Question

Two blocks are connected to a string, and the string is hung over a pulley connected to the ceiling, as shown in the figure below. Two blocks, labeled m1 and m2, are connected to a string which is hung over a pulley connected to the ceiling. The pulley is of mass M and radius R. A block labeled m1 hangs suspended off the surface on the left side of the pulley. A block m2 is on the right side of the pulley and rests on a surface below. m1 is larger than m2. The masses of the blocks are m1 = 18.0 kg and m2 = 10.0 kg, the mass of the pulley is M = 5.00 kg, and the radius of the pulley is R = 0.300 m. Block m2 is initially on the floor, and block m1 is initially 4.50 m above the floor when it is released from rest. The pulley's axis has negligible friction. The mass of the string is small enough to be ignored, and the string does not slip on the pulley, nor does it stretch. (a) How much time (in s) does it take block m1 to hit the floor after being released? Δt1 = Incorrect: Your answer is incorrect. Apply Newton's second law to each block. Consider the forces on each block and be careful of directions and signs of forces and acceleration. Then relate the net torque on the pulley to its angular acceleration. What is the moment of inertia of the pulley? How does the angular acceleration relate to the linear acceleration of the blocks? s (b) How would your answer to part (a) change if the mass of the pulley were neglected? (Enter the time, in seconds, it takes block m1 to hit the floor if the mass of the pulley were neglected.) Δt2 = s

Answer 1

solution L LU 1111 as (a) As shown in the figure zmU m Tita mig-Ti = mia I mig my Imag Tz - mag= maa. - Also the torque on the Pully by tensions Ti and Tz are t = TI R - T2 R = Ti-T2 DR = Id - = R a inertia of the and I = 2 m R² {disk] Pully. so (TI-T2 ) R = MR² a = TI-T2 = / Ma - By adding equations (1) , (2) and (3) (mi-m2 g = (m+ M2 + 1/2 m) a 8.0x 9.8 = (3015) a. & a = 2.57 m/s2 1. 19x4.50 = 1,87135 seconds - 3) so ti= v gos7 ~ 1,87 seconds ti= 1187 , seconds.