A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a horizontal shaft of radius 0.100 m along the axis of the cylinder. You wrap a light, nonstretching cable around the cylinder and tie the free end to a 0.500 kg block of cheese. You release the cheese from rest a distance h above the floor. If the cheese is moving downward at 4.00 m/s just before it hits the ground, what is the value of h?

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

[tex]\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f[/tex]

There is no kinetic energy in the initial state, nor potential energy in the end,

[tex]mgh+0=0+KE_f[/tex]

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

[tex]mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)[/tex]

The inertia of the bodies is given by the equation,

[tex]I=\frac{m(R_1^2+R^2_2)}{2}[/tex]

[tex]I=\frac{2(0.2^2+0.1^2)}{2}[/tex]

[tex]I=0.05Kgm^2[/tex]

On the other hand the angular velocity is given by

[tex]\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s[/tex]

Replacing these values in the equation,

[tex](0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2[/tex]

Solving for h,

[tex]h=2.86m[/tex]