## Great Formulas: V Squared Over r

All physics classes cover the formula for acceleration when the motion is in a circle: \(a=v^2 / r \). \(a\) is the acceleration of the mass, \(v\) is its velocity as it moves around the circle, and \(r\) is the radius of the circle. In an example that is always used, the formula lets us compute the tension in a string when a mass at the end of the string is being spun in a circle. For a given \(v\) and \(r\), we compute the acceleration \(a\). Then, Newton’s Second Law, \(F=ma\) gives the force; the tension in the string.

Although the textbooks do not make a fuss over the formula, its discovery by Isaac Newton in 1665 was a landmark in the history of science. Deriving the formula required a great deal of insight into the nature of motion, and it was the starting point for Newton’s work on gravity and the motion of planets. In an earlier post, I showed how Newton inferred the inverse square law of gravity by applying the formula to the Moon’s motion.

What Newton saw was that with circular motion, there are two simultaneous motions superimposed on one another. One of these is a straight-line motion, and the other is an accelerating motion toward the center.

The mass in the drawing would move, in the absence any external forces, from \(A\) to \(B\) in the time \(\Delta t\). Instead, the mass is at point \(C\) after time \(\Delta t\). The difference is that it has also “fallen” through the distance \(d\). The combination of straight line motion and inward motion is such that the mass follows the circular path. It is a bit tricky to depict or visualize because the two motions are going on continuously, and the \(\Delta t\) intervals are infinitely small.

The inward motion is due to a center-directed, constant force, which causes an acceleration toward the center. In the example of a mass being whirled around at the end of a string, the tension in the string provides the force. If it is a planet revolving around the Sun, then the inward directed force comes from the Sun’s gravity. In any case, the acceleration value is exactly such that the “falling” object stays on the circle.

#### \(v^2 / r\) Derived

College textbooks normally derive the formula using calculus (and high school books don’t include a derivation at all). Here, I will work out the formula using only elementary methematics.

As a preliminary step, we will need a little formula for the distance \(d\) in the drawing below.

The formula will actually apply only when the angle \(\theta\) is very small.

In the drawing, we have a right triangle whose hypotenuse is longer than \(r\) by the amount \(d\), so that, by the Pythagorean theorem:

\[r^2+s^2=(r+d)^2=r^2+2rd+d^2\]

We are seeking an expression for \(d\), and we write

\[d=\frac{s^2}{2r}-\frac{d^2}{2r}\]

This seems wrong; we have not solved for \(d\) because there is a \(d^2\) term on the right. But here’s the rub: we only care about the case where \(\theta\) is small. If \(\theta\) is small, then \(d\) is small. If \(d\) is small, then \(d^2\) is

*very*small, and we can ignore the second term on the right. For example, if \(d=0.001\), then \(d^2=0.000001\). More formally, we have,\[\lim_{\theta\rightarrow 0}d=\frac{s^2}{2r} \tag{1}\]

This reasoning may seem a bit suspect, especially to those who have not been exposed to calculus. However, the procedure and formula are sound, and when \(\theta\) is infinitely small, formula (1) is not an approximation; it is

*exact*.Now we return to the circular motion situation.

The diagram above shows what happens in a time period \(\Delta t\). The straight-line distance is \(v \cdot\Delta t\), and the mass has also fallen toward the center by the distance \(d\). The inward force, and therefore the acceleration, is constant. One of Galileo’s formulas gives us the distance fallen as a function of time:

\[d=\frac{1}{2}a (\Delta t)^2\mbox{ (from kinematics)} \tag{2}\]

where \(a\) is the (as yet unknown) acceleration. But from formula (1), we also have

\[d=\frac{(v\:\Delta t)^2}{2r}=\frac{1}{2}\frac{v^2}{r}(\Delta t)^2\mbox{ (from geometry)} \tag{3}\]

For these two expressions for \(d\) to be equal, we can see what \(a\) must be:

\[a=\frac{v^2}{r} \tag{4}\]

For a given \(v\) and \(r\), this is the acceleration value that will keep the mass “on the circle”.

The acceleration \(a\) in (4) is directed

*inward*- toward the center of the circle. The associated force on the moving mass (from \(F=ma\)) is also inward, and Newton invented the term*centripetal force*to describe it. This conception alone was a significant insight. Before Newton, others had used the term*centrifugal*for the force; meaning an outward directed force. Aristotle and Descartes believed that an object “wanted” to go in a straight line, and if it moved in a curve*the object itself*would exert a force pulling it toward the straight line.#### Another Way to Say It

It is worth pointing out an equivalent way to write the great formula. Recall that when an angle \(\theta\) is measured in radians, the length of the circle’s arc that it subtends is \(s=r \theta\). For example, when \(\theta\) is a full circle (that is, \(\theta = 2\pi\)), then \(s=r\cdot2\pi\), which is the circle’s circumference. Now imagine a point moving along the circle (i.e., \(\theta\) is changing). By convention, we denote the rate of change of \(\theta\) with the symbol \(\omega\) (omega), its units being

*radians per second.*It’s not hard to see that the speed \(v\) of the object is just \(v=r\omega\). Using this expression in place of \(v\) in (4), we get,\[a=\frac{(r\omega)^2}{r}=r\omega^2 \tag{5}\]

This gives us acceleration in terms of rotation rate, and that is often easier than using the speed \(v\).

#### Some Implications of the Formula

This fabulous formula does not apply to all motion that is not in a straight line. For example, the planets move in ellipses, not circles, so that r is not constant. Also, a planet’s velocity is not constant as it moves around its orbit. For these more complex situations, a fancier formula is required. Even so, the insight behind the \(v^2 / r \) was Newton’s starting point in working out the more general case.

Even if it does not apply all situations, the formula is central to many important physics topics. I plan to do future posts on at least two other uses of the formula:

- Kepler’s Third Law, which relates a planet’s orbital period with its distance from the Sun, is easy to derive using the formula. (perhaps you would like to try working out the math: the law says that, for all planets in a solar system, \( T^2 / r^3 \) is a constant, where \(T\) is the time for one orbit, and \(r\) is the orbit radius.)
- The invention of the cyclotron, the original “atom smasher” by Ernest Lawrence in 1929, centers around the \(v^2 / r \) formula.

July 13th, 2012 at 3:50 am

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October 26th, 2012 at 3:41 am

Thanks a bunch! Finally someone that explains this acceleration in a way I can understand it! For me this is very concrete and tangible. I have a hard time proving things with calculus. To me calculus was built upon summations and the Pythagorean theorem so why not prove everything using summations and the Pythagorean theorem?? So I thank you!

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