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Vinil7 [7]
3 years ago
7

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 ho

rizontal 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?
Physics
1 answer:
Aneli [31]3 years ago
5 0

Answer:

h=2.86m

Explanation:

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,

\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f

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

mgh+0=0+KE_f

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

mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)

The inertia of the bodies is given by the equation,

I=\frac{m(R_1^2+R^2_2)}{2}

I=\frac{2(0.2^2+0.1^2)}{2}

I=0.05Kgm^2

On the other hand the angular velocity is given by

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

Replacing these values in the equation,

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

Solving for h,

h=2.86m

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