Chapter 6, Problem 71P

(a)

To determine

Whether the magnitude of the force is $F_0$ and its direction is perpendicular to $\overrightarrow{r}=x\widehat{i}+y\widehat{j}$ .

Explanation of Solution

Given:

The force acting on the particle is $\left(\frac{F_0}{r}\right)\left(y\widehat{i}-x\widehat{j}\right)$ and $r$ is the distance of the particle from the origin.

Formula used:

The vector notation of vector $\overrightarrow{A}$ is $\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}$ .

Write the expression for magnitude of any vector.

  $\left|\overrightarrow{A}\right|=\sqrt{\left(A_x^2+A_y^2\right)}$

Here, $A_x$ is the $x-$ component of the vector and $A_y$ is the $y-$ component of the vector.

Calculation:

Write the expression for the magnitude of the force.

  $\begin{array}{c}\left|\overrightarrow{F}\right|=\sqrt{{\left(\frac{F_0}{r}y\right)}^2+{\left(\frac{F_0}{r}x\right)}^2}\\ =\left(\frac{F_0}{r}\right)\sqrt{\left(y^2+x^2\right)}\end{array}$

Substitute $r$ for $\sqrt{\left(y^2+x^2\right)}$ in the above expression.

  $\left|\overrightarrow{F}\right|=F_0$

Take the dot product between the force and $\left(x\widehat{i}+y\widehat{j}\right)$ .

  $\begin{array}{c}\overrightarrow{F}.\left(x\widehat{i}+y\widehat{j}\right)=\left(\left(\frac{F_0}{r}\right)\left(y\widehat{i}-x\widehat{j}\right)\right).\left(x\widehat{i}+y\widehat{j}\right)\\ =\left(\frac{F_0}{r}\right)\left(yx-xy\right)\\ =0\end{array}$

Conclusion:

Thus, the magnitude of the force is $F_0$ and its direction is perpendicular to $\overrightarrow{r}=x\widehat{i}+y\widehat{j}$ .

(b)

To determine

The work done by the force on the particle.

Explanation of Solution

Given:

The particle is moving around a circle of radius $5.0\text{m}$ centred at the origin. The force acting on the particle is $\left(\frac{F_0}{r}\right)\left(y\widehat{i}-x\widehat{j}\right)$ .

Formula Used:

Write the expression for the work done by a force on a particle.

  $W={\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\overrightarrow{F}.d\overrightarrow{r}}$   ........ (1)

Here, $W$ is the work done, $\overrightarrow{F}$ is the force applied, ${\theta }_1$ is the initial angular displacement of the particle and ${\theta }_2$ is the final angular displacement of the particle.

Calculation:

Write the expression for the instantaneous position of the particle on the circumference of the circle.

  $\begin{array}{c}\overrightarrow{r}=\left(x\text{}\widehat{i}+y\overrightarrow{j}\right)\\ =r\text{}\cos \theta \widehat{i}+r\sin \theta \widehat{j}\end{array}$

Here, $r$ is the magnitude of the radial vector and $\theta$ is the instantaneous value of the angular position of the paticle.

  $d\overrightarrow{r}=\left(-r\sin \theta \widehat{i}+r\cos \theta \widehat{j}\right)d\theta$

Rearrange the expression for force.

  $\overrightarrow{F}=\left(\frac{F_0}{r}\right)\left(r\sin \theta \widehat{i}-r\cos \text{}\theta \widehat{j}\right)$

Substitute $\left(\frac{F_0}{r}\right)\left(r\sin \theta \widehat{i}-r\cos \text{}\theta \widehat{j}\right)$ for $\overrightarrow{F}$ and $\left(-r\sin \theta \widehat{i}+r\cos \theta \widehat{j}\right)d\theta$ for $\overrightarrow{r}$ , 0 for ${\theta }_1$ and $2\pi$ for ${\theta }_2$ in expression (1).

  $\begin{array}{c}W={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(\frac{F_0}{r}\right)\left(rSin\theta \widehat{i}-r\mathrm{Cos}\text{}\theta \widehat{j}\right).}\left(-rSin\theta \widehat{i}+rCos\theta \widehat{j}\right)d\theta \\ =-F_{0}r{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\theta }\\ =-2\pi F_{0}r\end{array}$

Substitute $5.0\text{m}$ for $r$ in the above expression.

  $W=-\left(10\pi \right)F_0$

Conclusion:

Thus, the magnitude of the work done by the force on the particle is $\left(10\pi \right)F_0$ .