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A uniform Disk of radius

http://img210.imageshack.us/img210/6792/pulleywithmassquestionqy2.jpg [Broken]

[tex]\Sigma[/tex] F = ma

[tex]\Sigma[/tex] [tex]\tau[/tex] = I [tex]\alpha[/tex]

[tex]\alpha[/tex] = a / r

The Cord is massless and the Axis is frictionless, so that all equals to 0.

I set F_x = 0, and F_y = T + H - mg, T is the Tension force the pulley is pulling away from the weight attached to it by the string. So i defined the T as "Mg" where M is the mass of the pulley and "mg", where m is 0.57 kg. Finding the Tension yielding 5.586 N.

I set the sum of the torques equal to I[tex]\alpha[/tex], and substituted alpha as a/r. given the a causing rotational motion, because the system is being essentially rotated from the weight falling from rest attached to the pulley, making the 2.6 = a, and the r = 0.18 m. Given the substitution for [tex]\alpha[/tex], I can substitute the sums of torques as the magnitude of torques yielding rFsin(theta), since it's 90 degrees, the sin becomes 1, making the equation essentially F * radius. So, given the T from prior, 5.586, i multiplied by 0.18 m, attating 1.00548. I set 1.00548 yielding I x (2.6/0.18), and I solved for I, giving me a number of 0.069627. This is obviously incorrect though. I've been having trouble understanding the values of the application of how to SEE these torques in pictures and such. It's been giving me ALOT of trouble, so any help would be useful.

**0.18 m**is mounted on a frictionless, horizontal axis. A light cord wrapped around the disk supports a**0.57 kg**object. When released from rest, the object falls with a downward acceleration of**2.6 m/s^2****What is the Moment of Inertia of the disk?**http://img210.imageshack.us/img210/6792/pulleywithmassquestionqy2.jpg [Broken]

**Relevant Equations**[tex]\Sigma[/tex] F = ma

[tex]\Sigma[/tex] [tex]\tau[/tex] = I [tex]\alpha[/tex]

[tex]\alpha[/tex] = a / r

**My attempt**The Cord is massless and the Axis is frictionless, so that all equals to 0.

I set F_x = 0, and F_y = T + H - mg, T is the Tension force the pulley is pulling away from the weight attached to it by the string. So i defined the T as "Mg" where M is the mass of the pulley and "mg", where m is 0.57 kg. Finding the Tension yielding 5.586 N.

I set the sum of the torques equal to I[tex]\alpha[/tex], and substituted alpha as a/r. given the a causing rotational motion, because the system is being essentially rotated from the weight falling from rest attached to the pulley, making the 2.6 = a, and the r = 0.18 m. Given the substitution for [tex]\alpha[/tex], I can substitute the sums of torques as the magnitude of torques yielding rFsin(theta), since it's 90 degrees, the sin becomes 1, making the equation essentially F * radius. So, given the T from prior, 5.586, i multiplied by 0.18 m, attating 1.00548. I set 1.00548 yielding I x (2.6/0.18), and I solved for I, giving me a number of 0.069627. This is obviously incorrect though. I've been having trouble understanding the values of the application of how to SEE these torques in pictures and such. It's been giving me ALOT of trouble, so any help would be useful.

**1. Homework Statement****2. Homework Equations****3. The Attempt at a Solution**
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