공간에서의 운동

Displacement: $\left\langle -10,0\right\rangle$; distance traveled: $5\pi \approx 15.71$ft; average velocity: $\left\langle -10/\pi,0\right\rangle \approx \left\langle -3.18,0\right\rangle$; average speed: $5$ft/s

Displacement: $\left\langle 0,0\right\rangle$; distance traveled: $2\pi \approx 6.28$ft; average velocity: $\left\langle 0,0\right\rangle$; average speed: $1$ft/s

Displacement: $\left\langle 10,20,-20\right\rangle$; distance traveled: $30$ft; average velocity: $\left\langle 1,2,-2\right\rangle$; average speed: $3$ft/s

At $t$-values of $\sin^{-1}(9/30)/(4\pi) + n/2 \approx 0.024 + n/2$ seconds, where $n$ is an integer.

The stone, while whirling, can be modeled by $\vec r(t) = \left\langle 3\cos(8\pi t),3\sin (8\pi t)\right\rangle$.

(a) For $t$-values $t=\sin^{-1}(3/20)/(8\pi) + n/4 \approx 0.006 + n/4$, where $n$ is an integer. (b) $\left\|\vec r\,'(t)\right\| = 24\pi\approx 51.4$ft/s (c) At $t=0.006$, the stone is approximately $19.77$ft from Goliath. Using the formula for projectile motion, we want the angle of elevation that lets a projectile starting at $\left\langle 0,6\right\rangle$ with a initial velocity of $51.4$ft/s arrive at $\left\langle 19.77,9\right\rangle$. The desired angle is $0.27$ radians, or $15.69^\circ$.

(a) Holding the crossbow at an angle of $0.013$ radians, $\approx 0.745^\circ$ will hit the target $0.4$s later. (Another solution exists, with an angle of $89^\circ$, landing $18.75$s later, but this is impractical.) (b) In the .4 seconds the arrow travels, a deer, traveling at 20mph or 29.33ft/s, can travel 11.7ft. So she needs to lead the deer by 11.7ft.

The position function of the ball is $\vec r(t) = \left\langle (146.67\cos\theta)t,-16t^2+(146.67\sin\theta)t+3\right\rangle$, where $\theta$ is the angle of elevation.

(a) With $\theta=20^\circ$, the ball reaches 310ft from home plate in 2.25 seconds; at this time, the height of the ball is 34.9ft, not enough to clear the Green Monster. (b) With $\theta=21^\circ$, the ball reaches 310ft from home plate in 2.26s, with a height of 40ft, clearing the wall.

The position function is $\vec r(t) = \left\langle 220t,-16t^2+1000\right\rangle$. The $y$-component is 0 when $t=7.9$; $\vec r(7.9) = \left\langle 1739.25,0\right\rangle$, meaning the box will travel about 1740ft horizontally before it lands.

The position function of the ball is $\vec r(t) = \left\langle (v_0\cos\theta)t,-16t^2+(v_0\sin\theta)t+6\right\rangle$, where $\theta$ is the angle of elevation and $v_0$ is the initial ball speed.

(a) With $v_0 = 73.33$ft/s, there are two angles of elevation possible. An angle of $\theta = 9.47^\circ$ delivers the ball in 0.83s, while an angle of $79.57^\circ$ delivers the ball in $4.5$s. (b) With $\theta=8^\circ$, the initial speed must be 53.8mph$\approx 78.9$ft/s.

We don't have enough information to gauge the size of the vectors, but we can figure out their direction. Gravity pulls straight down, so the vector $-mg\hat{\boldsymbol{\jmath}}$ points straight down. The normal force will be normal to the curve.

This equation stems from $\mathbf{F}=m\mathbf{a}$. In that equation, $\mathbf{a}$ is acceleration — the second derivative of position with respect to time. So, $\mathbf{v}$ is the derivative of position with respect to time.

We previously used $\mathbf{v}$ as the derivative of position with respect to the parameter we use to define our position — which was often called $t$, but was not the necessarily time. So this is a good point to keep straight.

Solution 1:

For large, negative values of $x$, the wire is closer and closer to a vertical line. If the bead were sliding down a vertical wire, it could do so without even touching the wire, so the force exerted on the bead would be zero. As $x$ approaches 0 from the left, the wire approximates a horizontal line. If the bead were sitting on a horizontal line, the wire would be pushing up to counter gravity. So, we imagine the magnitude of the force exerted by the wire might increase as $x$ increases. That is, $\frac{\mathrm{d}W}{\mathrm{d}x}>0$.

Solution 2:

The net force exerted on the bead is

$$\begin{aligned} F=m\mathbf{a}&= W\hat{\mathbf{N}} -mg\hat{\boldsymbol{\jmath}} \\ \text{We dot both sides with $\hat{\mathbf{N}}$.} \\ W\hat{\mathbf{N}} \cdot \hat{\mathbf{N}} - mg\hat{\boldsymbol{\jmath}}\cdot\hat{\mathbf{N}}&=m\mathbf{a}\cdot\hat{\mathbf{N}} \\ \text{Using the equation $\mathbf{a}(t) = \frac{\mathrm{d}^2s}{\mathrm{d}{t}^2}\hat{\mathbf{T}} + \kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\hat{\mathbf{N}}$,} \\ W-mg\hat{\boldsymbol{\jmath}}\cdot\hat{\mathbf{N}}&=m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\\ W&=mg\hat{\mathbf{n}}\cdot\hat{\mathbf{N}}+m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\\ &=mg\cos\theta+m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2 \end{aligned}$$

where $\theta$ is the angle between $\hat{\boldsymbol{\jmath}}$ and $\hat{\mathbf{N}}$.

As $x$ moves from a highly negative number to zero, $\theta$ moves from nearly $\pi/2$ to nearly $0$. Therefore $\cos \theta$ increases from nearly zero to nearly one. Then $mg\cos\theta$ is increasing.

Furthermore, as $x$ increases, we see from the picture that the curvature $\kappa$ increases, and speed $\frac{\mathrm{d}s}{\mathrm{d}t}$ increases as well (kinetic energy is increasing as potential energy decreases).

So, $\frac{\mathrm{d}W}{\mathrm{d}x}>0$.

The skateboarder starts going back down at $y_S=\frac{E}{mg}$, so we solve $3\textrm{ m}=\frac{E}{100 \mathrm{ kg} \cdot 9.8~\frac{\mathrm{m}}{\mathrm{s}^2}}$ to find $E=2940 ~\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2}=2940 J$

Remark: we needed the diameter to be greater than 3m for the skateboarder to not be going all the way around the culvert, but choosing $r=5$ leads to an answer no different from, say, $r=50$.

From the text, the skateboarder will make it all the way around when $\frac{5}{2}(5) \le\frac{E}{mg}$. Energy $E$ is given by $E=\frac12m|\mathbf{v}|^2+mgy$, the sum of the kinetic and potential energy of the system. At $y=0$, all the energy is kinetic, so $E=\frac12m|\mathbf{v}|^2$, where $|\mathbf{v}|$ is the skater's velocity at the bottom of the culvert.

So, we solve:

$$\begin{aligned} \frac{25}{2}&\le\frac{E}{mg}=\frac{\frac12m|\mathbf{v}|^2}{m\cdot9.8}\\ |\mathbf{v}|&\ge5\sqrt{9.8} \end{aligned}$$

So, a speed of $5\sqrt{9.8}$ m/s or higher is needed. (That's about 56 kph.)

According to the equation in the text, the skier will become airborne when: $$|\mathbf{v}|>\sqrt{\frac{g}{\kappa}|\hat{\boldsymbol{\jmath}}\cdot\hat{\mathbf{N}}|}$$ We'll use the equation of the curve to find $\kappa$ and $\hat{\mathbf{N}}$.

Note that $g$ is given in metres per second, while the other quantities are in kilometres and hours. Converting, $9.8$ m/s$^2$ is the same as $\left(\frac{9.8\text{ m}}{1 \text{ s}^2}\right)\left( \frac{1 \text{ km}}{1000 \text{ m}}\right)\left(\frac{3600 \text{ s}}{1 \text{ hr}} \right)^2=98\cdot 6^4~ \frac{\text{km}}{\text{h}^2}=2^5\cdot 3^4\cdot 7^2~ \frac{\text{km}}{\text{h}^2}$.

$$\begin{aligned} \mathbf{r}(t)&=(\ln t,1-t)\\ \mathbf{r}'(t)&=\mathbf{v}(t)=(t^{-1}, -1)\qquad\qquad \frac{\mathrm{d}s}{\mathrm{d}t}=|\mathbf{v}(t)|=\sqrt{1+t^{-2}}\\ \mathbf{r}''(t)&=\mathbf{a}(t)=(-t^{-2},0)\\ \kappa (t)&=\frac{|\mathbf{v} \times \mathbf{a}|}{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^{3}}=\frac{t^{-2}}{\sqrt{1+t^{-2}}^3}=\frac{|t|}{(1+t^{2})^{3/2}}=\frac{t}{(1+t^{2})^{3/2}} \\ \text{Note $t$ is positive in the interval in question.} \\ \hat{\mathbf{T}}(t)&=\frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}=\frac{1}{\sqrt{1+t^{-2}}}(t^{-1},-1)=\left( \frac{1}{\sqrt{1+t^{2}}},\frac{-t}{\sqrt{1+t^{2}}}\right)\\ \hat{\mathbf{T}}'(t)&= \left( \frac{-t}{({1+t^2})^{3/2}},\frac{-1}{({1+t^2})^{3/2}}\right)\qquad\qquad |\hat{\mathbf{T}}'(t)|=\frac{1}{t^2+1}\\ \hat{\mathbf{N}}(t)&=\frac{\hat{\mathbf{T}}'(t)}{|\hat{\mathbf{T}}'(t)|}=\left( \frac{-t}{\sqrt{1+t^2}},\frac{-1}{\sqrt{1+t^2}}\right)\\ |\hat{\mathbf{N}} \cdot \hat{\boldsymbol{\jmath}}|&=\frac{1}{\sqrt{1+t^2}} \end{aligned}$$

Now, we have all the pieces we need to find the “escape velocity" of the ground.

$$\begin{aligned} |\mathbf{v}|&=\sqrt{\frac{g}{\kappa}|\hat{\mathbf{N}} \cdot \hat{\boldsymbol{\jmath}}|}=\sqrt{\frac{g\cdot (1+t^{2})^{3/2}}{t(1+t^2)^{1/2}}}=\sqrt{\frac{g(1+t^2)}{t}} \end{aligned}$$

Since the skier can take off anywhere on the hill, we just need their velocity to be larger than the smallest value of $\sqrt{\frac{g(1+t^2)}{t}}$ when $1/e \le t \le e$. To find that minimum, we find the location of the minimum of the simpler function $g(t)=\frac{1+t^2}{t}$. Using first-semester calculus, we find it to occur when $t=1$. So, the minimum value of $\sqrt{\frac{g(1+t^2)}{t}}$ (that is, smallest speed to achieve lift-off) occurs at $t=1$. We therefore need a minimum speed greater than:

$$\begin{aligned} \sqrt{\frac{g(1+t^2)}{|t|}}\Bigg|_{t=1}&=\sqrt{2g}=\sqrt{2^6\cdot3^4\cdot7^2}=2^3\cdot 3^2\cdot 7 = 504\text{ kph} \end{aligned}$$

(It seems unlikely that one could reach this speed on skis. The skier is probably earth-bound until they find a curvier hill.)

(a) By Newton's law of motion

$$\begin{aligned} E'(t)&=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{2} m |\mathbf{v}(t)|^2 +mg \mathbf{r}(t)\cdot\hat{\mathbf{k}}\right] =m\mathbf{v}(t)\cdot\mathbf{v}'(t)+mg\mathbf{v}(t)\cdot\hat{\mathbf{k}}\\ &=\mathbf{v}(t)\cdot\big[\mathbf{N}\big(\mathbf{r}(t)\big) - mg\hat{\mathbf{k}}\big]+mg\mathbf{v}(t)\cdot\hat{\mathbf{k}}\\ &=0 \end{aligned}$$

since $\mathbf{v}(t)\cdot\mathbf{N}\big(\mathbf{r}(t)\big)=0$. So $E(t)$ is a constant, independent of $t$.

(b) By part (a),

$$\begin{aligned} E(t)=E(0) \implies \tfrac{1}{2} m |\mathbf{v}(t)|^2 +mg b\theta(t) = mg(2\pi b) \implies|\mathbf{v}(t)|^2=2gb\big(2\pi-\theta(t)\big) \end{aligned}$$

(c) We wish to determine the time it takes to go from $\theta=2\pi$ to $\theta=0$. We'll first determine $\frac{\mathrm{d}\theta}{\mathrm{d}t}$.

$$\begin{aligned} &\mathbf{v}=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}t} =\big(-a\sin\theta,a\cos\theta,b\big)\frac{\mathrm{d}\theta}{\mathrm{d}t} \\ &\implies |\mathbf{v}|^2=\big[a^2+b^2]\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2\\ &\implies \frac{\mathrm{d}\theta}{\mathrm{d}t}=-\left[\frac{|\mathbf{v}|^2}{a^2+b^2}\right]^{1/2} =-\left[\frac{2gb(2\pi-\theta)}{a^2+b^2}\right]^{1/2} \end{aligned}$$

We have chosen the negative sign because $\theta$ must decrease from $2\pi$ to 0. The time required to do so is

$$\begin{aligned} \int dt&=\int_{2\pi}^0 \frac{dt}{d\theta}d\theta =-\left[\frac{a^2+b^2}{2gb}\right]^{1/2}\int_{2\pi}^0 \frac{1}{(2\pi-\theta)^{1/2}}d\theta \\ &=\left[\frac{a^2+b^2}{2gb}\right]^{1/2}\int^{2\pi}_0 \frac{1}{(2\pi-\theta)^{1/2}}d\theta\\ &=\left[\frac{a^2+b^2}{2gb}\right]^{1/2} \left[-2(2\pi-\theta)^{1/2}\right]_0^{2\pi} =2\left[\frac{a^2+b^2}{gb}\pi\right]^{1/2} \end{aligned}$$