Crank and slotted lever velocity analysis

Vector equation

Let B$_2$ be the crank pin (block) and B$_3$ the coincident point on the lever. The block slides along the lever, so

$$\vec v_{B_2} = \vec v_{B_3} + \vec v_{B_2B_3}$$

• $v_{B_2}$: magnitude $\omega\cdot O_1B = 10 \times 0.12 = 1.20$ m/s, perpendicular to $O_1B$ (horizontal, towards the left, since the crank is vertical and rotates anticlockwise).
• $v_{B_3}$: perpendicular to $O_2B$ (unknown magnitude).
• $v_{B_2B_3}$: along the slot $O_2B$ (unknown magnitude).

Geometry

$O_2B = \sqrt{300^2 + 120^2} = 323.1$ mm. The slot makes an angle $\alpha = \tan^{-1}(120/300) = 21.80^\circ$ with the horizontal, so $\cos\alpha = 0.9285$ and $\sin\alpha = 0.3714$.

Velocity polygon (resolving $v_{B_2}$)

         b2 <---- o      o: pole
          \ ' |
  along    \ ' | perp. to O2B
  slot      \  |
             b3

Draw $o b_2$ = 1.2 m/s horizontal. Through $o$ draw a line perpendicular to $O_2B$ (direction of $v_{B_3}$), and through $b_2$ draw a line parallel to $O_2B$ (direction of $v_{B_2B_3}$). They meet at $b_3$. Resolving $v_{B_2}$ in these two directions:

$$v_{B_2B_3} = v_{B_2}\cos\alpha = 1.2 \times 0.9285 = 1.114\ \text{m/s}$$ $$v_{B_3} = v_{B_2}\sin\alpha = 1.2 \times 0.3714 = 0.446\ \text{m/s}$$

Angular velocity of lever

$$\omega_{lever} = \frac{v_{B_3}}{O_2B} = \frac{0.446}{323.1/1000} = 1.379\ \text{rad/s (anticlockwise)}$$

Sense: B$_3$ lies up and to the left of $O_2$, and its velocity (perpendicular to the lever) points down and to the left, so the lever turns anticlockwise about $O_2$.

Velocity of D

$$v_D = \omega_{lever}\cdot O_2D = 1.379 \times 0.5 = 0.690\ \text{m/s}$$

perpendicular to the lever (D is on the opposite side of $O_2$, so it moves in the opposite direction to $B_3$).

Answer: $\omega_{lever} = 1.379$ rad/s anticlockwise, sliding velocity $= 1.114$ m/s, $v_D = 0.690$ m/s.