The minimum value of the coefficient of friction between the object and the floor of the truck which makes rolling of the object possible is μ = a/(K * g). What will be K?

To solve this problem, we need to consider the condition for rolling motion to occur, which means the object should roll without slipping.

The force of friction is responsible for causing the object to roll, and we need to relate the frictional force to the condition for rolling motion.

Key Concepts:
Rolling Motion Condition: For an object to roll without slipping, the frictional force must provide the necessary torque to rotate the object. This is given by the equation: F_friction = μ * N, where:

μ is the coefficient of friction.
N is the normal force, which is equal to the weight of the object (mg) for horizontal surfaces.
Translational and Rotational Motion: The rolling motion condition requires that the linear velocity (v) and angular velocity (ω) are related by v = ω * R, where R is the radius of the object. Additionally, the force causing rotation must satisfy the torque equation: Torque = I * α, where I is the moment of inertia and α is the angular acceleration.

Moment of Inertia: For a solid cylinder or disk, the moment of inertia is given by: I = (1/2) * m * R^2.

Torque Due to Friction: The frictional force creates a torque at the point of contact, and this torque is related to the angular acceleration: F_friction * R = I * α.

Newton's Second Law: For linear motion, the net force is related to the mass and acceleration: m * a = F_friction.

Now, let's set up the equations.

The frictional force provides the torque, so: F_friction * R = (1/2) * m * R^2 * α. Simplifying, we get: F_friction = (1/2) * m * R * α.

The frictional force also causes the linear acceleration: m * a = F_friction.

Since the object rolls without slipping, the linear acceleration is related to the angular acceleration by a = R * α.

Solution:
From the equation F_friction = (1/2) * m * R * α, and knowing that a = R * α, we can substitute α = a / R into the equation:

F_friction = (1/2) * m * R * (a / R).

This simplifies to: F_friction = (1/2) * m * a.

Now, using the equation for linear motion, m * a = F_friction, we get: m * a = (1/2) * m * a.

So, the frictional force needed is half of the force required for linear motion.

Now, since friction is also equal to μ * N, and the normal force N = m * g (where g is the acceleration due to gravity), we have: μ * m * g = (1/2) * m * a.

Simplifying, we get: μ = a / (2 * g).

Thus, comparing this with the given form of the equation, μ = a / (Kg), we can conclude that K = 2.

Final Answer:
K = 2.