A thin homogeneous wire of mass m_1 = 0.1 kg is bent into a…

A thin homogeneous wire of mass $m_1 = 0.1\ \mathrm{kg}$ is bent into a semicircle of radius $r = 0.25\ \mathrm{m}$. A small sphere of mass $m_2 = 0.15\ \mathrm{kg}$ is placed at the centre of the semicircle. Determine the gravitational force between the sphere and the wire.

Step 1: Idea

The wire is not a point, so Newton's law cannot be applied to it directly. We cut the wire into tiny pieces, each small enough to count as a point mass, find the force of each piece on the sphere and add all these forces as vectors. A useful simplification: every piece is at the same distance $r$ from the centre.

Step 2: Force from one piece

The wire has length $\pi r$, so its mass per unit length is $\lambda = m_1/(\pi r)$. Describe the position of a piece by the angle $\theta$ measured from the symmetry axis. A piece spanning the angle $d\theta$ has length $r\,d\theta$ and mass $dm = \lambda r\,d\theta$. It pulls the sphere toward itself with

\[ dF = \gamma\frac{m_2\,dm}{r^2} = \gamma\frac{m_2\lambda}{r}\,d\theta. \]

Step 3: Adding the components

Split each $dF$ into a component along the symmetry axis, $dF\cos\theta$, and one across it, $dF\sin\theta$. For every piece at angle $\theta$ there is a mirror piece at $-\theta$ whose cross component is equal and opposite, so all the cross components cancel. Only the components along the axis remain, and they all point the same way:

\[ F = \int_{-\pi/2}^{\pi/2}\gamma\frac{m_2\lambda}{r}\cos\theta\,d\theta = \gamma\frac{m_2\lambda}{r}\Big[\sin\theta\Big]_{-\pi/2}^{\pi/2}, \]

\[ F = \frac{2\gamma m_2\lambda}{r}. \]

Inserting $\lambda = m_1/(\pi r)$: $F = \frac{2\gamma m_1m_2}{\pi r^2}$.

Step 4: Numbers

\[ F = \frac{2 \cdot 6.67 \cdot 10^{-11} \cdot 0.1 \cdot 0.15}{\pi \cdot 0.0625}, \]

\[ F = \frac{2.001 \cdot 10^{-12}}{0.19635}. \]

\[F \approx 1.02 \cdot 10^{-11}\ \mathrm{N}\]

The force acts along the symmetry axis, pulling the sphere toward the middle of the arc.

Check

If the whole wire were at one point at distance $r$, the force would be $ \gamma m_1m_2/r^2 $ $= 1.6 \cdot 10^{-11}\ \mathrm{N} $. Spreading the mass round the semicircle reduces it by the factor $2/\pi \approx 0.64$, because the pieces pull partly sideways $\checkmark$