Atwoods Machine: The two masses (m1=5kg and m2=3kg) in the Atwoods machine shown
ID: 2189488 • Letter: A
Question
Atwoods Machine: The two masses (m1=5kg and m2=3kg) in the Atwoods machine shown in the figure released from rest, with m1 at hight og 0.78m above the floor. When when m1 hits the ground its speed is 1.6 m/s. A) Assuming that the pulley is a uniform disc with a radius of 12m, outline a strategy that allows you to find the mass of the pulley. B) Implement the strategy given in part A and determine the pulley's mass. The figure: there is a pulley that has a rope around it and mass 2 is sitting on the floor attached to the rope on left side, and mass 1 is hanging attached to the rope on right side. mass 1 is bigger than mass 2.Explanation / Answer
The initial energy in the system is just the potential energy of m1 which is:
Ei = PE1 = (m1)gh = (5.0 kg)(9.81 m/s^2)(0.78 m) = 38.22 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)(5.0 kg)(1.6 m/s)^2 =6.4 J
The kinetic energy in m2 just before m1 hits the ground is:
KE2 = (1/2)(m2)(v)^2 = (1/2)(3kg)(1.6 m/s)^2 =3.84 J
The potential energy in m2 is:
PE2 = (m2)gh = (3 kg)(9.81 m/s^2)(0.78 m) = 22.96 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 = (38.22 J) - (6.4 J) - (3.84 J) - (22.96 J) =5.02 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(5.02 J)/(1.6 m/s)^2 =7.844 kg
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