**Solution:**

Let the mass of the rod be m and the angle of the wedge be θ.

**Free-Body Diagrams:**

- Rod R:

There are two forces acting on the rod: the force of gravity (mg) and the tension force (T) from the string. Since the rod is constrained to move vertically, the tension force is in the same direction as the acceleration (upwards).

- Wedge W:

There are three forces acting on the wedge: the force of gravity (Mg), the normal force (N) from the ground, and the force (F) from the rod. Since the horizontal acceleration is in the same direction as the force from the rod, we need to resolve this force into its components.

**Equations of Motion:**

- For the rod:

Since the rod is moving vertically, we can use the equation:

ma = T - mg

- For the wedge:

Since the wedge is moving horizontally, we need to use the x-component of the equation:

Ma = Fsinθ

We also need to use the y-component of the equation to find the normal force:

N - Mg = 0

N = Mg

We can now substitute the value of F from the y-component equation into the x-component equation:

Ma = (mg + T)sinθ

**Relation between a1 and a2:**

- Acceleration of the rod:

We can solve for T in the equation of motion for the rod:

T = ma + mg

Substituting this value into the equation of motion for the wedge:

Ma = (ma + mg + Mg)sinθ

Dividing both sides by m:

a = (a + g)sinθ + gsinθ

a - asinθ = g(1 + sinθ)

a(1 - sinθ) = g(1 + sinθ)

a = g(1 + sinθ)/(1 - sinθ)

- Acceleration of the wedge:

Using the equation of motion for the wedge:

Ma = (mg + T)sinθ

Substituting the value of T:

Ma = (mg + ma + mg)sinθ

Dividing both sides by M:

a = (2g + a)sinθ

a - asinθ = 2gsinθ

a(1 - sinθ) = 2gsinθ

a = 2g sinθ/(1 - sinθ)

Dividing a1 by a2:

a1/a2 = (g(1 + sinθ)/(1 - sinθ))/(2g sinθ/(1 - sinθ))

a1/a2 = (1 + sinθ)/(2sinθ)

a1/a2 = (1/2)(1/sinθ + 1)

a1/a2 = (1/2)(cscθ + 1)

Therefore, the ratio of a1 to a2 is (1/2)(cscθ + 1).