so $\vec v_1$, $\vec v_2$ and their difference $\Delta\vec v$ form a second isosceles triangle with the same apex angle. Two isosceles triangles with equal apex angles are similar, so corresponding sides are in the same ratio:

$$\frac{|\Delta\vec v|}{v} = \frac{|\Delta\vec r|}{r}$$

Divide both sides by $\Delta t$:

$$\frac{|\Delta\vec v|}{\Delta t} = \frac{v}{r}\cdot\frac{|\Delta\vec r|}{\Delta t}$$

The left side becomes the magnitude of the acceleration as $\Delta t \to 0$. On the right, $|\Delta\vec r|$ is the straight chord between the two positions, and in that same limit the chord and the arc converge, so $|\Delta\vec r|/\Delta t \to v$. What is left is $a_c = v^2/r$. OpenStax runs the derivation this way in section 4.4 of University Physics Volume 1.

The direction falls out of the same picture. The velocity triangle is isosceles with apex $\Delta\theta$, so each base angle is $(180^\circ - \Delta\theta)/2$, which tends to $90^\circ$ as $\Delta\theta$ shrinks. In the limit $\Delta\vec v$ is perpendicular to $\vec v$, and it leans toward the inside of the turn. Perpendicular to the tangent and inward is exactly the radius, pointing at the centre.

If you would rather work in angular terms, $v = \omega r$ for angular velocity $\omega$, so

$$a_c = \frac{(\omega r)^2}{r} = \omega^2 r$$

The two forms disagree about how $a_c$ responds to radius, and both are right. Hold the speed fixed and a wider circle is gentler. Hold the angular rate fixed, as a point on a spinning disc must, and a wider circle is harsher - the rim of a record player is thrown around harder than the label.

Something has to supply it

$a_c$ is a kinematic requirement. Newton’s second law then says a real net force of magnitude $mv^2/r$ must point inward, and the interesting question in every problem is which force that is.

For a stone on a string it is tension, and the string breaks when $mv^2/r$ exceeds what it can carry. For a car on a level curve it is static friction between tyre and road, capped at $\mu_s mg$. For the Moon it is gravity, so $GMm/r^2 = mv^2/r$, which collapses to $v^2 = GM/r$ and is where orbital speed comes from. Three different agents, one job.

Take the car. A level curve of radius $50.0$ m with $\mu_s = 0.70$ between tyre and dry road. Friction is the only horizontal force, so the fastest cornering speed is set by $\mu_s mg = mv^2/r$. Mass cancels, as it must, since neither side knows about it:

$$v = \sqrt{\mu_s g r} = \sqrt{0.70 \times 9.8 \times 50.0} = \sqrt{343} = 18.5\ \text{m/s}$$

That is 66.7 km/h. The acceleration at that speed is $a_c = 343/50.0 = 6.86$ m/s$^2$, which is $0.70g$ - it has to be, because friction was capped at $0.70mg$ and friction is the entire net force. Check it through the angular form too: $\omega = 18.52/50.0 = 0.3704$ rad/s and $\omega^2 r = 0.1372 \times 50.0 = 6.86$ m/s$^2$. Same number, and a full lap of that circle takes $2\pi/\omega = 17.0$ s.

Bank the curve and friction stops being the only option. Tilt the road by $\theta$ and the normal force $N$ tips inward with it. Vertically $N\cos\theta = mg$, and horizontally the inward component supplies the centripetal force, $N\sin\theta = mv^2/r$. Divide the second by the first and $N$ and $m$ both vanish:

$$\tan\theta = \frac{v^2}{rg}$$

For the same 50.0 m curve at the same 18.5 m/s, $\tan\theta = 343/490 = 0.700$, so $\theta = 35.0^\circ$. The tangent of the ideal bank angle came out numerically equal to $\mu_s$, and that is not a coincidence: both conditions are the same sentence, “the road can push this car sideways with $0.70mg$”, once through friction and once through geometry. A curve banked at $35.0^\circ$ holds that speed on sheet ice.

Centripetal force is not a fourth arrow

Here is the mistake that costs the most marks. A student draws the car on the level curve, marks weight down, normal force up, friction inward, and then adds one more inward arrow labelled “centripetal force”. Newton’s second law along the inward direction now reads $f_s + F_c = ma_c$. With $F_c$ standing for $mv^2/r$ and $a_c$ for $v^2/r$, the two cancel and the equation collapses to $f_s = 0$: the curve can be taken on ice, at any speed. If instead the student keeps both terms as real inward pushes each capped at $\mu_s mg$, the answer goes the other way, $v = \sqrt{2 \times 343} = 26.2$ m/s, or 94 km/h on a corner that lets go at 67.

“Centripetal” is an adjective describing a direction, not a kind of force. $mv^2/r$ belongs on the right-hand side of $\sum F = ma$, never on the left. The left-hand side gets tension, friction, gravity, normal force - things with an agent that can be named. Write the free-body diagram, resolve along the inward radius, set the sum equal to $mv^2/r$, and the bookkeeping cannot double count.

Centrifugal force deserves an honest answer rather than a scolding. In the ground frame there is no outward force on the cornering passenger; they are travelling straight and the door is turning into them. Write Newton’s second law in the car’s rotating frame instead and an outward term $m\omega^2 r$ appears, which is legitimate arithmetic in that frame and not something any object exerts on anything.

Non-uniform circular motion adds a tangential piece and keeps this one intact. The acceleration becomes $\vec a = a_t\hat t + a_c\hat r$, where $a_t$ changes the speed and $a_c = v^2/r$ still uses whatever the instantaneous speed is. The circular motion problems mix the string, friction and gravity cases so the supplying force keeps moving, and the simulations let you vary $v$ and $r$ and watch the inward arrow rescale.

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