A solid cylinder is mounted above the ground with its axis of rotation oriented horizontally. A rope is wound around the cylinder and its free end is attached to a block of mass 54.0 kg that rests on a platform. The cylinder has a mass of 265 kg and a radius of 0.430 m. Assume that the cylinder can rotate about its axis without any friction and the rope is of negligible mass. The platform is suddenly removed from under the block. The block falls down toward the ground and as it does so, it causes the rope to unwind and the cylinder to rotate.

(a) What is the angular acceleration of the cylinder?

(b) How many revolutions does the cylinder make in 5 s?

(c) How much of the rope unwinds in this time interval?

To solve this problem, we can use the principles of rotational motion and conservation of angular momentum.

(a) To find the angular acceleration of the cylinder, we need to understand the forces acting on it. The only force acting on the cylinder is the tension in the rope, which causes the cylinder to rotate. Since there is no friction, there are no external torques acting on the cylinder. Therefore, the net torque on the cylinder is zero, and we can apply the principle of conservation of angular momentum.

The moment of inertia of a solid cylinder rotating about its axis is given by the equation I = 0.5 * m * r^2, where m is the mass of the cylinder and r is its radius.

The angular momentum of the system is given by the equation L = I * ω, where ω is the angular velocity of the cylinder.

At the beginning, when the block is released, the rope starts unwinding and the cylinder starts rotating. The angular momentum of the system is conserved, so we can equate the angular momentum of the cylinder at the beginning to the angular momentum of the cylinder when it has reached its final angular velocity.

L_initial = L_final

0.5 * m_cylinder * r^2 * ω_initial = 0.5 * m_cylinder * r^2 * ω_final

Since the cylinder starts from rest, ω_initial = 0, and ω_final can be denoted as ω.

0.5 * m_cylinder * r^2 * ω = 0.5 * m_cylinder * r^2 * ω_final

Simplifying the equation, we find that ω = ω_final = 0.5 * g / r, where g is the acceleration due to gravity.

The angular acceleration of the cylinder, α, is equal to the change in angular velocity per unit time, which can be expressed as α = ω / t, where t is the time.

Therefore, the angular acceleration of the cylinder is α = (0.5 * g / r) / t.

Substituting the given values of g, r, and t, we can calculate the angular acceleration.

(b) To find the number of revolutions the cylinder makes in 5 seconds, we need to calculate the final angular displacement, θ_final.

The angular displacement of an object can be calculated using the equation θ = ω_initial * t + 0.5 * α * t^2.

Since the cylinder starts from rest, ω_initial = 0, and the equation simplifies to θ_final = 0.5 * α * t^2.

To convert the angular displacement to revolutions, we need to divide it by 2π, since 1 revolution is equal to 2π radians.

The number of revolutions the cylinder makes in 5 seconds can be calculated as n = θ_final / (2π).

Substituting the calculated value of α and the given value of t, we can find the number of revolutions.

(c) Since the rope unwinds as the block falls, the length of the unwound rope can be calculated using the equation L_unwound = r * θ_final, where r is the radius of the cylinder.

Substituting the calculated value of θ_final and the given value of r, we can find the length of the unwound rope.

It is important to note that the given problem assumes that the rope does not slip on the cylinder, and the cylinder does not slip on the ground. However, in reality, there may be some slipping or sliding, which can affect the results.