Question

1. 12. [2pt]
   An Atwood Machine consists of two masses m₁ and m₂ attached to each other by a single massless string passing over a cylindrical pulley of mass M without slipping. The masses are released from rest, and allowed to accelerate until each has moved a distance h = 1.6 m. Assuming that upwards is positive, rank the velocity of m₂ (from most negative to most positive) at this moment for the following cases:
   
   Case A: m1 = 11 kg, m2 = 14 kg, M = 22 kg
   
   Case B: m1 = 12 kg, m2 = 14 kg, M = 25 kg
   
   Case C: m1 = 11 kg, m2 = 14 kg, M = 20 kg
   
   Case D: m1 = 12 kg, m2 = 14 kg, M = 28 kg
   
   Case E: m1 = 19 kg, m2 = 14 kg, M = 27 kg

Answer 1

Using Atwood machine formula for cylindrical pulley with mass M,

acceleration of m2, a = g*(m1-m2)/(m1+m2+0.5M)

caseA: a = 9.8*(11-14)/(11+14+0.5*22) = -0.8166 m/s^2

caseB: a = 9.8*(12-14)/(12+14+0.5*25) = -0.509 m/s^2

caseC: a = 9.8*(11-14)/(11+14+0.5*20) = -0.84 m/s^2

caseD: a = 9.8*(12-14)/(12+14+0.5*28) = -0.49 m/s^2

caseE: a = 9.8*(19-14)/(19+14+0.5*27) = 1.054 m/s^2

Also by third equation of motion, v = sqrt(2a*s) where s is the distance moved

the one with higher a will have higher speed v,

speed of C is most negative, then A, then B , then D, then E (most positive)