University Physics — Year 1 · Bachelor Year 1

12Newton’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 acting on it. This chapter states Newton’s three laws, catalogues the forces 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 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.

12.1 Mass, momentum, and the three laws

Definition 12.1 (Mass, momentum, force)

A point particle 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)

Principle of inertia. There exist frames, called inertial (or Galilean), in which a particle subject to no force, or to forces of zero resultant, moves in a straight line at constant velocity. Any frame in uniform rectilinear translation with respect to an inertial frame is inertial.
Law of momentum. In an inertial frame,

$\frac{\dd\vect p}{\dd t} = \sum\vect F, \qquad \text{i.e.}\qquad m\,\vect a = \sum\vect F \ \text{ for constant } m .$
Action and reaction. 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 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) is the first decision of every problem. The third law holds at each instant for the contact and gravitational forces 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 cancel in pairs (third law), and the total 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 vanish (Chapter 19 develops this).

12.2 The usual forces

Proposition 12.5 (Weight and gravitation)

Near the Earth’s surface a body of mass $m$ is pulled by its weight $\vect P = m\vect g$ , with $\vect g$ the local gravitational field, $g \approx 9.81\,\mathrm{m}/\mathrm{s}^{2}$ , directed downward (toward the center, very nearly). More generally a mass $M$ at distance $r$ attracts $m$ with $\vect F = -GMm/r^2\,\vect e_r$ (Chapter 16); 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; $g = GM/R^2 = 6.67\times10^{-11} \times 5.97\times10^{24}/(6.37\times10^6)^2 = 9.8\,\mathrm{m}/\mathrm{s}^{2}$ . ∎

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

\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 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 $\vect N$ , perpendicular to the surface and repulsive ( $N \geq 0$ : a surface cannot pull), and a tangential friction force $\vect T$ . 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. — 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 $\mu_sN$ ; 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 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) 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, derived in Chapter 21; the electric and magnetic forces on a charge are the subject of Chapter 17. All the forces 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)

System: name the body treated as a point.
Frame: name it and check it is inertial (or treat it as such to the accuracy required).
Forces: list every body touching or attracting the system, and draw each force on a diagram (the free-body diagram); nothing else enters.
Coordinates: choose axes adapted to the motion (Method 11.18) and express $\vect a$ .
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).
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 $m$ 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 $m$ on a string of length $\ell$ , angle $\theta$ from the downward vertical. Polar coordinates at the suspension point: $\vect a = -\ell\dot\theta^2\,\vect e_r + \ell\ddot\theta\,\vect e_\theta$ ; forces: 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 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). — 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 $v_\infty$ , 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 $m$ falling from rest under its weight and a drag force reaches a terminal velocity 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). 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 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 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. — 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 $N\sin\beta$ 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 $70\,\mathrm{kg}$ 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

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 $m_2 = 1.0\,\mathrm{kg}$ . Find the acceleration and the tension.

Solution

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?

Solution

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 in sight.

Exercise 12.4 ★

A $0.50\,\mathrm{kg}$ mass hangs from a spring of stiffness $200\,\mathrm{N}/\mathrm{m}$ . Find the static elongation; then the angular frequency and period of its vertical oscillations about equilibrium (show the equation of motion is harmonic).

Solution

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

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. Check with a Reynolds number that each regime is the right one.

Solution

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

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

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.

Exercise 12.9 ★★

A conical pendulum: a mass 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, the period and the tension (in units of $mg$ ).

Solution

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, 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

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

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 $m$ runs on the inside of a vertical circular loop of radius $R$ (Problem 11.1). Express the track’s normal reaction 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

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. — 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 $m = 80\,\mathrm{kg}$ . 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.

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

Part II — Falling through air.

Write the drag force for belly-down fall as a function of $v$ .
Write the equation of motion for $v(t)$ .
Find the terminal velocity $v_\infty$ (in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ ).
With $u = v/v_\infty$ and $\tau = v_\infty/g$ , show that $\dd u/\dd t = (1 - u^2)/\tau$ ; compute $\tau$ .
Solve by separation of variables and show that $v = v_\infty\tanh(t/\tau)$ .
After how long is $v = 0.9v_\infty$ ? $0.99v_\infty$ ?
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.
Estimate the Reynolds number at $v_\infty$ with $L = 0.5\,\mathrm{m}$ and justify the quadratic law.

Part III — The canopy opens.

Compute the new terminal velocity under the open canopy.
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?
Real canopies open over about $3\,\mathrm{s}$ . Estimate the mean deceleration and the mean force of the harness on the parachutist.
Once the canopy is fully open the speed decays toward the new $v_\infty$ : what is the new time constant $\tau' = v_\infty'/g$ , and what does it mean for the approach to steady descent?
In steady descent, what is the harness force? Justify with the second law.
Explain why the terminal velocity varies as $1/\sqrt{C_xS}$ : by what factor would a canopy of half the area raise it?
A heavier jumper ( $100\,\mathrm{kg}$ ) under the same canopy: terminal velocity?

Part IV — Landing.

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?
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 on the legs in each case, and conclude.
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.

A raindrop of radius $1.0\,\mathrm{mm}$ ( $C_x = 0.5$ ): compute its terminal velocity with the quadratic law, and check the Reynolds number.
A fog droplet of radius $10\,\text{µ}\mathrm{m}$ : use Stokes’s law for $v_\infty$ and check that the Reynolds number is indeed small.
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

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: 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) 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 $\approx m(g + a) \approx 1.7\,\mathrm{kN}$ .

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

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 $m(g + a) \approx 5\,\mathrm{kN}$ ; $17\,\mathrm{m}/\mathrm{s}^{2}$ ( $1.8g$ ) 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); $\mathrm{Re} \approx 0.02 \ll 1$ : 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.

Terms defined in this chapter

See all 393 terms in the glossary