Loading web-font TeX/Math/Italic

6.2 HW 2

PDF (letter size)

PDF (legal size)

6.2.1 Problem 1

$$\bar{v}=\dot{r}\hat{u}_{r}+r\dot{\theta }\hat{u}_{\theta }$$ Hence $$\begin{align*} \left \vert \bar{v}\right \vert & =\sqrt{\dot{r}^{2}+\left ( r\dot{\theta }\right ) ^{2}}\\ & =\sqrt{\left ( -22490\right ) ^{2}+\left ( 21000\times \left ( -2.933\right ) \right ) ^{2}}\\ & =65571\ \text{ft/sec} \end{align*}$$

Since

$$\begin{align*} \hat{u}_{r} & =\hat{\imath }\cos \theta +\hat{\jmath }\sin \theta \\ \hat{u}_{\theta } & =-\hat{\imath }\sin \theta +\hat{\jmath }\cos \theta \end{align*}$$

Then the velocity vector in Cartesian is

$$\begin{align*} \bar{v} & =\dot{r}\left ( \hat{\imath }\cos \theta +\hat{\jmath }\sin \theta \right ) +r\dot{\theta }\left ( -\hat{\imath }\sin \theta +\hat{\jmath }\cos \theta \right ) \\ & =\hat{\imath }\left ( \dot{r}\cos \theta -r\dot{\theta }\sin \theta \right ) +\hat{\jmath }\left ( \dot{r}\sin \theta +r\dot{\theta }\cos \theta \right ) \end{align*}$$

Plug-in numerical values

$$\begin{align*} \bar{v} & =\hat{\imath }\left ( -22490\cos \left ( 38\left ( \frac{\pi }{180}\right ) \right ) -\left ( 21000\right ) \left ( -2.933\right ) \sin \left ( 38\left ( \frac{\pi }{180}\right ) \right ) \right ) \\ & +\hat{\jmath }\left ( \left ( -22490\right ) \sin \left ( 38\left ( \frac{\pi }{180}\right ) \right ) +\left ( 21000\right ) \left ( -2.933\right ) \cos \left ( 38\left ( \frac{\pi }{180}\right ) \right ) \right ) \end{align*}$$

Or

$$\bar{v}=\left ( 20198.08\right ) \hat{\imath }-\hat{\jmath }\left ( 62382.17\right )$$ To check, we find magnitude of $\bar{v}$ in Cartesian $$\left \vert \bar{v}\right \vert =\sqrt{\left ( 20198.08\right ) ^{2}+\left ( 62382.17\right ) ^{2}}=65571$$ Which is the same as before. Hence the velocity vector makes angle $\tan ^{-1}\left ( \frac{-62382}{20198}\right ) =-1.2577$ rad or $-1.2577\left ( \frac{180}{\pi }\right ) =-72.061$ degrees with the x-axis.

6.2.2 Problem 2

$$\bar{a}=\left ( \ddot{r}-r\dot{\theta }^{2}\right ) \hat{u}_{r}+\left ( r\ddot{\theta }+2\dot{r}\dot{\theta }\right ) \hat{u}_{\theta }$$ Hence $$\begin{align*} \left \vert \bar{a}\right \vert & =\sqrt{\left ( \ddot{r}-r\dot{\theta }^{2}\right ) ^{2}+\left ( r\ddot{\theta }+2\dot{r}\dot{\theta }\right ) ^{2}}\\ & =\sqrt{\left ( 187300-\left ( 21400\right ) \left ( -2.944\right ) ^{2}\right ) ^{2}+\left ( \left ( 21400\right ) \left ( -5.407\right ) +2\left ( -22490\right ) \left ( -2.944\right ) \right ) ^{2}}\\ & =16810.49\ \text{ft/sec}^{2} \end{align*}$$

Since

$$\begin{align*} \hat{u}_{r} & =\hat{\imath }\cos \theta +\hat{\jmath }\sin \theta \\ \hat{u}_{\theta } & =-\hat{\imath }\sin \theta +\hat{\jmath }\cos \theta \end{align*}$$

Then the acceleration vector in Cartesian is

$$\begin{align} \bar{a} & =\left ( \ddot{r}-r\dot{\theta }^{2}\right ) \left ( \hat{\imath }\cos \theta +\hat{\jmath }\sin \theta \right ) +\left ( r\ddot{\theta }+2\dot{r}\dot{\theta }\right ) \left ( -\hat{\imath }\sin \theta +\hat{\jmath }\cos \theta \right ) \nonumber \\ & =\hat{\imath }\left ( \left ( \ddot{r}-r\dot{\theta }^{2}\right ) \cos \theta -\left ( r\ddot{\theta }+2\dot{r}\dot{\theta }\right ) \sin \theta \right ) +\hat{\jmath }\left ( \left ( \ddot{r}-r\dot{\theta }^{2}\right ) \sin \theta +\left ( r\ddot{\theta }+2\dot{r}\dot{\theta }\right ) \cos \theta \right ) \tag{1} \end{align}$$

But

$$\begin{align*} \left ( \ddot{r}-r\dot{\theta }^{2}\right ) & =\left ( 187300-\left ( 21400\right ) \left ( -2.944\right ) ^{2}\right ) =1823.290\\ \left ( r\ddot{\theta }+2\dot{r}\dot{\theta }\right ) & =\left ( \left ( 21400\right ) \left ( -5.407\right ) +2\left ( -22490\right ) \left ( -2.944\right ) \right ) =16711.32 \end{align*}$$

Hence (1) becomes

$$\begin{align*} \bar{a} & =\hat{\imath }\left ( 1823.290\cos \left ( 44\frac{\pi }{180}\right ) -\left ( 16711.32\right ) \sin \left ( 44\frac{\pi }{180}\right ) \right ) \\ & +\hat{\jmath }\left ( 1823.290\sin \left ( 44\frac{\pi }{180}\right ) +\left ( 16711.32\right ) \cos \left ( 44\frac{\pi }{180}\right ) \right ) \\ & \\ & =\hat{\imath }\left ( -10297.09\right ) +\hat{\jmath }\left ( 13287.68\right ) \end{align*}$$

To verify things, we check the magnitude of $\bar{a}$ is the same as found above (since the magnitude of vector does not depend on coordinates. We see that $\left \vert \bar{a}\right \vert =\sqrt{\left ( -10297.09\right ) ^{2}+\left ( 13287.68\right ) ^{2}}=16810.49$ which is the same as before.

Hence the acceleration vector makes angle $\tan ^{-1}\left ( \frac{13287.68}{-10297.09}\right ) =127.7\ $degrees with the x-axis.

6.2.3 Problem 3

$$\bar{a}=\dot{V}\hat{u}_{t}+\frac{V^{2}}{\rho }\hat{u}_{n}$$ But $\dot{V}=0$ since velocity is constant. Hence $\left \vert \bar{a}\right \vert =\frac{V^{2}}{\rho }$. Therefore $$\begin{align*} \frac{V^{2}}{\rho } & =9g\\ \frac{\left ( 750\left ( \frac{5280}{3600}\right ) \right ) ^{2}}{9\left ( 32.2\right ) } & =\rho \\ \rho & =4175.3\ \text{ft} \end{align*}$$

6.2.4 Problem 4

$$\bar{a}=\dot{V}\hat{u}_{t}+\frac{V^{2}}{\rho }\hat{u}_{n}$$ Maximum normal acceleration is $\frac{V^{2}}{\rho }$. Hence it occurs when $V$ is maximum, which is at start of turn. Then we want $$\begin{align*} \frac{V_{\max }^{2}}{\rho _{\min }} & =2g\\ \rho _{\min } & =\frac{V_{\max }^{2}}{g}=\frac{\left ( 85\left ( \frac{5280}{3600}\right ) \right ) ^{2}}{2\left ( 32.2\right ) }=241.33\text{ ft} \end{align*}$$

Now since

$$v_{f}^{2}=v_{o}^{2}+2a_{t}s$$ Where $a_{t}$ is tangential acceleration and $s$ is distance travelled which is $\frac{1}{4}$ of circumference of circle or $\frac{1}{4}\left ( 2\pi \rho _{\min }\right )$, therefore $$a_{t}=\frac{v_{f}^{2}-v_{o}^{2}}{2\left ( \frac{1}{2}\pi \rho _{\min }\right ) }=\frac{\left ( 80\left ( \frac{5280}{3600}\right ) \right ) ^{2}-\left ( 85\left ( \frac{5280}{3600}\right ) \right ) ^{2}}{\pi \left ( 241.33\right ) }=-2.341\text{ ft/sec}^{2}$$ Therefore to find time of travel, since acceleration is constant $$\begin{align*} v_{f} & =v_{0}+a_{t}t\\ t & =\frac{v_{f}-v_{0}}{a_{t}}=\frac{80\left ( \frac{5280}{3600}\right ) -85\left ( \frac{5280}{3600}\right ) }{-2.341}=3.133\text{ sec} \end{align*}$$

6.2.5 Problem 5

Since velocity is horizontal, then

$$\bar{v}=v_{0}\hat{\imath }$$ But $\hat{\imath }=\cos \theta \hat{u}_{r}-\sin \theta \hat{u}_{\theta }$, hence $$\begin{align*} \bar{v} & =v_{0}\cos \theta \hat{u}_{r}-v_{0}\sin \theta \hat{u}_{\theta }\\ & =\dot{r}\hat{u}_{r}+r\dot{\theta }\hat{u}_{\theta } \end{align*}$$

Therefore

$$\begin{align*} \dot{r} & =v_{0}\cos \theta \\ r\dot{\theta } & =-v_{0}\sin \theta \end{align*}$$

We are given that $v_{0}=560$ mph and $\theta =31$, hence solving gives

$$\begin{align*} \dot{r} & =560\cos \left ( 31\frac{\pi }{180}\right ) \\ \left ( 6.9\right ) \dot{\theta } & =-\left ( 560\right ) \sin \left ( 31\frac{\pi }{180}\right ) \end{align*}$$

Or

$$\begin{align*} \dot{r} & =480.014\text{ mph}\\ \dot{\theta } & =-41.8\text{ rad/h} \end{align*}$$

The acceleration vector is

$$\bar{a}=\left ( \ddot{r}-r\dot{\theta }^{2}\right ) \hat{u}_{r}+\left ( r\ddot{\theta }+2\dot{r}\dot{\theta }\right ) \hat{u}_{\theta }$$ Since constant velocity, then acceleration is zero. This gives us two equations to solve for $\ddot{\theta },\ddot{r}$ $$\begin{align*} \ddot{r}-r\dot{\theta }^{2} & =0\\ r\ddot{\theta }+2\dot{r}\dot{\theta } & =0 \end{align*}$$

Or

$$\begin{align*} \ddot{r}-\left ( 6.9\right ) \left ( -41.8\right ) ^{2} & =0\\ \left ( 6.9\right ) \ddot{\theta }+2\left ( 480.014\right ) \left ( -41.8\right ) & =0 \end{align*}$$

Solving gives

$$\begin{align*} \ddot{r} & =12054.23\text{ mph}\\ \ddot{\theta } & =5815.477\text{ rad/h}^{2} \end{align*}$$

6.2.6 Problem 6

$$\begin{align*} \bar{v} & =\dot{r}\hat{u}_{r}+r\dot{\theta }\hat{u}_{\theta }\\ & =\left ( 448.1\cos \theta +13.17\sin \theta \right ) \hat{u}_{r}+\left ( 13.17\cos \theta -448.1\sin \theta \right ) \hat{u}_{\theta } \end{align*}$$

Hence

$$\begin{align*} \left \vert \bar{v}\right \vert ^{2} & =\left ( 448.1\cos \theta +13.17\sin \theta \right ) ^{2}+\left ( 13.17\cos \theta -448.1\sin \theta \right ) ^{2}\\ & =\left ( 448.1\right ) ^{2}\cos ^{2}\theta +\left ( 13.17\right ) ^{2}\sin ^{2}\theta +2\left ( \left ( 448.1\right ) \left ( 13.17\right ) \cos \theta \sin \theta \right ) \\ & +\left ( 13.17\right ) ^{2}\cos ^{2}\theta +\left ( -448.1\right ) ^{2}\sin ^{2}\theta -2\left ( \left ( 448.1\right ) \left ( 13.17\right ) \cos \theta \sin \theta \right ) \end{align*}$$

Which simplifies to

$$\begin{align*} \left \vert \bar{v}\right \vert ^{2} & =\left ( 448.1\right ) ^{2}\cos ^{2}\theta +\left ( 13.17\right ) ^{2}\sin ^{2}\theta +\left ( 13.17\right ) ^{2}\cos ^{2}\theta +\left ( -448.1\right ) ^{2}\sin ^{2}\theta \\ & =\left ( 448.1\right ) ^{2}\left ( \cos ^{2}\theta +\sin ^{2}\theta \right ) +\left ( 13.17\right ) ^{2}\left ( \cos ^{2}\theta +\sin ^{2}\theta \right ) \\ & =\left ( 448.1\right ) ^{2}+\left ( 13.17\right ) ^{2}\\ & =200967.1 \end{align*}$$

Hence

$$\left \vert \bar{v}\right \vert =448.294\text{ mph}$$ Let $y$ be vertical distance. Hence $y=r\sin \theta$ and $$\begin{align*} \dot{y} & =\dot{r}\sin \theta +r\dot{\theta }\cos \theta \\ & =\left ( 448.1\cos \theta +13.17\sin \theta \right ) \sin \theta +\left ( 13.17\cos \theta -448.1\sin \theta \right ) \cos \theta \\ & =448.1\cos \theta \sin \theta +13.17\sin ^{2}\theta +13.17\cos ^{2}\theta -448.1\sin \theta \cos \theta \\ & =13.17\text{ mph} \end{align*}$$

Hence it is ascending. Convert to ft/sec

$$\begin{align*} \dot{y} & =13.17\frac{5280}{mile}\frac{hr}{3600}\\ & =13.17\frac{5280}{3600}\\ & =19.316\text{ ft/sec} \end{align*}$$