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

Explanation

To solve the problem of finding 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: Understand the Geometry
The semicircular wire has a radius \( R \). The length of the semicircular wire is given by:
\[ L = \pi R \]
From this, we can express the radius \( R \) in terms of the length \( L \):
\[ R = \frac{L}{\pi} \]

### Step 2: Consider a Small Element of the Wire
Let’s take a small element of the wire of length \( dx \) at an angle \( \theta \) from the vertical. The mass of this small element \( dm \) can be expressed as:
\[ dm = \frac{M}{L} \cdot dx \]

### Step 3: Relate \( dx \) to \( d\theta \)
The length of the small element \( dx \) can also be related to the angle \( d\theta \) as follows:
\[ dx = R \, d\theta \]
Thus, substituting for \( dx \) gives us:
\[ dm = \frac{M}{L} \cdot R \, d\theta \]

### Step 4: Calculate the Gravitational Force
The gravitational force \( dF \) between the small element \( dm \) and the mass \( m \) at the center is given by Newton's law of gravitation:
\[ dF = \frac{G \cdot dm \cdot m}{R^2} \]
Substituting for \( dm \):
\[ dF = \frac{G \cdot \left(\frac{M}{L} \cdot R \, d\theta\right) \cdot m}{R^2} = \frac{G \cdot M \cdot m}{L \cdot R} \, d\theta \]

### Step 5: Resolve the Force into Components
The force \( dF \) acts along the line connecting the mass \( m \) and the small element \( dm \). We need to consider only the vertical component of this force because the horizontal components will cancel out due to symmetry. The vertical component is given by:
\[ dF_y = dF \cdot \sin(\theta) \]
Thus:
\[ dF_y = \frac{G \cdot M \cdot m}{L \cdot R} \cdot \sin(\theta) \, d\theta \]

### Step 6: Integrate to Find the Total Force
To find the total gravitational force \( F \) on the mass \( m \), we integrate \( dF_y \) from \( \theta = 0 \) to \( \theta = \pi \):
\[ F = \int_0^\pi dF_y = \int_0^\pi \frac{G \cdot M \cdot m}{L \cdot R} \cdot \sin(\theta) \, d\theta \]
The integral of \( \sin(\theta) \) over \( 0 \) to \( \pi \) is \( 2 \):
\[ F = \frac{G \cdot M \cdot m}{L \cdot R} \cdot 2 \]

### Step 7: Substitute \( R \) in Terms of \( L \)
Substituting \( R = \frac{L}{\pi} \) into the expression for \( F \):
\[ F = \frac{G \cdot M \cdot m}{L \cdot \frac{L}{\pi}} \cdot 2 = \frac{2 \pi G \cdot M \cdot m}{L^2} \]

### Final Result
Thus, the gravitational attraction on the particle due to the wire is:
\[ F = \frac{2 \pi G \cdot M \cdot m}{L^2} \]