OBJECTIVES

When you finish this section, you should be able to:

1. Find the velocity, acceleration, and speed of a moving particle (p. 785)
2. Express the acceleration vector using tangential and normal components (p. 788)

spanDEFINITIONspan Velocity; Acceleration; Speed

The velocity, acceleration, and speed of a particle whose motion is along a smooth curve traced out by the twice differentiable vector function \(\mathbf{r}(t)=x( t) \mathbf{i}+y( t) \mathbf{j}+z( t) \mathbf{k}\), \(a\leq t\leq b\), are defined as

NOTE

Since \({\bf v}(t)=\mathbf{r}^{\prime}(t)\), the velocity vector is directed along the tangent vector.

Finding Velocity, Acceleration, and Speed

Find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(v\) of a particle that is moving along:

(a) the plane curve \(\mathbf{r}(t)=\left( \dfrac{1}{2}t^{2}+t\right) \mathbf{i}+t^{3}\mathbf{j}\) from \(t=0\) to \(t=2\).

(b) the space curve \(\mathbf{r}(t)=t\mathbf{i}+t^{2}\mathbf{j}+t^{3}\mathbf{k}\) from \(t=0\) to \(t=2.\)

For each curve, graph the motion of the particle and the vectors \(\mathbf{v}(1)\) and \(\mathbf{a}(1)\).

Figure 25 \(\mathbf{r}(t)=\left(\dfrac{1}{2}t^2+t\right)\mathbf{i}+ t^3 \mathbf{j}\), \\ \(0\le t\le 2\)

Solution (a) The velocity, acceleration, and speed are \begin{eqnarray*} \mathbf{v}(t)&\;=\;&\mathbf{r}^{\prime} (t)=\dfrac{d}{dt}\left( \dfrac{1}{2} t^{2}+t\right) \mathbf{i}+\dfrac{d}{dt}t^{3}\mathbf{j}=(t+1)\mathbf{i}+3t^{2} \mathbf{j} \\[6pt] \mathbf{a}(t)&\;=\;&\mathbf{r}^{\prime \prime} (t)=\dfrac{d}{dt}\mathbf{r^{\prime} } ( t)\;=\;\dfrac{d}{dt}(t+1)\mathbf{i}+\dfrac{d}{dt}3t^{2}\mathbf{j=i} +6t\mathbf{j} \\[6pt] v(t)&\;=\;&\left\Vert \mathbf{v}(t)\right\Vert\;=\;\sqrt{( t+1) ^{2}+( 3t^{2}) ^{2}}=\sqrt{9t^{4}+t^{2}+2t+1} \end{eqnarray*}

At \(t=1\), the velocity is \(\mathbf{v}(1)=\mathbf{r}^{\prime} (1)=2\mathbf{i}+3\mathbf{j}\) and the acceleration is \(\mathbf{a}(1)=\mathbf{r}^{\prime \prime}(1)=\mathbf{i}+6\mathbf{j}\). Figure 25 illustrates the graph of \(\mathbf{r}=\mathbf{r}( t)\), \(\mathbf{v}(1)\), and \(\mathbf{a}(1)\).

(b) The velocity, acceleration, and speed of the particle are \begin{eqnarray*} \mathbf{v}(t)&\;=\;&\mathbf{r}^{\prime} (t)=\dfrac{d}{dt}t\mathbf{i}+\dfrac{d}{dt} t^{2}\mathbf{j}+\dfrac{d}{dt}t^{3}\mathbf{k}=\mathbf{i}+2t\mathbf{j}+3t^{2} \mathbf{k} \\[12pt] \mathbf{a}(t)&\;=\;&\mathbf{r}^{\prime \prime} (t)=\dfrac{d}{dt}\mathbf{r^{\prime} } ( t)\;=\;\dfrac{d}{dt}\mathbf{i}+\dfrac{d}{dt}( 2t) \mathbf{j}+\dfrac{d}{dt}( 3t^{2}) \mathbf{k}=2\mathbf{j}+6t \mathbf{k} \\[12pt] v(t)&\;=\;&\left\Vert \mathbf{v}(t)\right\Vert\;=\;\sqrt{1^{2}+( 2t) ^{2}+( 3t^{2}) ^{2}}=\sqrt{1+4t^{2}+9t^{4}} \end{eqnarray*}

Figure 26 \(\mathbf{r}(t)= t \mathbf{i} + t^2 {\bf j} + t^3 \mathbf{k}\), \(0\le t \le 2\)

At \(t=1\), the velocity is \(\mathbf{v}(1)=\mathbf{i}+2\mathbf{j}+3\mathbf{k}\) and the acceleration is \(\mathbf{a}(1)=2\mathbf{j}+6\mathbf{k}\). See Figure 26.

NOW WORK

Problems 7 and 11.

Finding the Force Acting on a Particle Moving Along an Ellipse

The position of a particle of mass \(m\) that is moving along an ellipse with a constant angular speed \(\omega\) is given by the vector function \begin{equation*} \mathbf{r}(t)=A\cos ( \omega t) \mathbf{i}+B\sin ( \omega t) \mathbf{j}\qquad 0\leq t\leq 2\pi \end{equation*}

Find the force \(\mathbf{F}\) acting on the particle at any time \(t.\) Graph \(\mathbf{r}=\mathbf{r}( t)\) and \(\mathbf{F}=\mathbf{F}( t) .\)

Figure 27 \(\mathbf{F}(t)=-m\omega^2 \mathbf{r}(t)\)

Solution To find the force \(\mathbf{F}\), we use Newton's Second Law of Motion, \(\mathbf{F}=m\mathbf{a}\). We begin by finding the acceleration \(\mathbf{a}\) of the particle. \begin{eqnarray*} \mathbf{r}(t)&\;=\;&A\cos ( \omega t) \mathbf{i}+B\sin ( \omega t) \mathbf{j} \\[3pt] \mathbf{v}(t)&\;=\;&\mathbf{r^{\prime} }(t)=-A\omega \sin ( \omega t) \mathbf{i}+B\omega \cos ( \omega t) \mathbf{j} \\[3pt] \mathbf{a}(t)&\;=\;&\mathbf{r^{\prime \prime} }(t)=-A\omega ^{2}\cos ( \omega t) \mathbf{i}-B\omega ^{2}\sin ( \omega t) \mathbf{j} =-\omega ^{2}\mathbf{r}(t) \end{eqnarray*}

Then, by Newton's Law, \begin{equation*} \mathbf{F}(t)=m\mathbf{a}(t)=-m\hbox{ }\omega ^{2}\mathbf{r}(t) \end{equation*}

The direction of the force vector \(\mathbf{F}\) is opposite to that of the vector \(\mathbf{r}\) at any time \(t\), as shown in Figure 27.

NOTE

The force \(\mathbf{F}\) acting on the particle in Example 2 is directed toward the center of the ellipse at a time \(t\). Because of this, it is called a centripetal (center-seeking) force.

NOW WORK

Problem 23(a).

Finding the Position, Velocity, and Acceleration of a Particle Moving Along a Circle

(a) Find the position \(\mathbf{r}=\mathbf{r}( t)\) of a particle that moves counterclockwise along a circle of radius \(R\) with a constant speed \(v_{0}\).

(b) Find the velocity and acceleration of the particle.

(c) Find the magnitude of the acceleration.

Figure 28

Solution For convenience, we place the circle of radius \(R\) in the \(xy\)-plane, with its center at the origin, and assume that at time \(t=0\) the particle is on the positive \(x\)-axis.

(a) The particle is moving counterclockwise along the circle, as shown in Figure 28. If \(\theta (t)\) is the angle between the positive \(x\)-axis and the position vector of the particle \(\mathbf{r}=\) \(\mathbf{r}(t)\), then the vector \(\mathbf{r=r}( t) \) is \begin{equation*} \mathbf{r}(t)=R \cos [\theta (t)]{\bf i}+R\;\sin [\theta (t)]{\bf j}\tag{1} \end{equation*}

Notice that \(\mathbf{r}( 0)\;=\;R\mathbf{i}\), as required. Also for the motion to be counterclockwise, the function \(\theta\;=\;\theta ( t) \) must be increasing.

(b) The velocity \(\mathbf{v}\) of the particle is \begin{equation*} \mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=-R\;\sin [\theta (t)]\frac{d\theta }{dt} \mathbf{i}+R\;\cos [\theta (t)]\frac{d\theta }{dt}\mathbf{j}\qquad \color{#0066A7}{\hbox{Use the Chain Rule.}} \end{equation*}

and the speed \(v\) of the particle is \begin{eqnarray*} v(t)&=&\left\Vert \mathbf{v}(t)\right\Vert\;=\;\sqrt{R^{2}\sin ^{2}[\theta (t)]\left( \frac{ d\theta }{dt}\right) ^{2}+R^{2}\cos ^{2}[\theta (t)]\left( \frac{d\theta }{dt} \right) ^{2}}\\[6pt] &=&\sqrt{R^{2}\!\left( \frac{d\theta }{dt}\right) ^{2}}=R\left\vert \frac{d\theta }{dt}\right\vert \end{eqnarray*}

Since we know the speed is constant, \(v(t)=v_{0}\). Also, since the function \(\theta\;=\;\theta ( t) \) is increasing, \(\dfrac{d\theta }{dt}>0\). As a result, \begin{eqnarray*} v_{0}&=&R\frac{d\theta }{dt} \\[4pt] \frac{d\theta }{dt}&=&\frac{v_{0}}{R} \end{eqnarray*}

Since \(\dfrac{d\theta }{dt}\) is the rate at which the angle \(\theta \) is changing, the quantity \(\dfrac{v_{0}}{R}\) is the angular speed \(\omega \) of the particle. That is, \begin{equation*} \frac{d\theta }{dt}=\omega \end{equation*}

We solve this differential equation, using \(\theta ( 0)\;=\;0\) as the initial condition. \begin{eqnarray*} d\theta &=&\omega dt \\[3pt] \theta ( t) &=&\omega t+k \\[3pt] \theta ( 0) &=&k=0 \end{eqnarray*}

So, \(\theta (t)=\omega t\). Now substitute \(\theta (t)=\omega t\) into the vector function \(\mathbf{r=r}( t)\) [statement (1)] to obtain \begin{equation*} \mathbf{r}(t)=R\;\cos ( \omega t) \mathbf{i}+R\;\sin ( \omega t) \mathbf{j} \end{equation*}

Then the velocity \(\mathbf{v}\) and acceleration \(\mathbf{a}\) of the particle are \begin{eqnarray*} \mathbf{v}\;=\;\mathbf{v}(t) &=& -R\omega \sin ( \omega t) \mathbf{i} +R\omega \cos ( \omega t) \mathbf{j} \\[3pt] \mathbf{a}\;=\;\mathbf{a}(t) &=& -R\omega ^{2}\cos ( \omega t) \mathbf{i }-R\omega ^{2}\sin ( \omega t) \mathbf{j}=-\omega ^{2}\left[ R\;\cos ( \omega t) \mathbf{i}+R\;\sin ( \omega t) \mathbf{ j}\right]\\[3pt] &=&-\omega ^{2}\mathbf{r}(t) \end{eqnarray*}

(c) Since \(\omega=\dfrac{v_{0}}{R}\), the magnitude of the acceleration is \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\left\Vert \mathbf{a}(t)\right\Vert\;=\;\omega ^{2}\left\Vert \mathbf{r}(t)\right\Vert\;=\;\omega ^{2}R=\dfrac{v_{0}^{2}}{R}}\tag{2} \end{equation*}

NOW WORK

Problem 23(b).

NOTE

Near-Earth orbits are above \(100\) miles (out of Earth's atmosphere) up to an altitude of approximately \(15{,}000\) miles.

Finding the Speed Required for a Near-Earth Circular Orbit

Find the speed required to maintain a satellite in a near-Earth circular orbit. (The gravitational attraction of other bodies is ignored.)

Solution Let \(R\) be the distance of a satellite from the center of Earth. Then from (2), the magnitude of the acceleration of the satellite whose motion is circular is \begin{equation*} \left\Vert \mathbf{a}(t)\right\Vert\;=\;\dfrac{v_{0}^{2}}{R} \end{equation*}

For the satellite to remain in orbit, the magnitude of the acceleration \(\left\Vert \mathbf{a}(t)\right\Vert\) of the satellite at any time \(t\) must equal \(g\), the acceleration due to gravity for Earth. As a result, \begin{eqnarray*} \frac{v_{0}^{2}}{R}&\;=\;&g \\[3pt] v_{0}&\;=\;&\sqrt{gR} \end{eqnarray*}

The speed \(v_{0}\) required to maintain a near-Earth circular orbit is \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {v_{0}=\sqrt{gR}} \end{equation*}

where \(R\) is the distance of the satellite from the center of Earth and \(g\) is the acceleration due to gravity.

NOTE

The acceleration due to gravity at near-Earth orbits is somewhat less than \(g ≈ 32.2 \enspace \textrm{ft/s}^{2} ≈ 79{,}036 \enspace \textrm{mi/h}^{2}\), the acceleration due to gravity at Earth's surface.

NOW WORK

Problem 63.

Finding the Frictional Force Necessary to Prevent Skidding

A motorcycle with a mass of 150 kg is driven at a constant speed of 120 km/h on a circular track whose radius is 100 m. To keep the motorcycle from skidding, what frictional force must be exerted by the track on the tires?

Solution By Newton's Second Law of Motion, the force \(\mathbf{F}\) required to keep an object of mass \(m\) traveling along a curve traced out by \(\mathbf{r}=\mathbf{r}(t)\) is \(\mathbf{F}=m\mathbf{a}\). The magnitude of the frictional force exerted by the tires must therefore equal \begin{equation*} \left\Vert \mathbf{F}\right\Vert\;=\;m\left\Vert \mathbf{a}\right\Vert \end{equation*}

In this example, the motion is circular. So, using (2), we have \begin{eqnarray*} &&\left\Vert \mathbf{F}\right\Vert\;=\;m\left( \frac{v_{0}^{2}}{R}\right)\;=\;150\enspace \textrm{kg} \left[\dfrac{\left( 120 \enspace \textrm{km/h}\right)^{2}}{100\enspace \textrm{m}}\right]\quad\,\color{#0066A7}{\underset{{\hbox{$\uparrow$}} }{=}} ( 150) ( 144) \left(\dfrac{1000^{2}}{3600^{2}}\right) \dfrac{\textrm{kg}\,{\cdot}\,\textrm{m}}{\textrm{s}^{2}} \approx 1667\ \enspace \textrm{N}\\[-16.1pt] &&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \color{#0066A7}{\hbox{1 h = 3600s}} \end{eqnarray*}

The track must exert a force of magnitude 1667 \(N\) or greater to prevent skidding.

NOTE

The SI unit of force is the newton (\(\textrm{N}\)): \(\textrm{N}=\dfrac{\textrm{kg}\,{\cdot}\,\textrm{m}}{\textrm{s}^{2}}\).

NOW WORK

Problem 53.

RECALL

The unit tangent vector is \(\mathbf{T}( t)\;=\;\dfrac{\mathbf{r}^{\prime} ( t) }{\left\Vert \mathbf{r}^{\prime} ( t) \right\Vert }.\)

RECALL

The principal unit normal vector is \(\mathbf{N} ( t)\;=\;\dfrac{\mathbf{T}^{\prime} ( t) }{\left\Vert \mathbf{T}^{\prime} ( t) \right\Vert }\).

THEOREM The Acceleration Vector Expressed Using T and N

The acceleration vector \(\mathbf{a=a}(t)\) of a particle moving along a smooth curve \(C\) traced out by a twice differentiable vector function \(\mathbf{r}=\mathbf{r}(t)\) lies in the plane formed by the unit tangent vector \(\mathbf{T}(t)\) and the principal unit normal vector \(\mathbf{N}(t)\) to \(C\). Moreover, \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\mathbf{a}(t)=v^{\prime} (t)\mathbf{T}(t)+[ v(t)] ^{2}\kappa \mathbf{N}(t)}\tag{6} \end{equation*} where \(v( t)\) is the speed of the particle and \(\kappa \) is the curvature of the curve \(C\) traced out by \(\mathbf{r}=\mathbf{r}( t).\)

Figure 29 \(\mathbf{a}(t)=v^{\prime} (t)\mathbf{T}(t)+ [ v(t)]^{2}\kappa \mathbf{N}(t)\)

Finding the Tangential and Normal Components of Acceleration for an Elliptical Path

Find the tangential component \(a_{\mathbf{T}}\) and normal component \(a_{\mathbf{N}}\) of the acceleration of a particle moving along the ellipse \begin{equation*} \mathbf{r}(t)=3\sin t\mathbf{i}+4\cos t\mathbf{j}\quad 0\leq t\leq 2\pi \end{equation*}

Graph the ellipse, showing \(\mathbf{a}=\mathbf{a}( t), a_{\mathbf{T}}\), and \(a_{\mathbf{N}}\) when \(t=0,\) \(\dfrac{\pi }{4},\) and \( \dfrac{\pi }{2}\).

Solution We begin by finding the velocity, speed, and acceleration of the particle. \begin{eqnarray*} \mathbf{v}(t)&=&\mathbf{r}^{\prime} (t)=3\cos t\mathbf{i}-4\sin t\mathbf{j}\\[3pt] v(t)&=&\left\Vert \mathbf{v}(t)\right\Vert\;=\;\sqrt{(3\cos t)^{2}+(-4\sin t)^{2}}=\sqrt{9\cos ^{2}t+16\sin ^{2}t} \\[3pt] \mathbf{a}(t)&=&\mathbf{r}^{\prime \prime} (t)=-3\sin t\mathbf{i}-4\cos t \mathbf{j} \end{eqnarray*}

The tangential component of the acceleration is \begin{eqnarray*} a_{\mathbf{T}}=v^{\prime} ( t) &=&\dfrac{d}{dt}\sqrt{9\cos ^{2}t+16\sin ^{2}t}\\[3pt] &=&\dfrac{1}{2}( 9\cos ^{2}t+16\sin ^{2}t) ^{-1/2}[ 18\cos t ( -\!\sin t) +32\sin t (\cos t)]\\[3pt] &=&\frac{7\sin t\cos t}{\sqrt{9\cos ^{2}t+16\sin ^{2}t}} \end{eqnarray*}

To find the normal component of the acceleration, we need to find the curvature \(\kappa \) of \(\mathbf{r}=\mathbf{r}( t)\). \begin{eqnarray*} \kappa &=&\dfrac{\left\Vert \mathbf{r}^{\prime} ( t) \times \mathbf{r} ^{\prime \prime} ( t) \right\Vert }{\left\Vert \mathbf{r}^{\prime} ( t) \right\Vert ^{3}}=\dfrac{ \left\| \bbox[5px]{\begin{vmatrix} {\bf i} & {\bf j} & {\bf k}\\ {3 \cos t} & {-4 \sin t} & {0} \\ {-3 \sin t} & {-4 \cos t} & {0} \end{vmatrix}} \right\| }{( 9\cos ^{2}t+16\sin ^{2}t) ^{3/2}}=\dfrac{\left\vert\!-\!12\cos^{2}t-12\sin ^{2}t\right\vert }{( 9\cos ^{2}t+16\sin ^{2}t) ^{3/2} }\\[3pt] &=&\dfrac{12}{( 9\cos ^{2}t+16\sin ^{2}t) ^{3/2}} \end{eqnarray*}

Then \begin{equation*} a_{\mathbf{N}}=v^{2}( t) \kappa\;=\;( 9\cos ^{2}t+16\sin ^{2}t) \dfrac{12}{( 9\cos ^{2}t+16\sin ^{2}t) ^{3/2}}=\frac{ 12}{\sqrt{9\cos ^{2}t+16\sin ^{2}t}} \end{equation*}

The motion of the particle on the ellipse starts at \(\mathbf{r}(0)=4\mathbf{j}\). As \(t\) increases, \(x( t)\;=\;3 \sin t\) increases and \(y( t)\;=\;4\cos t\) decreases, so the motion of the particle is clockwise.

At \(t=0,\) \begin{equation*} \mathbf{a}( 0)\;=\;-3( 0) \mathbf{i}-4( 1) \mathbf{j}=-4\mathbf{j}\quad a_{\mathbf{T}}=\dfrac{7( 0) ( 1) }{3}=0\quad a_{\mathbf{N}}=\dfrac{12}{3}=4 \end{equation*}

At \(t=\dfrac{\pi }{4}\), \(\mathbf{r}\left( \dfrac{\pi }{4}\right)\;=\;\dfrac{3\sqrt{2}}{2}\mathbf{i}+2\sqrt{2}\mathbf{j}\), and \begin{equation*} \mathbf{a}\left( \dfrac{\pi }{4}\right)\;=\;-\dfrac{3\sqrt{2}}{2}\mathbf{i}-2 \sqrt{2}\mathbf{j}\quad a_{\mathbf{T}}=\dfrac{7\sqrt{2}}{10}\quad a_{\mathbf{N}}=\dfrac{12\sqrt{2}}{5} \end{equation*}

Figure 30 \(\mathbf{r}(t)\;=\;3 \sin t\mathbf{i} + 4 \cos t\mathbf{j}\)

At \(t=\dfrac{\pi }{2}\), \(\mathbf{r}\left( \dfrac{\pi }{2}\right)\;=\;3\mathbf{i}+4( 0) \mathbf{j}=3\mathbf{i}\), and \begin{equation*} \mathbf{a}\left( \dfrac{\pi }{2}\right)\;=\;-3( 1) \mathbf{i} -4( 0) \mathbf{j=}-\!3\mathbf{i}\quad a_{\mathbf{T} }=0\quad a_{\mathbf{N}}=\dfrac{12}{4}=3 \end{equation*}

See Figure 30.

NOW WORK

Problem 27.

Analyzing the Acceleration of a Car

A car on the ramp of a multistory parking garage travels along a curve traced out by \(\mathbf{r}( t)\;=\;10\cos t\mathbf{i}+10\sin t\mathbf{j}+3t\mathbf{k}\), where \(t\) is the time in hours and distance is in miles. Find the tangential component \(a_{\mathbf{T}}\) and normal component \(a_{\mathbf{N}}\) of the acceleration of the car. What are the magnitude and direction of the force on the driver?

Solution We begin by finding the velocity \(\mathbf{v}\), speed \(v \), and acceleration \(\mathbf{a}\) of the car. \begin{eqnarray*} \mathbf{v}(t)&\;=\;&\mathbf{r}^{\prime} (t)=\dfrac{d}{dt}( 10\cos t) \mathbf{i}+\dfrac{d}{dt}( 10\sin t) \mathbf{j}+\dfrac{d}{dt} ( 3t) \mathbf{k}=-10\sin t\mathbf{i}+10\cos t\mathbf{j+}3 \mathbf{k} \\[6pt] v(t)&\;=\;&\left\Vert \mathbf{v}(t)\right\Vert\;=\;\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;\sqrt{(-10\sin t)^{2}+(10\cos t)^{2}+9}=\sqrt{109}\approx 10.440 \enspace \textrm{mi/h}\\[6pt] \mathbf{a}(t)&\;=\;&\mathbf{r^{\prime\prime}} (t)=\dfrac{d}{dt}\mathbf{v}(t)= -10\cos t\mathbf{i}-10\sin t\mathbf{j } \end{eqnarray*}

Since the speed is constant, the tangential component of acceleration \(a_{ \mathbf{T}}\) is \begin{eqnarray*} a_{\mathbf{T}}=\dfrac{dv}{dt}=0 \end{eqnarray*}

The normal component of acceleration \(a_{\mathbf{N}}\) is

Since, \begin{equation*} \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -10\sin t & 10\cos t & 3 \\[3pt] -10\cos t & -10\sin t & 0 \end{array}\right| =30\sin t\mathbf{i}-30\cos t\mathbf{j}+100\mathbf{k} \end{equation*}

the normal component is \begin{eqnarray*} a_{\mathbf{N}}&=&\frac{\left\Vert 30\sin t\mathbf{i}-30\cos t\mathbf{j}+100 \mathbf{k}\right\Vert }{\sqrt{109}}=\dfrac{\sqrt{900\sin ^{2}t+900\cos ^{2}t+10000} }{\sqrt{109}}\\[4pt] &=&\dfrac{\sqrt{10900}}{\sqrt{109}}=10 \end{eqnarray*}

As the car travels on the ramp at a constant speed, the force \[ \mathbf{F}=m\mathbf{a}\;=\;ma_{\mathbf{N}} \mathbf{N}\;=\;10 m \mathbf{N} \]

pulls the car and driver toward the center of the ramp, with a magnitude about \(10\) times the mass of the car.

NOW WORK

Problem 69.

Finding the UVW-Axes of an Orbit

A satellite in a circular orbit intersects the \(x\)-axis and is inclined to the \(xy\)-plane at an angle of \(30{{}^\circ}\). Suppose the orbit has a radius \(a,\) and the motion has angular speed \(\omega \). Then the position of the satellite at time \(t\) is \begin{equation*} \mathbf{r}(t)=a\!\left( \cos ( \omega t) \mathbf{i}+\frac{\sqrt{3}}{ 2}\sin ( \omega t) \mathbf{j}+\frac{1}{2}\sin ( \omega t) \mathbf{k}\right) \end{equation*}

The velocity vector \(\mathbf{v}\) is \begin{equation*} \mathbf{v}(t)=\mathbf{r}^{\prime} (t)=a\omega \!\left( -\!\sin ( \omega t) \mathbf{i}+\frac{\sqrt{3}}{2}\cos ( \omega t) \mathbf{j}+ \frac{1}{2}\cos ( \omega t) \mathbf{k}\right) \end{equation*}

In this case, using (7), the UVW system of axes is \[ \begin{array}{rcl@{\qquad}rcl} \mathbf{i}_{\mathbf{U}}&=&\cos ( \omega t) \mathbf{i}+\frac{\sqrt{3}}{2}\sin ( \omega t) \mathbf{j}+\frac{1}{2}\sin ( \omega t) \mathbf{k} & \color{#0066A7}{\mathbf{i}_{\mathbf{U}}}&{\color{#0066A7}=}&\color{#0066A7}{\tfrac{\mathbf{r}}{\left\Vert \mathbf{r}\right\Vert }}\\ \mathbf{i}_{\mathbf{W}}&=&-\frac{1}{2}\mathbf{j}+\frac{\sqrt{3}}{2}\mathbf{k}& \color{#0066A7}{\mathbf{i}_{\mathbf{W}}}&{\color{#0066A7}=}& \color{#0066A7}{\tfrac{\mathbf{r}\times \mathbf{v}}{\left\Vert \mathbf{r}\times \mathbf{v}\right\Vert }}\\ \mathbf{i}_{\mathbf{V}}&\;=\;&-\sin ( \omega t) \mathbf{i}+\frac{ \sqrt{3}}{2}\cos ( \omega t) \mathbf{j}+\frac{1}{2}\cos ( \omega t) \mathbf{k} & \color{#0066A7}{\mathbf{i}_{\mathbf{V}}}&{\color{#0066A7}=}&\color{#0066A7}{\mathbf{i}_{\mathbf{W}}\times \mathbf{i}_{\mathbf{U}}} \end{array} \]

The directions of the onboard gyroscope axes UVW are now expressed in terms of \(\mathbf{i},\) \(\mathbf{j},\) and \(\mathbf{k}.\)