Chapter 11.5, Problem 11.179P

To determine

(a)

The magnitude of velocity and acceleration at $t=0$

Answer to Problem 11.179P

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

$a=\sqrt{8A^2-4A^2{B}^2+A^2{B}^4+4C^2}$

Explanation of Solution

Given information:

$\begin{array}{l}R=\frac{A}{\left(t+1\right)}\\ \theta =Bt\\ z=\frac{Ct}{\left(t+1\right)}\end{array}$

The cylindrical co-ordinates of velocity are defined as:

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

The cylindrical co-ordinates of acceleration are defined as :

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

Calculation:

According to given information,

$\begin{array}{l}R=\frac{A}{\left(t+1\right)}\\ \theta =Bt\\ z=\frac{Ct}{\left(t+1\right)}\end{array}$

Differentiate,

$\begin{array}{l}\dot{R}=-\frac{A}{{\left(t+1\right)}^2}\\ \ddot{R}=\frac{2A}{{\left(t+1\right)}^3}\end{array}$

$\begin{array}{l}\dot{\theta }=B\\ \ddot{\theta }=0\end{array}$

$\begin{array}{l}\dot{z}=\frac{C}{{\left(t+1\right)}^2}\\ \ddot{z}=-\frac{2C}{{\left(t+1\right)}^3}\end{array}$

Find the velocity,

$\begin{array}{l}v=\dot{R}{e}_R+R\dot{\theta }{e}_{\theta }+\dot{z}k\\ v=-\frac{A}{{\left(t+1\right)}^2}{e}_R+\frac{A}{\left(t+1\right)}B{e}_{\theta }+\frac{C}{{\left(t+1\right)}^2}k\end{array}$

Find the acceleration,

$\begin{array}{l}a=\left(\ddot{R}-R{\dot{\theta }}^2\right)e_R+\left(R\ddot{\theta }+2\dot{R}\dot{\theta }\right)e_{\theta }+\ddot{z}k\\ a=\left[\frac{2A}{{\left(t+1\right)}^3}-\frac{A}{\left(t+1\right)}{B}^2\right]e_R+2\left(-\frac{A}{{\left(t+1\right)}^2}\right)B{e}_{\theta }-\frac{2C}{{\left(t+1\right)}^3}k\end{array}$

For $t=0$, the velocity is equal to,

$\begin{array}{l}v=-A{e}_R+AB{e}_{\theta }+Ck\\ v=\sqrt{A^2+A^2{B}^2+C^2}\end{array}$

For $t=0$, the acceleration is equal to,

$\begin{array}{l}a=\left(2A-A{B}^2\right)e_R-2AB{e}_{\theta }-2Ck\\ a=\sqrt{{\left(2A-A{B}^2\right)}^2+4A^2+4C^2}\\ a=\sqrt{8A^2-4A^2{B}^2+A^2{B}^4+4C^2}\end{array}$

Conclusion:

The magnitude and velocity’s T = 0 according to the formula : $\begin{array}{l}R=\frac{A}{\left(t+1\right)}\\ \theta =Bt\\ z=\frac{Ct}{\left(t+1\right)}\end{array}$

To determine

(b)

The magnitude of velocity and acceleration at $t=\infty$ ?

Answer to Problem 11.179P

At $t=\infty$

$v=0$

$a=0$

Explanation of Solution

Given information:

$\begin{array}{l}R=\frac{A}{\left(t+1\right)}\\ \theta =Bt\\ z=\frac{Ct}{\left(t+1\right)}\end{array}$

The cylindrical co-ordinates of velocity is defined as,

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

The cylindrical co-ordinates of acceleration is defined as,

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

Calculation:

According to given information,

$\begin{array}{l}R=\frac{A}{\left(t+1\right)}\\ \theta =Bt\\ z=\frac{Ct}{\left(t+1\right)}\end{array}$

Differentiate,

$\begin{array}{l}\dot{R}=-\frac{A}{{\left(t+1\right)}^2}\\ \ddot{R}=\frac{2A}{{\left(t+1\right)}^3}\end{array}$

$\begin{array}{l}\dot{\theta }=B\\ \ddot{\theta }=0\end{array}$

$\begin{array}{l}\dot{z}=\frac{C}{{\left(t+1\right)}^2}\\ \ddot{z}=-\frac{2C}{{\left(t+1\right)}^3}\end{array}$

Find the velocity,

$\begin{array}{l}v=\dot{R}{e}_R+R\dot{\theta }{e}_{\theta }+\dot{z}k\\ v=-\frac{A}{{\left(t+1\right)}^2}{e}_R+\frac{A}{\left(t+1\right)}B{e}_{\theta }+\frac{C}{{\left(t+1\right)}^2}k\end{array}$

Find the acceleration,

$\begin{array}{l}a=\left(\ddot{R}-R{\dot{\theta }}^2\right)e_R+\left(R\ddot{\theta }+2\dot{R}\dot{\theta }\right)e_{\theta }+\ddot{z}k\\ a=\left[\frac{2A}{{\left(t+1\right)}^3}-\frac{A}{\left(t+1\right)}{B}^2\right]e_R+2\left(-\frac{A}{{\left(t+1\right)}^2}\right)B{e}_{\theta }-\frac{2C}{{\left(t+1\right)}^3}k\end{array}$

For $t=\infty$, the velocity is equal to,

$v=0$

For $t=\infty$, the acceleration is equal to,

$a=0$

Conclusion:

At $t=\infty$

The magnitude of velocity is equal to $v=0$

The magnitude of acceleration is equal to $a=0$