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University physics · Feynman Vol. I · Ch. 9

Newton’s Laws of Dynamics

This lesson follows chapter 9 of the Feynman Lectures, where Newton’s laws stop being a statement and become a tool for calculation. The first two simulations deal with force, momentum and velocity as a vector; the others put those ideas to work predicting a motion step by step, with a recipe that can be followed by hand, in a table, or handed over to a machine. The same recipe serves for a mass on a spring and for a planet going round the Sun. The original chapter can be read free of charge on the Caltech website.

  1. 1Force and momentum
  2. 2Velocity as a vector
  3. 3The spring step by step
  4. 4Precision
  5. 5The orbit
STEP 1

What does a force change in a body?

On the bench, two trolleys stand on a frictionless track, A with 1 kg and B with 4 kg. The Push button applies the same force to each, 2 N, for the same 2 s, and then lets them run on their own. While the force acts, A's velocity climbs four times as fast as B's; once it stops, A carries on at 4 m/s and B at 1 m/s.

From the end of the push onwards, both lines on the graph run flat. With no force, neither trolley gains or loses speed, and B would keep its 1 m/s for as much track as there was. The trolleys waiting before the button is pressed are the same case with zero velocity. This is inertia, which Galileo had already grasped.

The second law is about momentum, the product of mass and velocity, and it states that force is the rate at which momentum changes in time. With the mass constant, the derivative reaches only the velocity and the law becomes \(F = ma\): the same 2 N give the 1 kg trolley an acceleration of 2 m/s² and the 4 kg one an acceleration of 0.5 m/s². This is the sense in which mass measures inertia, a body's resistance to having its velocity changed, and the bench shows it in the slopes of the two lines during the push.

force, while it acts   velocity

The final momenta hold the most curious part. A ends with 1 kg times 4 m/s, B with 4 kg times 1 m/s, and both have the same 4 kg·m/s. Adding up the second law over the push, the change in momentum is the force times the time it acted for, the impulse \(F\,T\), and the mass never enters the sum. Any pair of masses on the controls ends with equal momenta; what the mass decides is how that momentum is split between a lot of mass and little velocity or the other way round. The secondary-school lesson on impulse uses the same relation to explain why a longer collision hurts less.

The law leaves one thing open, and it is worth saying which. It does not tell us what a force is or where one comes from; it tells us how much the momentum changes per second while the force acts. Predicting a motion takes a second piece of information, the force law of each case, and step 3 uses the spring's to work out a whole motion. The secondary-school lesson on the second law works with \(F = ma\) for given forces; what matters here is the form with momentum, the one Newton himself wrote down, under the name of quantity of motion.

\(\vec F = \dfrac{d\vec p}{dt},\quad \vec p = m\vec v\)\(F\,T = \Delta p\)\(\vec F\) is the force, \(\vec p\) the momentum, \(m\) the mass and \(\vec v\) the velocity; \(F\,T\) is the impulse of a constant force acting for a time \(T\), and \(\Delta p\) the change in momentum it produces.

In Feynman: §9-1 Momentum and force ↗

Let's discuss

  • Set A to 0.5 kg and B to 8 kg. How many times larger is A's final velocity than B's? And the final momenta?
  • Halve the force and double the duration of the push. What changes on the graph? And in the final momenta?
  • With the two masses equal, what happens to the two lines on the graph?
  • During the push the two momenta always read the same in the panels, although the velocities are quite different. Why? Compare with the F·t panel.
  • One to think through, since the bench has no friction: on a real track, what would the lines look like after the push? What does that say about friction?
Velocities A · B (m/s)
Momenta A · B (kg·m/s)
Impulse F·t (N·s)
STEP 2

Is a car on a bend with a steady speedometer accelerating?

The bench sets a point turning on a circle of radius 2 m, once round every 6.28 s and at 2 m/s throughout. A speedometer riding on it would sit at 2 m/s, and yet the blue velocity arrow is never the same at two instants, because it turns with the point. Physics keeps speed and velocity apart for exactly this reason. Speed is the size of the velocity, the figure on the speedometer; velocity is that figure together with a direction, and changing the direction alone already changes the velocity.

To see by how much it changes, the bench marks the point at two instants \(\Delta t\) apart and draws, alongside, the two velocities leaving a common origin. The red arrow running from the tip of the first to the tip of the second is \(\Delta\vec v\), what had to be added to the first to arrive at the second. With \(\Delta t = 0.1\) s the two arrows almost coincide and \(\Delta\vec v\) measures only 0.2 m/s, but divided by the 0.1 s it gives 1.999 m/s², an acceleration that does not fade away as the interval shrinks.

Copied onto the circle, at the point of the turn midway between the two instants, the same arrow points straight at the centre, whatever \(\Delta t\) is. It is this arrow that bends the path without touching the speed, and the ratio \(|\Delta\vec v|/\Delta t\) tends to \(v^2/R\), 2 m/s² with the starting values, the centripetal acceleration of the secondary-school lesson on circular motion. The car in the title, then, is accelerating, and the force that turns it comes from the road, through the grip of the tyres.

The bars on the right of the panel hold the same blue arrow as two numbers. Once two axes are chosen, \(v_x\) says how many metres per second the point travels in the x direction and \(v_y\) in the y direction, and the bars on the bench show the two trading values over the turn while \(\sqrt{v_x^2+v_y^2}\) stays at 2 m/s. Acceleration and force split up in the same way, and the second law holds on each axis separately. A force with no horizontal component cannot alter the horizontal velocity, which is why the ball in the Launch scenario, in step 6 of the chapter 8 lesson, keeps its 5 m/s horizontally from start to finish of the flight.

\(\vec v = (v_x, v_y),\ |\vec v| = \sqrt{v_x^2 + v_y^2}\)\(\vec a = \lim_{\Delta t\to0} \dfrac{\Delta \vec v}{\Delta t}\)\(F_x = m\,a_x,\ F_y = m\,a_y\)\(v_x\) and \(v_y\) are the components of the velocity along the two axes, \(|\vec v|\) is the speed and \(\vec a\) the acceleration; \(F_x\), \(F_y\), \(a_x\) and \(a_y\) are the components of the force and of the acceleration.

In Feynman: §9-2 Speed and velocity ↗ · §9-3 Components of velocity, acceleration, and force ↗

Let's discuss

  • Take \(\Delta t\) from 1 s down to 0.01 s. What value does \(|\Delta\vec v|/\Delta t\) approach? From which \(\Delta t\) on is the gap to \(v^2/R\) below 1 %?
  • With \(\Delta t = 1\) s the two velocities are already quite different. Does the red arrow copied onto the circle still point at the centre?
  • Double the angular velocity without changing the radius. By how many times does the speed grow? And the acceleration?
  • At which points of the turn is \(v_x\) zero? Where does the velocity point there?
Speed |v|
|Δv|/Δt
v²/R
STEP 3

How can we predict a spring's motion without solving an equation?

The bench hooks a 1 kg mass to a 2 N/m spring, draws the mass 0.6 m to the right and releases it. Nobody needs to know beforehand what motion will follow. The table under the graph is enough, filled in row by row with additions and multiplications, and each click on One more step adds a row to the table and a dot to the graph.

The second law on its own could not do this job. It ties force to the change in velocity, but someone has to say how large the force is at each position, and for the spring the answer is Hooke's law: stretched by \(x\) metres, it pulls back with \(kx\) newtons. Put the two together and the acceleration is \(-(k/m)\,x\) at every instant, a rule that says how the velocity is about to change from wherever the mass happens to be.

The table applies the rule with a step \(\varepsilon\) of 0.05 s. At t = 0 the mass is at 0.6 m and the acceleration is −2 × 0.6 = −1.2 m/s². The velocity in the first row is not the one at instant zero, which is nil, but the one in the middle of the first interval, at 0.025 s, and half a step of acceleration takes it to −0.03 m/s. With it the position at 0.05 s becomes 0.6 − 0.05 × 0.03 = 0.5985 m, the acceleration there turns into −1.197 m/s², and the velocity at 0.075 s comes out as −0.03 − 0.05 × 1.197 = −0.0899 m/s. Every new row repeats these three sums.

spring force on the mass   velocity

The columns of the table are out of step with one another: x and a hold at the instants 0, 0.05, 0.10 s, and v holds between them, at 0.025, 0.075, 0.125 s. Each velocity carries the position from one row to the next, and each acceleration updates the velocity from one mid-interval to the next. The offset is deliberate. A velocity taken at the start of an interval is already stale by its end, and travelling with it always errs to the same side, an error that builds up cycle after cycle, as Euler's method in step 4 makes visible; the mid-interval one sits close to the average over the interval, and the deviations either way nearly cancel. Feynman organises the calculation this way in the chapter, with other starting values and another table.

Compute a cycle fills the table for a whole period. The dots land on the dashed curve, \(x_0\cos\omega t\) with \(\omega = \sqrt{k/m}\), the exact solution the differential calculus provides and the table never used. The period comes from the crossings of x = 0, interpolated between the two neighbouring dots: the mass crosses zero heading left at 1.11 s and on its way back at 3.33 s, which makes a cycle of 4.44 s, equal to \(2\pi\sqrt{m/k}\) to the two decimal places on screen. The amplitude has no place in that formula, and the bench can confirm that pulling the spring further or less far leaves the period where it was.

\(m\,\dfrac{d^2x}{dt^2} = -k\,x\)\(x(t+\varepsilon) = x(t) + \varepsilon\,v(t+\tfrac{\varepsilon}{2})\)\(v(t+\tfrac{3\varepsilon}{2}) =\) \(v(t+\tfrac{\varepsilon}{2}) + \varepsilon\,a(t+\varepsilon)\)\(T = 2\pi\sqrt{m/k}\)\(k\) is the spring constant, \(m\) the mass, \(x\) the displacement from equilibrium and \(\varepsilon\) the time step; the acceleration at each instant is \(a = -(k/m)\,x\), and \(T\) is the period.

In Feynman: §9-4 What is the force? ↗ · §9-5 Meaning of the dynamical equations ↗ · §9-6 Numerical solution of the equations ↗

Let's discuss

  • In which row of the table does the acceleration change sign? Where is the mass at that instant, and what does the velocity do from then on?
  • Set k = 1 N/m, m = 1 kg and ε = 0.2 s and compute a cycle. What period comes out? Compare it with 2π.
  • Quadruple the mass without changing the spring. By how many times does the period grow? And if the amplitude goes from 0.6 m to 1 m?
  • Take ε to 0.5 s. How many dots fit into a cycle, and how far does the measured period stray from 2π√(m/k)?
t in s, x in m, v in m/s at mid-interval, a in m/s²
txv(t+ε/2)a
Measured period
2π√(m/k)
STEP 4

How small does the step have to be?

The bench takes the spring from step 3, with 1 kg, 2 N/m and 0.6 m of amplitude, and runs it for 5 cycles by two methods, both with a step of 0.05 s. The first, Euler's, crosses each interval with the velocity at its start and updates the velocity with the force at its start. The second is the half-step method, the one behind the table of step 3. The half step stays on the exact curve, while Euler's curve widens with every cycle until the amplitude has tripled.

The energy exposes the flaw without a close look at the graph. With no friction it ought to stay constant, and under Euler's method it grows by the same proportion at every step, 0.5 % with these numbers, and never comes down. By the end of the 5 cycles it is 9.16 times what it was at the start, energy that no force supplied. Under the half step it wobbles within a narrow band, at most 0.13 % below the starting value, and does not drift away from it cycle after cycle.

The small graph, with logarithmic scales on both axes, measures the error in the position at t = 2 s for eight different steps. Each method's dots fall almost on a straight line (Euler's bends at the larger steps), and the slope of that line is the exponent of \(\varepsilon\) in the error: 1 for Euler, 2 for the half step. Going from 0.05 s to 0.025 s takes the half-step error from 0.109 mm to 0.0272 mm, a quarter of what it was, while Euler's only falls from 58.9 mm to 29 mm. Twice the work buys four times the precision in one method and twice in the other.

More precision costs only more rows of the same calculation: the 5 cycles take 444, and a step ten times smaller would take over 4,400. Nothing in the recipe hinges on the force being the spring's: swap the force law and the same table works out a different motion, which step 5 does with the Sun's gravity.

half step   Euler   exact, dashed

\(E = \tfrac12 m v^2 + \tfrac12 k x^2\)\(\text{error} \propto \varepsilon^2\)\(E\) is the energy of the mass on the spring, kinetic plus the energy stored in the spring; the error is that of the half-step method, which falls with the square of the step \(\varepsilon\).

In Feynman: §9-6 Numerical solution of the equations ↗

Let's discuss

  • With ε = 0.025 s, by how many times does Euler's energy grow over 5 cycles? And with 0.01 s?
  • Compare the errors in the panels at ε = 0.1 s and at 0.05 s. What is the ratio for the half step? And for Euler?
  • With ε = 0.5 s, does the half step still follow the exact curve? What is left of Euler's curve?
  • How many steps of 0.01 s does each method need for the 5 cycles?
Half-step error at t = 2 s
Euler error at t = 2 s
Euler final energy / initial
STEP 5

Can a planet's orbit be worked out by hand?

The bench keeps the Sun fixed at the origin and launches a planet from x = 1, one unit away, with speed 1.2 pointing along y. The units are chosen to make GM equal to 1, the same simplification Feynman makes in the chapter. Holding the Sun still is an approximation, a sound one as long as its mass is far greater than the planet's. Each step of ε = 0.01 adds a dot to the curve, worked out with the recipe of step 3, now with two positions and two velocities instead of one of each.

The force law is the only new ingredient. Gravitation gives the planet an acceleration of size GM/r², aimed at the Sun, and the direction to the Sun is that of the vector (−x, −y) divided by its length r. Multiplying the two gives components −GMx/r³ and −GMy/r³, the ones in the box alongside. At every step of the calculation, then, r is found from x and y, the two accelerations from r, and the two velocities and two positions move on just as the spring's single pair did.

What the dots trace is an ellipse with the Sun at one focus, although the word ellipse appears nowhere in the calculation. The planet swings out to 2.57 from the Sun and gets there at 0.47, under 40 % of its starting speed. After some 1,500 steps it is back at x = 1, right on the dashed curve, which is Kepler's ellipse with the same energy. The period measured on this revolution is 14.994. The energy per unit mass, \(\tfrac12 v^2 - GM/r\), is −0.28 at the start and yields a semi-major axis of 1.786. With that axis, Kepler's third law, the same one step 2 of the chapter 7 lesson uses to reach 1/r², predicts 14.993, and with ε = 0.001 the two numbers agree to three places.

The shaded sectors of the first revolution each take in the same number of steps, 125, and therefore the same time. Near the Sun they come out short and wide, on the far side long and thin, and all of them sweep the same area, 0.750. This is the law of areas, which the calculation was never handed as a rule. It follows only from the force pointing at the Sun, as step 4 of the chapter 7 lesson shows, and that lesson already computed an orbit in short steps. The half-step method keeps the law to the letter for any ε, because each correction to the velocity points at the Sun and leaves the area of the next triangle untouched.

A large step exacts its price where the planet moves fastest. With v = 0.6 it starts from aphelion and dives to 0.22 from the Sun, where it swings round much faster. With ε = 0.05 the measured revolution already comes out 4.7 % longer than Kepler's, the computed orbit rotates a few degrees per revolution relative to the dashed one, and with ε = 0.2 it comes apart. The orbit for v = 1.2, which never comes within 1 of the Sun, survives the same ε = 0.2 with its period off by 1.8 %.

The planetary position tables used today by observatories and space missions come out of step-by-step numerical integrations, with methods more refined than the one on this bench. The underlying recipe is the same, and nothing in it calls for a fixed Sun and a lone planet. With several bodies, the acceleration of each becomes the sum of the pulls of all the others, each with its own term of the form −GMx/r³, and the Sun itself joins the calculation as one more body that moves. With steps small enough, the calculation tracks the deviations one planet causes in another, the perturbations.

\(a_x = -\dfrac{GM\,x}{r^3},\quad a_y = -\dfrac{GM\,y}{r^3}\)\(E = \tfrac12 v^2 - \dfrac{GM}{r}\)\(\Rightarrow a = -\dfrac{GM}{2E}\)\(T = 2\pi\sqrt{a^3/GM}\)\(x\) and \(y\) are the coordinates of the planet, with the Sun at the origin, and \(r = \sqrt{x^2+y^2}\) the distance between the two; \(GM\) is the gravitational constant times the mass of the Sun, \(E\) the planet's energy per unit mass, \(a\) the semi-major axis of the ellipse and \(T\) the period.

In Feynman: §9-7 Planetary motions ↗

Let's discuss

  • Set v = 1. What shape is the orbit, and what period appears in the panels? Compare it with 2π.
  • With v = 1.2, raise ε from 0.001 to 0.2. At which step does the measured period stray more than 1 % from Kepler's?
  • With v = 0.6, from which ε onwards does the panel warn that the step is too large? Why does the orbit for v = 1.2 tolerate larger steps?
  • Does the area swept per unit time change as ε grows? Divide a sector’s area by its duration, which the label on the bench shows.
  • Take v to 1.45. What sign does the energy have, and why does Kepler's period disappear from the panels?
Measured period
Kepler period
Semi-major axis a

Where we go next

Throughout the lesson the force came from outside, supplied by the spring or by a Sun that never moves, and only the second law was needed. Feynman's chapter 10, Conservation of Momentum, brings in the third law, action and reaction, and draws from it the conservation of momentum when two bodies push on each other, which would also free the Sun from its fixed place. It is also on the Caltech website, and the lesson in this track that goes with it comes next.