Two objects are connected to a rope, and the rope is hung over a pulley connected to the ceiling, as shown in the figure below. MR The masses of the objects are m, = 16.0 kg and m, = 11.0 kg, the mass of the pulley is M = 5.00 kg, and the radius of the pulley is R = 0.100 m. Object m, is initially on the floor, and object m, is initially 4.90 m above the floor when it is released from rest. The pulley's axis has negligible friction. The mass of the rope is small enough to be ignored, and the rope does not slip on the pulley, nor does it stretch. (a) How much time (in s) does it take object m, to hit the floor after being released? At, = (b) How would your answer to part (a) change if the mass of the pulley were neglected? (Enter the time, in seconds, it takes object m, to hit the floor if the mass of the pulley were neglected.) At, =

M = 5 kg m1 = 16 kg m2 = 11 kg R = 0.10 m h = 4.90 m Using newton's second law of motion for m1 : m1g - T1 = m1am1(g-a) = T1 ...(1) Using newton's second law of motion for m2 : T2 - m2g = m2 aT2 = m2(g+a) ...(2) A)For Pulley : Torque = I ×αR×(T1-T2) = 12×M×R2×aR ...(3)Put value from (1) , (2) : m1(g-a) - m2(g+a) = 12×M×a16×(9.8-a) - 11×(9.8+a) = 12×5×a49 - 27a = 2.5 a49 = 29.5×a1.66 ms2 = a using equation of motion : h = (ut) + (12at2)4.9 = 0 + (1.662×t2)5.90 = t22.43 seconds = t Thus, ∆t1 = 2.43 seconds B) Put M = 0 in equation (3) m1(g-a) = m2(g+a) 16×(9.8-a) = 11×(9.8+a) 1.4545×(9.8-a) = 9.8 + a14.25 - 1.4545a = 9.8 + a4.45 = 2.4545 a 1.81 ms2 = a Using equation of motion : h = ut + (12at2) 4.9 = 0 + (1.812×t2) 5.4 = t22.32 seconds = t Thus, ∆t2 = 2.32 seconds