---
title: "Newton’s Laws of Dynamics"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 12
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics
---

# Chapter 12 — Newton’s Laws of Dynamics

A skydiver steps out of the plane and, within a dozen seconds, stops accelerating: at two hundred kilometers per hour the air pushes back exactly as hard as the Earth pulls. A car takes a bend too fast and the tires give way; a truck on a mountain road crawls in low gear, held by the same friction that lets it stop on the flat. Kinematics described motions; dynamics explains them, from a single law linking the acceleration of a body to the [forces](#def-b1-newton-dynamics-momentum) acting on it. This chapter states Newton’s three laws, catalogues the [forces](#def-b1-newton-dynamics-momentum) one meets at the level of this volume — weight, springs, strings, contacts, friction dry and fluid — and shows how to go from a drawing of the [forces](#def-b1-newton-dynamics-momentum) to a differential equation and its solution.

![The International Space Station seen from the departing shuttle (NASA): everything aboard is in free fall together, which is why nothing aboard seems to weigh anything.](https://one-course.com/images/onecourse/chapters/physics-3/b1-newton-dynamics/img-58b3db574027.jpg)

*The International Space Station seen from the departing shuttle (NASA): everything aboard is in free fall together, which is why nothing aboard seems to weigh anything.*

## 12.1 Mass, momentum, and the three laws

**Definition 12.1 (Mass, momentum, force).**

A [point particle](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-frame) has an *inertial mass* $m > 0$ (kilograms), a measure of its resistance to changes of velocity, independent of the frame. Its *momentum* in a frame is

$$
\vect p = m\,\vect v \qquad (\mathrm{kg}\,\mathrm{m}/\mathrm{s}).
$$

A *force* $\vect F$ (newtons) is the action of another body on the particle; forces add as vectors, and the *resultant* is their sum $\sum\vect F$.

**Theorem 12.2 (Newton’s laws).**

1. *Principle of inertia.* There exist frames, called *inertial* (or Galilean), in which a particle subject to no [force](#def-b1-newton-dynamics-momentum) , or to [forces](#def-b1-newton-dynamics-momentum) of zero resultant, moves in a straight line at constant velocity. Any frame in uniform rectilinear translation with respect to an [inertial frame](#thm-b1-newton-dynamics-laws) is inertial.
2. *Law of [momentum](#def-b1-newton-dynamics-momentum).* In an [inertial frame](#thm-b1-newton-dynamics-laws), $$\frac{\dd\vect p}{\dd t} = \sum\vect F, \qquad \text{i.e.}\qquad  m\,\vect a = \sum\vect F \ \text{ for constant } m .$$
3. *[Action and reaction](#thm-b1-newton-dynamics-laws).* If body $A$ exerts $\vect F_{A\to B}$ on body $B$ , then $B$ exerts $\vect F_{B\to A} = -\vect F_{A\to B}$ on $A$ , along the line joining them.

**Proof.** *Admitted at this level.* ∎

**Remark 12.3 (What the laws say).**

The first law is not a special case of the second: it *defines* the frames in which the second holds. In practice a laboratory on the Earth is inertial to excellent accuracy for experiments lasting minutes ([Chapter 18](https://one-course.com/books/physics/3/en/chapter/18-non-inertial-frames-dynamics-on-earth#ch-b1-non-inertial-frames) quantifies the error); a frame attached to the Sun and the distant stars is better still. The second law is a vector equation: three scalar equations, one per axis, and the choice of axes ([Chapter 11](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#ch-b1-point-kinematics)) is the first decision of every problem. The third law holds at each instant for the contact and gravitational [forces](#def-b1-newton-dynamics-momentum) of this volume.

**Definition 12.4 (Closed system; momentum conservation).**

A *closed system* exchanges no matter with the outside. For a system of particles, the internal [forces](#def-b1-newton-dynamics-momentum) cancel in pairs (third law), and the total [momentum](#def-b1-newton-dynamics-momentum) $\vect P = \sum m_i\vect v_i$ obeys $\dd\vect P/\dd t = \sum\vect F_{\mathrm{ext}}$: it is conserved when the external [forces](#def-b1-newton-dynamics-momentum) vanish ([Chapter 19](https://one-course.com/books/physics/3/en/chapter/19-systems-of-points-introduction-to-rigid-bodies#ch-b1-systems-of-points) develops this).

## 12.2 The usual forces

**Proposition 12.5 (Weight and gravitation).**

Near the Earth’s surface a body of [mass](#def-b1-newton-dynamics-momentum) $m$ is pulled by its *weight* $\vect P = m\vect g$, with $\vect g$ the local [gravitational field](#prop-b1-newton-dynamics-weight), $g \approx 9.81\,\mathrm{m}/\mathrm{s}^{2}$, directed downward (toward the center, very nearly). More generally a [mass](#def-b1-newton-dynamics-momentum) $M$ at distance $r$ attracts $m$ with $\vect F = -GMm/r^2\,\vect e_r$ ([Chapter 16](https://one-course.com/books/physics/3/en/chapter/16-central-forces-planets-and-satellites#ch-b1-central-forces)); at the surface of a sphere of radius $R$, $g = GM/R^2$. Note the unit: $\mathrm{N}/\mathrm{kg}$ and $\mathrm{m}/\mathrm{s}^{2}$ are the same, which is why all bodies fall alike.

**Proof.** Newton’s law of gravitation is admitted and explored in [Chapter 16](https://one-course.com/books/physics/3/en/chapter/16-central-forces-planets-and-satellites#ch-b1-central-forces); $g = GM/R^2 = 6.67\times10^{-11} \times
5.97\times10^{24}/(6.37\times10^6)^2 = 9.8\,\mathrm{m}/\mathrm{s}^{2}$. ∎

**Proposition 12.6 (Spring; string).**

An ideal spring of stiffness $k$ and natural length $\ell_0$, stretched to length $\ell$ along the unit vector $\vect u$ pointing from its fixed end to the body, exerts on the body the *restoring [force](#def-b1-newton-dynamics-momentum)*

$$
\vect F = -k(\ell - \ell_0)\,\vect u \qquad (\text{Hooke's law}),
$$

toward its natural length. An ideal string (massless, inextensible) exerts a *tension* $\vect T$ along itself, pulling, of the same magnitude at both ends and all along it (even over a massless, frictionless pulley).

**Proof.** [Hooke’s law](#prop-b1-newton-dynamics-springstring) is an empirical model valid for small deformations. The string: a massless element obeys $0 = \vect T_1 + \vect T_2$ from Newton’s second law, so the tension is transmitted unchanged. ∎

**Proposition 12.7 (Contact forces: normal reaction and dry friction).**

A solid surface exerts on a body touching it a *[normal reaction](#prop-b1-newton-dynamics-friction)* $\vect N$, perpendicular to the surface and repulsive ($N \geq 0$: a surface cannot pull), and a tangential *friction [force](#def-b1-newton-dynamics-momentum)* $\vect T$. Coulomb’s laws: if the body does not slide, $\abs{\vect T} \leq \mu_s N$ (*static* friction adjusts to what is needed, up to that limit); if it slides at velocity $\vect v_g$ relative to the surface, $\vect T = -\mu_d N\,\vect v_g/\abs{\vect v_g}$ (*kinetic* friction, opposing the sliding), with $\mu_d \leq \mu_s$ coefficients of order $0.1$ to $1$ depending on the materials, independent of the contact area and of the speed.

**Proof.** *Admitted at this level.* ∎

![Left: the forces on a block on an incline — weight, normal reaction, friction along the surface — and the natural axes. Right: Coulomb’s laws: at rest the friction balances the applied force up to _sN; once sliding starts it drops to _dN and stays there.](https://one-course.com/images/onecourse/chapters/physics-3/b1-newton-dynamics/fig-73cfb08faca4.svg)

![Left: the forces on a block on an incline — weight, normal reaction, friction along the surface — and the natural axes. Right: Coulomb’s laws: at rest the friction balances the applied force up to _sN; once sliding starts it drops to _dN and stays there.](https://one-course.com/images/onecourse/chapters/physics-3/b1-newton-dynamics/fig-5588c1d0c4b1.svg)

*Left: the [forces](#def-b1-newton-dynamics-momentum) on a block on an incline — weight, [normal reaction](#prop-b1-newton-dynamics-friction), friction along the surface — and the natural axes. Right: Coulomb’s laws: at rest the friction balances the applied [force](#def-b1-newton-dynamics-momentum) up to $\mu_sN$; once sliding starts it drops to $\mu_dN$ and stays there.*

**Proposition 12.8 (Fluid friction).**

A body moving at velocity $\vect v$ through a fluid at rest feels a *drag* opposite to $\vect v$: at low speed (small size, viscous fluid) it is linear, $\vect F = -\alpha\vect v$ — for a sphere of radius $r$ in a fluid of viscosity $\eta$, $\alpha = 6\pi\eta r$ (Stokes); at high speed (large, fast bodies in air or water) it is quadratic,

$$
\vect F = -\tfrac12\,\rho\,C_xS\,v\,\vect v ,
$$

with $\rho$ the fluid’s density, $S$ the body’s cross-section facing the flow and $C_x$ a [dimensionless](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#def-b1-units-dimensions-dimension) *drag coefficient* ($0.5$ for a sphere, $0.3$ for a car, $1$ for a flat disk). The Reynolds number $\rho vL/\eta$ ([Remark 1.10](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#rem-b1-units-dimensions-pi)) decides: linear below $1$, quadratic above about $1000$.

**Proof.** *Admitted at this level.* ∎

**Remark 12.9 (Other forces).**

Buoyancy (Archimedes) is a resultant of pressure [forces](#def-b1-newton-dynamics-momentum), derived in [Chapter 21](https://one-course.com/books/physics/3/en/chapter/21-fluid-statics#ch-b1-fluid-statics); the electric and magnetic [forces](#def-b1-newton-dynamics-momentum) on a charge are the subject of [Chapter 17](https://one-course.com/books/physics/3/en/chapter/17-charged-particles-in-e-and-b-fields#ch-b1-charged-particles). All the [forces](#def-b1-newton-dynamics-momentum) of this chapter are *models* of contact or field interactions, valid within stated limits — Hooke’s spring breaks, Coulomb’s friction fails at high pressure, Stokes’s drag fails at high speed. Part of every solution is checking that the model used applies.

## 12.3 Solving a problem of dynamics

**Method 12.10 (From the situation to the equation).**

1. *System* : name the body treated as a point.
2. *Frame* : name it and check it is inertial (or treat it as such to the accuracy required).
3. *[Forces](#def-b1-newton-dynamics-momentum)* : list every body touching or attracting the system, and draw each [force](#def-b1-newton-dynamics-momentum) on a diagram (the free-body diagram); nothing else enters.
4. *Coordinates* : choose axes adapted to the motion ( [Method 11.18](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#met-b1-point-kinematics-choose) ) and express $\vect a$ .
5. *Project* $m\vect a = \sum\vect F$ on the axes; use the constraints (a body on a surface stays on it: $a_\perp = 0$ ; a string of fixed length: equal speeds at both ends).
6. *Solve* and *check* : dimensions, limiting cases, signs ( $N \geq 0$ , $\abs T \leq \mu_sN$ if at rest).

**Example 12.11 (Block on an incline).**

Block of [mass](#def-b1-newton-dynamics-momentum) $m$ on a plane inclined at $\alpha$, coefficients $\mu_s$, $\mu_d$. Axes: $x$ down the slope, $y$ along the normal. Weight $(mg\sin\alpha, -mg\cos\alpha)$, reaction $(0, N)$, friction $(-T, 0)$ if sliding down. Along $y$: $N = mg\cos\alpha$. At rest along $x$: $T = mg\sin\alpha \leq \mu_sN = \mu_smg\cos\alpha$, i.e. $\tan\alpha
\leq \mu_s$: the block holds up to the angle $\arctan\mu_s$ — a measurement of $\mu_s$ needing no balance. Sliding: $ma = mg\sin\alpha -
\mu_dmg\cos\alpha$, so $a = g(\sin\alpha - \mu_d\cos\alpha)$, independent of $m$: $3.2\,\mathrm{m}/\mathrm{s}^{2}$ for $\alpha = 30^\circ$, $\mu_d = 0.2$.

**Example 12.12 (The simple pendulum).**

[Mass](#def-b1-newton-dynamics-momentum) $m$ on a string of length $\ell$, angle $\theta$ from the downward vertical. [Polar coordinates](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-cylindrical) at the suspension point: $\vect a =
-\ell\dot\theta^2\,\vect e_r + \ell\ddot\theta\,\vect e_\theta$; [forces](#def-b1-newton-dynamics-momentum): weight $mg(\cos\theta\,\vect e_r - \sin\theta\,\vect e_\theta)$ and tension $-T\vect e_r$. Along $\vect e_\theta$: $m\ell\ddot\theta =
-mg\sin\theta$,

$$
\ddot\theta + \frac{g}{\ell}\sin\theta = 0 ,
$$

which for small angles ($\sin\theta \approx \theta$) is the harmonic equation with $\omega_0 = \sqrt{g/\ell}$: period $T_0 = 2\pi\sqrt{\ell/g}$, the $k = 2\pi$ that [Example 1.8](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#ex-b1-units-dimensions-pendulum) could not supply. Along $\vect e_r$: $T = mg\cos\theta + m\ell\dot\theta^2$ — the string pulls hardest at the bottom, where the speed is greatest.

![Left: the simple pendulum, its two forces and the polar basis used to project Newton’s law. Right: a body falling from rest under its weight and a drag force: the speed approaches the terminal velocity v_∈fty, exponentially for linear drag (= m/), as a hyperbolic tangent for quadratic drag (= v_∈fty/g).](https://one-course.com/images/onecourse/chapters/physics-3/b1-newton-dynamics/fig-42aff5404e26.svg)

![Left: the simple pendulum, its two forces and the polar basis used to project Newton’s law. Right: a body falling from rest under its weight and a drag force: the speed approaches the terminal velocity v_∈fty, exponentially for linear drag (= m/), as a hyperbolic tangent for quadratic drag (= v_∈fty/g).](https://one-course.com/images/onecourse/chapters/physics-3/b1-newton-dynamics/fig-017cdd74ae49.svg)

*Left: the [simple pendulum](#ex-b1-newton-dynamics-pendulum), its two [forces](#def-b1-newton-dynamics-momentum) and the polar basis used to project Newton’s law. Right: a body falling from rest under its weight and a drag [force](#def-b1-newton-dynamics-momentum): the speed approaches the [terminal velocity](#prop-b1-newton-dynamics-terminal) $v_\infty$, exponentially for linear drag ($\tau = m/\alpha$), as a hyperbolic tangent for quadratic drag ($\tau = v_\infty/g$).*

**Proposition 12.13 (Fall with drag; terminal velocity).**

A body of [mass](#def-b1-newton-dynamics-momentum) $m$ falling from rest under its weight and a drag [force](#def-b1-newton-dynamics-momentum) reaches a *[terminal velocity](#prop-b1-newton-dynamics-terminal)* at which the two balance. Linear drag: $m\dot v = mg - \alpha v$, $v = v_\infty(1 - \eu^{-t/\tau})$ with $v_\infty = mg/\alpha$, $\tau = m/\alpha$. Quadratic drag: $m\dot v = mg -
\tfrac12\rho C_xSv^2$, $v = v_\infty\tanh(t/\tau)$ with

$$
v_\infty = \sqrt{\frac{2mg}{\rho C_xS}}, \qquad \tau = \frac{v_\infty}{g} .
$$

**Proof.** Linear: a first-order equation, $\tau\dot v + v = v_\infty$ ([Theorem 7.2](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#thm-b1-transient-regimes-firstorder)). Quadratic: with $u = v/v_\infty$, $\dot u = (1 - u^2)/\tau$; separating variables, $\int\dd u/(1 - u^2) = \operatorname{artanh}u = t/\tau$ (the rational fraction $1/(1 - u^2) = \tfrac12[1/(1 - u) + 1/(1 + u)]$ integrates to $\tfrac12\ln\frac{1 + u}{1 - u}$), so $u = \tanh(t/\tau)$. ∎

**Example 12.14 (Two terminal velocities).**

A skydiver, $m = 80\,\mathrm{kg}$, belly down, $C_xS \approx 0.8\,\mathrm{m}^{2}$ in air ($\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}$): $v_\infty = \sqrt{2 \times 785/(1.2
\times 0.8)} = 40\,\mathrm{m}/\mathrm{s}$, reached (to $90\%$) in about $1.5\tau =
6\,\mathrm{s}$. A fog droplet, $r = 10\,\text{µ}\mathrm{m}$, in Stokes’s regime ($\eta_{\text{air}} = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}$): $v_\infty = mg/6\pi\eta r =
1.2\,\mathrm{cm}/\mathrm{s}$ — which is why fog hangs in the air.

## 12.4 Motion on a circle: constraint forces

**Proposition 12.15 (Centripetal requirement).**

A particle moving on a circle of radius $R$ at speed $v$ must receive from the resultant of the [forces](#def-b1-newton-dynamics-momentum) a component $mv^2/R$ directed toward the center. Whatever provides it — a string’s tension, friction, a banked road’s reaction, gravity — sets the maximum speed or the minimum speed at which the motion is possible.

**Proof.** Newton’s second law with the Frenet acceleration: the normal component of $\sum\vect F$ equals $mv^2/\rho$. ∎

**Example 12.16 (Three circles).**

*Flat bend*: friction must supply $mv^2/R \leq \mu_sN = \mu_smg$, so $v \leq \sqrt{\mu_sgR}$: $20\,\mathrm{m}/\mathrm{s}$ for $R = 50\,\mathrm{m}$, $\mu_s = 0.8$. *Banked bend* at angle $\beta$, no friction: the reaction $\vect N$ is normal to the road; vertically $N\cos\beta = mg$, horizontally $N\sin\beta = mv^2/R$, so $v^2 = gR\tan\beta$ — the one speed at which the bend is taken without any friction. *Loop*: at the top, with $\vect N$ pointing down toward the center, $N + mg =
mv^2/R$, and $N \geq 0$ requires $v \geq \sqrt{gR}$: below that, the car leaves the track ([Problem 11.1](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#pb-b1-point-kinematics-1) again, now explained).

![A banked bend seen in cross-section: without friction the road’s normal reaction must both hold the car up and push it toward the center; the horizontal component N provides mv2/R, which happens at exactly one speed, v = √gR.](https://one-course.com/images/onecourse/chapters/physics-3/b1-newton-dynamics/fig-6eb28464a9ff.svg)

*A banked bend seen in cross-section: without friction the road’s [normal reaction](#prop-b1-newton-dynamics-friction) must both hold the car up and push it toward the center; the horizontal component $N\sin\beta$ provides $mv^2/R$, which happens at exactly one speed, $v = \sqrt{gR\tan\beta}$.*

## 12.5 Exercises

**Exercise 12.1 ★.**

A person of [mass](#def-b1-newton-dynamics-momentum) $70\,\mathrm{kg}$ stands on a scale in an elevator. What does the scale read (in newtons and in “kilograms”) when the elevator accelerates upward at $2.0\,\mathrm{m}/\mathrm{s}^{2}$, downward at $2.0\,\mathrm{m}/\mathrm{s}^{2}$, moves at constant velocity, and falls freely?

**Solution of Exercise 12.1.**

Scale reading $N = m(g + a)$ with $a$ the upward acceleration: up, $70 \times 11.81 = 827\,\mathrm{N}$ (“$84\,\mathrm{kg}$”); down, $70 \times 7.81 =
547\,\mathrm{N}$ (“$56\,\mathrm{kg}$”); constant velocity, $687\,\mathrm{N}$ ($70\,\mathrm{kg}$); free fall, $0$.

**Exercise 12.2 ★.**

A block $m_1 = 2.0\,\mathrm{kg}$ on a frictionless table is tied by a string over a frictionless pulley to a hanging [mass](#def-b1-newton-dynamics-momentum) $m_2 = 1.0\,\mathrm{kg}$. Find the acceleration and the tension.

**Solution of Exercise 12.2.**

Same acceleration $a$ for both (inextensible string): $m_1a = T$, $m_2a = m_2g - T$, so $a = m_2g/(m_1 + m_2) = 3.3\,\mathrm{m}/\mathrm{s}^{2}$, $T = m_1a = 6.5\,\mathrm{N}$.

**Exercise 12.3 ★.**

A block slides down a $30^\circ$ incline with $\mu_d = 0.20$. Compute its acceleration and its speed after $5.0\,\mathrm{m}$ from rest. Does the answer depend on its [mass](#def-b1-newton-dynamics-momentum)?

**Solution of Exercise 12.3.**

$a = g(\sin\alpha - \mu_d\cos\alpha) = 9.81(0.50 - 0.17) = 3.2\,\mathrm{m}/\mathrm{s}^{2}$; $v = \sqrt{2 \times 3.2 \times 5.0} = 5.7\,\mathrm{m}/\mathrm{s}$. No [mass](#def-b1-newton-dynamics-momentum) in sight.

**Exercise 12.4 ★.**

A $0.50\,\mathrm{kg}$ [mass](#def-b1-newton-dynamics-momentum) hangs from a spring of stiffness $200\,\mathrm{N}/\mathrm{m}$. Find the static elongation; then the [angular frequency](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) and period of its vertical oscillations about equilibrium (show the equation of motion is harmonic).

**Solution of Exercise 12.4.**

$k\Delta\ell = mg$: $\Delta\ell = 0.50 \times 9.81/200 = 2.5\,\mathrm{cm}$. With $x$ the displacement from equilibrium, $m\ddot x = -kx$ (the weight and the static stretch cancel): $\omega_0 = \sqrt{k/m} = 20\,\mathrm{rad}/\mathrm{s}$, $T = 0.31\,\mathrm{s}$.

**Exercise 12.5 ★★.**

A crate sits on a tilting platform; it starts to slide at $22^\circ$. Find $\mu_s$. A car on a flat road with $\mu_s = 0.80$: maximum speed in a bend of radius $50\,\mathrm{m}$, and on a wet road with $\mu_s = 0.40$.

**Solution of Exercise 12.5.**

$\mu_s = \tan 22^\circ = 0.40$. Bend: $v \leq \sqrt{\mu_sgR}$: $\sqrt{0.8
\times 9.81 \times 50} = 20\,\mathrm{m}/\mathrm{s}$ ($71\,\mathrm{km}/\mathrm{h}$); wet, $14\,\mathrm{m}/\mathrm{s}$ ($50\,\mathrm{km}/\mathrm{h}$).

**Exercise 12.6 ★★.**

Terminal velocities: a skydiver ($m = 80\,\mathrm{kg}$, $C_xS = 0.70\,\mathrm{m}^{2}$, $\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}$); a fog droplet of radius $10\,\text{µ}\mathrm{m}$ in air ($\eta = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}$), using [Stokes’s law](#prop-b1-newton-dynamics-drag). Check with a Reynolds number that each regime is the right one.

**Solution of Exercise 12.6.**

Skydiver: $v_\infty = \sqrt{2 \times 785/(1.2 \times 0.70)} = 43\,\mathrm{m}/\mathrm{s}$; $\mathrm{Re} = \rho vL/\eta \approx 1.2 \times 43 \times 0.5/1.8\times10^{-5}
\approx 10^6$: quadratic. Droplet: $m = \tfrac43\pi r^3\rho_w = 4.2 \times 10^{-12}\,\mathrm{kg}$, $v_\infty = mg/6\pi\eta r = 4.1\times10^{-11}/3.4\times10^{-9} = 1.2\,\mathrm{cm}/\mathrm{s}$; $\mathrm{Re} = 1.2 \times 0.012 \times 2\times10^{-5}/1.8\times10^{-5} = 0.02$: linear.

**Exercise 12.7 ★★.**

A pendulum of length $\ell$ is released from rest at $\theta_0 = 60^\circ$. Given that its speed at the bottom is $\sqrt{2g\ell(1 - \cos\theta_0)}$ (energy, next chapter), find the string tension at the bottom in units of $mg$, and at the release point.

**Solution of Exercise 12.7.**

Bottom: $v^2 = 2g\ell(1 - \cos 60^\circ) = g\ell$; $T = mg + mv^2/\ell =
2mg$. Release point: $v = 0$, $T = mg\cos\theta_0 = 0.5mg$.

**Exercise 12.8 ★★.**

A bend of radius $200\,\mathrm{m}$ is banked at $15^\circ$. At what speed can it be taken with no friction at all? What must friction do at lower and at higher speeds?

**Solution of Exercise 12.8.**

$v = \sqrt{gR\tan\beta} = \sqrt{9.81 \times 200 \times 0.268} = 23\,\mathrm{m}/\mathrm{s}$ ($82\,\mathrm{km}/\mathrm{h}$). Slower: the car tends to slide down the bank, friction must point up-slope; faster: friction points down-slope to add [centripetal force](#prop-b1-newton-dynamics-centripetal).

**Exercise 12.9 ★★.**

A conical pendulum: a [mass](#def-b1-newton-dynamics-momentum) on a string of length $\ell = 1.0\,\mathrm{m}$ describes a horizontal circle, the string making $30^\circ$ with the vertical. Find the [angular velocity](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#cor-b1-point-kinematics-circular), the period and the tension (in units of $mg$).

**Solution of Exercise 12.9.**

$T\cos\alpha = mg$, $T\sin\alpha = m\omega^2\ell\sin\alpha$: $\omega =
\sqrt{g/\ell\cos\alpha} = \sqrt{9.81/0.866} = 3.4\,\mathrm{rad}/\mathrm{s}$, period $1.9\,\mathrm{s}$; $T = mg/\cos\alpha = 1.15mg$.

**Exercise 12.10 ★★★.**

A $1000\,\mathrm{kg}$ car reaches $100\,\mathrm{km}/\mathrm{h}$ from rest in $8.0\,\mathrm{s}$ with constant acceleration. Find the resultant [force](#def-b1-newton-dynamics-momentum), and the minimum friction coefficient between driving tires and road if the driving wheels carry the whole weight. What limits a sports car on a wet road?

**Solution of Exercise 12.10.**

$a = 27.8/8.0 = 3.5\,\mathrm{m}/\mathrm{s}^{2}$, $F = 3.5\,\mathrm{kN}$; $F \leq \mu_smg$ gives $\mu_s \geq a/g = 0.35$. On a wet road $\mu_s \approx 0.4$ caps the acceleration near $0.4g$ whatever the engine: grip, not power.

**Exercise 12.11 ★★★.**

A body falls from rest with linear drag, $\tau = m/\alpha = 0.50\,\mathrm{s}$. Write $v(t)$ and derive the distance fallen $z(t)$; compute $v_\infty$, and $z$ after $2.0\,\mathrm{s}$; compare with free fall.

**Solution of Exercise 12.11.**

$v = v_\infty(1 - \eu^{-t/\tau})$, $v_\infty = g\tau = 4.9\,\mathrm{m}/\mathrm{s}$; $z = \int v = v_\infty[t - \tau(1 - \eu^{-t/\tau})]$. At $2.0\,\mathrm{s}$: $4.9 \times (2.0 - 0.5 \times 0.98) = 7.4\,\mathrm{m}$, against $\tfrac12 gt^2
= 20\,\mathrm{m}$ in vacuum.

**Exercise 12.12 ★★★.**

A car of [mass](#def-b1-newton-dynamics-momentum) $m$ runs on the inside of a vertical circular loop of radius $R$ ([Problem 11.1](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#pb-b1-point-kinematics-1)). Express the track’s [normal reaction](#prop-b1-newton-dynamics-friction) as a function of the angle $\theta$ from the bottom and of the speed $v(\theta)$; deduce the minimum speed at the top; using $v^2 = v_0^2 - 2gR(1 - \cos\theta)$, find the minimum entry speed and the reaction at the bottom in that case.

**Solution of Exercise 12.12.**

Radial projection (toward the center): $N - mg\cos\theta = mv^2/R$, so $N = m(v^2/R + g\cos\theta)$. Top ($\theta = \pi$): $N = m(v^2/R - g)
\geq 0$ iff $v \geq \sqrt{gR}$. With the speed law, $v_{\text{top}}^2 =
v_0^2 - 4gR \geq gR$: $v_0 \geq \sqrt{5gR}$; then at the bottom $N =
m(5g + g) = 6mg$.

![Free fall with air: after a few seconds the quadratic drag balances the weight and the skydiver falls at a terminal speed near 200\, km/ h — the weekend problem.](https://one-course.com/images/onecourse/chapters/physics-3/b1-newton-dynamics/img-56eb68ace300.jpg)

*Free fall with air: after a few seconds the quadratic drag balances the weight and the skydiver falls at a terminal speed near $200\,\mathrm{km}/\mathrm{h}$ — the weekend problem.*

## 12.6 Problem: The parachutist

**Problem 12.1.**

Weekend problem — from the door of the plane to the ground: free fall, the air that pushes back, a canopy that opens too fast, and the landing — one law of motion, two drag laws

A parachutist with gear has [mass](#def-b1-newton-dynamics-momentum) $m = 80\,\mathrm{kg}$. Air: $\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}$, $\eta = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}$. Falling belly down, $C_xS = 0.80\,\mathrm{m}^{2}$; under the open canopy, $C_xS =
25\,\mathrm{m}^{2}$. $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$; $z$ is counted downward from the jump point.

**Part I — Without air.**

1. Write Newton’s second law and give $v(t)$ and $z(t)$ ; compute both at $t = 10\,\mathrm{s}$ .
2. Compute the weight.
3. The parachutist feels “weightless” during this phase although gravity acts fully. Explain with the [forces](#def-b1-newton-dynamics-momentum) (what is missing compared with standing on the ground).
4. Is the Earth’s frame inertial enough for this problem? Name what is neglected.

**Part II — Falling through air.**

5. Write the drag [force](#def-b1-newton-dynamics-momentum) for belly-down fall as a function of $v$ .
6. Write the equation of motion for $v(t)$ .
7. Find the [terminal velocity](#prop-b1-newton-dynamics-terminal) $v_\infty$ (in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ ).
8. With $u = v/v_\infty$ and $\tau = v_\infty/g$ , show that $\dd u/\dd t = (1 - u^2)/\tau$ ; compute $\tau$ .
9. Solve by separation of variables and show that $v = v_\infty\tanh(t/\tau)$ .
10. After how long is $v = 0.9v_\infty$ ? $0.99v_\infty$ ?
11. Show that $z(t) = v_\infty\tau\ln\cosh(t/\tau)$ and compute the distance fallen at $t = 10\,\mathrm{s}$ ; compare with Part I.
12. Estimate the Reynolds number at $v_\infty$ with $L = 0.5\,\mathrm{m}$ and justify the quadratic law.

**Part III — The canopy opens.**

13. Compute the new [terminal velocity](#prop-b1-newton-dynamics-terminal) under the open canopy.
14. If the canopy opened instantaneously at $v = 40\,\mathrm{m}/\mathrm{s}$ , what would the deceleration be, in $\mathrm{m}/\mathrm{s}^{2}$ and in $g$ ? Why is this unacceptable?
15. Real canopies open over about $3\,\mathrm{s}$ . Estimate the mean deceleration and the mean [force](#def-b1-newton-dynamics-momentum) of the harness on the parachutist.
16. Once the canopy is fully open the speed decays toward the new $v_\infty$ : what is the new [time constant](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-firstorder) $\tau' = v_\infty'/g$ , and what does it mean for the approach to steady descent?
17. In steady descent, what is the harness [force](#def-b1-newton-dynamics-momentum) ? Justify with the second law.
18. Explain why the [terminal velocity](#prop-b1-newton-dynamics-terminal) varies as $1/\sqrt{C_xS}$ : by what factor would a canopy of half the area raise it?
19. A heavier jumper ( $100\,\mathrm{kg}$ ) under the same canopy: [terminal velocity](#prop-b1-newton-dynamics-terminal) ?

**Part IV — Landing.**

20. The landing speed is the canopy’s $v_\infty$ . From what height would one have to jump, with no air, to hit the ground at that speed?
21. A stiff-legged landing stops the body over $0.5\,\mathrm{m}$ ; a roll over $1.5\,\mathrm{m}$ . Estimate the mean deceleration (in $g$ ) and the mean [force](#def-b1-newton-dynamics-momentum) on the legs in each case, and conclude.
22. Just before touching down the parachutist pulls both toggles, which briefly increases the canopy’s drag. Explain qualitatively the effect on the speed.

**Part V — Small things fall differently.**

23. A raindrop of radius $1.0\,\mathrm{mm}$ ( $C_x = 0.5$ ): compute its [terminal velocity](#prop-b1-newton-dynamics-terminal) with the quadratic law, and check the Reynolds number.
24. A fog droplet of radius $10\,\text{µ}\mathrm{m}$ : use [Stokes’s law](#prop-b1-newton-dynamics-drag) for $v_\infty$ and check that the Reynolds number is indeed small.
25. Summarize: one equation of motion, two drag regimes, and the three numbers ( $\mathrm{m}/\mathrm{s}$ ) that decide whether something falls like a stone, like a leaf or not at all.

**Solution of Problem 12.1.**

**1.** $m\dot v = mg$: $v = gt$, $z = \tfrac12 gt^2$; at $10\,\mathrm{s}$: $98\,\mathrm{m}/\mathrm{s}$, $490\,\mathrm{m}$.

**2.** $mg = 785\,\mathrm{N}$.

**3.** On the ground the floor pushes up with $N = mg$ — that push is what one feels as weight. In free fall there is no contact [force](#def-b1-newton-dynamics-momentum): gravity acts on every part alike and nothing presses.

**4.** Yes to a fraction of a percent over a minute; neglected: the Earth’s rotation (centrifugal and Coriolis terms, [Chapter 18](https://one-course.com/books/physics/3/en/chapter/18-non-inertial-frames-dynamics-on-earth#ch-b1-non-inertial-frames)) and the wind.

**5.** $F_d = \tfrac12\rho C_xSv^2 = 0.48\,v^2$ (SI).

**6.** $m\,\dd v/\dd t = mg - 0.48v^2$.

**7.** $v_\infty = \sqrt{mg/0.48} = \sqrt{1635} = 40\,\mathrm{m}/\mathrm{s} =
146\,\mathrm{km}/\mathrm{h}$.

**8.** Divide by $mg$: $\dd u/\dd t\,(v_\infty/g) = 1 - u^2$; $\tau = 40.4/9.81 = 4.1\,\mathrm{s}$.

**9.** $\int_0^u\dd u'/(1 - u'^2) = \operatorname{artanh}u = t/\tau$, so $u = \tanh(t/\tau)$.

**10.** $t = \tau\operatorname{artanh}(0.9) = 1.47\tau = 6.1\,\mathrm{s}$; $0.99$: $2.65\tau = 11\,\mathrm{s}$.

**11.** $z = \int v_\infty\tanh(t/\tau)\dd t = v_\infty\tau\ln\cosh(t/\tau)$; at $10\,\mathrm{s}$: $166 \times \ln\cosh(2.43) = 166 \times 1.74 = 290\,\mathrm{m}$, against $490\,\mathrm{m}$.

**12.** $\mathrm{Re} = 1.2 \times 40 \times 0.5/1.8\times10^{-5} \approx
1.3 \times 10^{6}$: far above $1000$, quadratic.

**13.** $v_\infty' = \sqrt{785/(0.6 \times 25)} = \sqrt{52} = 7.2\,\mathrm{m}/\mathrm{s}$.

**14.** Drag $= 15 \times 1600 = 24\,\mathrm{kN}$; $a = (24000 - 785)/80
= 290\,\mathrm{m}/\mathrm{s}^{2} \approx 30g$: the harness would break, or the jumper.

**15.** Mean $a \approx (40 - 7)/3 = 11\,\mathrm{m}/\mathrm{s}^{2} \approx 1.1g$; harness [force](#def-b1-newton-dynamics-momentum) $\approx m(g + a) \approx 1.7\,\mathrm{kN}$.

**16.** $\tau' = 7.2/9.81 = 0.73\,\mathrm{s}$: within a few seconds of full opening the descent is steady.

**17.** $\vect a = \vect 0$: harness [force](#def-b1-newton-dynamics-momentum) $= mg = 785\,\mathrm{N}$.

**18.** Balance $mg = \tfrac12\rho C_xSv^2$ gives $v \propto (C_xS)^{-1/2}$: half the area, $\sqrt2$ times the speed ($10\,\mathrm{m}/\mathrm{s}$).

**19.** $v \propto \sqrt m$: $7.2 \times \sqrt{100/80} = 8.1\,\mathrm{m}/\mathrm{s}$.

**20.** $h = v^2/2g = 52/19.6 = 2.6\,\mathrm{m}$: a jump from a first floor.

**21.** $a = v^2/2d$: $52\,\mathrm{m}/\mathrm{s}^{2}$ ($5.3g$) over $0.5\,\mathrm{m}$, [force](#def-b1-newton-dynamics-momentum) $m(g + a) \approx 5\,\mathrm{kN}$; $17\,\mathrm{m}/\mathrm{s}^{2}$ ($1.8g$) over $1.5\,\mathrm{m}$, $2.2\,\mathrm{kN}$. Roll.

**22.** A larger drag (and some lift) momentarily exceeds the weight: the jumper decelerates and touches down below $v_\infty'$.

**23.** $m = \tfrac43\pi r^3\rho_w = 4.2 \times 10^{-6}\,\mathrm{kg}$, $S = \pi r^2 =
3.1 \times 10^{-6}\,\mathrm{m}^{2}$: $v_\infty = \sqrt{2mg/\rho C_xS} = \sqrt{8.2\times10^{-5}/
1.9\times10^{-6}} = 6.6\,\mathrm{m}/\mathrm{s}$; $\mathrm{Re} = 1.2 \times 6.6 \times
2\times10^{-3}/1.8\times10^{-5} \approx 900$: quadratic (roughly).

**24.** $v_\infty = 1.2\,\mathrm{cm}/\mathrm{s}$ ([Exercise 12.6](#exo-b1-newton-dynamics-6)); $\mathrm{Re} \approx 0.02 \ll 1$: Stokes applies, consistently.

**25.** $m\dot{\vect v} = m\vect g + \vect F_{\text{drag}}$; drag linear in $v$ for small, slow things, quadratic for large, fast ones. The numbers: $40\,\mathrm{m}/\mathrm{s}$ for a falling body, $7\,\mathrm{m}/\mathrm{s}$ under a canopy (or a raindrop), $1\,\mathrm{cm}/\mathrm{s}$ for fog — the same law spanning four orders of magnitude.