Question

instead of answering question calculate distance from instantaneous center to center of mass

end A moves to the right

Answer 1

Since, you do not want the answer to the asked questions, I'm going to skip the procedure, but the velocity of end B is required to find out the Instantaneous center therefore, I'll just use the results.
V_B = 1.869 in/s at an angle of 59.1^o with horizontal.

Now, we know, the instantaneous center of rotation can be found out by drawing perpendicular lines to the velocity vectors at the ends of the rod and locating the intersection point of the two lines. Velocity of end A is horizontal, therefore, perpendicular to it will be a vertical line. Similarly, a perpendicular line is drawn to the velocity vector at end B, which will be at an angle of 30.9^o with horizontal as shown below:

1.363 in/s 30.g 59.1 com 35 l O

In the figure, point C represents the Center of mass and O represents the Instantaneous Center of Rotation.
Now, We need to find out the length OC.
AD = AB Sin 35^o = 20 x 0.574 = 11.47"
BD = AB Cos 35⁰ = 16.38"
OD = BD Tan 30.9^o = 9.805"
OA = AD + OD = 11.47 + 9.805 = 21.275"
Now, AE = 10 Sin 35^o = 5.735"
Therefore, OE = OA - AE = 21.275 - 5.735 = 15.54"
And EC = 10 Cos 35^o = 8.19"
Using Pythagoras theroem,
OC = (OE² + EC²)^1/2 = (15.54² + 8.19²)^1/2 = 17.56"

Therefore, the distance between the Instantaneous center of rotation and the center of mass = 17.56 inches.

Thank you! Upvotes are appreciated.