A frictionless pulley has the shape of a uniform solid disk of mass 2.60 kg and
ID: 1469863 • Letter: A
Question
A frictionless pulley has the shape of a uniform solid disk of mass 2.60 kg and radius 10 cm. A 1.00 kg 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 5.50 J of kinetic energy? Part B What percent of the total kinetic energy does the pulley have? C is there anything else we can know about this system? how could I evaluate these answers, in other words how could I assume this makes sense.
Explanation / Answer
Moment of Inertia of Solid disk, I = 1/2 * MR^2
w = v/r
K.E of System -
5.50 = (1/2) I ^2
5.50 = (1/2) (1/2)MR^2 * ^2
5.50 * 4 = Mv^2,
22.0 = 2.60 * v^2
v = 2.91
(a)
Let d = distance by which the stone must fall
loss of PE of the stone = gain of KE of the stone + pulley
(1.0) * 9.81 * d = (1/2) * (1.0) * (2.91)^2 + 5.50
d = 0.992 m.
(b)
% of total KE that the pulley has
= 5.5 / (5.5 + 4.23) x 100
% of total KE that the pulley has = 56.53 %