Answer: 8.38 revolutions per second
Step-by-step explanation:
Before the second disk is dropped, the initial angular momentum of the system is given by:
L = I1 * w1
where I1 is the moment of inertia of the first disk, and w1 is its angular velocity.
Substituting the given values, we have:
L = (25 kg m^2) * (2 rev/sec * 2π rad/rev) = 100π kg m^2/s
When the second disk is dropped onto the rotating disk, the total moment of inertia of the system will be the sum of the moment of inertia of the first disk and the moment of inertia of the second disk:
Itotal = I1 + I2/2
where I2/2 is the moment of inertia of the second disk, which is half as large as that of the first disk.
Substituting the given values, we have:
Itotal = (25 kg m^2) + (12.5 kg m^2) = 37.5 kg m^2
Conservation of angular momentum requires that the initial angular momentum of the system be equal to its final angular momentum, so:
L = Itotal * wf
where wf is the final angular velocity of the combined disk.
Solving for wf, we get:
wf = L / Itotal = (100π kg m^2/s) / (37.5 kg m^2) ≈ 8.38 rev/sec
Therefore, the new rotational speed of the combined rotating object is approximately 8.38 revolutions per second.