A light rope is wrapped several times around a large wheel with a radius of 0.435 m. The wheel?

Question

A light rope is wrapped several times around a large wheel with a radius of 0.435 m. The wheel rotates in frictionless bearings about a stationary horizontal axis. The free end of the rope is tied to a suitcase with a mass of 18.0 kg. The suitcase is released from rest at a height of 4.00 m above the ground. The suitcase has a speed of 3.60 m/s when it reaches the ground.
(a) the angular velocity of the wheel when the suitcase reaches the ground
my correct anwser is 8.28 rad/s
(b) the moment of inertia of the wheel.
I=? kg*m^2
im having trouble with just part b

Answer 1

a. ω = v/R = 3.6/0.435 = 8.28 rad/s2

b. From kinematics:
s = v0 - a•t^2/2
v = v0 + a•t.
Since v0 = 0, a = v^2/2•s =
= ( 3.6)^2/2•4 = 1.62 m/s2.
Second law:
m•a = m•g – T,
T = m•g - m•a= 18(9.8-1.62) =147.24 N.
The second law for rotation:
I•ε = M
I• (a/R)=T•R,
I=T•R^2/a=147.24•(0.435)^2/1.62=
=17.2 kg•m2.