A thin rod with mass$\mathbf{M}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$ M=is bent in a semicircle of radius$\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{650}\mathbf{}\mathbf{m}$. (Fig. 13-56). (a) What is its gravitational force (both magnitude and direction on a particle with mass$\mathbf{m}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{kg}$at P, the center of curvature? (b) What would be the force on the particle the rod were a complete circle?

The mass of the rod $\mathrm{M}=5.00\mathrm{kg}$.

The mass of the particle$\mathrm{m}=3.00\times {10}^{-3}\mathrm{kg}$and is placed at P, the centre of the semicircle.

The radius of the semicircle $\mathrm{R}=0.650\mathrm{m}$.

We can use Newton's law of gravitation and the concept of integration to find the gravitational force.

Formula:

$\mathrm{F}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$

To determine the force exerted by the rod, we will have to calculate its mass per unit length since the rod is in the form of a semicircle.

Mass per unit length$\mathrm{\mu }=\frac{\mathrm{M}}{\mathrm{L}}$, where L = length of wire =$\mathrm{\pi R}$

Now, consider a small section dl of the rod at an angle θ.

The mass of this section is$\mathrm{\mu dl}$.

The force exerted by the section dl on the particle at P is given by .

The total force by all such elements will be directed radially. Hence the net force will be sum of the vertical (sin) components of these forces. All the horizontal (cos) components get cancelled in pairs as they will be directed opposite to each other.

Hence, the net force on particle at P due to semicircular wire is

$\begin{array}{rcl}\mathrm{F}& =& \int F^{\text{'}}\mathrm{sin\theta }\\ & =& \int \frac{\mathrm{GM\mu dlsin\theta }}{{\mathrm{R}}^2}\end{array}$

From the figure, we have$\mathrm{dl}=\mathrm{Rd\theta }$.

Hence we write

$\begin{array}{rcl}\mathrm{F}& =& \int \frac{\mathrm{GM\mu Rd\theta sin\theta }}{{\mathrm{R}}^2}\\ & =& \frac{\mathrm{GmM}}{{\mathrm{\pi R}}^2}{\int }_0^{\mathrm{\pi }}\mathrm{sin\theta }d\mathrm{\theta }\end{array}$

On integrating, we get

$\begin{array}{rcl}\mathrm{F}& =& \frac{2\mathrm{GmM}}{{\mathrm{\pi R}}^2}\\ & =& \frac{2\times 5\times 3\times {10}^{-3}\times 6.67\times {10}^{-11}}{3.14\times {\left(0.650\right)}^2}\\ & =& 1.51\times {10}^{-12}\mathrm{N}\end{array}$

This force will be directed upwards along the + Y axis

If the rod is now made in the form of a complete circle, the net force acting on the particle at P will be zero. As we have seen in part (a), all the horizontal components get cancelled in pairs. So, all the vertical components will also get cancelled.