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 ?

In the Atwood's machine shown in Fig.m_1 = 7 kg and m_2 = 12 kg are co

1.	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 ?
Answer» Solution :`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)`