A semicircular wire has a length L and mass M. A particle of mass m is placed at the center of the circle. Find the gravitational attraction on the particle due to the wire.

Explanation

To find the gravitational attraction on a particle of mass \( m \) placed at the center of a semicircular wire of length \( L \) and mass \( M \), we can follow these steps:

### Step 1: Define the Geometry
Assume the semicircular wire is centered at the origin of a coordinate system, with its ends touching the x-axis. The radius \( R \) of the semicircle can be expressed in terms of the length \( L \) as:
\[ R = \frac{L}{\pi} \]

### Step 2: Determine the Incremental Mass
Consider an infinitesimal element of the wire at an angle \( \theta \) with respect to the x-axis. The length of this infinitesimal element is given by:
\[ dL = R \, d\theta \]
The mass of this infinitesimal element \( dm \) can be expressed as:
\[ dm = \frac{M}{L} \cdot dL = \frac{M}{L} \cdot R \, d\theta = \frac{M}{L} \cdot \frac{L}{\pi} \, d\theta = \frac{M}{\pi} \, d\theta \]

### Step 3: Calculate the Gravitational Force by the Element
The gravitational force \( dF \) exerted by the mass element \( dm \) on the particle of mass \( m \) at the center is given by Newton's law of gravitation:
\[ dF = \frac{G m \, dm}{R^2} \]
Substituting \( dm \):
\[ dF = \frac{G m \cdot \frac{M}{\pi} \, d\theta}{\left(\frac{L}{\pi}\right)^2} = \frac{G m M \pi}{L^2} \, d\theta \]

### Step 4: Resolve the Force into Components
The force \( dF \) can be resolved into its x and y components. The x-component is \( dF_x = dF \cos \theta \) and the y-component is \( dF_y = dF \sin \theta \):
\[ dF_x = \frac{G m M \pi}{L^2} \cos \theta \, d\theta \]
\[ dF_y = \frac{G m M \pi}{L^2} \sin \theta \, d\theta \]

### Step 5: Integrate to Find Total Force
To find the total gravitational force, we integrate \( dF_x \) and \( dF_y \) over the limits \( 0 \) to \( \pi \):
\[ F_x = \int_0^\pi dF_x = \frac{G m M \pi}{L^2} \int_0^\pi \cos \theta \, d\theta \]
\[ F_y = \int_0^\pi dF_y = \frac{G m M \pi}{L^2} \int_0^\pi \sin \theta \, d\theta \]

Calculating these integrals:
- The integral of \( \cos \theta \) from \( 0 \) to \( \pi \) is \( 0 \).
- The integral of \( \sin \theta \) from \( 0 \) to \( \pi \) is \( 2 \).

Thus, we have:
\[ F_x = 0 \]
\[ F_y = \frac{G m M \pi}{L^2} \cdot 2 = \frac{2 G m M \pi}{L^2} \]

### Step 6: Final Result
The total gravitational force \( F \) acting on the particle is directed in the positive y-direction (towards the wire) and has a magnitude of:
\[ F = 0 \hat{i} + \frac{2 G m M \pi}{L^2} \hat{j} \]

### Conclusion
The gravitational attraction on the particle due to the semicircular wire is:
\[ F = \frac{2 G m M \pi}{L^2} \hat{j} \]