Given information:
- Mass of the cylinder (m) = 100 gm = 0.1 kg
- Radius of the cylinder (r) = 10 cm = 0.1 m
- Initial speed of the cylinder (v) = 15 m/s

Initial conditions:
- The cylinder is released without any rotation, so the initial angular velocity (ω) is zero.
- There is no external torque acting on the cylinder.

Analysis:
When the cylinder is released, it starts moving forward due to its initial linear velocity. However, since there is no rotation initially, there is no rolling friction to oppose its motion. As a result, the cylinder continues to slide without rolling.

In order for the cylinder to start rolling, there must be a non-zero angular velocity. This can be achieved if there is some external torque acting on the cylinder. In this case, the only external torque that can act on the cylinder is due to the friction force between the cylinder and the surface.

When the cylinder starts rolling, the point of contact between the cylinder and the surface is at rest. This means that the linear velocity of the point of contact is zero. At the same time, the topmost point of the cylinder has the maximum linear velocity.

Explanation:
- The linear velocity of the topmost point of the cylinder is equal to the sum of the linear velocity due to rotation and the linear velocity due to translation.
- The linear velocity due to rotation is given by v_rot = ω * r, where ω is the angular velocity and r is the radius of the cylinder.
- The linear velocity due to translation is equal to the linear velocity of the center of mass, which is given by v_trans = v.
- At the point of contact, the sum of these linear velocities is zero, so we can write:
v_rot + v_trans = 0
ω * r + v = 0
- Since the initial angular velocity is zero, we can solve for the final angular velocity:
ω = -v / r
- The torque acting on the cylinder is given by τ = I * α, where I is the moment of inertia and α is the angular acceleration.
- Since the cylinder is a uniform solid cylinder, the moment of inertia is given by I = (1/2) * m * r^2.
- The torque can also be written as τ = r * F_friction, where F_friction is the frictional force between the cylinder and the surface.
- Equating the two expressions for torque, we have: r * F_friction = I * α
- Substituting the values of I and α, we get: r * F_friction = (1/2) * m * r^2 * α
- The frictional force can be written as F_friction = μ * N, where μ is the coefficient of friction and N is the normal force.
- The normal force is equal to the weight of the cylinder, which is given by N = m * g, where g is the acceleration due to gravity.
- Substituting the values of F_friction and N, we get: r * μ * m * g = (1/2) * m * r^2 * α
- Simplifying the equation, we find: α = 2 * μ * g