An object suspended by a spool of string. (Angular momentum problem)

[letsfailsafe]

Homework Statement

An object with a mass of m = 4.80 kg is attached to the end of a string wrapped around a reel of radius R = 0.230 m and mass M = 3.00 kg. The reel is a solid disc. The suspended object is released from rest 1.90 m above the floor.

http://imgur.com/ElfqPSX

a) Determine the tension in the string.
b) Determine the acceleration of the object.
c) Determine the speed with which the object hits the floor.

Homework Equations

\tau = I \alpha
T = m(g - a)
E = (0.5)(I)(\varpi)²

The Attempt at a Solution

mg = \tau = I \alpha
mg = I \alpha
mg = (1/2)(MR²)(\varpi
mg = (1/2)(MR²)(a/R)
mg = (1/2)(MR)(a)

a = (2mg)/(MR)
a = ((2)(4.8)(9.8))/((3)(0.23))
a = 136.34 m/s²

There is no way this is right. This is one of the many methods I have tried and all the values were ridiculously high so I am posting a question here as a last resort.

 

[collinsmark]

Homework Statement

An object with a mass of m = 4.80 kg is attached to the end of a string wrapped around a reel of radius R = 0.230 m and mass M = 3.00 kg. The reel is a solid disc. The suspended object is released from rest 1.90 m above the floor.

http://imgur.com/ElfqPSX

a) Determine the tension in the string.
b) Determine the acceleration of the object.
c) Determine the speed with which the object hits the floor.

Homework Equations

\tau = I \alpha
T = m(g - a)
E = (0.5)(I)(\varpi)²

The Attempt at a Solution

mg = \tau = I \alpha
mg = I \alpha

I think here is where the error starts.

For starters, you're mixing force and torque. mg is a force, not a torque.

Secondly, the force applied to the disk is not mg, rather it's the tension of the string T. Putting that in other words, the tension on the string is not mg, it's something less than that. [Edit: btw, you already have this in your relevant equations.]

To solve this problem, I suggest using Newton's second law of motion in two different ways. First, use the more conventional form of the law applied to the mass, m. The sum of all forces applied to the mass m is equal to ma, the mass times its acceleration, a.

\sum_n \vec F_n = m \vec a

Then do the same, but for the angular version with torques and moment of inertia, and apply that to the disk.

\sum_n \vec \tau_n = I \vec \alpha

Combine that with your knowledge of the relationship betwee \alpha and a, and you have enough information to solve the problem.