Question 19
A uniform disk of mass $2 \mathrm{~kg}$ is spinning on its (vertical) axis of symmetry (remember the rotational inertia of a disk is given by $0.5 \mathrm{mr}^{2}$ ), with an angular velocity of $\mathrm{h}=5 \mathrm{rad} / \mathrm{s}$. A piece of mud if mass 1 $\mathrm{kg}$ is dropped on the disk near its edge, and the disk slows down. What is its angular velocity now? (you don't need to know the radius of the disk to solve this). $\mathrm{rad} / \mathrm{s}$

Step 1 ：The problem involves the conservation of angular momentum. The initial angular momentum of the system is the angular momentum of the disk, and the final angular momentum of the system is the sum of the angular momenta of the disk and the mud. Since no external torques are acting on the system, the total angular momentum of the system remains constant. Therefore, the initial angular momentum equals the final angular momentum.

Step 2 ：The angular momentum of a rotating object is given by the product of its moment of inertia and its angular velocity. The moment of inertia of a disk is given by \(0.5 \times \text{mass} \times \text{radius}^2\), and the moment of inertia of a point mass (like the mud) is given by \(\text{mass} \times \text{radius}^2\).

Step 3 ：Since we don't know the radius of the disk, we can't calculate the moments of inertia directly. However, we can express the moment of inertia of the mud in terms of the moment of inertia of the disk, since they are both at the same radius from the axis of rotation.

Step 4 ：Let's denote the initial angular velocity of the disk as \(w1\), the final angular velocity as \(w2\), the mass of the disk as \(m1\), the mass of the mud as \(m2\), and the moment of inertia of the disk as \(I1\). Then the initial angular momentum is \(I1 \times w1\), and the final angular momentum is \((I1+m2 \times r^2) \times w2\). Setting these equal to each other and solving for \(w2\) gives us the final angular velocity of the disk.

Step 5 ：Given that \(m1 = 2\), \(m2 = 1\), and \(w1 = 5\), we can calculate \(w2\) to be approximately 3.33.

Step 6 ：The final angular velocity of the disk, after the mud is dropped on it, is approximately 3.33 rad/s. This is less than the initial angular velocity, as expected, because the mud adds to the moment of inertia of the system, causing the disk to slow down.

Step 7 ：Final Answer: The final angular velocity of the disk is approximately \(\boxed{3.33}\) rad/s.