Physics help?

Question

Two masses are suspended from a pulley as shown in the figure below (the Atwood machine). The pulley itself has a mass of 0.30 kg, a radius of 0.15 m, and a constant torque of 0.35 m·N due to the friction between the rotating pulley and its axle. What is the magnitude of the acceleration of the suspended masses if m1 = 0.30 kg and m2 = 0.70 kg? (Neglect the mass of the string.)

Ash · Answer

Since m₂>m₁, the direction of acceleration, a ,  will be towards m₂

The tension in the strings will be 
T₁ = m₁g + m₁a....................(1)
T₂ = m₂g - m₂a.....................(2)

Net Torque, τ(net) = I α..............(I = moment of Inertia of pulley and α is angular acceleration)
Since I = ½m₃r²......where m₃ is mass of pulley and r = radius.............and angular acceleration, α = a/r
T₂r -T₁r - τ(friction) = (½m₃r²) (a/r)
r(T₂ -T₁) - 0.35 = ½m₃ra
T₂ -T₁ - (0.35/r) = ½m₃a
 From (1) & (2), we get
m₂g - m₂a - m₁g - m₁a - (0.35/0.15) = ½(0.30)a
(m₂-m₁)g - (m₂+m₁)a - (7/3) = (3/20)a
(m₂+m₁+ (3/20))a = (m₂-m₁)g - (7/3)
a =  [(m₂-m₁)g - (7/3)] / [(m₂+m₁+ (3/20)]
a = [0.70-0.30)9.81 - (7/3)] / [(0.70+0.30)+(3/20)]
a = 1.38 m/s²

? · Answer

mass m1 and m2 can be seen by the pulley system as equivalent rotaing masses me1 and me2 worth, respectively 0.60 and 1.40 kg 
motive force Fm = g(m2-m1) = 9.8*0.4 N
friction torque  Tf = 0.35 N*m
accelerating torque Ta = Fm*r-0.35 = 9.8*0.4*0.15-0.35 = 0.2380 N*m
total moment of inertia J = (mp+me1+me2)/2*r^2 = (0.30+0.60+1.4)/2*0.15^2 = 0.0259 kg*m^2
angular acceleration α = Δω/Δt = Ta/J = 0.2380*10^4/259 = 2380/259 = 9.20 rad/sec^2
tangential acceleration a = α*r = 9.20*0.15 = 1.380 m/sec^2

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Physics help?

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