Acceleration of Cylinder's Center of Mass (Due 9AM)

[bmoore509]

Homework Statement

A 4.02 kg hollow cylinder with inner radius
0.19 m and outer radius 0.42 m rolls with-
out slipping when it is pulled by a horizontal
string with a force of 47 N, as shown in the
diagram below.
What is the acceleration of the cylinder’s
center of mass? Its moment of inertia about
the center of mass is .5m(r(out)^2 + r(in)^2).
Answer in units of m/s2

Homework Equations

T=F(r)
T=I*alpha
alpha=r*a

The Attempt at a Solution

F(r)=(.5m(r(out)^2 + r(in)^2))(r*a)
a=F(r)/(.5m(r(out)^2 + r(in)^2))(r)

 

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