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Given that resultant force is at a right angle to F, what is the acceleration of the roller? Express in terms of radius a, alpha, and radius of gyration k.

My set-up (and failure):

Let H be resultant force, f_s is frictional force

Forces in x-dir:

[tex]Hsin( \alpha) = Fcos( \alpha) - f_s[/tex]

Forces in y-dir:

[tex]-Hcos( \alpha) = Fsin( \alpha) - mg[/tex]

Rotational equation of motion:

[tex]af_s = \frac{mk^2}{a} \ddot{x}[/tex]

Assume no slipping:

[tex]\ddot{x} = a \ddot{ \theta}[/tex]

Solve for x-acceleration:

[tex]\ddot{x} = \frac{a^2 F - a^2 mgsin( \alpha)}{mk^2 cos( \alpha)}[/tex]

Correct answer:

[tex]\ddot{x} = \frac{ga^2 sin( \alpha) cos( \alpha)}{k^2 + a^2 sin^2 (\alpha)}[/tex]

Where am I going wrong? I don't see why F isn't in the answer...