In the Atwood's machine shown in Fig. `m_1 = 7` kg and `m_2 = 12` kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley of radius 5 cm. When the system is released from rest the heavier mass is found to fall 1 m in 2 s. Calculate the moment of inertia of the pulley ?

Explanation

`m_(1) = 7 kg, m_(2) = 12 kg`
Radius of the pulley `=r = 5cm = 0.05` m
Moment of intertia of the pulley = l = ?
Displacement of the block =s =1 m
Time =t =2s
Linear acceleration a= ?
`s= ut + 1/2 at^(2)`
`a = (2s)/t^(2) (2 xx 1)/22 = 0.5 m//s^(2)`
Decrease in P.E. of the system = Increase in K.E. of the system
`(m_(2)gs ~ m_(1)s) =1/2m_(1)v_(1)^(2) + 1/2m_(2)v^(2) + 1/2 I omega^(2)`
`v_(1)^(2) = v_(2)^(2) =2 as`
Final angular velocity of the pulley `omega = omega_(0) + alpha l =0 + a/r t`
`m_(2)gs - m_(1)gs = 1/2 m_(1) xx 2as + 1/2 m_(2) xx 2as +1/2I xx (a^(2)t^(2))/r^(2)`
`(m_(2)-m_(1))gs = (m_(1) + m_(2))as + (I a^(2)t^(2))/(2r^(2))`
`I =[(m_(2) - m_(1))gs -(m_(1) + m_(2))as] =(2r^(2))/(a^(2)t^(2))`
`=[(12-7)9.8 xx 1 -(12 + 7) xx 0.5 xx 1] xx (2 xx 0.0^(2))/(0.5^(2) xx 2^(2)) = 3.95 kg m^(2)`