Solution 63GP Introduction We shall have to calculate net torque first followed by linear acceleration and time.

Step 1: Given that, diameter = 12 cm Radius R = 6 cm R = 0.06 m Mass of the pulley M = 2.0 kg 1 2 2 Therefore, its moment of inertia I = ×2.0 kg×(0.06) kg.m 2 I = 0.0036 kg.m Let T be the tension of the rope attached to the 4.0 kg mass and T be the tension of the 1 2 rope attached to the 2.0 kg mass. For the 4.0 kg mass, the equation of motion can be written as follows. 4gT = 41 ( a is the acceleration) T =14g4a…..(1) For the 2.0 kg mass, the equation of motion can be written as follows. T 2g = 2a 2 T =22g+2a…..(2) Thus, torque on the pulley = (T T1)0.062N.m (from equations (1) and (2)) Now, the net torque excluding the frictional torque = (T T )0.06 N.m0.50 N.m 1 2 …..(3) Step 2: We know that, torque = I Therefore, substituting the values of I = 0.0036 kg.m and from equation (3) 2 (T 1 )0.26 N.m0.50 N.m = 0.0036 kg.m × Now, substituting the values of T an1 T from2equations (1) and (2) and using a = R 2 (4g4a2g2a)0.06 N.m0.50 N.m = 0.0036 kg.m ×a/R 2 (2g6a)0.06 N.m0.50 N.m = 0.0036 kg.m ×a/0.06 2×9.8×0.06 N.m0.36a N.m0.50 N.m = 0.06a N.m 1.18 N.m0.50 N.m0.36a N.m = 0.06a N.m 0.68 N.m = 0.42a N.m a = 1.62 m/s 2