Chapter 11.5, Problem 11.179P

To determine

(a)

The magnitude of the velocity and acceleration when $t=0$ at given condition.

Answer to Problem 11.179P

The required values are:

Velocity: $v=\sqrt{A^2+B^2}$

Acceleration: $a=\sqrt{\left(1+16{\pi }^2\right)A^2+B^2}$

Explanation of Solution

Given Information:

$\begin{array}{l}R=A\left(1-e^{-t}\right)\\ \theta =2\pi t\\ z=B\left(1-e^{-t}\right)\\ t=0\end{array}$

Formula used:

$v=\dot{R}{e}_R+R\dot{\theta }{e}_{\theta }+\dot{z}k$

$a=\left(\ddot{R}-R{\dot{\theta }}^1\right)e_R+\left(R\ddot{\theta }+2\dot{R}\dot{\theta }\right)e_{\theta }+\ddot{z}k$

Calculation:

In order to solve this problem, we will use the double differentiation method;

$\Rightarrow R=A\left(1-e^{-t}\right)\theta =2\pi tz=B\left(1-e^{-t}\right)$ ;

$\Rightarrow \dot{R}=A{e}^{-t}\dot{\theta }=2\pi \dot{z}=B{e}^{-t}$ ;

$\Rightarrow \ddot{R}=-A{e}^{-t}\ddot{\theta }=0\ddot{z}=-B{e}^{-t}$ ;

We know that, $v=\dot{R}{e}_R+R\dot{\theta }{e}_{\theta }+\dot{z}k$

(eq. $11.49$ )

Put the values in the above equation;

$\Rightarrow v=A{e}^{-t}{e}_R+2\pi A\left(1-e^{-t}\right)e_{\theta }+B{e}^{-t}k$

When, $t=0:e^{-t}=e^0=1;v=A{e}_R+Bk$ ;

$\Rightarrow v=\sqrt{A^2+B^2}$ ;

We know that, $a=\left(\ddot{R}-R{\dot{\theta }}^1\right)e_R+\left(R\ddot{\theta }+2\dot{R}\dot{\theta }\right)e_{\theta }+\ddot{z}k$ ;

(eq. $11.50$ )

$\Rightarrow \left[-A{e}^{-t}-A(1-e^{-t})4{\pi }^2\right]e_R+\left[0+2A{e}^{-t}\left(2\pi \right)\right]e_{\theta }-B{e}^{-t}k$

When, $t=0:e^{-t}=e^0=1$ ;

$\Rightarrow a=-A{e}_R+4\pi A{e}_{\theta }-Bk$ ;

$\Rightarrow a=\sqrt{A^2+{\left(4\pi A\right)}^2+B^2}$ ;

$\Rightarrow a=\sqrt{\left(1+16{\pi }^2\right)A^2+B^2}$.

To determine

(b)

The magnitude of the velocity and acceleration when $t=\infty$ at given condition.

Answer to Problem 11.179P

The required values are:

Velocity: $v=2\pi A$

Acceleration: $a=4{\pi }^{2}A$

Explanation of Solution

Given information:

$\begin{array}{l}R=A\left(1-e^{-t}\right)\\ \theta =2\pi t\\ z=B\left(1-e^{-t}\right)\\ t=\infty \end{array}$

Formula used:

$v=\dot{R}{e}_R+R\dot{\theta }{e}_{\theta }+\dot{z}k$

$a=\left(\ddot{R}-R{\dot{\theta }}^1\right)e_R+\left(R\ddot{\theta }+2\dot{R}\dot{\theta }\right)e_{\theta }+\ddot{z}k$

Calculation:

We know that, $v=A{e}^{-t}{e}_R+2\pi A\left(1-e^{-t}\right)e_{\theta }+B{e}^{-t}k$ ;

(eq. $11.49$ )

When, $\Rightarrow t=\infty :e^{-t}=e^{-\infty }=0v=2\pi A{e}_{\theta }$ ;

$\Rightarrow v=2\pi A$ ;

We know that, $a=\left(\ddot{R}-R{\dot{\theta }}^1\right)e_R+\left(R\ddot{\theta }+2\dot{R}\dot{\theta }\right)e_{\theta }+\ddot{z}k$ ;

(eq. $11.50$ )

$\Rightarrow \left[-A{e}^{-t}-A(1-e^{-t})4{\pi }^2\right]e_R+\left[0+2A{e}^{-t}\left(2\pi \right)\right]e_{\theta }-B{e}^{-t}k$

When, $t=\infty :e^{-t}=e^{-\infty }=0$ ;

$\Rightarrow a=-4{\pi }^{2}A{e}_R$ ;

$\Rightarrow a=4{\pi }^{2}A$.