Chapter 11.4, Problem 11.93P

(a)

To determine

The position, velocity, and acceleration of the particle at $t=0$.

Answer to Problem 11.93P

The position (r), velocity (v), and acceleration of the particle (a) at $t=0$ is

$\underset{\_}{20\text{mm}(\uparrow )}$, $\underset{\_}{43.4\text{mm/s}\left(46.3°-\text{IVQuad}\right)}$, and $\underset{\_}{743{\text{mm/s}}^2\left(85.4°-\text{IIIQuad}\right)}$ respectively.

Explanation of Solution

Given information:

The given position vector of vibrating particle (r) is $x_1\left[1-\frac{1}{t+1}\right]\text{i}+\left[y_1{e}^{\frac{-\pi t}{2}}\cos 2\pi t\right]\text{j}$.

The value of $x_1$ is 30 mm.

The value of $y_1$ is 20 mm.

Calculation:

Determine the position vector of the particle using the relation.

$\text{r}=x_1\left[1-\frac{1}{t+1}\right]\text{i}+\left[y_1{e}^{\frac{-\pi t}{2}}\cos 2\pi t\right]\text{j}$

Substitute 30 mm for $x_1$ and 20 mm for $y_1$.

$\text{r}=30\left[1-\frac{1}{t+1}\right]\text{i}+\left[20e^{\frac{-\pi t}{2}}\cos 2\pi t\right]\text{j}$ (1)

Determine the velocity value $\left(\text{v}\right)$.

Differentiate the Equation (1) with respect to t.

$\begin{array}{c}\text{v}=\frac{d\text{r}}{dt}\\ =\frac{d}{dt}\left(30\left[1-\frac{1}{t+1}\right]\text{i}+\left[20e^{\frac{-\pi t}{2}}\cos 2\pi t\right]\text{j}\right)\\ =30\left[\frac{1}{{\left(t+1\right)}^2}\right]\text{i}-20\pi \left[e^{\frac{-\pi t}{2}}\left(\frac{1}{2}\cos 2\pi t+2\sin 2\pi t\right)\right]\text{j}\end{array}$ (2)

Determine the acceleration value (a).

Differentiate the Equation (2) with respect to t.

$\begin{array}{c}\text{a}=\frac{d\text{v}}{dt}\\ =\frac{d}{dt}\left(30\left[\frac{1}{{\left(t+1\right)}^2}\right]\text{i}-20\pi \left[e^{\frac{-\pi t}{2}}\left(\frac{1}{2}\cos 2\pi t+2\sin 2\pi t\right)\right]\text{j}\right)\\ =30\left[\frac{-2}{{\left(t+1\right)}^3}\right]\text{i}-\left\{20\pi \left[\begin{array}{l}-\frac{\pi }{2}{e}^{\frac{-\pi t}{2}}\left(\frac{1}{2}\cos 2\pi t+2\sin 2\pi t\right)+\\ e^{\frac{-\pi t}{2}}\left(-\pi \sin 2\pi t+4\cos 2\pi t\right)\end{array}\right]\text{j}\right\}\\ =-\frac{60}{{\left(t+1\right)}^3}\text{i}+10{\pi }^2{e}^{\frac{-\pi t}{2}}\left(0.5\cos 2\pi t+4\cos 2\pi t-8\cos 2\pi t\right)\text{j}\end{array}$

$\text{a}=-\frac{60}{{\left(t+1\right)}^3}\text{i}+10{\pi }^2{e}^{\frac{-\pi t}{2}}\left(4\sin 2\pi t-7.5\cos 2\pi t\right)\text{j}$ (3)

Determine the position of the particle at $t=0$.

Substitute 0 for t in Equation (1).

$\begin{array}{c}\text{r}=30\left[1-\frac{1}{0+1}\right]\text{i}+\left[20e^{\frac{-\pi \left(0\right)}{2}}\cos 2\pi \left(0\right)\right]\text{j}\\ =\left(\text{30}\text{mm}\right)\text{i}+\left(20\text{mm}\right)\text{j}\end{array}$

Determine the velocity of the particle at $t=0$.

Substitute 0 for t in Equation (2).

$\begin{array}{c}\text{v}=30\left[\frac{1}{{\left(0+1\right)}^2}\right]\text{i}-20\pi \left[e^{\frac{-\pi \left(0\right)}{2}}\left(\frac{1}{2}\cos 2\pi \left(0\right)+2\sin 2\pi \left(0\right)\right)\right]\text{j}\\ =\left(30\text{mm/s}\right)\text{i}-\left(31.42\text{mm/s}\right)\end{array}$

Determine the magnitude of velocity of the particle at $t=0$.

$\text{v}=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}$

Substitute 30 mm/s for $v_x$ and 31.42 mm/s for $v_y$.

$\begin{array}{c}\text{v}=\sqrt{{30}^2+{\left(31.42\right)}^2}\\ =43.4\text{mm/s}\end{array}$

Determine the direction of the velocity using the relation.

$\tan \theta =\frac{v_y}{v_x}$

Substitute 31.42 mm/s for $v_y$ and 30 mm/s for $v_x$.

$\begin{array}{l}\tan \theta =\frac{31.42}{30}\\ \theta ={\tan }^{-1}\left(1.047\right)\\ \theta =46.3°\left(\text{IV-Quad}\right)\end{array}$

Determine the acceleration of the particle at $t=0$.

Substitute 0 for t in Equation (3).

$\begin{array}{c}\text{a}=-\frac{60}{{\left(0+1\right)}^3}\text{i}+10{\pi }^2{e}^{\frac{-\pi \left(0\right)}{2}}\left(4\sin 2\pi \left(0\right)-7.5\cos 2\pi \left(0\right)\right)\text{j}\\ =-\left(60{\text{mm/s}}^2\right)\text{i}+\left(740.22{\text{mm/s}}^2\right)\text{j}\end{array}$

Determine the magnitude of acceleration of the particle at $t=0$.

$\text{a}=\sqrt{{\left(a_x\right)}^2+{\left(a_y\right)}^2}$

Substitute $-\left(60{\text{mm/s}}^2\right)$ for $a_x$ and $\left(740.22{\text{mm/s}}^2\right)$ for $a_y$.

$\begin{array}{c}\text{a}=\sqrt{{\left(-60\right)}^2+{\left(-740.22\right)}^2}\\ =743{\text{mm/s}}^2\end{array}$

Determine the direction of the velocity using the relation.

$\tan \theta =\frac{a_y}{a_x}$

Substitute $\left(740.22{\text{mm/s}}^2\right)$ for $a_y$ and $\left(60{\text{mm/s}}^2\right)$ for $a_x$.

$\begin{array}{l}\tan \theta =\frac{740.22}{60}\\ \theta ={\tan }^{-1}\left(12.34\right)\\ \theta =85.4°\left(\text{III-Quad}\right)\end{array}$

Therefore, the position (r), velocity (v), and acceleration of the particle (a) at $t=0$ is

$\underset{\_}{20\text{mm}(\uparrow )}$, $\underset{\_}{43.4\text{mm/s}\left(46.3°-\text{IVQuad}\right)}$, and $\underset{\_}{743{\text{mm/s}}^2\left(85.4°-\text{IIIQuad}\right)}$ respectively.

(b)

To determine

The position, velocity, and acceleration of the particle at $t=1.5\text{s}$.

Answer to Problem 11.93P

The position (r), velocity (v), and acceleration of the particle (a) at $1.5\text{s}$ is

$\underset{\_}{18.10\text{mm}\left(6.01°-\text{IVQuad}\right)}$, $\underset{\_}{5.63\text{mm/s}\left(31.5°-\text{IQuad}\right)}$, and $\underset{\_}{69.3{\text{mm/s}}^2\left(86.8°-\text{IIQuad}\right)}$ respectively.

Explanation of Solution

Given information:

The given position vector of vibrating particle (r) is $x_1\left[1-\frac{1}{t+1}\right]\text{i}+\left[y_1{e}^{\frac{-\pi t}{2}}\cos 2\pi t\right]\text{j}$.

The value of $x_1$ is 30 mm.

The value of $y_1$ is 20 mm.

Calculation:

Determine the position of the particle at $t=1.5\text{s}$.

Substitute 1.5 s for t in Equation (1).

$\begin{array}{c}\text{r}=30\left[1-\frac{1}{1.5+1}\right]\text{i}+\left[20e^{\frac{-\pi \left(1.5\right)}{2}}\cos 2\pi \left(1.5\right)\right]\text{j}\\ =\left(\text{18}\text{mm}\right)\text{i}+\left(1.87\text{mm}\right)\text{j}\end{array}$

Determine the magnitude of position of the particle at $t=1.5\text{s}$.

$\text{r}=\sqrt{{\left(r_x\right)}^2+{\left(r_y\right)}^2}$

Substitute 18 mm for $r_x$ and 1.87 mm for $r_y$.

$\begin{array}{c}\text{r}=\sqrt{{18}^2+{1.87}^2}\\ =18.1\text{mm}\end{array}$

Determine the direction of the position using the relation.

$\tan \theta =\frac{r_y}{r_x}$

Substitute 1.87 mm for $r_y$ and 18 mm for $r_x$.

$\begin{array}{l}\tan \theta =\frac{1.87}{18}\\ \theta ={\tan }^{-1}\left(0.104\right)\\ \theta =6.01°\left(\text{IV-Quad}\right)\end{array}$

Determine the velocity of the particle at $t=1.5\text{s}$.

Substitute 0 for t in Equation (2).

$\begin{array}{c}\text{v}=30\left[\frac{1}{{\left(1.5+1\right)}^2}\right]\text{i}-20\pi \left[e^{\frac{-\pi \left(1.5\right)}{2}}\left(\frac{1}{2}\cos 2\pi \left(1.5\right)+2\sin 2\pi \left(1.5\right)\right)\right]\text{j}\\ =\left(4.8\text{mm/s}\right)\text{i}-\left(2.94\text{mm/s}\right)\text{j}\end{array}$

Determine the magnitude of velocity of the particle at $t=1.5\text{s}$.

$\text{v}=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}$

Substitute 4.8 mm/s for $v_x$ and –2.94 mm/s for $v_y$.

$\begin{array}{c}\text{v}=\sqrt{{4.8}^2+{\left(2.94\right)}^2}\\ =5.63\text{mm/s}\end{array}$

Determine the direction of the velocity using the relation.

$\tan \theta =\frac{v_y}{v_x}$

Substitute 2.94 mm/s for $v_y$ and 4.8 mm/s for $v_x$.

$\begin{array}{l}\tan \theta =\frac{2.94}{4.8}\\ \theta ={\tan }^{-1}\left(0.6119\right)\\ \theta =31.5°\left(\text{I-Quad}\right)\end{array}$

Determine the acceleration of the particle at $t=1.5\text{s}$.

Substitute 1.5 s for t in Equation (3).

$\begin{array}{c}\text{a}=-\frac{60}{{\left(1.5+1\right)}^3}\text{i}+10{\pi }^2{e}^{\frac{-\pi \left(1.5\right)}{2}}\left(4\sin 2\pi \left(1.5\right)-7.5\cos 2\pi \left(1.5\right)\right)\text{j}\\ =-\left(3.84{\text{mm/s}}^2\right)\text{i}+\left(69.21{\text{mm/s}}^2\right)\text{j}\end{array}$

Determine the magnitude of acceleration of the particle at $t=1.5\text{s}$.

$\text{a}=\sqrt{{\left(a_x\right)}^2+{\left(a_y\right)}^2}$

Substitute $-\left(3.84{\text{mm/s}}^2\right)$ for $a_x$ and $\left(69.21{\text{mm/s}}^2\right)$ for $a_y$.

$\begin{array}{c}\text{a}=\sqrt{{3.84}^2+{\left(69.21\right)}^2}\\ =69.3{\text{mm/s}}^2\end{array}$

Determine the direction of the velocity using the relation.

$\tan \theta =\frac{a_y}{a_x}$

Substitute $\left(69.21{\text{mm/s}}^2\right)$ for $a_y$ and $\left(3.84{\text{mm/s}}^2\right)$ for $a_x$.

$\begin{array}{l}\tan \theta =\frac{69.21}{3.84}\\ \theta ={\tan }^{-1}\left(18.02\right)\\ \theta =86.8°\left(\text{II-Quad}\right)\end{array}$

Therefore, the position (r), velocity (v), and acceleration of the particle (a) at $1.5\text{s}$ is

$\underset{\_}{18.10\text{mm}\left(6.01°-\text{IVQuad}\right)}$, $\underset{\_}{5.63\text{mm/s}\left(31.5°-\text{IQuad}\right)}$, and $\underset{\_}{69.3{\text{mm/s}}^2\left(86.8°-\text{IIQuad}\right)}$ respectively.