Chapter 11.5, Problem 11.162P

The path of a particle P is a limacon. The motion of the particle is defined by the relations $r=b\left(2+\cos \pi t\right)$ and $\theta =\pi t$ where t and $\theta$ are expressed in seconds and radians, respectively. Determine (a) the velocity and the acceleration of the particle when $t=2s$ , (b) the value of $\theta$ for which the magnitude of the velocity is maximum.

To determine

(a)

The velocity and the acceleration of the particle at $t=2s$.

Answer to Problem 11.162P

Velocity of the particle: $v=3b\pi e_{\theta }$

Acceleration of the particle: $a=-4b{\pi }^2{e}_r$

Explanation of Solution

Given information:

The path of the particle is a limacon.

The motion of the particle is defined by the relation: $r=b\left(2+\cos \pi t\right)\text{ and }\theta =\pi t$.

Calculations:

From the given relation for the distance of the particle: $r=b\left(2+\cos \pi t\right)$

Taking the time derivative:

$\begin{array}{l}\dot{r}=-b\pi \sin \pi t\mathrm{.........}\left(1\right)\\ and,\\ \ddot{r}=-b{\pi }^2\cos \pi t\mathrm{................}\left(2\right)\end{array}$

And from another given relation for the angle of the particle: $\theta =\pi t$

The time derivatives;

$\begin{array}{l}\dot{\theta }=\pi \mathrm{..................}\left(3\right)\\ and,\\ \ddot{\theta }=0\mathrm{.................}\left(4\right)\end{array}$

Hence, from the relation for the velocity of particle in polar coordinates:

$\begin{array}{l}v=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }\\ \left(\text{where }{e}_r\text{ and }{e}_{\theta }\text{ are unit vectors in radial and transverse directions}\text{.}\right)\\ \text{Substituting from eq}\text{.}\left(\text{1}\right)\text{ and }\left(\text{3}\right):\\ v=\left(-b\pi \sin \pi t\right)e_r+\left(b\left(2+\cos \pi t\right)\pi \right)e_{\theta }\\ at,t=2s\\ v=\left(-b\pi \sin 2\pi \right)e_r+\left(b\left(2+\cos 2\pi \right)\pi \right)e_{\theta }\\ \left\{\sin ce,\sin 2\pi =0and\cos 2\pi =1\right\}\\ \therefore v=\left(-b\pi \times 0\right)e_r+\left(b\left(2+1\right)\pi \right)e_{\theta }\\ or,\\ \Rightarrow v=3b\pi e_{\theta }\end{array}$

Now, using the relation for the acceleration of the particle in polar coordinates:

$\begin{array}{l}a=\left(\ddot{r}-e{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }\\ \text{Substituting values from eq}\text{. }\left(\text{1}\right)\text{, }\left(\text{2}\right)\text{, }\left(\text{3}\right)\text{ and }\left(\text{4}\right):\\ a=\left(-b{\pi }^2\cos \pi t-b\left(2+\cos \pi t\right){\pi }^2\right)e_r+\left(\left[b\left(2+\cos \pi t\right)\times 0\right]+\left[2\left(-b\pi \sin \pi t\right)\times \pi \right]\right)e_{\theta }\\ or,\\ a=\left(-2b{\pi }^2\cos \pi t-2b{\pi }^2\right)e_r+\left(-2b{\pi }^2\sin \pi t\right)e_{\theta }\\ At,t=2s\\ a=\left(-2b{\pi }^2\cos 2\pi -2b{\pi }^2\right)e_r+\left(-2b{\pi }^2\sin 2\pi \right)e_{\theta }\\ \left\{\sin ce,\sin 2\pi =0and\cos 2\pi =1\right\}\\ \therefore a=\left(-4b{\pi }^2\right)e_r+0\\ \Rightarrow a=-4b{\pi }^2{e}_r\end{array}$

Conclusion:

The velocity of the particle at $t=2s$ is $v=3b\pi e_{\theta }$ and the acceleration is $a=-4b{\pi }^2{e}_r$.

To determine

(b)

The value of θ for which the magnitude of the velocity is maximum.

Answer to Problem 11.162P

The value of θ : $\theta =2n\pi ,\text{ For }n=0,1,\mathrm{2......}$

Explanation of Solution

Given information:

The path of the particle is a limacon.

The motion of the particle is defined by the relation: $r=b\left(2+\cos \pi t\right)\text{ and }\theta =\pi t$.

Calculations:

The magnitude of the velocity is calculated as:

$\begin{array}{l}v=\sqrt{v_r^2+v_{\theta }^2}\\ where,v_r=\dot{r}\text{ and }{v}_{\theta }=r\dot{\theta }\\ \therefore v_r=-b\pi \left(\sin \pi t\right)\\ v_{\theta }=\pi b\left(2+\cos \pi t\right)\\ substitut\text{i}\text{n}g;\\ v=\sqrt{{\left(-b\pi \left(\sin \pi t\right)\right)}^2+{\left(\pi b\left(2+\cos \pi t\right)\right)}^2}\\ \text{To find the maximum magnitude of velocity, set }\frac{d}{dt}\left(v\right)=0\\ \frac{d}{dt}\sqrt{{\left(-b\pi \left(\sin \pi t\right)\right)}^2+{\left(\pi b\left(2+\cos \pi t\right)\right)}^2}=0\\ \frac{1}{2\sqrt{{\left(-b\pi \left(\sin \pi t\right)\right)}^2+{\left(\pi b\left(2+\cos \pi t\right)\right)}^2}}\times \frac{d}{dt}\left[{\left(-b\pi \left(\sin \pi t\right)\right)}^2+{\left(\pi b\left(2+\cos \pi t\right)\right)}^2\right]=0\\ or,\\ \frac{d}{dt}\left[{\left(-b\pi \left(\sin \pi t\right)\right)}^2+{\left(\pi b\left(2+\cos \pi t\right)\right)}^2\right]=0\\ \frac{d}{dt}\left(b^2{\pi }^2{\sin }^2\pi t+b^2{\pi }^2\left(4+{\cos }^2\pi t+\left(2\times \cos \pi t\times 2\right)\right)\right)=0\end{array}$

$\begin{array}{l}\frac{d}{dt}\left(b^2{\pi }^2\left({\sin }^2\pi t+{\cos }^2\pi t\right)+4b^2{\pi }^2+4b^2{\pi }^2\cos \pi t\right)=0\\ or,\\ \frac{d}{dt}\left(5b^2{\pi }^2+4b^2{\pi }^2\cos \pi t\right)=0\\ 0+4b^2{\pi }^2\left(-\sin \pi t\right)=0\\ or,\\ \sin \pi t=0\\ \sin ce,\theta =\pi t\\ \therefore \sin \theta =0\\ \text{The equation is satisfied for:}\\ \Rightarrow \theta =2n\pi ,where,n=0,1,2,\mathrm{........}\end{array}$

Conclusion:

The value of θ for which the magnitude of the velocity is maximum is 2np, where, $n=\left(0,1,2,\mathrm{........}\right)$.