Answer:
a) the angular acceleration of the bar AB is 
b) the acceleration
of point B = 
Explanation:
The sketched diagram below shows an illustration of what the question comprises of:
Now, from the diagram ; we can deduce the following relations;


Taking point H as the instantaneous center of rotation of line HD. The distance between H and D is represented as:

The angular velocity of the bar AB from the diagram can be determined by using the relation:

where:

and
= velocity of point D = 18.5 ft/s
L = length of the bar = 8 ft
Then;


Using vector approach to acceleration analysis;
Acceleration about point B can be expressed as;
---equation(1)
The y - component of
⇒
= 
where;
= 0
Then
0 = 
Making
the subject of the formula; we have:
------- equation (2)
Acceleration about point D is expressed as follows:

The y - component of
⇒ 
Replacing - 23 ft/s² for
; we have:

Also; replacing equation (2) for
in the above expression; we have


Making
the subject of the formula ; we have:

Replacing 8 ft for L; 27° for θ; 4.625 rad/s for 
Then;




Thus, the angular acceleration of the bar AB is 
b)
To calculate the acceleration of point B using equation (1); we have:

Replacing

L = 8 ft
θ = 27°
= 4.625 rad/s
= y = 0
Then;
![a_B = [-29.31(8)cos 27^0- (4.625^2)(8sin27^0)] i +0j](https://tex.z-dn.net/?f=a_B%20%3D%20%5B-29.31%288%29cos%2027%5E0-%20%284.625%5E2%29%288sin27%5E0%29%5D%20%20%20i%20%2B0j)

Thus, the acceleration
of point B = 