University Physics — Year 2 · Bachelor Year 2

1Rigid-Body Mechanics

Watch a bicycle wheel from the kerb: the spokes near the road are almost still and sharp, the spokes at the top are a blur moving at twice the speed of the cyclist. Every point of the wheel has a different velocity, yet the wheel is one object — its points keep their distances, and that single constraint organizes all their motions into two vectors, a translation and a rotation. The Year 1 volume treated the special case of a solid turning about a fixed axis; this chapter gives the general kinematics of a rigid body, its kinetic energy and angular momentum, the laws of its motion, and the law that governs every wheel, tyre, brake and ladder: Coulomb’s law of dry friction.

1.1 Kinematics of a rigid body

Definition 1.1 (Rigid body)

A rigid body (or solid) is a system of points whose mutual distances stay constant: $\|\vect{AB}\|$ is independent of time for every pair $A$ , $B$ of its points. A frame in which every point of the solid is at rest is a frame attached to the solid.

Theorem 1.2 (Velocity field of a rigid body)

At every instant there is a vector $\vect\Omega$ , the rotation vector of the solid (unit $\mathrm{rad}/\mathrm{s}$ ), such that the velocities of any two of its points $A$ and $M$ in a frame $\mathcal R$ satisfy

\vect v_M = \vect v_A + \vect\Omega\wedge\vect{AM} .

$\vect\Omega$ does not depend on the choice of $A$ ; a vector $\vect u$ fixed in the solid evolves as $\dd\vect u/\dd t = \vect\Omega\wedge\vect u$ .

Proof. Let $(\vect e_1, \vect e_2, \vect e_3)$ be an orthonormal basis attached to the solid. From $\vect e_i\cdot\vect e_j = \delta_{ij}$ , the coefficients $a_{ij} = \dot{\vect e}_i\cdot\vect e_j$ satisfy $a_{ij} + a_{ji} = 0$ : the matrix of the map $\vect u \mapsto \dot{\vect u}$ (for $\vect u$ fixed in the solid, $\vect u = \sum u_i\vect e_i$ with constant $u_i$ ) is antisymmetric, and an antisymmetric $3\times3$ matrix is the matrix of $\vect u \mapsto \vect\Omega\wedge\vect u$ with $\Omega_1 = a_{23}$ , $\Omega_2 = a_{31}$ , $\Omega_3 = a_{12}$ . Apply it to $\vect u = \vect{AM}$ : $\vect v_M - \vect v_A = \vect\Omega\wedge\vect{AM}$ . If $\vect\Omega'$ did the same job, $(\vect\Omega - \vect\Omega')\wedge \vect{AM} = \vect 0$ for every $M$ , so $\vect\Omega' = \vect\Omega$ . ∎

Remark 1.3 (Three motions)

Translation: $\vect\Omega = \vect 0$ , all points share one velocity (not necessarily a straight-line motion: a Ferris-wheel cabin translates on a circle). Rotation about a fixed axis $\Delta$ : the points of $\Delta$ are at rest, $\vect\Omega = \omega\,\vect e_\Delta$ , and a point at distance $r$ from the axis moves on a circle at speed $r|\omega|$ — the case of the Year 1 volume. The general motion combines both: $\vect v_M = \vect v_G + \vect\Omega\wedge\vect{GM}$ , a translation at the velocity of the centre of mass plus a rotation about $G$ .

Definition 1.4 (Rolling without slipping)

Two solids $S_1$ and $S_2$ touch at a point $I$ . The slip velocity of $S_1$ on $S_2$ is $\vect v_{\text{slip}} = \vect v_{I \in S_1} - \vect v_{I \in S_2}$ , the difference of the velocities of the two material points that coincide at $I$ at this instant. $S_1$ rolls without slipping on $S_2$ when $\vect v_{\text{slip}} = \vect 0$ . For a wheel of radius $R$ rolling without slipping on a fixed ground, with centre velocity $v_G$ and angular velocity $\omega$ , the condition reads $v_G = R\omega$ (signs chosen so that a wheel rolling to the right turns clockwise).

Proof. Apply Theorem 1.2 to the wheel between $G$ and the contact point $I$ : $\vect v_{I\in S_1} = \vect v_G + \vect\Omega \wedge\vect{GI}$ ; with $\vect v_G = v_G\vect e_x$ , $\vect\Omega = -\omega \vect e_z$ and $\vect{GI} = -R\vect e_y$ , this is $(v_G - R\omega)\vect e_x$ , which must vanish. ∎

Example 1.5 (Velocity field of a rolling wheel)

For the rolling wheel, $\vect v_M = \vect\Omega\wedge\vect{IM}$ : at each instant the wheel rotates about its contact point. The contact point is at rest, the centre moves at $v = R\omega$ , the top at $2v$ , and a point of the rim at height $h$ moves at $\omega\sqrt{2Rh}$ , perpendicular to $\vect{IM}$ . A spoke valve traces a cycloid; a stone flung from the tread leaves at the speed of the rim point, up to $2v$ — which is why a lorry’s mudflaps matter.

A wheel caught by a slow shutter: the spokes near the road, almost at rest, stay sharp; those at the top, moving at twice the bicycle’s speed, blur.

A wheel rolling without slipping. Left: the velocities of a few points — zero at the contact point I, v_G = R at the centre, 2v_G at the top. Right: every velocity is perpendicular to the segment joining the point to I and proportional to its length: at each instant the wheel turns about its contact point. — A wheel rolling without slipping. Left: the velocities of a few points — zero at the contact point $I$ , $v_G = R\omega$ at the centre, $2v_G$ at the top. Right: every velocity is perpendicular to the segment joining the point to $I$ and proportional to its length: at each instant the wheel turns about its contact point.

1.2 Kinetic quantities of a solid

Proposition 1.6 (Momentum, angular momentum, kinetic energy)

For a solid of mass $M$ , centre of mass $G$ , in a frame $\mathcal R$ :

its momentum is $\vect p = M\vect v_G$ ;
if it rotates about an axis $\Delta$ (fixed in $\mathcal R$ , or passing through $G$ and fixed in direction) at angular velocity $\omega$ , its angular momentum along that axis is $L_\Delta = J_\Delta\,\omega$ , where

$J_\Delta = \sum_i m_i r_i^2 = \int r^2\,\dd m$

is its moment of inertia about $\Delta$ ( $r$ the distance to the axis; unit $\mathrm{kg}\,\mathrm{m}^{2}$ );
its kinetic energy is (Koenig) $E_k = \tfrac12Mv_G^2 + E_k^*$ , where the kinetic energy in the barycentric frame is $E_k^* = \tfrac12J_{G}\omega^2$ for a rotation at $\omega$ about an axis through $G$ — and $E_k = \tfrac12J_\Delta\omega^2$ for a rotation about a fixed axis $\Delta$ .

Proof. The momentum and Koenig’s theorems were proved in the Year 1 volume for any system. For the rotation about $\Delta$ through $O$ , a point at distance $r_i$ from the axis has speed $r_i\omega$ on a circle around the axis; its angular momentum about $O$ projected on $\vect e_\Delta$ is $m_ir_i^2\omega$ (the other components, which cancel for a symmetric solid, are not needed here); sum. Kinetic energy: $\sum\tfrac12m_ir_i^2 \omega^2$ . ∎

Remark 1.7 (What is admitted)

For an arbitrary rotation $\vect\Omega$ the full vector $\vect L_G$ is not parallel to $\vect\Omega$ in general: $\vect L_G = \mathbf J\,\vect\Omega$ with $\mathbf J$ the inertia tensor, a symmetric matrix whose eigenvectors are the principal axes. Every solid of this chapter rotates about a fixed axis or about a symmetry axis through $G$ , for which $\vect L = J\vect\Omega$ exactly; the general case is the Year 3 volume’s.

Proposition 1.8 (Moments of inertia)

For homogeneous solids of mass $M$ , about an axis through $G$ : thin rod of length $\ell$ (axis perpendicular), $J = \tfrac1{12}M\ell^2$ ; thin hoop or cylindrical shell of radius $R$ (own axis), $J = MR^2$ ; disk or full cylinder, $J = \tfrac12MR^2$ ; full sphere, $J = \tfrac25MR^2$ ; spherical shell, $J = \tfrac23MR^2$ . Huygens’ theorem (parallel axes): about an axis $\Delta$ parallel to an axis $\Delta_G$ through $G$ at distance $d$ ,

J_\Delta = J_{\Delta_G} + Md^2 .

Proof. Rod: $\int_{-\ell/2}^{\ell/2}x^2\,(M/\ell)\dd x = M\ell^2/12$ . Disk: rings of mass $(2M/R^2)r\,\dd r$ , $\int_0^R r^2(2M/R^2)r\,\dd r = MR^2/2$ . Sphere: by symmetry $J = \tfrac23\int r^2\dd m$ ( $r$ the distance to the centre, since $x^2 + y^2 = \tfrac23(x^2 + y^2 + z^2)$ on average over the three axes) and $\int r^2\dd m = \int_0^R r^2(3M/R^3)r^2\dd r = \tfrac35MR^2$ . Huygens: with $\vect r_i$ the vector from $\Delta$ to $m_i$ perpendicular to the axis and $\vect r_i = \vect d + \vect r_i^*$ , $\sum m_ir_i^2 = Md^2 + 2\vect d\cdot\sum m_i\vect r_i^* + \sum m_ir_i^{*2}$ , and the middle sum vanishes by definition of $G$ . ∎

Example 1.9 (A flywheel)

A steel disk of $50\,\mathrm{kg}$ and radius $30\,\mathrm{cm}$ : $J = \tfrac12 \times 50 \times 0.09 = 2.25\,\mathrm{kg}\,\mathrm{m}^{2}$ ; at $3000\,\mathrm{rpm}$ ( $\omega = 314\,\mathrm{rad}/\mathrm{s}$ ) it stores $\tfrac12J\omega^2 = 111\,\mathrm{kJ}$ , the kinetic energy of a small car at $50\,\mathrm{km}/\mathrm{h}$ . Energy-storage flywheels spin carbon-fibre rotors at $50\,000\,\mathrm{rpm}$ in vacuum on magnetic bearings: the energy grows as $\omega^2$ , the limit is the tensile strength of the rim.

1.3 Dynamics of a solid

Theorem 1.10 (Laws of motion of a solid)

In a Galilean frame, for a solid subject to external forces $\vect F_i$ applied at points $A_i$ :

(momentum) $M\,\dd\vect v_G/\dd t = \sum\vect F_i$ ;
(angular momentum about $G$ , or about a fixed point $O$ ) $\dd\vect L_G/\dd t = \sum\vect{GA_i}\wedge\vect F_i$ ; projected on a fixed axis $\Delta$ or on an axis through $G$ of fixed direction along which the solid rotates at $\omega$ : $J_\Delta\,\dd\omega/\dd t = \mathcal M_\Delta$ , the sum of the moments of the external forces about the axis;
(power) the power of a set of forces on a solid is $\mathcal P = \vect R\cdot\vect v_A + \vect{\mathcal M}_A\cdot \vect\Omega$ for any point $A$ of the solid, with $\vect R$ the sum of the forces and $\vect{\mathcal M}_A$ their total moment about $A$ ; the internal forces of a solid have zero power, and $\dd E_k/\dd t = \mathcal P_{\text{ext}}$ .

Proof. (1) and (2) are the theorems of the Year 1 volume for systems of points (the angular momentum theorem about $G$ holds even when $G$ accelerates, because the inertial forces of the barycentric frame have zero moment about $G$ ). (3) With $\vect v_{A_i} = \vect v_A + \vect\Omega \wedge\vect{AA_i}$ : $\sum\vect F_i\cdot\vect v_{A_i} = \vect R\cdot\vect v_A + \sum\vect F_i\cdot(\vect\Omega\wedge\vect{AA_i}) = \vect R\cdot\vect v_A + \vect\Omega\cdot\sum\vect{AA_i}\wedge\vect F_i$ (mixed product). Internal forces: their sum and their total moment vanish (action–reaction, same line), hence zero power on a solid — not on a deformable system. The kinetic energy theorem then keeps only the external power. ∎

Definition 1.11 (Perfect pivot)

A solid rotates about a fixed axis $\Delta$ in a perfect pivot (an ideal bearing) when the contact forces of the bearing have zero moment about $\Delta$ ; they then develop no power ( $\vect v_A = \vect 0$ on the axis, $\mathcal M_\Delta = 0$ ). A real bearing adds a friction torque $-\Gamma_f\operatorname{sgn}\omega$ , or a viscous one $-h\omega$ .

Proposition 1.12 (The physical pendulum)

A solid of mass $M$ swings about a horizontal fixed axis $\Delta$ in a perfect pivot, its centre of mass at distance $d$ from the axis. With $\theta$ the angle of $\vect{OG}$ from the downward vertical,

J_\Delta\ddot\theta = -Mgd\sin\theta ,

so that small oscillations have the period $T = 2\pi\sqrt{J_\Delta/Mgd}$ — that of a simple pendulum of length $\ell_{\text{eq}} = J_\Delta/Md$ .

Proof. Angular momentum theorem about $\Delta$ : the weight’s moment is $-Mgd\sin\theta$ , the pivot’s is zero. Linearize. ∎

Example 1.13 (The metre rule)

A metre rule pivoted at one end: $J = \tfrac13M\ell^2$ (Huygens from $\tfrac1{12}M\ell^2$ ), $d = \ell/2$ , $\ell_{\text{eq}} = \tfrac23\ell = 0.667\,\mathrm{m}$ and $T = 1.64\,\mathrm{s}$ — a whole-rule pendulum beats slower than a point mass hung at its centre ( $1.42\,\mathrm{s}$ ) and faster than one at its end ( $2.0\,\mathrm{s}$ ).

Left: the physical pendulum — the weight’s moment about the pivot drives the rotation, the pivot’s force has no moment. Right: the forces on a wheel rolling on a horizontal ground: weight, normal reaction N and tangential friction T at the contact point, whose magnitude is bounded by Coulomb’s law. — Left: the physical pendulum — the weight’s moment about the pivot drives the rotation, the pivot’s force has no moment. Right: the forces on a wheel rolling on a horizontal ground: weight, normal reaction $\vect N$ and tangential friction $\vect T$ at the contact point, whose magnitude is bounded by Coulomb’s law.

1.4 Contact forces and Coulomb friction

Definition 1.14 (Contact force, normal and tangential)

The force exerted by a support on a solid at a contact point $I$ splits into a normal reaction $\vect N$ , perpendicular to the common tangent plane and directed toward the solid ( $N \ge 0$ : a support can only push), and a tangential component $\vect T$ in the tangent plane, the friction force.

Theorem 1.15 (Coulomb’s laws of dry friction)

For two dry solids in contact, with a coefficient of friction $f$ depending on the materials and their state of surface, but not on the area of contact nor (to a good approximation) on the speed:

if the solids slip on each other ( $\vect v_{\text{slip}} \ne \vect 0$ ), the friction force is opposite to the slip velocity and of magnitude $\|\vect T\| = f\|\vect N\|$ (kinetic friction);
if they do not slip, $\|\vect T\| \le f\|\vect N\|$ (static friction): the friction force is then whatever the other laws of motion require, within that bound, and its direction is not known in advance.

A perfect contact is the idealization $f = 0$ : $\vect T = \vect 0$ .

Proof. Admitted at this level. ∎

Remark 1.16 (Reading Coulomb’s laws)

The first law is an equation, the second an inequality — and that is the whole difficulty of friction problems. The method is to assume no slipping, solve for $\vect T$ and $\vect N$ , and check $\|\vect T\| \le f\|\vect N\|$ ; if the check fails, the solid slips, $\|\vect T\| = f\|\vect N\|$ opposite to the slip, and the problem is solved again. Geometrically, the contact force must stay inside the friction cone of half-angle $\varphi$ with $\tan\varphi = f$ . Typical values: rubber on dry asphalt $f \approx 0.8$ – $1$ , on wet asphalt $0.4$ – $0.6$ , on ice $0.1$ ; steel on steel $0.15$ dry, $0.05$ greased; the static coefficient is in reality slightly larger than the kinetic one, which is why a skid is hard to stop once started.

Proposition 1.17 (Power of the contact forces; rolling without slipping)

The power of the contact forces on a solid moving on a fixed support is $\mathcal P = \vect T\cdot\vect v_{I\in S}$ (the normal force is perpendicular to the velocity of the contact point, which is in the tangent plane). In rolling without slipping this power is zero: friction then does no work, though it may be essential to the motion. In sliding it is $-f N v_{\text{slip}} < 0$ : heat.

Proof. $\mathcal P = (\vect N + \vect T)\cdot\vect v_{I\in S}$ and $\vect v_{I\in S} = \vect v_{\text{slip}}$ lies in the tangent plane, so only $\vect T$ contributes; it is zero when the slip velocity is zero, and $-fNv_{\text{slip}}$ when the solids slip ( $\vect T$ opposite to $\vect v_{\text{slip}}$ ). Real rolling dissipates a little through the deformation of tyre and ground (rolling friction, modelled by a small torque $\Gamma_r = \mu_rNR$ or an equivalent force $\mu_rN$ , with $\mu_r \approx 0.01$ for a tyre on a road): a different, much smaller effect. ∎

Method 1.18 (A solid rolling down an incline)

A homogeneous solid of radius $R$ , mass $M$ and moment of inertia $J = kMR^2$ about its axis ( $k = 1$ hoop, $\tfrac12$ disk, $\tfrac25$ sphere) is released on an incline of angle $\alpha$ with friction coefficient $f$ . Assume rolling without slipping:

momentum along the slope: $M\dot v = Mg\sin\alpha - T$ ; normal: $N = Mg\cos\alpha$ ;
angular momentum about $G$ : $kMR^2\dot\omega = TR$ ;
no slip: $v = R\omega$ , hence $T = kM\dot v$ and

$\dot v = \frac{g\sin\alpha}{1 + k} , \qquad T = \frac{k}{1 + k}\, Mg\sin\alpha ;$
check: $T \le fN \iff \tan\alpha \le f\,\dfrac{1 + k}{k}$ . Beyond that slope the solid slips, $T = fMg\cos\alpha$ , $\dot v = g(\sin\alpha - f\cos\alpha)$ and $\dot\omega = fg\cos\alpha/kR$ .

Energy check (no slip, friction powerless): $\tfrac12(1 + k)Mv^2 = Mgh$ gives $v = \sqrt{2gh/(1 + k)}$ — a sphere beats a disk, which beats a hoop, independently of mass and radius.

Left: forces on a solid rolling down an incline — weight at G, normal reaction and friction at the contact point. Right: the acceleration of a disk (k = 1/2) as a function of the slope for f = 0.5: it rolls without slipping while 3f, then slips and its acceleration follows g( - f ), always below the frictionless g. — Left: forces on a solid rolling down an incline — weight at $G$ , normal reaction and friction at the contact point. Right: the acceleration of a disk ( $k = \tfrac12$ ) as a function of the slope for $f = 0.5$ : it rolls without slipping while $\tan\alpha \le 3f$ , then slips and its acceleration follows $g(\sin\alpha - f\cos\alpha)$ , always below the frictionless $g\sin\alpha$ .

Example 1.19 (Why a car needs friction to accelerate)

A front-wheel-drive car: the engine applies a torque to the front wheels; the only external horizontal force on the car is the friction of the road on the tyres, and it is this force that accelerates the car. With the whole weight on the driven wheels the maximum acceleration is $fg \approx 8\,\mathrm{m}/\mathrm{s}^{2}$ on dry asphalt — and half of that if only half the weight rests on the driven axle; on ice, $1\,\mathrm{m}/\mathrm{s}^{2}$ . The engine’s work is not done by the road (whose contact point is at rest): it is internal work, converted by the transmission into kinetic energy — the same way the internal forces of a walker’s muscles, not the ground, supply the walker’s energy.

Example 1.20 (The sliding-to-rolling transition)

A bowling ball is released sliding at $v_0$ without spin. Kinetic friction $fMg$ slows the centre ( $\dot v = -fg$ ) and spins the ball up ( $\tfrac25MR^2\dot\omega = fMgR$ , $\dot\omega = 5fg/2R$ ) until $v = R\omega$ : at $t_1 = 2v_0/7fg$ , with $v_1 = \tfrac57v_0$ . From then on it rolls without slipping at constant speed. Two sevenths of the speed, and $1 - (\tfrac57)^2\tfrac75 = \tfrac27$ of the kinetic energy, are lost to heat in the slide — whatever $f$ is; $f$ only sets how fast.

A ball released sliding without spin: kinetic friction brakes the centre and spins the ball up, linearly in time, until R catches v_G at 5/7v_0; from then on the ball rolls without slipping and friction stops acting. — A ball released sliding without spin: kinetic friction brakes the centre and spins the ball up, linearly in time, until $R\omega$ catches $v_G$ at $\tfrac57v_0$ ; from then on the ball rolls without slipping and friction stops acting.

1.5 Exercises

Exercise 1.1 ★

Moments of inertia: (a) a $1.0\,\mathrm{m}$ , $0.50\,\mathrm{kg}$ rod about a perpendicular axis through its centre, then through one end; (b) a $2.0\,\mathrm{kg}$ disk of radius $20\,\mathrm{cm}$ about its axis, then about a parallel axis through its rim; (c) a $0.16\,\mathrm{kg}$ billiard ball of diameter $57\,\mathrm{mm}$ . (d) A thin hoop and a full disk of the same mass and radius: which resists spin-up more, and by what factor?

Solution

Solution of Exercise 1.1.

(a) $\tfrac1{12}M\ell^2 = 4.2 \times 10^{-2}\,\mathrm{kg}\,\mathrm{m}^{2}$ ; through one end (Huygens, $d = \ell/2$ ): $\tfrac13M\ell^2 = 0.167\,\mathrm{kg}\,\mathrm{m}^{2}$ . (b) $\tfrac12MR^2 = 4.0 \times 10^{-2}\,\mathrm{kg}\,\mathrm{m}^{2}$ ; at the rim $\tfrac32MR^2 = 0.12\,\mathrm{kg}\,\mathrm{m}^{2}$ . (c) $\tfrac25 \times 0.16 \times (0.0285)^2 = 5.2 \times 10^{-5}\,\mathrm{kg}\,\mathrm{m}^{2}$ . (d) The hoop, $MR^2$ against $\tfrac12MR^2$ : twice.

Exercise 1.2 ★

A bicycle rides at $18\,\mathrm{km}/\mathrm{h}$ on wheels of diameter $70\,\mathrm{cm}$ . Angular velocity of the wheels; speed, in the ground frame, of the valve when it is at the top, at the bottom, at the front of the wheel (height of the axle); speed of the tread relative to the frame of the bicycle. A stone stuck in the tread comes loose at the top: at what speed does it leave?

Solution

Solution of Exercise 1.2.

$v = 5.0\,\mathrm{m}/\mathrm{s}$ , $\omega = v/R = 5/0.35 = 14.3\,\mathrm{rad}/\mathrm{s}$ . Top: $2v = 10\,\mathrm{m}/\mathrm{s}$ forward; bottom: $0$ ; front: $\vect v_G + \vect\Omega \wedge\vect{GM} = (v, -v)$ , i.e. $v\sqrt2 = 7.1\,\mathrm{m}/\mathrm{s}$ at $45{}^{\circ}$ downward. Relative to the bicycle the tread moves at $R\omega = v = 5\,\mathrm{m}/\mathrm{s}$ everywhere. The stone leaves at $10\,\mathrm{m}/\mathrm{s}$ , horizontally forward.

Exercise 1.3 ★

A flywheel is a $40\,\mathrm{kg}$ steel disk of radius $25\,\mathrm{cm}$ . (a) Energy stored at $6000\,\mathrm{rpm}$ . (b) Constant torque needed to spin it up from rest in $2.0\,\mathrm{min}$ ; power at the end of the spin-up. (c) In a bus, it must deliver $20\,\mathrm{kW}$ for $10\,\mathrm{s}$ : to what speed does it slow down? (d) Why are high-speed flywheels run in vacuum?

Solution

Solution of Exercise 1.3.

$J = \tfrac12 \times 40 \times 0.25^2 = 1.25\,\mathrm{kg}\,\mathrm{m}^{2}$ ; $\omega = 628\,\mathrm{rad}/\mathrm{s}$ . (a) $E = \tfrac12J\omega^2 = 246\,\mathrm{kJ}$ . (b) $\Gamma = J\omega/\Delta t = 1.25 \times 628/120 = 6.5\,\mathrm{N}\,\mathrm{m}$ ; $P = \Gamma\omega = 4.1\,\mathrm{kW}$ at the end. (c) $200\,\mathrm{kJ}$ taken out leave $46\,\mathrm{kJ}$ : $\omega = \sqrt{2 \times 46\,000/1.25} = 271\,\mathrm{rad}/\mathrm{s} = 2600\,\mathrm{rpm}$ . (d) Air drag on the rim grows as $\omega^2$ (power as $\omega^3$ ) and would heat and slow the rotor; in vacuum on magnetic bearings the losses fall to a few watts.

Exercise 1.4 ★

A uniform disk of radius $R$ swings in a vertical plane about a horizontal axis through a point of its rim. Moment of inertia about the axis; period of small oscillations for $R = 20\,\mathrm{cm}$ ; length of the equivalent simple pendulum. Same questions for the axis at a distance $R/2$ from the centre. Where should the axis be for the shortest period?

Solution

Solution of Exercise 1.4.

$J = \tfrac12MR^2 + MR^2 = \tfrac32MR^2$ , $d = R$ : $\ell_{\text{eq}} = \tfrac32R = 0.30\,\mathrm{m}$ , $T = 2\pi\sqrt{0.30/9.81} = 1.10\,\mathrm{s}$ . At $R/2$ : $J = \tfrac12MR^2 + \tfrac14MR^2 = \tfrac34MR^2$ , $d = R/2$ , $\ell_{\text{eq}} = \tfrac32R$ again, same period. In general $\ell_{\text{eq}}(d) = (R^2/2 + d^2)/d$ , minimal at $d = R/\sqrt2$ : $\ell_{\text{eq}} = R\sqrt2 = 0.283\,\mathrm{m}$ , $T = 1.07\,\mathrm{s}$ .

Exercise 1.5 ★★

A hoop, a disk and a sphere of the same radius are released together at the top of a $2.0\,\mathrm{m}$ high incline of slope $20{}^{\circ}$ , $f = 0.30$ . (a) Check that all three roll without slipping. (b) Speed at the bottom for each; order of arrival. (c) Time taken by each. (d) Show that the friction force on the sphere is $\tfrac27Mg\sin\alpha$ and that it does no work. (e) At what slope would the hoop start to slip? The sphere?

Solution

Solution of Exercise 1.5.

(a) $\tan20^\circ = 0.36 \le f(1 + k)/k$ , i.e. $\le 0.6$ (hoop), $0.9$ (disk), $1.05$ (sphere): all roll. (b) $v = \sqrt{2gh/(1 + k)}$ : hoop $4.4\,\mathrm{m}/\mathrm{s}$ , disk $5.1\,\mathrm{m}/\mathrm{s}$ , sphere $5.3\,\mathrm{m}/\mathrm{s}$ ; sphere first, hoop last. (c) Length $h/\sin\alpha = 5.85\,\mathrm{m}$ , $a = g\sin\alpha/(1 + k) = 1.68$ , $2.24$ , $2.40\,\mathrm{m}/\mathrm{s}^{2}$ , $t = \sqrt{2d/a} = 2.6$ , $2.3$ , $2.2\,\mathrm{s}$ . (d) $T = \dfrac{k}{1 + k}Mg\sin\alpha = \tfrac27Mg\sin\alpha$ ; the contact point is at rest, so $\vect T\cdot\vect v_I = 0$ . (e) Hoop: $\tan\alpha > 2f = 0.6$ , $\alpha > 31{}^{\circ}$ ; sphere: $\tan\alpha > 3.5f = 1.05$ , $\alpha > 46{}^{\circ}$ .

Exercise 1.6 ★★

Disk brake. A wheel and its brake disk have $J = 1.2\,\mathrm{kg}\,\mathrm{m}^{2}$ and spin at $900\,\mathrm{rpm}$ ; two pads press on the disk at a mean radius of $12\,\mathrm{cm}$ , each with a normal force of $800\,\mathrm{N}$ , $f = 0.40$ . Braking torque; angular deceleration; time and number of turns to stop; heat released, and temperature rise of the $1.5\,\mathrm{kg}$ steel disk ( $c = 470\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$ ) if it keeps it all.

Solution

Solution of Exercise 1.6.

$\Gamma = 2fNr = 2 \times 0.4 \times 800 \times 0.12 = 77\,\mathrm{N}\,\mathrm{m}$ ; $\dot\omega = \Gamma/J = 64\,\mathrm{rad}/\mathrm{s}^{2}$ ; $\omega_0 = 94\,\mathrm{rad}/\mathrm{s}$ , $t = 1.5\,\mathrm{s}$ ; $\theta = \omega_0^2/2\dot\omega = 69\,\mathrm{rad} = 11$ turns; heat $= \tfrac12J \omega_0^2 = 5.3\,\mathrm{kJ}$ , $\Delta T = 5300/(1.5 \times 470) = 7.6\,\mathrm{K}$ .

Exercise 1.7 ★★

Atwood’s machine with a real pulley. Masses $m_1 = 2.0\,\mathrm{kg}$ and $m_2 = 1.5\,\mathrm{kg}$ hang on a rope over a pulley of radius $R = 10\,\mathrm{cm}$ and moment of inertia $J = 5.0 \times 10^{-3}\,\mathrm{kg}\,\mathrm{m}^{2}$ on a perfect pivot; the rope does not slip on the pulley. (a) Why are the two rope tensions different? (b) Equations for the two masses and the pulley; acceleration. (c) The two tensions. (d) Compare with the massless pulley; by energy, recover the acceleration.

Solution

Solution of Exercise 1.7.

(a) The rope must exert a net torque to spin the pulley up: $T_1 \ne T_2$ . (b) $m_1a = m_1g - T_1$ , $m_2a = T_2 - m_2g$ , $J\dot\omega = (T_1 - T_2)R$ with $a = R\dot\omega$ : $a = (m_1 - m_2)g/(m_1 + m_2 + J/R^2) = 0.5 \times 9.81/4.0 = 1.23\,\mathrm{m}/\mathrm{s}^{2}$ . (c) $T_1 = m_1(g - a) = 17.2\,\mathrm{N}$ , $T_2 = m_2(g + a) = 16.6\,\mathrm{N}$ . (d) Massless: $1.40\,\mathrm{m}/\mathrm{s}^{2}$ . Energy: $(m_1 - m_2)gx = \tfrac12(m_1 + m_2 + J/R^2)v^2$ ; differentiate.

Exercise 1.8 ★★

The ladder. A uniform ladder of length $\ell$ and mass $m$ leans at angle $\theta$ from the horizontal against a smooth vertical wall; the ground has friction coefficient $f$ . (a) Forces and their moments about the foot; equilibrium equations. (b) Show that equilibrium requires $\tan\theta \ge 1/2f$ . (c) Minimum angle for $f = 0.40$ . (d) A person of mass $3m$ climbs the ladder set at $60{}^{\circ}$ : how far up (fraction of $\ell$ ) can they go?

Solution

Solution of Exercise 1.8.

(a) Weight $mg$ at the middle; wall: horizontal $N_w$ at the top; ground: vertical $N = mg$ and horizontal friction $T = N_w$ . Moments about the foot: $mg\,\tfrac{\ell}2\cos\theta = N_w\ell\sin\theta$ . (b) $T = N_w = mg/(2\tan\theta) \le fmg \iff \tan\theta \ge 1/2f$ . (c) $\tan\theta \ge 1.25$ , $\theta \ge 51{}^{\circ}$ . (d) Person at $x$ : $N = 4mg$ , $N_w = mg(\tfrac12 + 3x/\ell)/\tan60^\circ \le 4fmg = 1.6mg$ : $x/\ell \le (2.77 - 0.5)/3 = 0.76$ .

Exercise 1.9 ★★

Back-spin. A billiard ball is struck so that it leaves with centre speed $v_0$ and back-spin $\omega_0$ (rotation opposite to rolling). Friction coefficient $f$ . (a) Equations for $v$ and $\omega$ while sliding. (b) Time at which it starts rolling without slipping and its speed then, as functions of $v_0$ , $R\omega_0$ . (c) Condition on $R\omega_0$ for the ball to come back toward the player. (d) Numbers: $v_0 = 2.0\,\mathrm{m}/\mathrm{s}$ , $R\omega_0 = 6.0\,\mathrm{m}/\mathrm{s}$ , $f = 0.2$ , $R = 2.9\,\mathrm{cm}$ .

Solution

Solution of Exercise 1.9.

(a) The contact point moves forward at $v + R\omega$ ( $\omega > 0$ for back-spin): friction $-fmg$ on the centre, $\dot v = -fg$ ; torque $-fmgR$ on the spin: $\dot\omega = -5fg/2R$ . (b) Slip $v + R\omega = v_0 + R\omega_0 - \tfrac72fgt$ vanishes at $t_1 = 2(v_0 + R\omega_0)/7fg$ , then $v_1 = v_0 - fgt_1 = (5v_0 - 2R\omega_0)/7$ . (c) $v_1 < 0 \iff R\omega_0 > \tfrac52v_0$ . (d) $t_1 = 16/(7 \times 1.96) = 1.2\,\mathrm{s}$ ; $v_1 = (10 - 12)/7 = -0.29\,\mathrm{m}/\mathrm{s}$ : it comes back.

Exercise 1.10 ★★★

The spool. A spool (mass $M$ , moment of inertia $J = kMR^2$ about its axis, outer radius $R$ ) rests on a horizontal table; a thread wound on its inner hub of radius $r < R$ is pulled horizontally with a force $F$ , the thread coming off the bottom of the hub. (a) Assuming rolling without slipping, write the momentum and angular-momentum equations and find the acceleration of the centre; which way does the spool roll? (b) Friction force required, and the condition on $F$ for no slipping. (c) Same questions if the thread is pulled at an angle $\beta$ above the horizontal; show that the spool stays still when $\cos\beta = r/R$ , and explain geometrically (instantaneous axis through the contact point). (d) What happens for $\cos\beta < r/R$ ?

Solution

Solution of Exercise 1.10.

(a) With $x$ toward the pull, friction $T$ along $x$ : $Ma = F + T$ ; moments about $G$ (counterclockwise): $rF + RT = -kMRa$ (rolling, $\omega_z = -a/R$ ). Hence $a = F(R - r)/(1 + k)MR > 0$ : the spool rolls toward the pull, winding the thread up. (b) $T = -F(kR + r)/(1 + k)R$ (backward); no slip if $F \le fMg(1 + k)R/(kR + r)$ . (c) The thread’s line of action is tangent to the hub; its moment about the contact point $I$ is $F(r - R\cos\beta)$ , clockwise (forward roll) when $\cos\beta > r/R$ , zero when the line passes through $I$ — the instantaneous axis — and then $a = F(R\cos\beta - r)/(1 + k)MR$ , $T = -F(kR\cos\beta + r)/(1 + k)R$ , $N = Mg - F\sin\beta$ . (d) For $\cos\beta < r/R$ the moment about $I$ is counterclockwise: the spool rolls away from the puller.

Exercise 1.11 ★★★

Where to hit a billiard ball. A cue delivers a horizontal impulse $P$ (very short force) to a ball of radius $R$ at rest, at height $h$ above the table. (a) Momentum and angular momentum just after the hit (friction negligible during the hit). (b) Show that the ball rolls without slipping at once if $h = \tfrac75R$ , has top-spin above, back-spin below. (c) For $h = R$ (centre hit), how far does the ball slide before rolling, with $f = 0.2$ and $v_0 = 3\,\mathrm{m}/\mathrm{s}$ ? (d) Why is the cushion of a billiard table built at height $\tfrac75R$ ?

Solution

Solution of Exercise 1.11.

(a) $Mv_0 = P$ ; about $G$ : $\tfrac25MR^2\omega_0 = P(h - R)$ (top-spin sense for $h > R$ ). (b) $v_0 = R\omega_0 \iff 2R = 5(h - R) \iff h = \tfrac75R$ ; above it $R\omega_0 > v_0$ (top-spin), below it back-spin or under-spin. (c) $\omega_0 = 0$ : slides for $t_1 = 2v_0/7fg = 0.44\,\mathrm{s}$ , distance $v_0t_1 - \tfrac12fgt_1^2 = 1.1\,\mathrm{m}$ . (d) A cushion at $\tfrac75R$ returns the ball rolling without slipping: no sliding phase, no speed lost to the cloth, a predictable rebound.

Exercise 1.12 ★★★

Cylinder in a bowl. A full cylinder of radius $r$ rolls without slipping inside a fixed cylindrical surface of radius $R > r$ , axes horizontal and parallel. Let $\theta$ be the angle of the line of centres from the vertical. (a) Express the angular velocity of the cylinder about its own axis in terms of $\dot\theta$ (hint: the contact point is at rest). (b) Kinetic energy and potential energy; show the period of small oscillations is $T = 2\pi\sqrt{3(R - r)/2g}$ . (c) Compare with a point sliding without friction in the bowl. (d) Minimum friction coefficient for rolling without slipping at amplitude $\theta_0$ (small angles).

Solution

Solution of Exercise 1.12.

(a) $v_G = (R - r)\dot\theta$ ; the contact point is at rest, so the cylinder spins at $\omega = v_G/r = (R - r)\dot\theta/r$ about its axis. (b) $E_k = \tfrac12Mv_G^2 + \tfrac12\cdot\tfrac12Mr^2\omega^2 = \tfrac34M(R - r)^2 \dot\theta^2$ , $E_p = -Mg(R - r)\cos\theta$ ; $\dd E/\dd t = 0$ gives $\ddot\theta + \dfrac{2g}{3(R - r)}\sin\theta = 0$ , $T = 2\pi\sqrt{3(R - r)/2g}$ . (c) A sliding point: $2\pi\sqrt{(R - r)/g}$ — rolling is slower by $\sqrt{3/2}$ , a third of the energy being rotation. (d) Tangential: $M(R - r)\ddot\theta = -Mg\sin\theta + T$ , so $T = \tfrac13Mg\sin\theta$ ; at the turning point $N = Mg\cos\theta_0$ : $f \ge \tfrac13\tan\theta_0 \approx \theta_0/3$ .

1.6 Problem: A vehicle on wheels

Problem 1.1

Weekend problem — what a car asks of its tyres: rolling, accelerating, braking, and the one force that does it all

A car of mass $M = 1200\,\mathrm{kg}$ has a wheelbase $L = 2.6\,\mathrm{m}$ ; its centre of mass is at height $h = 0.55\,\mathrm{m}$ and at horizontal distances $a = 1.1\,\mathrm{m}$ from the front axle and $b = 1.5\,\mathrm{m}$ from the rear axle. Each of the four wheels has radius $R = 0.31\,\mathrm{m}$ , mass $m = 15\,\mathrm{kg}$ and moment of inertia $J = 1.0\,\mathrm{kg}\,\mathrm{m}^{2}$ about its axle. Tyre–road friction coefficient $f = 0.80$ (dry); rolling-friction coefficient $\mu_r = 0.012$ . $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$ .

Part I — Rolling.

The car drives at $v = 90\,\mathrm{km}/\mathrm{h}$ . Angular velocity of the wheels; velocity, in the road frame, of the top, bottom and front points of a tyre.
Total kinetic energy of the car, separating the translation of the whole and the rotation of the four wheels; what fraction do the wheels’ rotations represent?
At rest on level ground, normal forces $N_f$ (both front wheels together) and $N_r$ (both rear) by the momentum and angular-momentum equations — which axle carries more?
The rolling-friction torque on each wheel is $\mu_rNR$ . Power dissipated by rolling friction at $90\,\mathrm{km}/\mathrm{h}$ ; compare with the aerodynamic drag $\tfrac12\rho C_xSv^2$ with $C_xS = 0.70\,\mathrm{m}^{2}$ , $\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}$ .
The engine is cut and the car coasts on level ground: write the equation for $v(t)$ (include the wheels’ inertia as an effective mass) and estimate the distance to slow from $90\,\mathrm{km}/\mathrm{h}$ to $45\,\mathrm{km}/\mathrm{h}$ , neglecting drag; then including only drag.

Part II — Accelerating. The car is front-wheel drive; the engine applies a torque $\Gamma$ to the two front wheels together. Neglect drag and rolling friction here.

Draw the external forces on the whole car. Which force accelerates it? What is the power of the road on the car, and where does the kinetic energy come from?
Write the momentum equation of the car, the angular-momentum equation of the front wheels about their axle, and the no-slip condition; show that the acceleration is $\dot v = \Gamma/R\,/\,(M + 4J/R^2)$ and compute the effective mass.
Angular momentum about the centre of mass for the whole car (treat the wheels’ angular momenta as negligible): show that the load transfer under acceleration $\dot v$ is $N_r - N_r^0 = Mh\dot v/L$ — the front axle is unloaded, the rear loaded.
Maximum acceleration without the driven (front) wheels spinning: show that $\dot v_{\max} = fgb/(L + fh)$ and compute it; the same for a rear-wheel-drive car, $\dot v_{\max} = fga/(L - fh)$ . Which layout accelerates harder, and why do dragsters put the weight at the back?
Engine torque at the wheels needed for $\dot v_{\max}$ ; the corresponding power at $50\,\mathrm{km}/\mathrm{h}$ .
The same car on ice ( $f = 0.10$ ): maximum acceleration and the time to reach $50\,\mathrm{km}/\mathrm{h}$ .
What does the friction force on the front tyres do to the tyre — sliding or not — and why does a spinning wheel on ice accelerate the car less than a gripping one?

Part III — Braking.

The brakes apply a torque to each wheel. Show that, for a wheel that keeps rolling, the braking force on the car is the road’s friction on the tyre, and that the brake torque is limited by $fN_iR$ per wheel.
Load transfer under deceleration $\gamma$ : front axle load $N_f = Mg\,b/L + Mh\gamma/L$ ; numbers at $\gamma = fg$ .
Maximum deceleration with all four wheels at the limit of slipping; stopping distance from $90\,\mathrm{km}/\mathrm{h}$ on dry road, and on wet road ( $f = 0.45$ ).
If the brakes are set so that front and rear torques are equal, which axle locks first at hard braking? Why is a locked rear axle dangerous (think of the direction of the friction force on a sliding tyre)?
A wheel locks and slides: deceleration from the road, and what fraction of the peak braking is lost with $f_{\text{kinetic}} = 0.7$ vs $f_{\text{static}} = 0.8$ . Explain what an anti-lock system does and why it pulses the brakes.
Heat released in the brakes from $90\,\mathrm{km}/\mathrm{h}$ ; temperature rise of four $6\,\mathrm{kg}$ steel disks ( $c = 470\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$ ); why do mountain roads have runaway-truck ramps?
An electric car recovers braking energy through its motor at $70\,\%$ efficiency: energy recovered per stop from $90\,\mathrm{km}/\mathrm{h}$ ; number of such stops a $50\,\mathrm{kWh}$ battery is worth.

Part IV — Cornering and a wheel off the ground.

The car turns on a flat road on a circle of radius $\rho = 50\,\mathrm{m}$ at $v$ : the friction forces must supply $Mv^2/\rho$ . Maximum cornering speed on dry and on wet road.
Angular-momentum about the centre of mass along the direction of motion: show that the outer wheels are loaded by $\Delta N = Mv^2h/\rho w$ where $w = 1.5\,\mathrm{m}$ is the track; speed at which the inner wheels lift ( $\Delta N = Mg/2$ ), and compare with the skidding speed: does this car roll over or skid first?
The bend is banked at an angle $\beta$ . Show that at the speed $v = \sqrt{g\rho\tan\beta}$ no friction is needed at all; value for $\beta = 10{}^{\circ}$ .
A wheel is jacked off the ground and spun by hand to $60\,\mathrm{rpm}$ ; its bearing exerts a friction torque of $0.05\,\mathrm{N}\,\mathrm{m}$ . How long does it spin? Kinetic energy lost.
The spinning wheel is dropped onto the road (car at rest, wheel at $60\,\mathrm{rpm}$ , $N = 3\,\mathrm{kN}$ ): sliding time before it stops, and the distance the car would be pushed if it were free to roll (treat the car as a mass $M$ on free wheels, neglect the other wheels’ inertia).
Sum up in one table: the four regimes (rolling, accelerating, braking, cornering), the force that acts in each and the Coulomb bound it must respect.

Solution

Solution of Problem 1.1.

1. $v = 25\,\mathrm{m}/\mathrm{s}$ , $\omega = 80.6\,\mathrm{rad}/\mathrm{s}$ ; top $50\,\mathrm{m}/\mathrm{s}$ , bottom $0$ , front $(25, -25)$ : $35\,\mathrm{m}/\mathrm{s}$ at $45{}^{\circ}$ downward.

2. $\tfrac12Mv^2 = 375\,\mathrm{kJ}$ ; $4 \times \tfrac12J\omega^2 = 13\,\mathrm{kJ}$ ; total $388\,\mathrm{kJ}$ , wheels $3.3\%$ .

3. $N_f + N_r = Mg = 11.8\,\mathrm{kN}$ , $N_fa = N_rb$ : $N_f = Mgb/L = 6.8\,\mathrm{kN}$ , $N_r = Mga/L = 5.0\,\mathrm{kN}$ — the front (engine) axle.

4. $P_r = \sum\mu_rN_iR\omega = \mu_rMgv = 0.012 \times 11772 \times 25 = 3.5\,\mathrm{kW}$ ; drag $\tfrac12 \times 1.2 \times 0.7 \times 625 = 262\,\mathrm{N}$ , $6.6\,\mathrm{kW}$ .

5. $(M + 4J/R^2)\dot v = -\mu_rMg - \tfrac12\rho C_xSv^2$ , $M_{\text{eff}} = 1200 + 41.6 = 1242\,\mathrm{kg}$ . Rolling only: $\dot v = -0.114\,\mathrm{m}/\mathrm{s}^{2}$ , $d = (25^2 - 12.5^2)/0.228 = 2.1\,\mathrm{km}$ . Drag only: $v = v_0\eu^{-x/\lambda}$ , $\lambda = M_{\text{eff}}/\tfrac12\rho C_xS = 2.96\,\mathrm{km}$ , $d = \lambda\ln2 = 2.0\,\mathrm{km}$ (both together: $1.0\,\mathrm{km}$ ).

6. Weight, normal reactions, friction of the road on the tyres. The forward friction on the front tyres is the only horizontal external force, so it accelerates the car; its power is zero (contact point at rest); the kinetic energy comes from the engine — internal work.

7. $M\dot v = T_f - T_r$ ; front wheels $2J\dot v/R = \Gamma - T_fR$ ; rear wheels $2J\dot v/R = T_rR$ . Adding: $(M + 4J/R^2)\dot v = \Gamma/R$ ; $M_{\text{eff}} = 1242\,\mathrm{kg}$ .

8. Moments about $G$ (horizontal ground forces at depth $h$ , total $M\dot v$ ): $aN_f - bN_r + hM\dot v = 0$ with $N_f + N_r = Mg$ , so $N_r = Mga/L + Mh\dot v/L$ : the rear gains $Mh\dot v/L$ , the front loses it.

9. Front drive: $M\dot v \le fN_f = f(Mgb/L - Mh\dot v/L)$ , so $\dot v_{\max} = fgb/(L + fh) = 11.77/3.04 = 3.9\,\mathrm{m}/\mathrm{s}^{2}$ . Rear drive: $M\dot v \le f(Mga/L + Mh\dot v/L)$ , $\dot v_{\max} = fga/(L - fh) = 4.0\,\mathrm{m}/\mathrm{s}^{2}$ . Load transfer helps the rear axle: rear drive, and weight at the back, accelerate harder.

10. $\Gamma = RM_{\text{eff}}\dot v_{\max} = 0.31 \times 1242 \times 3.87 = 1.5\,\mathrm{kN}\,\mathrm{m}$ ; $P = \Gamma v/R = 1490 \times 13.9/0.31 = 67\,\mathrm{kW}$ .

11. $\dot v_{\max} = 0.1 \times 9.81 \times 1.5/2.655 = 0.55\,\mathrm{m}/\mathrm{s}^{2}$ ; $25\,\mathrm{s}$ to $50\,\mathrm{km}/\mathrm{h}$ .

12. Gripping: static friction, no slip, no dissipation at the contact. Spinning: kinetic friction (slightly smaller coefficient), its work heats the tyre and polishes or melts the ice, and the engine’s power goes into spinning the wheel, not moving the car.

13. The brake torque is internal (wheel–body); the car is slowed only by the road’s backward friction on the tyre. For the wheel (negligible $J\dot\omega$ ): $\Gamma_b \approx T_iR \le fN_iR$ .

14. Same balance with $\dot v = -\gamma$ : $N_f = Mgb/L + Mh\gamma/L$ ; at $\gamma = fg = 7.85\,\mathrm{m}/\mathrm{s}^{2}$ : $N_f = 6792 + 1992 = 8.8\,\mathrm{kN}$ , $N_r = 3.0\,\mathrm{kN}$ .

15. $\gamma_{\max} = fg = 7.85\,\mathrm{m}/\mathrm{s}^{2}$ ; $d = v^2/2\gamma = 40\,\mathrm{m}$ ; wet $f = 0.45$ : $4.4\,\mathrm{m}/\mathrm{s}^{2}$ , $71\,\mathrm{m}$ .

16. Equal torques give equal forces, but the rear axle carries only $3.0\,\mathrm{kN}$ : it locks first. A sliding tyre’s friction is opposite to its slip velocity and can no longer supply a lateral force: a locked rear axle loses directional stability and the car spins.

17. Locked: $\gamma = 0.7g = 6.9\,\mathrm{m}/\mathrm{s}^{2}$ , $12\%$ less, stopping distance $45\,\mathrm{m}$ instead of $40\,\mathrm{m}$ — and no steering. An anti-lock system senses a wheel decelerating toward lock, releases and reapplies the brake many times a second, keeping the tyre near the static peak and rolling.

18. $388\,\mathrm{kJ}$ into $4 \times 6 \times 470 = 11.3\,\mathrm{kJ}/\mathrm{K}$ : $\Delta T = 34\,\mathrm{K}$ per stop. A $40\,\mathrm{t}$ truck descending $1000\,\mathrm{m}$ must dissipate $Mgh = 390\,\mathrm{MJ}$ : brakes overheat and fade; a gravel ramp stops the truck by the friction of the gravel.

19. $0.7 \times 388 = 272\,\mathrm{kJ} = 0.075\,\mathrm{kWh}$ ; a $50\,\mathrm{kWh}$ battery is about $660$ such stops.

20. $Mv^2/\rho \le fMg$ : $v \le \sqrt{fg\rho} = 19.8\,\mathrm{m}/\mathrm{s} = 71\,\mathrm{km}/\mathrm{h}$ ; wet $53\,\mathrm{km}/\mathrm{h}$ .

21. Lateral friction $Mv^2/\rho$ at depth $h$ , normal forces at $\pm w/2$ : $(N_{\text{out}} - N_{\text{in}})\,w/2 = Mv^2h/\rho$ , so $\Delta N = Mv^2h/\rho w$ . Inner wheels lift at $\Delta N = Mg/2$ : $v = \sqrt{g\rho w/2h} = 25.9\,\mathrm{m}/\mathrm{s} = 93\,\mathrm{km}/\mathrm{h} > 71\,\mathrm{km}/\mathrm{h}$ : it skids first (since $f < w/2h = 1.36$ ).

22. On a bank of angle $\beta$ with no friction, $N\sin\beta = Mv^2/\rho$ and $N\cos\beta = Mg$ : $v = \sqrt{g\rho\tan\beta} = 9.3\,\mathrm{m}/\mathrm{s} = 33\,\mathrm{km}/\mathrm{h}$ for $10{}^{\circ}$ .

23. $\omega_0 = 6.28\,\mathrm{rad}/\mathrm{s}$ , $J\dot\omega = -0.05\,\mathrm{N}\,\mathrm{m}$ : $t = 126\,\mathrm{s}$ ; $\tfrac12J\omega_0^2 = 20\,\mathrm{J}$ lost.

24. Kinetic friction $fN = 2.4\,\mathrm{kN}$ pushes the car forward: $\dot v = 2400/1200 = 2.0\,\mathrm{m}/\mathrm{s}^{2}$ ; on the wheel $J\dot\omega = -fNR$ , $\dot\omega = -744\,\mathrm{rad}/\mathrm{s}^{2}$ . The slip $R\omega - v$ starts at $1.95\,\mathrm{m}/\mathrm{s}$ and falls at $233\,\mathrm{m}/\mathrm{s}^{2}$ : it vanishes in $8.4\,\mathrm{ms}$ , the car having gained $1.7\,\mathrm{cm}/\mathrm{s}$ and moved $70\,\text{µ}\mathrm{m}$ — a jolt, not a push.

25. Rolling: friction $\approx 0$ , rolling resistance $\mu_rN$ , $|T| \le fN$ trivially. Accelerating: forward friction on the driven tyres, $T \le fN_{\text{driven}}$ , $\dot v \le fgb/(L + fh)$ . Braking: backward friction on all tyres, $T_i \le fN_i$ , $\gamma \le fg$ . Cornering: lateral friction $Mv^2/\rho \le fMg$ . In every case one force — the road’s friction, bounded by Coulomb’s law — does the whole job.

Terms defined in this chapter

See all 393 terms in the glossary