Question

An Atwood's machine consists of two weights, m 3.4kg and m2 1.1 kg, connected by a by a string over a pulley of mass mp-2.3 kg and radius r-0.28 m. Assume the pulley is a uniform solid cylinder (disk) 1m7 177 The system is released from rest when mj is 2.7 m above the floor, and m2 begins on the floor. Assume there is no friction in the pulley. Take the ground to be the location of zero GPE kg m What is the rotational inertia of the pulley? I-09016 What is the total mechanical energy of the system when it is at rest? Esotal 89.96 J 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 KE. v-727 m/s Immediately after m, hits the floor is m2 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

mi 2.3.4.19 m2 119 mp. -2.3 kg rp = 0.28m V=w.dp 523 = 2.7m me that from longebra bion of energy ME; = M*El > migh =maght I mi v?t 1 mav²+ £ 10? -> (m.-m2) gh= limitm2) v²+ + mpx? v? -> (314-).1)x10x 2.7 = $(3.4+).)) v?+ * *2.3 x Geef v? 4 2.3 x10 x 2.7 = v2( 4.5+ 373) Tv=4.64 m3/ greatest height of m2 = 12th = 2.7+(4.6932 2800 = 308m maxm height attained by ma from ground = 3.8m