Solution: Torque

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First, we pick the positive direction of the rotation, and we will arbitrarily pick clockwise as the positive direction.

The resulting torques on the cue is as follow:
t_tot = FR(cos q - sin q)
The y-component of the force subtracts because that torque would cause a negative spin on the cue.

For angular acceleration:
FR(cos q - sin q) = (2/5)MR²a

For linear acceleration:
F = Ma

a(cos q - sin q) = (2/5)Ra
a = 5a(cos q - sin q)/2R

Answer: 5a(cos theta - sin theta)/2R