Chapter 11.5, Problem 11.161P

To determine

(a)

Velocity and acceleration of robot tip A.

Answer to Problem 11.161P

We got,

Final velocity, $v=3.927\text{ ft/s}$

Final acceleration, $a=5.994{\text{ ft/s}}^{\text{2}}$

Explanation of Solution

Given information:

Time $t=3\text{ s}$

$r=3+0.5\cos (4\theta )$

$\theta =-\frac{\pi }{4}{t}^2+\pi t$

Concept used:

Velocity $v_r=\dot{r},v_{\theta }=r\dot{\theta }$

Acceleration $a_r=\ddot{r}-r{\dot{\theta }}^2,a_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }$

Calculation:

Velocity,

$\begin{array}{l}{v}_r=\dot{r}\\ v_r=\frac{dr}{dt}=\frac{dr}{d\theta }\times \frac{d\theta }{dt}\\ v_r=\frac{d}{d\theta }\left(3+0.5\cos (4\theta )\right)\times \frac{d}{dt}\left(-\frac{\pi }{4}{t}^2+\pi t\right)\end{array}$

$\begin{array}{l}{v}_r=\left(-2\sin (4\theta )\right)\times \left(-\frac{\pi }{2}t+\pi \right)\\ v_r=\left(-2\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times \left(-\frac{\pi }{2}t+\pi \right)\\ \text{Put t}=3\text{ s}\\ v_r=\left(-2\sin \left(4\left(-\frac{\pi }{4}{3}^2+3\pi \right)\right)\right)\times \left(-\frac{3\pi }{2}+\pi \right)\\ v_r=0\end{array}$

And,

$\begin{array}{l}{v}_{\theta }=r\dot{\theta }\\ v_{\theta }=r\times \frac{d\theta }{dt}\\ v_{\theta }=\left(3+0.5\cos (4\theta )\right)\times \frac{d}{dt}\left(-\frac{\pi }{4}{t}^2+\pi t\right)\end{array}$

$\begin{array}{l}{v}_{\theta }=\left(3+0.5\cos (4\theta )\right)\times \left(-\frac{\pi }{2}t+\pi \right)\\ v_{\theta }=\left(3+0.5\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times \left(-\frac{\pi }{2}t+\pi \right)\\ \text{Put t}=3\text{ s}\\ v_{\theta }=\left(3+0.5\cos \left(4\left(-\frac{\pi }{4}{3}^2+3\pi \right)\right)\right)\times \left(-\frac{3\pi }{2}+\pi \right)\\ v_{\theta }=-3.927\end{array}$

Final velocity,

$\begin{array}{l}v=\sqrt{v_r{}^2+v_{\theta }{}^2}\\ v=\sqrt{0^2+{(-3.927)}^2}\\ v=3.927\end{array}$

Radial acceleration,

$\begin{array}{l}{a}_r=\ddot{r}-r{\dot{\theta }}^2\\ a_r=\frac{d}{dt}\dot{r}-r{\dot{\theta }}^2\end{array}$

$a_r={\left\{\begin{array}{l}\frac{d}{dt}\left[\left(-2\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times \left(-\frac{\pi }{2}t+\pi \right)\right]\\ -\left[\left(3+0.5\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times {\left(-\frac{\pi }{2}t+\pi \right)}^2\right]\end{array}\right\}}_{t=3}$

$a_r={\left\{\begin{array}{l}\left[\left(-2\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times \frac{d}{dt}\left(-\frac{\pi }{2}t+\pi \right)\right]\\ +\left[\frac{d}{dt}\left(-2\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times \left(-\frac{\pi }{2}t+\pi \right)\right]\\ -\left[\left(3+0.5\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times {\left(-\frac{\pi }{2}t+\pi \right)}^2\right]\end{array}\right\}}_{t=3}$

$a_r={\left\{\begin{array}{l}\left[\pi \cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right]+\left[\left(-8\sin \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times {\left(-\frac{\pi }{2}t+\pi \right)}^2\right]\\ -\left[\left(3+0.5\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times {\left(-\frac{\pi }{2}t+\pi \right)}^2\right]\end{array}\right\}}_{t=3}$

$a_r=\left\{\begin{array}{l}\left[\pi \cos \left(4\left(-\frac{\pi }{4}{3}^2+3\pi \right)\right)\right]+\left[\left(-8\sin \left(4\left(-\frac{\pi }{4}{3}^2+3\pi \right)\right)\right)\times {\left(-\frac{\pi }{2}3+\pi \right)}^2\right]\\ -\left[\left(3+0.5\cos \left(4\left(-\frac{\pi }{4}{3}^2+3\pi \right)\right)\right)\times {\left(-\frac{\pi }{2}3+\pi \right)}^2\right]\end{array}\right\}$

$a_r=0.78{\text{ ft/s}}^{\text{2}}$

Transverse acceleration,

$a_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }$

$\begin{array}{l}{a}_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }\\ a_{\theta }={\left[\begin{array}{l}\left\{\left(3+0.5\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times \left(-\frac{\pi }{2}\right)\right\}\\ +2\left\{\left(-2\cos \left(4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right)\right)\times \left(-\frac{\pi }{2}t+\pi \right)\times \left(-\frac{\pi }{2}t+\pi \right)\right\}\end{array}\right]}_{t=3}\\ a_{\theta }={\left[\begin{array}{l}\left\{\left(3+0.5\cos \left(4\left(-\frac{\pi }{4}{3}^2+3\pi \right)\right)\right)\times \left(-\frac{\pi }{2}\right)\right\}\\ +2\left\{\left(-2\cos \left(4\left(-\frac{\pi }{4}{3}^2+3\pi \right)\right)\right)\times \left(-\frac{\pi }{2}3+\pi \right)\times \left(-\frac{\pi }{2}3+\pi \right)\right\}\end{array}\right]}_{t=3}\\ a_{\theta }=5.9426{\text{ ft/s}}^{\text{2}}\end{array}$

Final acceleration,

$\begin{array}{l}a=\sqrt{a_r{}^2+a_{\theta }{}^2}\\ a=\sqrt{{0.78}^2+{(5.9426)}^2}\\ a=5.994{\text{ ft/s}}^{\text{2}}\end{array}$

Conclusion:

We got,

Final velocity, $v=3.927\text{ ft/s}$

Final acceleration, $a=5.994{\text{ ft/s}}^{\text{2}}$

To determine

(b)

To plot: the path of tip.

Explanation of Solution

Given information:

Time $0\le t\le 4\text{ s}$

$r=3+0.5\cos (4\theta )$

$\theta =-\frac{\pi }{4}{t}^2+\pi t$

Concept used:

MS Excel shall be used to plot the path.

Calculation:

We know

$\begin{array}{l}r=3+0.5\cos (4\theta )\\ r=3+0.5\cos \left\{4\left(-\frac{\pi }{4}{t}^2+\pi t\right)\right\}\end{array}$

Following table is being made,

Conclusion:

Plot has been shown above.