Why Does the Mass Take 3.54s to Reach the Floor in This Torque Problem?

[AMacias2008]

Homework Statement

A mass of 40.98kg is attached to a light string which is wrapped around a cylindrical spool of radius 10cm and moment of inertia 4.00kg * m^2. The spool is suspended from the ceiling and the mass is then released from rest a distance 5.70m above the floor. How long does it take to reach the floor?

A) 1.13s
B) 3.54 s
C) 4.98s
D)3.37s
E)7.02

Homework Equations

$\sum$$\tau$=I$\alpha$ (Equation 1)

a$_{tangential}$=R$\alpha$ (equation 2)

y$_{f}$=y$_{i}$+v$_{i}$t-1/2gt$^{2}$ (equation 3)

The Attempt at a Solution

Found the torque by using equation 1.
$\tau$=I$\alpha$=Fl=(40.98kg)(9.80m/s^2)(.1m)=40.1604N*m

Used the above result to find the acceleration along the y-axis of the mass using equation 2.
a$_{tangential}$=R$\alpha$

Solve for $\alpha$, substitute into equation 1, solve for acceleration,

a$_{tangential}$=($\tau$*R)/I = 1.00401m/s^2

Substitue the acceleration into equation 3. Since it starts from rest, initial velocity is zero we end up with...

Δy=1/2at$^{2}$

Δy=5.70m, a=1.00401m/s^2, solve for t, I get...

t=√(Δy*2)/a = 3.37s.

However, the answer is B, 3.54s. Been trying to figure this out the past 2 hours and I've come here because I really don't know what I did wrong. Thank you for your time and help.

 

[haruspex]

Found the torque by using equation 1.
$\tau$=I$\alpha$=Fl=(40.98kg)(9.80m/s^2)(.1m)=40.1604N*m

You are wrongly assuming that the tension will be equal to the weight of he mass. but if that were the case the mass would not descend!
Introduce an unknown for the tension and write separate equations for the two objects.