Let's go through the solution for each part of the problem.

Given Data:

• Block mass $m = 30 \, \text{kg}$
• Distance to fall $h = 2 \, \text{m}$
• Pulley mass $M = 5 \, \text{kg}$
• Pulley radius $R = 0.1 \, \text{m}$ (20 cm diameter)
• Pulley moment of inertia $I = \frac{1}{2}MR^2$
• We assume the rope doesn't slip on the pulley.

Solution Steps

Part (a): Velocity of the 30 kg block just before it hits the ground

Using energy conservation, the gravitational potential energy of the block will be converted into:

1. Translational kinetic energy of the block.

2. Rotational kinetic energy of the pulley.

3. Gravitational potential energy (initial): $U = mgh = 30 \times 9.8 \times 2 = 588 \, \text{J}$

4. Kinetic energy (final): At the point just before impact, the block has kinetic energy: $KE_{\text{block}} = \frac{1}{2}mv^2$ The pulley has rotational kinetic energy: $KE_{\text{pulley}} = \frac{1}{2}I\omega^2$ where $\omega$ is the angular velocity of the pulley. Since $v = \omega R$, we can substitute $\omega = \frac{v}{R}$.

   The total energy conservation equation is: $mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2$

5. Simplify and solve for $v$: Substitute values and solve for $v$: $588 = \frac{1}{2}(30)v^2 + \frac{1}{2} \cdot \frac{1}{2}(5)(0.1)^2 \cdot \frac{v^2}{(0.1)^2}$ $588 = 15v^2 + \frac{1}{4} \cdot 5 \cdot v^2$ $588 = 15v^2 + 1.25v^2$ $588 = 16.25v^2$ $v^2 = \frac{588}{16.25}$ $v \approx 2.73 \, \text{m/s}$

So, the velocity of the block just before it touches the ground is approximately 2.73 m/s.

Part (b): Angular velocity of the pulley at that instant

Since $v = \omega R$: $\omega = \frac{v}{R} = \frac{2.73}{0.1} = 27.3 \, \text{rad/s}$

The angular velocity of the pulley at that instant is 27.3 rad/s.

Part (c): Tension in the rope

To find the tension, consider the forces acting on the 30 kg block. Let $T$ be the tension in the rope.

From Newton's second law for the block: $mg - T = ma$ where $a$ is the acceleration of the block. Also, using the rotational dynamics of the pulley, we get: $T \cdot R = I \cdot \alpha$ where $\alpha = \frac{a}{R}$ is the angular acceleration of the pulley. Substituting $I = \frac{1}{2}MR^2$: $T \cdot R = \frac{1}{2}MR^2 \cdot \frac{a}{R}$ $T = \frac{1}{2}Ma$

Now we can solve for $a$ and $T$ by substituting values.

Part (d): Time for the block to reach the ground

Using $v = u + at$ and $v = 2.73 \, \text{m/s}$