Question

2. Rod ABD is guided by wheels at A and B that roll in horizontal and vertical tracks. Knowing that at the instant shown, B = 60 degrees, and the velocity of wheel B is 800 mm/s downward, determine: a. Angular velocity of the rod b. Velocity of Point D (10 pts) (7 pts) (direction not required) A 300 mm

AB =300mm , BD=300mm.

AB = 300mm, BD=300mm.

Answer 1

Given

AB = BD = 300 mm VB = 800 mm/s B = 60°

I'm taking the angle $\beta$ as angle between the rod and vertical(I can't find it in the figure and that is only one that looks )

09

The motion of the rod will like in the figure below

0 B VA 300 MAN

a)

The perpendicular drawn to the velocity at A and B intersect at O, therefore O is the center of rotation of the rod.

Angular velocity of the rod

$\\ \omega_{ABD}=\frac{V_B}{OB}=\frac{V_B}{AB\sin\beta}=\frac{800}{300 \sin60\degree}= 3.0792\ rad/s$

b)

D O VO C 30 A 60 300 Congruent Triangle

The velocity of D can be written as

$\\ V_D=\omega_{ABD}*OD=\omega_{ABD}*\sqrt{OC^2+CD^2}=\omega_{ABD}\sqrt{(2*AB \sin60\degree)^2+(AB \cos60\degree)^2} \\ V_D=3.0792 *540.83= 1665.3\ mm/s$