let mA = 250 g = 0.25 kg
mB = 100 g = 0.1 kg
R = 0.05 m
let alfa is the angular acceleration of the pulley and a is the acceleration of the bodies.
Net force acting on mA, FnetA = mA*g - TA
mA*a = mA*g - TA
==> TA = mA*g - mA*a -------(1)
Net force acting on mB, FnetB = TB - mB*g
mB*a = TB - mB*g
==> TB = mB*g + mB*a --------(2)
Net Torque acting on pulley, Tnet = I*alfa
(TA - TB)*R = I*a/R
TA - TB = I*a/R^2
mA*g - mA*a - mB*g - mB*a = I*a/R^2
(mA - mB)*g = a*(I/R^2 + mA + mB)
==> a = (mA - mB)*g/(I/R^2 + mA + mB)
= (0.25 - 0.1)*9.8/(6.25*10^-4/0.05^2 + 0.25 + 0.1)
= 2.45 m/s^2
alfa = a/R
= 2.45/0.05
= 49 rad/s^2 <<<<<<<-----------Answer
b) TA = mA*g - mA*a
= 0.25*9.8 - 0.25*2.45
= 1.8375 N <<<<<<<-----------Answer
TB = mB*g + mB*a
= 0.1*9.8 + 0.1*2.45
= 1.225 N <<<<<<<-----------Answer
c) w = wo + alfa*t
= 0 + 49*3
= 147 rad/s <<<<<<<-----------Answer
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