University Physics — Year 1 · Bachelor Year 1

11Kinematics of a Point

The car on a roller coaster enters the loop at $90\,\mathrm{km}/\mathrm{h}$ , slows as it climbs, hangs for an instant at the top, and roars out again; its riders feel crushed into their seats at the bottom and almost weightless at the top. Describing that ride — where the car is, how fast it moves, how its velocity turns and changes — is kinematics, and it needs more than the $x$ , $y$ , $z$ of a straight road: polar coordinates for anything that turns, and a frame that travels with the car itself. This chapter sets up those tools, proves the formulas for velocity and acceleration in each of them, and applies them to the motions that every later chapter will call upon: circular, helical, harmonic, and the motion along any curve.

11.1 Position, velocity, acceleration

Definition 11.1 (Reference frame; position; trajectory)

A reference frame is a rigid body taken as fixed (the laboratory, the Earth, a train), together with a clock. A point particle $M$ — a body whose size is irrelevant to the motion studied — is located at each instant by its position vector $\vect{OM}(t)$ from an origin $O$ attached to the frame; the curve described by $M$ is its trajectory. Motion is always motion relative to a frame: the same passenger is at rest in the train and moving in the station.

Definition 11.2 (Velocity and acceleration)

The velocity and acceleration of $M$ in the frame are

\vect v = \frac{\dd\vect{OM}}{\dd t}, \qquad \vect a = \frac{\dd\vect v}{\dd t} = \frac{\dd^2\vect{OM}}{\dd t^2},

derivatives of vector functions: the derivative of a vector is obtained by differentiating its components in a basis that is fixed in the frame. The velocity is tangent to the trajectory and points along the motion; its norm $v$ is the speed. Units: $\mathrm{m}/\mathrm{s}$ , $\mathrm{m}/\mathrm{s}^{2}$ .

Proposition 11.3 (Cartesian coordinates)

With a fixed orthonormal basis $(\vect e_x, \vect e_y, \vect e_z)$ and $\vect{OM} = x\,\vect e_x + y\,\vect e_y + z\,\vect e_z$ ,

\vect v = \dot x\,\vect e_x + \dot y\,\vect e_y + \dot z\,\vect e_z, \qquad \vect a = \ddot x\,\vect e_x + \ddot y\,\vect e_y + \ddot z\,\vect e_z,

where the dot denotes $\dd/\dd t$ .

Proof. The basis vectors are constant; only the components vary. ∎

Example 11.4 (Projectile)

Launched from $O$ at speed $v_0$ and angle $\alpha$ above the horizontal, a ball’s coordinates (dynamics, next chapter) are $x = v_0\cos\alpha\,t$ , $z = v_0\sin\alpha\,t - \tfrac12 gt^2$ . Then $\vect v = v_0\cos\alpha\,\vect e_x + (v_0\sin\alpha - gt)\vect e_z$ , $\vect a = -g\vect e_z$ : the acceleration is constant while the velocity turns; at the apex ( $\dot z = 0$ , $t = v_0\sin\alpha/g$ ) the velocity is horizontal and the acceleration perpendicular to it. Eliminating $t$ : $z = x\tan\alpha - gx^2/(2v_0^2\cos^2\alpha)$ , a parabola.

A projectile’s parabola: the velocity is tangent to the trajectory and turns, the acceleration g is constant. At the apex velocity and acceleration are perpendicular: the speed is momentarily stationary while the direction still changes. — A projectile’s parabola: the velocity is tangent to the trajectory and turns, the acceleration $\vect g$ is constant. At the apex velocity and acceleration are perpendicular: the speed is momentarily stationary while the direction still changes.

11.2 Cylindrical coordinates

Definition 11.5 (Cylindrical coordinates and basis)

Point $M$ is located by $r \geq 0$ (distance to the $z$ axis), the angle $\theta$ of its projection on the $xy$ plane with $\vect e_x$ , and $z$ : $\vect{OM} = r\,\vect e_r + z\,\vect e_z$ , where

\vect e_r = \cos\theta\,\vect e_x + \sin\theta\,\vect e_y, \qquad \vect e_\theta = -\sin\theta\,\vect e_x + \cos\theta\,\vect e_y,

and $\vect e_z$ form the local basis $(\vect e_r, \vect e_\theta, \vect e_z)$ , orthonormal and direct, which moves with $M$ . In the plane ( $z$ fixed) these are the polar coordinates $(r, \theta)$ .

Lemma 11.6 (Derivatives of the local basis)

\frac{\dd\vect e_r}{\dd t} = \dot\theta\,\vect e_\theta, \qquad \frac{\dd\vect e_\theta}{\dd t} = -\dot\theta\,\vect e_r, \qquad \frac{\dd\vect e_z}{\dd t} = \vect 0 .

Proof. Differentiate the components: $\dd\vect e_r/\dd t = \dot\theta(-\sin\theta\, \vect e_x + \cos\theta\,\vect e_y) = \dot\theta\vect e_\theta$ , and similarly for $\vect e_\theta$ . A rotating unit vector’s derivative is perpendicular to it, of norm the angular rate. ∎

Theorem 11.7 (Velocity and acceleration in cylindrical coordinates)

\vect v = \dot r\,\vect e_r + r\dot\theta\,\vect e_\theta + \dot z\,\vect e_z, \qquad \vect a = \big(\ddot r - r\dot\theta^2\big)\vect e_r + \big(r\ddot\theta + 2\dot r\dot\theta\big)\vect e_\theta + \ddot z\,\vect e_z .

Proof. Differentiate $\vect{OM} = r\vect e_r + z\vect e_z$ with the lemma: $\vect v = \dot r\vect e_r + r\dot\theta\vect e_\theta + \dot z\vect e_z$ . Differentiate again: $\ddot r\vect e_r + \dot r\dot\theta\vect e_\theta + \dot r\dot\theta\vect e_\theta + r\ddot\theta\vect e_\theta - r\dot\theta^2 \vect e_r + \ddot z\vect e_z$ ; collect. ∎

Polar coordinates: the local basis ( e_r, e_ ) turns with M; the velocity has a radial part r and an orthoradial part r. Here the trajectory (red) spirals outward, so r > 0 and v leans outward from the tangent to the circle. — Polar coordinates: the local basis $(\vect e_r, \vect e_\theta)$ turns with $M$ ; the velocity has a radial part $\dot r$ and an orthoradial part $r\dot\theta$ . Here the trajectory (red) spirals outward, so $\dot r > 0$ and $\vect v$ leans outward from the tangent to the circle.

Corollary 11.8 (Circular motion)

On a circle of radius $R$ ( $r = R$ , $z = 0$ ), with angular velocity $\omega = \dot\theta$ :

\vect v = R\omega\,\vect e_\theta, \qquad \vect a = -R\omega^2\,\vect e_r + R\dot\omega\,\vect e_\theta = -\frac{v^2}{R}\,\vect e_r + \frac{\dd v}{\dd t}\,\vect e_\theta .

For uniform circular motion ( $\omega$ constant) the acceleration is purely centripetal, of norm $v^2/R = R\omega^2$ , although the speed is constant.

Proof. Set $\dot r = \ddot r = 0$ in the theorem; $v = R\omega$ . ∎

Example 11.9 (Two circles)

A car rounds a roundabout of radius $20\,\mathrm{m}$ at $36\,\mathrm{km}/\mathrm{h}$ ( $10\,\mathrm{m}/\mathrm{s}$ ): $a = v^2/R = 5.0\,\mathrm{m}/\mathrm{s}^{2}$ , half a $g$ , directed toward the center — what the tires must supply. The Moon ( $R = 3.84 \times 10^{8}\,\mathrm{m}$ , $T = 27.3\,\mathrm{d}$ ): $\omega = 2\pi/T = 2.66 \times 10^{-6}\,\mathrm{rad}/\mathrm{s}$ , $a = R\omega^2 = 2.7 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ , one $3600$ th of $g$ at the Earth’s surface, at sixty Earth radii — the comparison Newton made (Chapter 16).

11.3 Spherical coordinates

Definition 11.10 (Spherical coordinates)

Point $M$ is located by its distance $r = OM$ , the colatitude $\theta \in [0, \pi]$ between $\vect{OM}$ and $\vect e_z$ , and the azimuth $\varphi$ of its projection on the $xy$ plane: $\vect{OM} = r\,\vect e_r$ with

\vect e_r = \sin\theta\cos\varphi\,\vect e_x + \sin\theta\sin\varphi\,\vect e_y + \cos\theta\,\vect e_z ;

$\vect e_\theta$ (tangent to the meridian, toward increasing $\theta$ ) and $\vect e_\varphi$ (tangent to the parallel) complete the local orthonormal basis. On the Earth, $\theta$ is $90^\circ$ minus the latitude and $\varphi$ the longitude.

Proposition 11.11 (Velocity in spherical coordinates)

\vect v = \dot r\,\vect e_r + r\dot\theta\,\vect e_\theta + r\sin\theta\,\dot\varphi\,\vect e_\varphi .

Proof. $\dd\vect e_r/\dd t = \dot\theta\,\vect e_\theta + \sin\theta\,\dot\varphi\, \vect e_\varphi$ : the first term is the rotation in the meridian plane (as in the polar case), the second the rotation about $\vect e_z$ at rate $\dot\varphi$ of a vector whose distance to the axis is $\sin\theta$ . Then $\vect v = \dd(r\vect e_r)/\dd t$ . The acceleration is not needed this year. ∎

Spherical coordinates (r, , ) and their local basis: e_r radial, e_ along the meridian (toward the south on a globe), e_ along the parallel (toward the east). — Spherical coordinates $(r, \theta, \varphi)$ and their local basis: $\vect e_r$ radial, $\vect e_\theta$ along the meridian (toward the south on a globe), $\vect e_\varphi$ along the parallel (toward the east).

11.4 The Frenet frame

Definition 11.12 (Curvilinear abscissa; Frenet basis)

Along a trajectory, the curvilinear abscissa $s(t)$ is the arc length from a reference point, counted algebraically; $v = \dot s$ . At $M$ , the unit tangent $\vect T = \dd\vect{OM}/\dd s$ points along increasing $s$ ; the radius of curvature $\rho > 0$ and the unit normal $\vect N$ , pointing toward the concave side (the center of curvature), are defined by

\frac{\dd\vect T}{\dd s} = \frac{1}{\rho}\,\vect N .

$(\vect T, \vect N)$ is the Frenet basis; $\rho = \infty$ on a straight line, $\rho = R$ on a circle of radius $R$ .

Theorem 11.13 (Acceleration in the Frenet basis)

\vect v = v\,\vect T, \qquad \vect a = \frac{\dd v}{\dd t}\,\vect T + \frac{v^2}{\rho}\,\vect N .

The tangential component $\dd v/\dd t$ measures how fast the speed changes; the normal component $v^2/\rho$ , always toward the center of curvature, how fast the direction turns. The motion is uniform iff $\vect a \perp \vect v$ ; it is rectilinear iff $\vect a \parallel \vect v$ .

Proof. $\vect v = (\dd\vect{OM}/\dd s)(\dd s/\dd t) = v\vect T$ . Then $\vect a = \dot v\vect T + v\,\dd\vect T/\dd t = \dot v\vect T + v(\dd\vect T/\dd s)\dot s = \dot v\vect T + (v^2/\rho)\vect N$ . That $\dd\vect T/\dd s \perp \vect T$ follows from $\vect T\cdot\vect T = 1$ differentiated. ∎

The Frenet basis at a point of a curve: T along the motion, N toward the center of curvature C. The acceleration splits into a tangential part v T (speeding up or braking) and a normal part (v2/ ) N (turning). — The Frenet basis at a point of a curve: $\vect T$ along the motion, $\vect N$ toward the center of curvature $C$ . The acceleration splits into a tangential part $\dot v\vect T$ (speeding up or braking) and a normal part $(v^2/\rho)\vect N$ (turning).

Example 11.14 (Braking in a bend)

A car at $90\,\mathrm{km}/\mathrm{h}$ ( $25\,\mathrm{m}/\mathrm{s}$ ) in a bend of radius $100\,\mathrm{m}$ brakes at $2.0\,\mathrm{m}/\mathrm{s}^{2}$ : $a_T = -2.0\,\mathrm{m}/\mathrm{s}^{2}$ , $a_N = 625/100 = 6.25\,\mathrm{m}/\mathrm{s}^{2}$ , $a = \sqrt{4 + 39} = 6.6\,\mathrm{m}/\mathrm{s}^{2}$ . The normal part dominates: a bend is first of all a turning, and the tires’ grip (next chapter) limits $\sqrt{a_T^2 + a_N^2}$ , which is why one brakes before the bend.

11.5 Standard motions

Proposition 11.15 (Uniformly accelerated rectilinear motion)

On an axis, $\ddot x = a$ constant, from $x_0$ and $v_0$ : $v = v_0 + at$ , $x = x_0 + v_0t + \tfrac12 at^2$ , and $v^2 - v_0^2 = 2a(x - x_0)$ .

Proof. Integrate twice; eliminate $t$ between the first two relations. ∎

Proposition 11.16 (Helical motion)

$r = R$ , $\theta = \omega t$ , $z = v_zt$ (constants $R$ , $\omega$ , $v_z$ ): the speed $v = \sqrt{R^2\omega^2 + v_z^2}$ is constant, the acceleration $\vect a = -R\omega^2\vect e_r$ is centripetal of constant norm, and the radius of curvature is $\rho = v^2/a = R(1 + v_z^2/R^2\omega^2) > R$ .

Proof. Theorem 11.7 with $\dot r = 0$ , $\ddot\theta = 0$ , $\ddot z = 0$ ; uniform motion, so $\vect a$ is purely normal and $v^2/\rho = R\omega^2$ . ∎

Proposition 11.17 (Harmonic motion)

$x = A\cos(\omega t + \varphi)$ : $\dot x = -A\omega\sin(\omega t + \varphi)$ , $\ddot x = -\omega^2 x$ . The velocity leads the position by a quarter period and the acceleration is opposite to the position; $v_{\max} = A\omega$ , $a_{\max} = A\omega^2$ . This is the projection on a diameter of a uniform circular motion of radius $A$ and angular velocity $\omega$ .

Proof. Differentiate; for the projection, take $x = A\cos\theta$ with $\theta = \omega t + \varphi$ . ∎

Method 11.18 (Choosing coordinates)

Straight-line or parabolic motion, constant acceleration: Cartesian.
Motion around an axis or a point (circles, spirals, orbits, turntables): cylindrical/polar; the basis moves, so use Theorem 11.7, never naive component differentiation.
A given curve (track, bend, loop) when the speed law is known: Frenet, which separates “how fast” from “which way”.

Whatever the choice, $\vect v$ and $\vect a$ are frame-dependent vectors, and a result is complete only with the frame named.

11.6 Exercises

Exercise 11.1 ★

A car starts from rest with $x(t) = 2.0\,t^2$ (SI units). Velocity and acceleration at $t = 3.0\,\mathrm{s}$ ; distance covered in the first $5\,\mathrm{s}$ ; time to reach $100\,\mathrm{km}/\mathrm{h}$ .

Solution

Solution of Exercise 11.1.

$v = 4.0t$ : $12\,\mathrm{m}/\mathrm{s}$ at $3.0\,\mathrm{s}$ ; $a = 4.0\,\mathrm{m}/\mathrm{s}^{2}$ ; $x(5) = 50\,\mathrm{m}$ ; $100\,\mathrm{km}/\mathrm{h} = 27.8\,\mathrm{m}/\mathrm{s}$ at $t = 27.8/4.0 = 6.9\,\mathrm{s}$ .

Exercise 11.2 ★

A point moves in a plane with $r = 2.0\,\mathrm{m}$ and $\theta = 3.0\,t$ (rad, SI). Give $\vect v$ and $\vect a$ in the polar basis, their norms, and the nature of the motion.

Solution

Solution of Exercise 11.2.

$\dot r = 0$ , $\dot\theta = 3.0\,\mathrm{rad}/\mathrm{s}$ : $\vect v = r\dot\theta\,\vect e_\theta = 6.0\,\vect e_\theta$ ( $\mathrm{m}/\mathrm{s}$ ); $\vect a = -r\dot\theta^2\vect e_r = -18\,\vect e_r$ ( $\mathrm{m}/\mathrm{s}^{2}$ ). Uniform circular motion: constant speed, centripetal acceleration.

Exercise 11.3 ★

A point on the equator ( $R = 6.37 \times 10^{6}\,\mathrm{m}$ , sidereal day $86\,164\,\mathrm{s}$ ): speed and acceleration in the geocentric frame. Compare the acceleration with $g$ .

Solution

Solution of Exercise 11.3.

$v = 2\pi R/T = 2\pi \times 6.37\times10^6/86164 = 465\,\mathrm{m}/\mathrm{s}$ ; $a = v^2/R = 0.034\,\mathrm{m}/\mathrm{s}^{2}$ , $g/290$ .

Exercise 11.4 ★

A ball is thrown with $v_0 = 12\,\mathrm{m}/\mathrm{s}$ at $\alpha = 40^\circ$ ; its coordinates are those of Example 11.4. Find the time to the apex, the maximum height, the range, and the velocity at landing.

Solution

Solution of Exercise 11.4.

$v_0\sin\alpha = 7.71\,\mathrm{m}/\mathrm{s}$ , $v_0\cos\alpha = 9.19\,\mathrm{m}/\mathrm{s}$ . Apex at $t = 7.71/9.81 = 0.79\,\mathrm{s}$ , height $7.71^2/(2 \times 9.81) = 3.0\,\mathrm{m}$ . Range $2 \times 9.19 \times 0.786 = 14.5\,\mathrm{m}$ . Landing: $(9.19, -7.71)$ , $12\,\mathrm{m}/\mathrm{s}$ at $40^\circ$ below the horizontal.

Exercise 11.5 ★★

A cyclist at $10\,\mathrm{m}/\mathrm{s}$ enters a bend of radius $25\,\mathrm{m}$ and accelerates at $1.0\,\mathrm{m}/\mathrm{s}^{2}$ . Give $\vect a$ in the Frenet basis and its norm; after $5\,\mathrm{s}$ of the same acceleration on the same bend?

Solution

Solution of Exercise 11.5.

$a_T = 1.0\,\mathrm{m}/\mathrm{s}^{2}$ , $a_N = 10^2/25 = 4.0\,\mathrm{m}/\mathrm{s}^{2}$ , $a = 4.1\,\mathrm{m}/\mathrm{s}^{2}$ . After $5\,\mathrm{s}$ : $v = 15\,\mathrm{m}/\mathrm{s}$ , $a_N = 225/25 = 9.0\,\mathrm{m}/\mathrm{s}^{2}$ , $a = 9.1\,\mathrm{m}/\mathrm{s}^{2}$ .

Exercise 11.6 ★★

A point follows the spiral $r = b\theta$ , $\theta = \omega t$ ( $b$ , $\omega$ constants). Compute $\vect v$ , the speed, and $\vect a$ in the polar basis. Where does the $2\dot r\dot\theta$ term come from?

Solution

Solution of Exercise 11.6.

$r = b\omega t$ , $\dot r = b\omega$ , $\ddot r = 0$ , $\dot\theta = \omega$ : $\vect v = b\omega\,\vect e_r + b\omega^2 t\,\vect e_\theta$ , $v = b\omega \sqrt{1 + \omega^2t^2}$ ; $\vect a = -b\omega^3 t\,\vect e_r + 2b\omega^2\, \vect e_\theta$ . The $2\dot r\dot\theta$ term has two equal halves: the radial velocity’s direction turns at rate $\dot\theta$ (giving $\dot r\dot\theta\,\vect e_\theta$ ), and moving outward increases the orthoradial speed $r\dot\theta$ (another $\dot r\dot\theta$ ).

Exercise 11.7 ★★

Helix: $R = 2.0\,\mathrm{m}$ , $\omega = 1.0\,\mathrm{rad}/\mathrm{s}$ , $v_z = 0.50\,\mathrm{m}/\mathrm{s}$ . Speed, acceleration, pitch of the helix, radius of curvature.

Solution

Solution of Exercise 11.7.

$v = \sqrt{(2.0 \times 1.0)^2 + 0.5^2} = 2.1\,\mathrm{m}/\mathrm{s}$ ; $a = R\omega^2 = 2.0\,\mathrm{m}/\mathrm{s}^{2}$ toward the axis; pitch $v_z \times 2\pi/\omega = 3.1\,\mathrm{m}$ ; $\rho = v^2/a = 4.25/2.0 = 2.1\,\mathrm{m} > R$ .

Exercise 11.8 ★★

A piston moves as $x = 0.10\cos(10t)$ (SI). Give $v_{\max}$ , $a_{\max}$ , the phase relations between $x$ , $v$ , $a$ , and the positions where $\abs v$ and $\abs a$ are largest.

Solution

Solution of Exercise 11.8.

$v_{\max} = A\omega = 1.0\,\mathrm{m}/\mathrm{s}$ ; $a_{\max} = A\omega^2 = 10\,\mathrm{m}/\mathrm{s}^{2}$ ; $v$ leads $x$ by a quarter period, $a = -\omega^2x$ is opposite to $x$ . $\abs v$ is largest at $x = 0$ , $\abs a$ at the extremes $x = \pm A$ .

Exercise 11.9 ★★

Starting from $\vect e_r = \cos\theta\,\vect e_x + \sin\theta\,\vect e_y$ , derive $\dd\vect e_r/\dd t$ and $\dd\vect e_\theta/\dd t$ , and then the polar velocity and acceleration of Theorem 11.7.

Solution

Solution of Exercise 11.9.

$\dot{\vect e}_r = \dot\theta(-\sin\theta\,\vect e_x + \cos\theta\,\vect e_y) = \dot\theta\,\vect e_\theta$ ; $\dot{\vect e}_\theta = \dot\theta(-\cos\theta\, \vect e_x - \sin\theta\,\vect e_y) = -\dot\theta\,\vect e_r$ . Then $\vect v = \dot r\vect e_r + r\dot\theta\vect e_\theta$ , and differentiating once more with the product rule gives the acceleration of Theorem 11.7.

Exercise 11.10 ★★★

For the projectile of Example 11.4, find the radius of curvature of the trajectory at the apex and at the launch point (use the normal component of $\vect g$ ). Which is smaller, and why?

Solution

Solution of Exercise 11.10.

Apex: $v = v_0\cos\alpha$ , $a_N = g$ : $\rho = v_0^2\cos^2\alpha/g$ . Launch: $v = v_0$ , $a_N = g\cos\alpha$ : $\rho = v_0^2/(g\cos\alpha)$ . The apex radius is smaller by $\cos^3\alpha$ : the parabola bends most where it is slowest and where all of $\vect g$ is normal.

Exercise 11.11 ★★★

A wheel of radius $R$ rolls without slipping at speed $v$ ; a point of its rim has coordinates $x = vt - R\sin(vt/R)$ , $y = R - R\cos(vt/R)$ (a cycloid). Compute its velocity and acceleration; show that when it touches the ground its velocity vanishes and its acceleration is $v^2/R$ upward; find its speed at the top.

Solution

Solution of Exercise 11.11.

$\dot x = v - v\cos(vt/R)$ , $\dot y = v\sin(vt/R)$ ; $\ddot x = (v^2/R) \sin(vt/R)$ , $\ddot y = (v^2/R)\cos(vt/R)$ . At the ground ( $vt/R = 2\pi n$ ): $\dot x = \dot y = 0$ and $\vect a = (0, v^2/R)$ , upward. At the top ( $vt/R = \pi$ ): $\dot x = 2v$ , $\dot y = 0$ : speed $2v$ .

Exercise 11.12 ★★★

A person walks outward along a radius of a turntable turning at constant $\omega$ , at constant speed $u$ relative to the turntable: $r = ut$ , $\theta = \omega t$ . Compute $\vect a$ in the polar basis, identify the two terms, and evaluate them for $u = 1.0\,\mathrm{m}/\mathrm{s}$ , $\omega = 1.0\,\mathrm{rad}/\mathrm{s}$ at $t = 2.0\,\mathrm{s}$ . Which term has no counterpart for a person walking on the ground?

Solution

Solution of Exercise 11.12.

$\ddot r = 0$ , $\dot r = u$ , $\dot\theta = \omega$ : $\vect a = -ut\omega^2\, \vect e_r + 2u\omega\,\vect e_\theta$ . At $t = 2.0\,\mathrm{s}$ : $-2.0\,\vect e_r + 2.0\,\vect e_\theta$ ( $\mathrm{m}/\mathrm{s}^{2}$ ). The first is the centripetal term of the rotation; the second, $2\dot r\dot\theta$ , exists only because the radial direction itself rotates — on the ground, walking straight at constant speed gives no acceleration at all.

A vertical loop (Yomiuriland, Tokyo): a teardrop rather than a circle, so that the curvature is largest at the top where the speed is least — the geometry of the weekend problem. Photograph: Jeremy Thompson, CC BY 2.0.

11.7 Problem: The roller-coaster loop

Problem 11.1

Weekend problem — a car enters a vertical loop at ninety kilometers per hour: where it slows, where its acceleration points, how many $g$ the riders feel, and why modern loops are not circles

A car, treated as a point $M$ , runs on a vertical circular loop of radius $R = 10\,\mathrm{m}$ , center $C$ . Its position is given by the angle $\theta$ between $\vect{CM}$ and the downward vertical ( $\theta = 0$ at the bottom, $\pi$ at the top). Friction is neglected, and the speed law (derived in Chapter 13) is

v^2 = v_0^2 - 2gR(1 - \cos\theta),

with $v_0 = 25\,\mathrm{m}/\mathrm{s}$ the speed at the bottom; $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$ .

Part I — Kinematics on the circle.

In polar coordinates centered at $C$ (with $\vect e_r$ from $C$ to $M$ ), write $\vect v$ and $\vect a$ for motion on the circle.
Identify the Frenet vectors $\vect T$ and $\vect N$ in terms of $\vect e_r$ and $\vect e_\theta$ , and the radius of curvature.
Compute the speed at the top, and at the side points $\theta = \pi/2$ and $3\pi/2$ .
Differentiate the speed law with respect to time to show that the tangential acceleration is $a_T = -g\sin\theta$ . Interpret its sign on the way up and on the way down.
Express the normal acceleration $a_N(\theta)$ and compute it at the bottom, the sides and the top.
Give the norm of $\vect a$ at the four points and the angle it makes with the vertical at the top.
At which points is $\vect a$ perpendicular to $\vect v$ ? Parallel?

Part II — Angular rates and timing.

Express $\dot\theta$ as a function of $\theta$ and compute it at the bottom and the top.
Show that $\ddot\theta = -(g/R)\sin\theta$ (the equation of a pendulum) and comment.
For comparison, compute the period of small oscillations of a simple pendulum of length $R$ (Example 1.8 gave its form; $k = 2\pi$ ).
Estimate the time to go round the loop, using the average of the speeds at the four reference points.
The minimum speed at the top for the car to stay on the track (next chapter) is $\sqrt{gR}$ . Is it satisfied? What minimum $v_0$ would be needed?

Part III — What the riders feel. The apparent weight per unit mass felt by a rider is $\vect a - \vect g$ (admitted here, derived in Chapter 12); its norm in units of $g$ is the “ $g$ -load”.

Compute the $g$ -load at the bottom.
Compute it at the top, and say which way the rider is pressed (seat or restraint).
Compute it at the side points.
Riders tolerate about $4g$ briefly. Does this loop pass? Where is the problem?
Show that, for a circular loop entered just fast enough to clear the top (zero apparent weight there), the bottom $g$ -load is necessarily $6g$ , whatever $R$ .

Part IV — The clothoid loop. Modern loops are “teardrops”: the radius of curvature is large at the bottom and small at the top.

Using the Frenet form of $\vect a$ , explain why varying $\rho$ along the track changes $a_N$ without changing the speed law (which depends only on height).
Choose $\rho_{\text{bottom}}$ so that the $g$ -load at the bottom is $3g$ with $v_0 = 25\,\mathrm{m}/\mathrm{s}$ .
At the top, the height is still $2R = 20\,\mathrm{m}$ above the bottom. Choose $\rho_{\text{top}}$ so that the $g$ -load there is $1.5g$ .
A clothoid has curvature proportional to arc length, $1/\rho = s/A^2$ . Along such an entry, with the speed nearly constant, how does $a_N$ grow with time, and why is that gentler on the neck than a circle (where $a_N$ jumps at the entry)?
Sketch the resulting loop shape qualitatively and summarize in two sentences why it replaced the circle.

Part V — Seen from the ground. A camera at ground level, $50\,\mathrm{m}$ from the loop’s center in its plane, follows the car.

When the car is at the top, moving horizontally at $15\,\mathrm{m}/\mathrm{s}$ , at what angular rate (rad/s) must the camera turn? (Polar coordinates from the camera.)
Why is a camera that turns at constant rate a poor tracker of uniform circular motion?
Summarize the chapter’s lesson from the loop: which coordinate system answered which question.

Solution

Solution of Problem 11.1.

1. $\vect v = R\dot\theta\,\vect e_\theta$ ; $\vect a = -R\dot\theta^2 \,\vect e_r + R\ddot\theta\,\vect e_\theta$ .

2. $\vect T = \vect e_\theta$ (motion with $\theta$ increasing), $\vect N = -\vect e_r$ (toward $C$ ), $\rho = R$ .

3. Top: $v^2 = 625 - 392 = 233$ , $v = 15.3\,\mathrm{m}/\mathrm{s}$ . Sides: $625 - 196 = 429$ , $v = 20.7\,\mathrm{m}/\mathrm{s}$ .

4. $2v\dot v = -2gR\sin\theta\,\dot\theta$ and $v = R\dot\theta$ : $\dot v = -g\sin\theta$ . Negative for $0 < \theta < \pi$ (slowing on the way up), positive on the way down.

5. $a_N = v^2/R = v_0^2/R - 2g(1 - \cos\theta)$ : bottom $62.5\,\mathrm{m}/\mathrm{s}^{2}$ ; sides $62.5 - 19.6 = 42.9\,\mathrm{m}/\mathrm{s}^{2}$ ; top $62.5 - 39.2 = 23.3\,\mathrm{m}/\mathrm{s}^{2}$ .

6. Bottom: $62.5\,\mathrm{m}/\mathrm{s}^{2}$ (no tangential part); sides: $\sqrt{42.9^2 + 9.81^2} = 44.0\,\mathrm{m}/\mathrm{s}^{2}$ ; top: $23.3\,\mathrm{m}/\mathrm{s}^{2}$ , pointing straight down toward $C$ : angle $0$ with the vertical.

7. Perpendicular at the bottom and top ( $a_T = 0$ ); never parallel ( $a_N > 0$ everywhere).

8. $\dot\theta = v/R = \sqrt{v_0^2 - 2gR(1 - \cos\theta)}/R$ : bottom $2.50\,\mathrm{rad}/\mathrm{s}$ , top $1.53\,\mathrm{rad}/\mathrm{s}$ .

9. $\ddot\theta = \dot v/R = -(g/R)\sin\theta$ : the equation of a simple pendulum of length $R$ — the car on its track is a pendulum bob that goes over the top.

10. $2\pi\sqrt{R/g} = 2\pi\sqrt{10/9.81} = 6.3\,\mathrm{s}$ .

11. Mean of $25$ , $20.7$ , $15.3$ , $20.7$ : about $20\,\mathrm{m}/\mathrm{s}$ ; length $2\pi R = 63\,\mathrm{m}$ : roughly $3\,\mathrm{s}$ .

12. $\sqrt{gR} = 9.9\,\mathrm{m}/\mathrm{s} < 15.3$ : yes. Minimum: $v_0^2 \geq gR + 4gR = 5gR = 490$ , $v_0 \geq 22.1\,\mathrm{m}/\mathrm{s}$ .

13. Bottom: $\vect a = 62.5$ upward, $\vect g$ downward: $\abs{\vect a - \vect g} = 62.5 + 9.8 = 72.3\,\mathrm{m}/\mathrm{s}^{2} = 7.4g$ .

14. Top: both downward, $23.3 - 9.8 = 13.5\,\mathrm{m}/\mathrm{s}^{2} = 1.4g$ , pressing the rider into the seat (toward the center, i.e. upward for the inverted rider).

15. Sides: $\vect a = 42.9$ horizontal toward $C$ plus $9.8$ downward (tangential); $\vect a - \vect g$ is $42.9$ horizontal: $4.4g$ sideways, toward the center.

16. Fails: $7.4g$ at the bottom (the top and sides are fine).

17. Zero apparent weight at the top: $v_{\text{top}}^2 = gR$ ; then $v_0^2 = gR + 4gR = 5gR$ , $a_N = 5g$ at the bottom and the load is $5g + g = 6g$ , independent of $R$ .

18. $a_N = v^2/\rho$ : $v$ is fixed by the height (energy), $\rho$ by the track’s geometry; a different $\rho$ at the same height changes $a_N$ alone.

19. $a_N + g = 3g$ : $a_N = 2g$ , $\rho = v_0^2/2g = 625/19.6 = 32\,\mathrm{m}$ .

20. $v_{\text{top}} = 15.3\,\mathrm{m}/\mathrm{s}$ ; $a_N - g = 1.5g$ : $a_N = 24.5\,\mathrm{m}/\mathrm{s}^{2}$ , $\rho = 233/24.5 = 9.5\,\mathrm{m}$ .

21. $a_N = v^2/\rho = v^2s/A^2 = v^3t/A^2$ : it grows linearly from zero — a constant rate of change of acceleration — instead of jumping from $0$ to $v^2/R$ at the entry of a circle, which the neck feels as a blow.

22. A teardrop: wide and gently curved at the bottom where the car is fast, tight at the top where it is slow. It keeps the load near $3g$ throughout and brings it on gradually.

23. Line of sight: horizontal distance $50\,\mathrm{m}$ , height $20\,\mathrm{m}$ : $r = 53.9\,\mathrm{m}$ , elevation $\beta = 21.8^\circ$ . The horizontal velocity’s transverse component is $v\sin\beta = 15 \times 0.371 = 5.6\,\mathrm{m}/\mathrm{s}$ ; $\dot\theta = 5.6/53.9 = 0.10\,\mathrm{rad}/\mathrm{s}$ .

24. Seen from an off-center point the angular rate of the line of sight is not constant (fast when the car is near, slow when far): a constant-rate camera drifts off the car.

25. Polar coordinates at $C$ for velocity and acceleration on the circle; the Frenet split for “speeding up” versus “turning” and for the clothoid; polar coordinates at the camera for the tracking rate; energy (next chapters) for the speed law.

Terms defined in this chapter

See all 393 terms in the glossary