In the figure, the pulley (which is in the form of a disk) has a mass of 7.5 kg

Question

In the figure, the pulley (which is in the form of a disk) has a mass of 7.5 kg and a radius of 35 cm. The smaller block is initially 3.8 m above the floor and is made of iron; it has a mass of 9.2 kg. The larger block, which is initially on the floor, is made of wood and has a mass of 3.40 kg. The string connecting the elements of the system together is mass less and does not stretch. The pulley axle is assumed to be friction-less. The system is released from rest. What is the speed of the iron block prior to its hitting the floor?

m1=9.2 kg

m2=3.4 kg

h=3.8 m

Pulley

Explanation / Answer

FOLLOW THIS The initial energy in the system is just the potential energy of m1 which is: Ei = PE1 = (m1)gh = (5.09 kg)(9.81 m/s^2)(0.791 m) = 39.50 J Just before m1 hits the ground, the total energy in the system is equal to the kinetic energy in m1 and m2 plus the potential energy in m2 plus the kinetic energy in the pulley. The kinetic energy in m1 just before it hits the ground is: KE1 = (1/2)(m1)(v)^2 = (1/2)(5.09 kg)(0.231 m/s)^2 = 0.1358 J The kinetic energy in m2 just before m1 hits the ground is: KE2 = (1/2)(m2)(v)^2 = (1/2)(3.15 kg)(0.231 m/s)^2 = 0.0840 J The potential energy in m2 is: PE2 = (m2)gh = (3.15 kg)(9.81 m/s^2)(0.791 m) = 24.44 J Use conservation of energy to find the kinetic energy in the pulley (KEr): Ei = Ef PE1 = KE1 + KE2 + PE2 + KEr KEr = PE1 - KE1 - KE2 - PE2 = (39.50 J) - (0.1358 J) - (0.0840 J) - (24.44 J) = 14.83 J Note that the moment of inertia of a uniform disk is: I = (1/2)MR^2 So, the kinetic energy of the disk is: KEr = (1/2)Iw^2 = (1/2)[(1/2)MR^2]w^2 Using the fact that v = Rw (since the string is not slipping on the pulley): KEr = (1/4)Mv^2 Now solve for M, and plug in values: M = 4*KEr/v^2 = 4(14.83 J)/(0.231 m/s)^2 = 1112 kg This is an extremely heavy pulley. This might seem unreasonable, however, if the pulley was frictionless and massless, then the masses would be traveling at 1.91 m/s instead of 0.231 m/s. With no friction, the pulley would have to have that large of a mass to slow down the masses by that much. (I have checked my calculations several times and found no error. If anyone finds an error, feel free to point it out.)

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