Final answer:
The first disk's kinetic energy before it is dropped is 18.75 J. The final angular velocity of the combined disks after they stick together is 3.33 rad/s. The final kinetic energy of the combined disks is 24.97 J.
Step-by-step explanation:
Given a solid disk of mass 3 kg, radius 0.5 m, and initial angular velocity 10 rad/sec, and a second disk with the same radius but with double the mass (6 kg), which is initially at rest, we need to find the following:
- Kinetic energy of the first disk before it is dropped
- Final angular velocity of the combined disks after they stick together
- Final kinetic energy of the combined disks
(a) The kinetic energy (KE) of the first disk before it is dropped is calculated using the formula for rotational kinetic energy, KE = (1/2)Iω², where I is the moment of inertia for a solid disk (I = (1/2)mr²) and ω is the angular velocity. Therefore, KE = (1/2)((1/2)mr²)ω² = (1/2)((1/2)(3 kg)(0.5 m)²)(10 rad/s)² = 18.75 J.
(b) To find the final angular velocity (ω'), we conserve angular momentum since no external torques are acting. L_initial = L_final, which means Iω = (I1 + I2)ω'. Calculating, we get (1/2)(3 kg)(0.5 m)²(10 rad/s) = ((1/2)(3 kg)(0.5 m)² + (1/2)(6 kg)(0.5 m)²)ω', solving for ω' gives us 3.33 rad/s.
(c) The final kinetic energy (KE_final) is calculated with the combined moment of inertia and the final angular velocity using KE_final = (1/2)(I1 + I2)ω'^2. Using the values from (b), we have KE_final = (1/2)((1/2)(3 kg)(0.5 m)² + (1/2)(6 kg)(0.5 m)²)(3.33 rad/s)² = 24.97 J.