What is the tension in the cord while the mass is falling?

[jg156]

Recently stumbled on this problem:

A uniform disk (I =½ MR2
) of mass 8.0 kg can rotate without
friction on a fixed axis. A string is wrapped around its
circumference and is attached to a 6.0 kg mass. The string does
not slip. What is the tension in the cord while the mass is
falling?

The answer is 24 Newtons, but i would like to know why.
If anyone can provide a solution/explanation, that would be greatly appreciated.
I am having trouble figuring out the effect the disk's inertia has on the system's acceleration.

 

[member 392791]

post it in the Homework section

 

[MariaResnick]

Hello! To do that problem, you have to, as you know, understand how the rotation of the pulley affects the tension in the cord. Because Tension T is pulling down on the point A where the string is perpendicular to the center of the pulley, and pulling up further down the rope to counteract the mg of the block, we have that:
1) T - mg = ma
Since we know what mg is, we have to find a in terms of T, and solve for T
We know that the Torque about the pulley is Torque = r x F, and since Mg passes through the pulley's center, we don't have to consider it. We only consider T.
2) Torque = r x F = TR
Now, remember that Torque = I(alpha), and alpha = a/R?
TR = I(alpha) [Let's do some substitution]
TR = MR^2/2 * a/R
So, a = 2T/M.
Now let's solve for T!
Go back to 1) and rearrange it. T + ma = mg. Now plug in a.
So, we have that T = mg/(1+2m/M). T = 23.52N which is 24.