Velocity of center of mass of spinning disk

[atlarge]

A solid uniform disk of mass 19.0 kg and radius 70.0 cm (.7 m) is at rest FLAT on a frictionless surface. A string is wrapped around the rim of the disk and a constant force of 35.0 N is applied to the string. The string does not slip on the rim. The string is being pulled. When the disk has moved a distance of 6 m, what is the speed of the center of mass?



Relevant equations (I think...):
I=1/2MR^2
FR=Iα
α=a/R


So far, I have:
FR=Iα
FR=1/2MR^2(a/R)
F=1/2Ma
a=2*F/M=3.33333 m/s^2

and then,
v^2=2*a*6m, since initial velocity is 0.
v=7.26 m/s

But this is wrong, and I'm not sure why. Please help!

 

[gneill]

Newton's second law holds for translational motion, even if the net force results in a torque. The angular and linear components can be dealt with separately when they are not "coupled" via some friction or other mechanical mechanism (like a rolling object or a pulley and string).

 

[atlarge]

Thank you. I'm not sure how I would use Newton's Second Law, due to the fact that the normal force and mg are perpendicular to the movement and therefore not contributing to the velocity.

I'm so frustrated :)

 

[gneill]

Thank you. I'm not sure how I would use Newton's Second Law, due to the fact that the normal force and mg are perpendicular to the movement and therefore not contributing to the velocity.

I'm so frustrated :)

mg and the normal force are equal and opposite, so their contribution vanishes. You only have to deal with the force applied via the string.

 

[atlarge]

OH, thank you gneill!

Answer is:

F=ma
35N=(21kg)a
a=1.667 m/s^2

v^2=0 + 2(1.667)(7.9), 0 being the initial velocity
v=5.1 m/s