The problem involves rotational kinematics and dynamics. First, we find the angular acceleration and displacement from the given kinematic information. Then, using the moment of inertia for a solid cylinder and Newton's second law for rotation, we find the required torque and the pulling force. Finally, the work done is calculated from the torque and angular displacement.

The final angular velocity is $\omega_f = 2\pi f$. Since it starts from rest, the angular acceleration is $\alpha = \omega_f / t$. The angular displacement is $\Delta\theta = \frac{1}{2}\omega_f t$, and the number of revolutions is $N = \Delta\theta / (2\pi)$.

$$ \alpha = \frac{2\pi f}{t} $$ $$ N = \frac{(1/2)(2\pi f)t}{2\pi} = \frac{ft}{2} $$

The moment of inertia for a solid cylinder is $I = \frac{1}{2}MR^2$. The torque is $\tau = I\alpha$, which is also given by $\tau = TR$, where $T$ is the pulling force. The work done is $W = \tau\Delta\theta$.

$$ T = \frac{I\alpha}{R} = \frac{(\frac{1}{2}MR^2)(\frac{2\pi f}{t})}{R} = \frac{MR\pi f}{t} $$ $$ W = \tau\Delta\theta = (TR)(\pi ft) = (\frac{MR\pi f}{t})R(\pi ft) = MR^2\pi^2 f^2 $$

Substituting the given values: $M = 5.0$ kg, $D = 0.30$ m ($R=0.15$ m), $t = 0.50$ s, $f = 10$ r/s.

$$ \alpha = \frac{2\pi (10 \text{ r/s})}{0.50 \text{ s}} = 40\pi \text{ rad/s}^2 \approx 1.3 \times 10^2 \text{ rad/s}^2 $$ $$ N = \frac{(10 \text{ r/s})(0.50 \text{ s})}{2} = 2.5 \text{ rev} $$ $$ T = \frac{(5.0 \text{ kg})(0.15 \text{ m})\pi(10 \text{ r/s})}{0.50 \text{ s}} = 15\pi \text{ N} \approx 47 \text{ N} $$ $$ W = (5.0 \text{ kg})(0.15 \text{ m})^2\pi^2(10 \text{ r/s})^2 = 11.25\pi^2 \text{ J} \approx 1.1 \times 10^2 \text{ J} $$