A block of mass 2.5 kg hangs vertically from a frictionless pulley of mass 6.5 kg and radius 16 cm. Treat the pulley as a disk. Find:

a) the acceleration of the block
b) the tension in the rope

I don't even know where to begin,

1 answer

assume I = (1/2) m r^2 for disk
= (1/2)(6.5)(r)^2

Torque = I alpha where alpha = ang acceleration = a/r

T = tension in rope
so torque = T * r = I (a/r)
T = I a/r^2 = (.5 m )a=.5(6.5)a

now the block
force down = m g
force up = T
ma = mg - T
T = mg-ma = 2.5(g-a)

so putting them together
2.5(g-a) = .5(6.25)a