Chapter 11, Problem 11.192RP

The end point B of a boom is originally 5 m from fixed point A when the driver starts to retract the boom with a constant radial acceleration of r = −1.0 m/s² and lower it with a constant angular acceleration θ = −0.5 rad/ s². At t = 2 s, determine (a) the velocity of point B,(b) the acceleration of point B, (c) the radius of curvature of the path.

Fig. P11.192

To determine

The velocity $\left(v_0\right)$ of the point B.

Answer to Problem 11.192RP

The velocity $\left(v_0\right)$ of the point B is $\underset{\_}{3.61\text{m/s}}$ at an angle $\left(\varphi \right)$ of $\underset{\_}{59°}$.

Explanation of Solution

Given information:

The distance $\left(r_0\right)$ between the point A and B is 5 m.

The boom with a constant radial velocity $\left({\dot{r}}_0\right)$ of $0\text{m/s}$.

The boom with a constant radial acceleration $\left(\ddot{r}\right)$ of $-1.0{\text{m/s}}^{\text{2}}$.

The boom with a constant angular acceleration $\left(\ddot{\theta }\right)$ of $-0.5{\text{rad/s}}^{\text{2}}$.

The time (t) is 2 sec.

Calculation:

Write the expression for rectangular position coordinate (r) of point B using equation of motion:

$r=r_0+{\dot{r}}_{0}t+\frac{1}{2}\ddot{r}{t}^2$ (1)

Here, $r_0$ is initial position of point B in rectangular coordinates and ${\dot{r}}_0$ is initial velocity of point B in rectangular coordinate.

The radial initial velocity of point B is 0.

${\dot{r}}_0=0$

Calculate the radial coordinate (r) of point B:

Substitute 5m for $r_0$, 0 for ${\dot{r}}_0$, $-1\text{m}/{\text{s}}^{\text{2}}$ for $\ddot{r}$, and 2s for t in Equation (1).

$\begin{array}{c}r=5+0+\frac{1}{2}\left(-1\right){\left(2\right)}^2\\ =3\text{m}\end{array}$

Calculate the radial velocity $\left(\dot{r}\right)$ of point B using Equation of motion.

$\dot{r}={\dot{r}}_0+\ddot{r}t$

Substitute 0 for ${\dot{r}}_0$, $-1\text{m}/{\text{s}}^{\text{2}}$ for $\ddot{r}$, and 2s for t.

$\begin{array}{c}\dot{r}=\left(-1\right)\left(2\right)\\ =-2\text{m}/\text{s}\end{array}$

Calculate the angular coordinate $\left(\theta \right)$ of point B using Equation of motion:

$\theta ={\theta }_0+{\dot{\theta }}_{0}t+\frac{1}{2}\ddot{\theta }{t}^2$

Here, ${\theta }_0$ is initial position of point B in angular coordinate, ${\dot{\theta }}_0$ is initial velocity of point B in angular coordinates, and $\ddot{\theta }$ is acceleration of point B in angular coordinate.

Angular coordinate of initial velocity of point B is 0. Thus,

${\dot{\theta }}_0=0$

Substitute $60°$ for ${\theta }_0$, 0 for ${\dot{\theta }}_0$, $-0.5\text{rad}/{\text{s}}^{\text{2}}$ for $\ddot{\theta }$, and 2s for t.

$\begin{array}{c}\theta =\left(60°\times \frac{\pi }{180°}\right)+0+\frac{1}{2}\left(-0.5\right){\left(2\right)}^2\\ =\left(0.047198\text{rad}\right)\left(\frac{180°}{\pi }\right)\\ =2.70°\end{array}$

Calculate the $\left(\dot{\theta }\right)$ of velocity point B in angular coordinate using equation of motion.

$\dot{\theta }={\dot{\theta }}_0+\ddot{\theta }t$

Substitute zero for ${\dot{\theta }}_0$, $-0.5\text{rad}/{\text{s}}^{\text{2}}$ for $\ddot{\theta }$, and 2s for t.

$\begin{array}{c}\dot{\theta }=0+\left(-0.5\right)\left(2\right)\\ =-1\text{rad}/\text{s}\end{array}$

Calculate velocity $\left(v_B\right)$ of point B in polar coordinates.

$v_B=v_r{e}_r+v_{\theta }{e}_{\theta }$ (2)

Here, $v_r$ is magnitude of velocity of point B in radial direction and $v_{\theta }$ is magnitude of velocity of point B in transverse direction.

Rewrite Equation (2) in terms of r, $\dot{r}$, and $\dot{\theta }$.

$v_B=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }$

Substitute $-2\text{m}/\text{s}$ for $\dot{r}$, 3m for r, and $-1\text{rad}/\text{s}$ for $\dot{\theta }$.

$\begin{array}{c}{v}_B=\left(-2\text{m}/\text{s}\right)e_r+\left(3\text{m}\right)\left(-1\text{rad}/\text{s}\right)e_{\theta }\\ =\left(-2\text{m}/\text{s}\right)e_r+\left(-3\text{m}/\text{s}\right)e_{\theta }\end{array}$ (3)

Calculate the magnitude of $\left(v_B\right)$ using dot product of $\left(v_B\right)$:

$v_B=v_B\cdot v_B$

Substitute $\left(-2\text{m}/\text{s}\right)e_r+\left(-3\text{m}/\text{s}\right)e_{\theta }$ for $v_B$.

$\begin{array}{l}{v}_B\cdot v_B=\left[\left(-2\text{m}/\text{s}\right)e_r+\left(-3\text{m}/\text{s}\right)e_{\theta }\right]\cdot \left[\left(-2\text{m}/\text{s}\right)e_r+\left(-3\text{m}/\text{s}\right)e_{\theta }\right]\\ v_B=\sqrt{4+9}\\ v_B=3.6055\text{m}/\text{s}\\ v_B\approx 3.61\text{m}/\text{s}\end{array}$

Calculate unit vector $\left(e_t\right)$ in terms of $v_B$ and $v_B$.

$e_t=\frac{v_B}{v_B}$

Substitute $\left[\left(-2\text{m}/\text{s}\right)e_r+\left(-3\text{m}/\text{s}\right)e_{\theta }\right]$ for $v_B$ and $3.6055\text{m}/\text{s}$ for $v_B$.

$\begin{array}{c}{e}_t=\frac{\left(-2\text{m}/\text{s}\right)e_r+\left(-3\text{m}/\text{s}\right)e_{\theta }}{3.6055}\\ =\frac{\left(-2\text{m}/\text{s}\right)e_r}{3.6055}+\frac{\left(-3\text{m}/\text{s}\right)e_{\theta }}{3.6055}\\ =-0.5547e_r-0.8320e_{\theta }\end{array}$

Calculate angle $\left(\alpha \right)$ between the components of $v_B$.

$\tan \alpha =\frac{v_{\theta }}{v_r}$

Substitute $-2\text{m}/\text{s}$ for $v_r$ and $-3\text{m}/\text{s}$ for $v_{\theta }$.

$\begin{array}{l}\tan \alpha =\frac{-3}{-2}\\ \alpha ={\tan }^{-1}\left(1.5\right)\\ \alpha =56.3099°\\ \alpha \approx 56.31°\end{array}$

Calculate the angle $\left(\varphi \right)$ using the relation:

$\varphi =\theta +\alpha$

Substitute $2.70°$ for $\theta$ and $56.31°$ for $\alpha$.

$\begin{array}{c}\varphi =2.70°+56.31°\\ =59.01°\\ \simeq 59°\end{array}$

Therefore, the velocity $\left(v_0\right)$ of the point B is $\underset{\_}{3.61\text{m/s}}$ at an angle $\left(\varphi \right)$ of $\underset{\_}{59°}$.

To determine

The acceleration $\left(a_B\right)$ of point B.

Answer to Problem 11.192RP

The acceleration $\left(a_B\right)$ of point B is $\underset{\_}{4.72{\text{m/s}}^{\text{2}}}$ at an angle $\left(\phi \right)$ of $\underset{\_}{29.3°}$.

Explanation of Solution

Given information:

The distance $\left(r_0\right)$ between the point A and B is 5 m.

The boom with a constant radial acceleration $\left({\dot{r}}_0\right)$ of $0\text{m/s}$.

The boom with a constant radial acceleration $\left(\ddot{r}\right)$ of $-1.0{\text{m/s}}^{\text{2}}$.

The boom with a constant angular acceleration $\left(\ddot{\theta }\right)$ of $-0.5{\text{rad/s}}^{\text{2}}$.

The time (t) is 2 sec.

Calculation:

Show the values of ${\theta }_0$ and $\theta$ as in Figure (1).

Write acceleration $\left(a_B\right)$ of point B vector in polar coordinates:

$a_B=a_r{e}_r+a_{\theta }{e}_{\theta }$

Here, $a_r$ is magnitude of acceleration of point B in radial direction and $a_{\theta }$ is magnitude of acceleration of point B in transverse direction.

Rewrite the above equation in term of r, $\dot{r}$, $\ddot{r}$, $\dot{\theta }$, and $\ddot{\theta }$.

$a_B=\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }$

Substitute $-1\text{m}/{\text{s}}^{\text{2}}$ for $\ddot{r}$, 3m for r, $-1\text{rad}/\text{s}$ for $\dot{\theta }$, $-0.5\text{rad}/{\text{s}}^{\text{2}}$ for $\ddot{\theta }$, and $-2\text{m}/\text{s}$ for $\dot{r}$.

$\begin{array}{c}{a}_B=\left(-1-\left(3\right){\left(-1\right)}^2\right)e_r+\left(3\left(-0.5\right)+2\left(-2\right)\left(-1\right)\right)e_{\theta }\\ =\left(-4\text{m}/{\text{s}}^{\text{2}}\right)e_r+\left(2.5\text{m}/{\text{s}}^{\text{2}}\right)e_{\theta }\end{array}$

Here, $-4\text{m}/{\text{s}}^{\text{2}}$ is $a_r$ and $2.5\text{m}/{\text{s}}^{\text{2}}$ is $a_{\theta }$.

Calculate the magnitude of $\left(a_B\right)$ using dot product:

$a_B=a_B\cdot a_B$

Substitute $\left(-4\text{m}/{\text{s}}^{\text{2}}\right)e_r+\left(2.5\text{m}/{\text{s}}^{\text{2}}\right)e_{\theta }$ for $a_B$.

$\begin{array}{c}{a}_B=\left[\left(-4\text{m}/{\text{s}}^{\text{2}}\right)e_r+\left(2.5\text{m}/{\text{s}}^{\text{2}}\right)e_{\theta }\right]\cdot \left[\left(-4\text{m}/{\text{s}}^{\text{2}}\right)e_r+\left(2.5\text{m}/{\text{s}}^{\text{2}}\right)e_{\theta }\right]\\ =\sqrt{16+6.25}\\ =4.7169\text{m}/{\text{s}}^{\text{2}}\\ \approx 4.72\text{m}/{\text{s}}^{\text{2}}\end{array}$

Calculate the angle $\left(\beta \right)$ between the components of $a_B$:

$\tan \beta =\frac{a_{\theta }}{a_r}$

Substitute $-4{\text{m}/\text{s}}^2$ for $a_r$ and $2.5\text{m}/{\text{s}}^2$ for $a_{\theta }$.

$\begin{array}{l}\tan \beta =\frac{2.5}{-4}\\ \beta ={\tan }^{-1}\left(0.625\right)\\ \beta =-32°\end{array}$

Calculate the angle $\left(\phi \right)$ using the relation:

$\phi =\theta +\beta$

Substitute $2.70°$ for $\theta$ and $32°$ for $\beta$.

$\begin{array}{c}\phi =2.70°-32°\\ =29.30°\end{array}$

Therefore, the acceleration $\left(a_B\right)$ at point B is $\underset{\_}{4.72{\text{m/s}}^{\text{2}}}$ at an angle $\left(\phi \right)$ of $\underset{\_}{29.3°}$.

To determine

The radius of curvature $\left(\rho \right)$ of the path.

Answer to Problem 11.192RP

The radius of curvature $\left(\rho \right)$ of the path is $\underset{\_}{2.76\text{m}}$.

Explanation of Solution

Given information:

The distance $\left(r_0\right)$ between the point A and B is 5 m.

The boom with a constant radial acceleration $\left({\dot{r}}_0\right)$ of $0\text{m/s}$.

The boom with a constant radial acceleration $\left(\ddot{r}\right)$ of $-1.0{\text{m/s}}^{\text{2}}$.

The boom with a constant angular acceleration $\left(\ddot{\theta }\right)$ of $-0.5{\text{rad/s}}^{\text{2}}$.

The time (t) is 2 sec.

Calculation:

Calculate the tangential component of acceleration $\left(a_t\right)$ vector using unit vector $e_t$:

$a_t=\left(a_B\cdot e_t\right)e_t$

Substitute $\left(-4\text{m}/{\text{s}}^{\text{2}}\right)e_r+\left(2.5\text{m}/{\text{s}}^{\text{2}}\right)e_{\theta }$ for $a_B$ and $-0.5547e_r-0.8320e_{\theta }$ for $e_t$.

$\begin{array}{c}{a}_t=\left[\left(-4e_r+2.5e_{\theta }\right)\cdot \left(-0.5547e_r-0.8320e_{\theta }\right)\right]e_t\\ =\left(2.2188-2.08\right)e_t\\ =\left(0.1388\text{m}/{\text{s}}^{\text{2}}\right)e_t\end{array}$

Write normal component $\left(a_n\right)$ of acceleration vector:

$a_n=a_B-a_t$

Substitute $\left(-4\text{m}/{\text{s}}^{\text{2}}\right)e_r+\left(2.5\text{m}/{\text{s}}^{\text{2}}\right)e_{\theta }$ for $a_B$ and $\left(0.1388\text{m}/{\text{s}}^{\text{2}}\right)e_t$ for $a_t$.

$a_n=\left(-4e_r+2.5e_{\theta }\right)-\left(0.1388\text{m}/{\text{s}}^{\text{2}}\right)e_t$

Substitute $-0.5547e_r-0.8320e_{\theta }$ for $e_t$.

$\begin{array}{c}{a}_n=\left(-4e_r+2.5e_{\theta }\right)-\left(0.1388\right)\left(-0.5547e_r-0.8320e_{\theta }\right)\\ =\left(-4e_r+2.5e_{\theta }\right)-\left(0.1388\right)\left(-0.5547e_r-0.8320e_{\theta }\right)\\ =-3.9231e_r+2.6184e_{\theta }\end{array}$

Calculate the normal acceleration $\left(a_n\right)$ using dot product of $a_n$:

$a_n=a_n\cdot a_n$

Substitute $-3.9231e_r+2.6184e_{\theta }$ for $a_n$.

$\begin{array}{c}{a}_n=\left(-3.9231e_r+2.6184e_{\theta }\right)\cdot \left(-3.9231e_r+2.6184e_{\theta }\right)\\ =\sqrt{22.23103}\\ =4.7149\text{m}/{\text{s}}^{\text{2}}\end{array}$

Calculate the radius of curvature $\left(\rho \right)$ using the normal component of acceleration formula:

$a_n=\frac{v^2}{\rho }$

Rewrite Equation for radius of curvature.

$\rho =\frac{v^2}{a_n}$

Substitute $4.7149\text{m}/{\text{s}}^2$ for $a_n$ and $3.6055\text{m}/\text{s}$ for v.

$\begin{array}{c}\rho =\frac{{\left(3.6055\right)}^2}{4.7149}\\ =2.757\text{m}\\ \approx 2.76\text{m}\end{array}$

Therefore, the radius of curvature $\left(\rho \right)$ of the path is $\underset{\_}{2.76\text{m}}$.