Question

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?

Answer 1

The final kinetic energy Kf of the two-disk system after reaching a common angular speed will be I0 * ω0², which is less than the initial total kinetic energy E0 due to energy loss in the inelastic interaction.

Step-by-step explanation:

When the top disk, spinning counterclockwise at an angular speed of ω0, is dropped on the bottom disk, spinning clockwise at an angular speed of 3ω0, angular momentum is conserved but kinetic energy is not because of the inelastic nature of the coupling. The combined final angular speed ωf can be found using the conservation of angular momentum. Assuming the total moment of inertia for the system is I (sum of the individual moments of inertia, which are both I0), the final angular speed ωf can be calculated from I0 * 3ω0 + (-I0 * ω0) = (I0 + I0) * ωf. Solving this yields ωf = ω0. The final kinetic energy Kf is given by Kf = 1/2 * 2I0 * ωf², which simplifies to Kf = 1/2 * 2I0 * ω0² = I0 * ω0². Since energy is not conserved, we expect a loss due to the inelastic interaction; hence, the final kinetic energy Kf is less than the initial total kinetic energy E0.