Chapter 5, Problem 130P

(a)

To determine

To Show:

Path of the particle is circle of radius R.

Explanation of Solution

Given:

Mass of a particle $\text{= }m\text{ = 0}\text{.80 kg}$

Position of a particle is given by $\overrightarrow{r}=x\widehat{i}+y\widehat{j}=\left(R\sin \omega t\right)\widehat{i}+\left(R\cos \omega t\right)\widehat{j}$

Radius of circle $=4.0\text{m}$

  $\omega =2\pi {\text{ s}}^{-1}$

Calculation:

To show that particle is moving in a circle take a dot product of $\overrightarrow{r}$

  $\begin{array}{l}\overrightarrow{r}\cdot \overrightarrow{r}=\left(x\widehat{i}+y\widehat{j}\right)\cdot \left(x\widehat{i}+y\widehat{j}\right)\\ \therefore r^2=x^2+y^2={\left(R\sin \omega t\right)}^2+{\left(R\cos \omega t\right)}^2\\ \therefore x^2+y^2=R^2\left({\sin }^2\omega t+{\cos }^2\omega t\right)\\ =x^2+y^2=R^2\end{array}$

  $$

It is the equation of the circle with radius R and center at the origin. It is given, R $=4.0\text{m}$ hence equation of the circle is

  $=x^2+y^2=16$

Conclusion:

It is shown that the path of this particle is the circle of radius R and center at the origin.

(b)

To determine

To Compute:

The velocity vectors.

To Show

: $\frac{v_x}{v_y}=-\frac{y}{x}$

Explanation of Solution

Given:

  $\begin{array}{l}x=R\sin \left(\omega t\right)\\ y=R\cos \left(\omega t\right)\end{array}$

Mass of a particle $\text{= }m\text{ = 0}\text{.80 kg}$

Position of a particle is given by $\overrightarrow{r}=x\widehat{i}+y\widehat{j}=\left(R\sin \omega t\right)\widehat{i}+\left(R\cos \omega t\right)\widehat{j}$

Radius of circle $=4.0\text{m}$

  $\omega =2\pi {\text{ s}}^{-1}$

Calculation:

  $v=\frac{dx}{dt}$

  $\begin{array}{l}\therefore v_x=\frac{dx}{dt}=\frac{d\left(R\sin \left(\omega t\right)\right)}{dt}=\omega R\cos \left(\omega t\right)=8\pi \cos \left(2\pi t\right)\\ v_y=\frac{dy}{dt}=\frac{d\left(R\cos \left(\omega t\right)\right)}{dt}=-\omega R\sin \left(\omega t\right)=-8\pi \sin \left(2\pi t\right)\end{array}$

Hence,

  $\begin{array}{l}\frac{v_x}{v_y}=\frac{\omega R\sin \left(\omega t\right)}{-\omega R\cos \left(\omega t\right)}\\ \frac{v_x}{v_y}=-\frac{y}{x}\end{array}$

Conclusion:

It is shown that $\frac{v_x}{v_y}=-\frac{y}{x}$ .

(c)

To determine

To compute:

The acceleration vector

To Show:

The acceleration is directed towards the origin and has magnitude $\frac{v^2}{R}$ .

Explanation of Solution

Given:

  $\begin{array}{l}{v}_x=\omega R\cos \left(\omega t\right)\\ v_y=-\omega R\sin \left(\omega t\right)\end{array}$

Calculation:

  $a=\frac{dv}{dt}$

  $\begin{array}{l}\therefore a_x=\frac{d{v}_x}{dt}=\frac{d\left(\omega R\cos \left(\omega t\right)\right)}{dt}=-{\omega }^{2}R\sin \left(\omega t\right)=-\left(16{\pi }^2\sin \left(2\pi t\right)\right)\\ a_y=\frac{d{v}_y}{dt}=\frac{d\left(-\omega R\sin \left(\omega t\right)\right)}{dt}=-{\omega }^{2}R\cos \left(\omega t\right)=-\left(16{\pi }^2\cos \left(2\pi t\right)\right)\end{array}$

  $\begin{array}{l}\overrightarrow{a}=a_x\widehat{i}+a_y\widehat{j}=-{\omega }^{2}R\sin \left(\omega t\right)\widehat{i}-{\omega }^{2}R\cos \left(\omega t\right)\widehat{j}\\ \therefore \overrightarrow{a}=-{\omega }^2\left(R\sin \left(\omega t\right)\widehat{i}+R\cos \left(\omega t\right)\widehat{j}\right)\\ \therefore \overrightarrow{a}=-{\omega }^2\left(\widehat{r}\right)={\omega }^2\left(-\widehat{r}\right)=4{\pi }^2\left(-\widehat{r}\right)\end{array}$

Hence,acceleration is directed towards the center.

Magnitude of acceleration is

  $\begin{array}{l}\overrightarrow{a}=a_x\widehat{i}+a_y\widehat{j}=-{\omega }^{2}R\sin \left(\omega t\right)\widehat{i}-{\omega }^{2}R\cos \left(\omega t\right)\widehat{j}\\ \left|\overrightarrow{a}\right|=\sqrt{{\left(-{\omega }^{2}R\sin \left(\omega t\right)\right)}^2+{\left(-{\omega }^{2}R\cos \left(\omega t\right)\right)}^2}\\ \left|\overrightarrow{a}\right|=\sqrt{{\omega }^4{R}^2\left(\sin {\left(\omega t\right)}^2+\cos {\left(\omega t\right)}^2\right)}\\ \left|\overrightarrow{a}\right|=\sqrt{{\omega }^4{R}^2}=R{\omega }^2=16{\pi }^2\end{array}$

But $v=R\omega$ hence,

  $\begin{array}{l}\left|\overrightarrow{a}\right|=R{\left(\frac{v}{R}\right)}^2\\ \left|\overrightarrow{a}\right|=\frac{v^2}{R}\end{array}$

Conclusion:

It is shown that the acceleration vector is directed towards the origin and has magnitude $\frac{v^2}{R}=16{\pi }^2$ .

(d)

To determine

To Find:

The magnitude and direction of net force acting on the particle.

Answer to Problem 130P

Magnitude of net force acting on the particle is $F=\frac{m{v}^2}{r}$ and its direction is towards the center.

Explanation of Solution

Given:

Mass of a particle $\text{= }m\text{ = 0}\text{.80 kg}$

Radius of circle $R=4.0\text{m}$

  $\omega =2\pi {\text{ s}}^{-1}$

Formula Used:

Centripetal force:

  $\begin{array}{l}F=m\frac{v^2}{r}\\ \omega =\frac{v}{r}\end{array}$

Calculation:

The only force acting on the particle in a circular motion is the centripetal force. Hence the magnitude of the total force on the particle is

  $\begin{array}{l}F=ma\\ F=m\left(\frac{v^2}{r}\right)=\frac{m{v}^2}{r}=\left(0.80\right)\left(16{\pi }^2\right)\\ F=12.8{\pi }^2=126.2\text{N}\end{array}$

The direction of force is equal to the direction of centripetal acceleration i.e. towards the center, along the radius.

Conclusion:

The magnitude of the net force acting on the particle is $126.2\text{N}$ and its direction is towards the center.