Question:

A uniform solid disc of mass 2 kg and radius 0.5 m is initially at rest on a frictionless horizontal surface. A block of mass 1 kg slides towards the edge of the disc with a speed of 4 m/s and lands on the disc exactly at its edge. The block sticks to the disc upon impact. What is the angular speed of the disc-block system after the collision?

Note: The moment of inertia of a solid disc about its central axis is given by I = 1/2 * m * r^2, where m is the mass of the disc and r is its radius.

Answer:

To solve this problem, we will use the principle of conservation of angular momentum. According to this principle, the total angular momentum of a system remains constant if no external torque is acting on it.

We start by calculating the initial angular momentum of the system before the block lands on the disc. Since the disc is at rest initially, its initial angular momentum is zero. The angular momentum of the block is given by L_block = I_block * ω_block, where I_block is the moment of inertia of the block and ω_block is its initial angular velocity.

As the block slides towards the edge of the disc, it will acquire an angular velocity due to its linear velocity. Given that the block has a mass of 1 kg and a speed of 4 m/s, we can calculate its initial angular velocity as follows:

v_block = ω_block * r 4 = ω_block * 0.5 ω_block = 8 rad/s

Now, the block lands on the disc at its edge and sticks to it. Since they stick together, the moment of inertia of the system changes. We can calculate the new moment of inertia of the disc-block system using the formula provided:

I_system = I_disc + I_block = (1/2 * m_disc * r^2) + (1/2 * m_block * r^2) = (1/2 * 2 kg * (0.5 m)^2) + (1/2 * 1 kg * (0.5 m)^2) = 0.75 kg·m^2

Now, using the conservation of angular momentum principle, we equate the initial angular momentum of the block to the final angular momentum of the disc-block system. We assume the angular velocity of the disc-block system after the collision is ω_final.

L_initial = L_final I_block * ω_block = I_system * ω_final (1 kg·m^2) * (8 rad/s) = (0.75 kg·m^2) * ω_final 8 kg·m^2/s = 0.75 kg·m^2 * ω_final ω_final = 8 kg·m^2/s / 0.75 kg·m^2 ω_final ≈ 10.67 rad/s

Hence, the angular speed of the disc-block system after the collision is approximately 10.67 rad/s.