An Atwood machine consists of two masses, mA = 61 kg and mB = 75 kg , connected

Question

An Atwood machine consists of two masses, mA = 61 kg and mB = 75 kg , connected by a massless inelastic cord that passes over a pulley free to rotate (Figure 1) . The pulley is a solid cylinder of radius R = 0.42 m and mass 7.0 kg . [Hint: The tensions FTA and FTB are not equal.] Part A Determine the acceleration of each mass. Express your answer to two significant figures and include the appropriate units Part B What % error would be made if the moment of inertia of the pulley is ignored? Express your answer using two significant figures. *CANT FIGURE OUT PLEASE SHOW STEPS AND EXPLAIN *BHARATHI B answered it previously but its wrong? please help

Explanation / Answer

T1 = tension in the cord above m1

T2 = tension in the cord above m2

I = moment of inertia of pulley

Now we know that totrque is given by

torque = rF = I*alpha

net torque on the pulley will be

net torque = R*T1 - R*T2 = I*alpha

alpha = a/R

R*T1 - R*T2 = I*a/R

T1 - T2 = I*a/R^2

I = MR^2/2

T1 - T2 = M*a/2(1)

now using force equation

on mass m1:

m1*g - T1 = m1*a (2)

on mass m2:

T2 - m2*g = m2*a (3)

solving the above three equations:

a = g*(m1 - m2)/(M/2 + m1 + m2)

a = 9.81*(61 - 75)/(7/2 + 61 + 75) = -0.984 m/sec^2

a = 0.98 m/sec^2 (-ve sign means direction will be downford for mass 1)

Case 2: when moment of inertia of pulley ignored

force balance for mass 1

m1*g - T1 = m1*a

for mass 2:

T2 - m2*g = m2*a

solving these two equation

a = g*(m1 - m2)/(m1 + m2)

a = 9.81*(61 - 75)/(61 + 75)

a = -1.0098 m/sec^2 = -1.01 m/sec^2

error% will be

%error = (a1- a2)/a1 = [(-0.98) - (-1.01)]/(-0.98)

%error = 3.06%

Let me know if you have any doubt.

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