Question

9. By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with mı = 10.6 kg and m2 = 42.6 kg. The pulley can be treated as a uniform, solid, cylindrical disk. The downward acceleration of the 42.6 kg block is observed to be exactly one-half the acceleration due to gravity. Noting that the tension in the rope is not the same on each side of the pulley, find the mass of the pulley. kg m m2

Answer 1

In Order to find the mass of the pulley we have to find the net force acting on the system .The tension of the string will act upward direction while the weight of each block will acts in downward direction.
let the mass of the pulley be m. Here as the block of mass my is moving in downward director, the block my will more in upward direction. The tension on each block will ads in upward direction while the weight of brooks will alts in doorward direction. If the acceleration of the blocks is ä and it wire fatal upward direction as the sign, For block m Zhamar Ti-mg. Team at mag am lang) for blouh me, I formy oten mag T2 Emo 19-a): Now, as Tir m (11) a 3 mg Tia 155.820. 3* 10.6kg * 5.4 ms 5155.62 N -T2-426.kg *** 9.8 ms) 208.74 N. Net torce acting on fur sushem, CT2 T = (208.74 -155.82) N a $2.92 N

Now it r ne the radius of the pulley, now, the set torowe acting on the system, Eraar-Tara tart? 5 52.92*R. again, Ir? E9, a to the argurar accernation. whore, arpa . 20 .- a= moment of meria of the pulong a 2 me. sa max 52.92*R mr. 12x52.92 N. 3.8 ms m= 21.6 kg so, mass of the pullen) is 21.6 kg.