Final answer:
The angular acceleration of the disc when the 3.0 kg mass is released is 33.65 rad/s², calculated using the relationship τ = Iα and the moment of inertia for a thin uniform disc.
Step-by-step explanation:
The problem involves finding the angular acceleration of a disc when a mass is released from rest. As the mass falls, it will cause the disc to rotate. To find the angular acceleration, we use Newton's Second Law for rotation, τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
First, we'll calculate the torque (τ) produced by the weight of the mass as it falls. The torque due to the weight (mg) is given by τ = r * F₀ where r is the radius of the disc and F₀ is the force applied at the edge of the disc, which is the weight of the mass. So, τ = 0.25 m * 3.0 kg * 9.81 m/s² = 7.3575 N·m.
The moment of inertia (I) of a solid disc about its central axis is given by I = 1/2 * M * R², where M is the mass of the disc and R is its radius. For our disc, I = 1/2 * 7.0 kg * (0.25 m)² = 0.21875 kg·m².
Now we can calculate the angular acceleration using α = τ / I. So, α = 7.3575 N·m / 0.21875 kg·m² = 33.65 rad/s².
The angular acceleration of the disc when the mass is released is therefore 33.65 rad/s².