Sol. We, know that the angular velocity and angular acceleration of the wheel are

ω = v O /r and α = α O /r

The acceleration of A is written in terms of the given acceleration of O. Thus

a A = a O + A A/O = a O + (a A/O ) n + (a A/O ) t

The relative-acceleration terms are viewed as though O were fixed, and for the relative circular motion they have the magnitudes

(a A/O ) n = r 0 ω 2 = r 0

(a A/O ) t = r 0 α = r 0

and the directions shown.

Adding the vectors head-to-tail gives a A as shown. In a numerical problem, we may obtain the combination algebraically or graphically. The algebraic expression for the magnitude of a A is found for the square root of the sum of the squares of its components. If we use n- and t-directions, we have

a A =

=

= Ans.

The direction of a A can be computed if desired.

The acceleration of the instantaneous center C of zero velocity, considered a point on the wheel, is obtained from the expression

a C = a O + a C/O

where the components of the relative-acceleration term are (a C/O ) n = r ω 2 directed from C to O and (a C/O ) t = r α directed to the right because of the counterclockwise angular acceleration of line CO about O. The term are added together in the lower diagram, and it is seen that

a C = r ω 2 Ans.

Helpful Hints :

(i) The counterclockwise angular acceleration α of OA determines the positive direction of (a A /O) t . The normal component (a A/O ) n is, of course, directed towards the reference
center O.

(ii) If the wheel were following to the right with the same velocity v O but still had an acceleration a O to the left, note that the solution for a A would be unchanged.

(iii) We note that the acceleration of the instantaneous center of zero velocity is independent of α and is directed toward the center of the wheel. This conclusion is a useful result to remember.