A horizontal disk is rotating counter-clockwise about its axis of symmetry at `14 rps` . Its moment of inertia with respect to its axis of symmetry...

The system has some angular momentum `L_0` at its initial state. To calculate its value, we first calculate the angular momentum of each disk and then we add them together. The formula for calculating the angular momentum of a rotating body is given by (1):

`(1) L = Iw `

Where `I` is the moment of the inertia with respect to the axis of rotation, and `w` is the angular frequency of the rotation. Note that the angular momentum is a vector, pointing in the direction given by the vector cross product `v^^ r` , where `v` is the velocity of the mass element located at position `r` relative to the axis of rotation. For our disks, this direction is orthogonal to the disk. We will adopt the following: for counter-clockwise rotation this vector points "up" and for clockwise rotation the vector points "down".

Let's calculate the angular momentum `L_a` of the first disk.

`L_a = (8 kg.m^2)(14 rps) = 112 kg.m^2/s`

And `L_b` for the second disk:

`L_b = (2 kg.m^2)(-7 rps) = -14 kg.m^2/s`

Note the minus sign, since the disk is rotating in the opposite direction relative to the first disk.

Now, the initial angular momentum of the system is simply:

`L_0 = L_a + L_b = (112 - 14) kg.m^2/s = 98 kg.m^2/s`

After we drop the second disk over the first, they stick together with final angular velocity `w_f`  and moment of inertia `I_a + I_b`  (the moment of inertia of a body is the sum of the moment of inertia of its parts - relative to the correct axis of rotation). Since no external torque is being applied to the system, the total angular momentum is conserved, so:

`L_f = (I_a + I_b)w_f = L_0`

Rearranging for `w_f` :

`w_f = L_0/(I_a + I_b) = (98/10) s^(-1) = 9.8 rps`

Since this value is positive, the system is rotating counter-clockwise about its axis of symmetry.