In this problem, the net force on the cylinder is the tension $$T$$ in the rope, which is equal to the force applied to the rope:

$$F=T=52.5~\mathrm{N}$$

The force causes the cylinder to accelerate with an angular acceleration $$\alpha$$, which is related to its linear acceleration $$a$$ and the radius of the cylinder $$R$$ by the equation:

$$\alpha=\frac{a}{R}$$

The linear acceleration $$a$$ of the cylinder can be found using the formula $$F=ma$$:

$$ma=F=52.5~\mathrm{N}$$

where $$m=5~\mathrm{kg}$$ is the mass of the cylinder. Solving for $$a$$, we get:

$$a=\frac{F}{m}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}}=10.5~\mathrm{m/s^2}$$

Substituting this value of $$a$$ into the equation for $$\alpha$$, we get:

$$\alpha=\frac{a}{R}=\frac{10.5~\mathrm{m/s^2}}{0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$$

Therefore, the angular acceleration of the cylinder is $$15~\mathrm{rad/s^2}$$.

Alternate Method:

Let's first draw a free body diagram of the cylinder. The force $$F$$ applied to the rope creates a tension in the rope, which in turn exerts a force on the cylinder in the opposite direction. This force is given by:

$$T=F$$

where $$T$$ is the tension in the rope. The cylinder also experiences a torque due to the tension in the rope, which causes it to rotate. The torque is given by:

$$\tau=TR$$

where $$R$$ is the radius of the cylinder.

The net torque on the cylinder is equal to the product of the moment of inertia $$I$$ of the cylinder and its angular acceleration $$\alpha$$:

$$\tau=I\alpha$$

The moment of inertia of a hollow cylinder about its geometrical axis which is parallel to its length is given by:

$$I=MR^2$$

where $$M$$ is the mass of the cylinder.

Substituting the given values, we get:

$$\tau=TR=I\alpha=MR^2\alpha$$

Solving for $$\alpha$$, we get:

$$\alpha=\frac{T}{MR}=\frac{F}{MR}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}\cdot0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$$

Therefore, the angular acceleration of the cylinder is $$15~\mathrm{rad/s^2}$$.

Rotational Motion

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