Two blocks are connected by a massless rope. This rope moves without slipping ov

Question

Two blocks are connected by a massless rope. This rope moves without slipping over a pulley. The mass and radius of the pulley are 2 kg and 6 cm, respectively. The mass of block A is 6 kg and the mass of block Bis 4 kg. When the system is released from rest block B will move down, mass A will move to the right, and the pulley will rotate clockwise. During this process the change in the gravitational potential energy of block B will be partitioned into changes in the kinetic energies of the moving objects. What fraction (as a decimal) of this partitioned energy is converted into the rotational kinetic energy of the pulley?

Explanation / Answer

4(9.8)h = 1/2 ( 6) v^2 + 1/2 I w^2

I = mr^2 = 2 ( 0.0036) = 0.0072 kg m^2

w = v/r

39.2 h = 3v^2 + 2v^2

h = 5v^2 / 39.2

5v^ 2= 3v^2 + Rotational KE

therefore rotattional KE must be = 2 v^2

fraction = 2/5

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