Rotational Dynamics: Pulley and mass system

[henryli78]

Homework Statement

A 8.0-cm radius disk with a rotational inertia of 0.12 kg*m^2 is free to rotate on a horizontal axis. A string is fastened to the surface of the disk and a 10-kg mass hangs from the other end. The mass is raised by a using a crank to apply a 9.0-N*m torque to the disk. The acceleration of the mass is:
A. 0.50 m/s^2
B. 1.7 m/s^2
C. 6.2 m/s^2
D. 12 m/s^2
E. 20 m/s^2
Answer: A

Homework Equations

∑$\tau$ = I$\alpha$
∑F = ma
a = $\alpha$r

The Attempt at a Solution

The net torque of the system is:
I*$\alpha$ = I*a/r = 9.0 N*m - $F_{T}$*r
Thus, $F_{T}$ = (Ia/r-9)/(-r)
By N2L:
ma = $F_{T}$ - mg = (Ia/r-9)/(-r) - mg
Rearranging gives:
a = (mg)/((I/r-9/(-r))-m) and substituting in values gives me an answer of about 1.17 m/s^2.

Can someone direct me on where I have gone wrong in my calculations? I would very much appreciate it :)

 

[grzz]

...ma = $F_{T}$ - mg = (Ia/r-9)/(-r) - mg...

Open up the RHS and carefully collect the terms containing a.

 

[henryli78]

Ok thank you. I guess my mistake was just in bad algebra :P