The disk on the bottom has a rotational inertia of i0 and is spinning clockwise on a horizontal surface having negligible friction at an angular speed of 3ω0. the top disk is then dropped directly on top of the bottom disk. the top disk has a rotational inertia of i0 and was initially spinning counterclockwise at an angular speed of ω0. if the initial total kinetic energy of the two-disk system was e0, what will the kinetic energy of the system be once the objects reach a common angular speed?

  K = I₀ w₀²

Explanation:

This is a problem of conservation of angular momentum,

We take the counterclockwise turn as positive

Initial. Before the crash

       L₀ = -I₀ 3w₀ + I₀ w₀

Final. The two albums together

       = (I₀ + I₀) wf

The moment is preserved

      L₀ =  

      -I₀ 2w₀ = (2I₀)  

         = -w₀

Let's look for kinetic energy

       K = ½ I w²

initial

       K₀ = ½ I₀ (-3w₀)² + ½ I₀ w₀²

       K₀ = ½ I₀ w₀² (9-1)

       K₀ = 4 I₀ w₀² = E₀

Final

        K = ½ 2 I₀ (-w₀)²

         K = I₀ w₀²

unbalanced force

explanation:

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y