The pulley system represented in the figure, of radii R1 = 0.25 m and R2 = 1 m and masses m1 = 20 kg and m2 = 60 kg is lifting an object of mass M = 1000 kg. At a certain moment, when the object is at a height of 2 m above the ground, the brake is released and the mass falls from rest. Calculate the speed of the mass when it reaches the ground.
Givens: The moment of inertia of a disc with respect to an axis that passes through its center of mass is: ICM = (1/2)MR2
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Solution:
We will solve this problem using the principle of conservation of energy. For this, we choose the initial (A) and final (B) states for the system consisting of the two pulleys and the mass M. In the following figure both states have been represented, as well as the origin of heights that we will use to calculate the gravitational energy:
In state A the three objects that make up the system are at rest. Therefore, the mechanical energy of the system at that instant is equal to the gravitational energy of the mass M:
In state B the mass M hit the ground, it has no gravitational energy but it has a certain speed; on the other hand the two pulleys are rotating. The total energy in state B will therefore be the sum of the translational kinetic energy of the mass and the rotational energy of the pulleys:
As there is no non-conservative force (friction) acting on the system, its mechanical energy is preserved:
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On the other hand, if we assume that the rope does not slide on the pulleys, the linear velocity of a point at the periphery of the pulleys must be equal to the velocity of the mass M. Therefore the angular velocity of each pulley can be related to the linear velocity of the mass M by means of the following equation:
And after substituting in the energy conservation equation we get:
When we replace the moment of inertia of the pulleys we get:
Finally we find v and we substitute the givens to get:
In the problem we have used g = 10 m/s2.
Do not forget to include the units in the results.
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