Answer:

Introduction: In this problem, we are given a spiracle shell of mass M and radius R, which is pulled on a horizontal plane by a horizontal force F acting on it's center of mass. We need to find the angular acceleration of the spiracle shell if it is rolling without slipping.

Rolling without slipping: When an object is rolling without slipping, the point of contact between the object and the surface it is rolling on is stationary. This means that the velocity of the point of contact is zero.

Newton's Second Law: The net force acting on an object is equal to its mass times its acceleration. We can apply this law to find the acceleration of the spiracle shell.

Torque: When a force is applied to an object, it can cause the object to rotate around its center of mass. This rotational motion is called torque. The torque acting on an object is equal to the force applied multiplied by the distance from the center of mass to the point where the force is applied.

Angular acceleration: The angular acceleration of an object is the rate at which its angular velocity changes with time. In other words, it is the rate at which the object is rotating.

Solution:
The net force acting on the spiracle shell is F, which is the force that is pulling the shell.
The acceleration of the shell is a, which is the linear acceleration of the center of mass.
The torque acting on the shell is τ, which is the torque due to the force F.
The moment of inertia of the shell is I, which is given by:

I = (2/5)MR^2

where M is the mass of the shell and R is the radius of the shell.

From Newton's Second Law, we have:

F - f = Ma

where f is the force of friction acting on the shell.
Since the shell is rolling without slipping, we have:

f = (1/2)Ma

From the torque equation, we have:

τ = FR

where R is the radius of the shell.

The angular acceleration of the shell is given by:

α = τ/I

Substituting the values of τ and I, we get:

α = FR/(2/5)MR^2

Simplifying the equation, we get:

α = (5/2)(F/MR)

Therefore, the angular acceleration of the spiracle shell is (5/2)(F/MR).

Conclusion: In this problem, we have shown that the angular acceleration of the spiracle shell is (5/2)(F/MR) when it is rolling without slipping.