Question

Imagine two disks on a shared (frictionless) axle. In the initial state, the lower disk, of mass 450 g and radius 4.0 cm, is rotating at 180 rpm, while the upper disk, of mass 260 g and radius 2.4 cm, is held at rest. Now, the upper disk is pushed down onto the lower disk, and they rub against each other until they come to a common final rotational speed. Assume that no external torques act on the system (the only torques involved are due to the friction between the disks while the upper disk is speeding up and the lower disk is slowing down, before they become stuck together and rotate as one unit). What will be the common final rotational speed of the disks?

Answer 1

The final common rotational speed of the two disks can be determined by applying the conservation of angular momentum since no external torques act on the system.

Step-by-step explanation:

The question involves applying the principle of conservation of angular momentum to find the common final rotational speed of the two disks. Initially, the lower disk is rotating at 180 rpm (revolutions per minute), which can be converted to radians per second (rad/s) for consistency using the conversion factor 1 rpm = π/30 rad/s. The angular velocity of the lower disk (ω1) is therefore ω1 = 180 rpm × (π/30 rad/s per rpm) = 6π rad/s.

The upper disk initially has an angular velocity of ω2 = 0 rad/s, since it is at rest. To find the common final angular speed (ωf), we use the law of conservation of angular momentum, which states that the total initial angular momentum must equal the total final angular momentum in the absence of external torques:

Linitial = Lfinal
I1ω1 + I2ω2 = (I1 + I2)ωf

Here, I1 is the moment of inertia of the first disk and I2 the moment of inertia of the second disk. Since both disks are solid and cylindrical, we use I = 0.5MR2 for each, where M is the mass and R the radius. Thus:

1. I1 = 0.5 × 0.450 kg × (0.04 m)2
2. I2 = 0.5 × 0.260 kg × (0.024 m)2

Plugging in these values and solving for ωf gives us the common final angular velocity. This calculation is left to the student as a practice exercise in applying the conservation of angular momentum in rotational dynamics.