A frictionless pulley has the shape of a uniform solid disk of mass 2.50 and rad
ID: 2189049 • Letter: A
Question
A frictionless pulley has the shape of a uniform solid disk of mass 2.50 and radius 10 . A 1.40 stone is attached to a very light wire that is wrapped around the rim of the pulley (the figure (Figure 1) ), and the system is released from rest. Part A How far must the stone fall so that the pulley has 4.30 of kinetic energy? Part B What percent of the total kinetic energy does the pulley have?Explanation / Answer
KE = (1/2) I ?^2 = (1/2) (1/2)MR^2 * ?^2 = (1/4) Mv^2, where v = linear velocity of the stone => 4.5 = (1/4) * 2.50 * v^2 => v^2 = 18/2.50 = 7.2 (a) Let d = distance by which the stone must fall => loss of PE of the stone = gain of KE of the stone + pulley => (1.50) * 9.81 * d = (1/2) * (1.50) * (7.2) + 4.50 => d = (5.4 + 4.50) / (1.50 * 9.81) => d = 0.673 m. (b) % of total KE that the pulley has = 4.5 / (4.5 + 5.4) x 100 = 45.45.