Acceleration of a point on an accelerating rolling wheel

[thenewbosco]

hello i am not sure how to even begin this problem (i.e. how to set it up)

A wheel of radius b, rolls along the ground with constant forward acceleration a0. Show that, at any given instant, the magnitude of the acceleration of any point of the wheel is $$(a_{0}^2 + \frac{v^4}{b^2})^\frac{1}{2}$$

The second part asks to show the magnitude of the acceleration relative to the ground is $$a_{0}[2 + 2cos\theta + \frac{v^4}{a_{0}^2 b^2} - (\frac{2v^2}{a_{0}b})sin\theta]^\frac{1}{2}$$
Here v is the instantaneous forward speed and theta defines the location of the point on the wheel, measured forward from the highest point.

Could someone help me get started on this as well as explain what theta is, since i do not understand based on the question.

 

[Astronuc]

The problem is one of finding the resultant acceleration for two components, the translational acceleration $a_0$ and the angular acceleration $v^2/r$.

The formula $$(a_{0}^2 + \frac{v^4}{b^2})^\frac{1}{2}$$
implies that the two accelerations are mutually perpendicular, so one would use the square root of the sum of the squares, but the second part implies r = b, i.e. it applies to the points on the circumference.