The two masses (m1 = 5.13 kg and m2 = 2.87 kg) in the Atwood\'s machine shown in
ID: 2186274 • Letter: T
Question
The two masses (m1 = 5.13 kg and m2 = 2.87 kg) in the Atwood's machine shown in the figure below are released from rest, with m1 at a height of 0.739 m above the floor. When m1 hits the ground its speed is 0.207 m/s. Assuming that the pulley is a uniform disk with a radius of 10.1 cm, calculate the pulley's mass.
The two masses (m1 = 5.13 kg and m2 = 2.87 kg) in the Atwood's machine shown in the figure below are released from rest, with m1 at a height of 0.739 m above the floor. When m1 hits the ground its speed is 0.207 m/s. Assuming that the pulley is a uniform disk with a radius of 10.1 cm, calculate the pulley's mass.Explanation / Answer
The initial energy in the system is just the potential energy of m1 which is: Ei = PE1 = (m1)gh = (5.13 kg)(9.81 m/s^2)(0.739 m) = 37.15 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)(5.13 kg)(0.207m/s)^2 = 0.11 J The kinetic energy in m2 just before m1 hits the ground is: KE2 = (1/2)(m2)(v)^2 = (1/2)(2.87kg)(0.207 m/s)^2 = 0.06 J The potential energy in m2 is: PE2 = (m2)gh = (2.87 kg)(9.81 m/s^2)(0.739 m) = 20.79J 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 = (37.15 J) - (0.11 J) - (0.06 J) - (20.79J) = 16.19 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 M = 4*KEr/v^2 = 4(16.19 J)/(0.207 m/s)^2 = 1511 kg
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