Question

An Atwood's machine consists of two weights, m, = 3.2 kg and m2 = 2 kg, connected by a by a string over a pulley of mass m, -1.7 kg and radius r = 0.2 m. Assume the pulley is a uniform solid cylinder (disk). The system is released from rest when my is 1.6 m above the floor, and my begins on the floor. Assume there is no friction in the pulley. Take the ground to be the location of zero GPE. What is the rotational inertia of the pulley? I-L w kg ma What is the total mechanical energy of the system when it is at rest? E l- Use conservation of energy to determine how fast is m, moving just before hitting the ground. You do not need to consider the GPE of the pulley since its location will not change; however, it will have rotational KEY= m/s Immediately after m, hits the floor is m, still moving? yes, up What is the greatest height that my will reach? Hint: This part can be solved with basic linear kinematics. h=

Answer 1

Rotational inertial of the pulley Ip ² Mp op I = ² x 17X (0.2)2 = 0.0341g-m2 ( Ip = 0.034 Kgam? -3.2x9.8X1.6 2 Initiall, system is atrest, EFEttal = mghas Etotal a 50. 176 v 4 = Efinal = 2m,u2+ 2 mg V2 + m2 gh + 1/ I w2 = almu2 +4 m2 V2 + magh+FCmpre) 6.)? -(mt£ me +£mp) v2+ mogh. using conservations of evez gy Eszet migh = (££m, & 2 m, t2 mp 4²+ migh [21 m.-ma) gh va V T mitmat mp) (21 3.2-2)*9.881-6 (3.2+2 +1.7) V = 2-34 m/s ✓ 2:34 m)s - E L yes, up m2 moving up. ~ -). have you = 0.278 hmax a 1.6+0.278= 1.878 m v