Coriolis meaning; engine crank 2000 rpm

Coriolis acceleration is the extra acceleration a point has when it moves relative to a rotating (moving) frame or body. If a point $P$ moves with relative velocity $\vec v_{rel}$ on a body rotating with angular velocity $\vec\omega$, then

$$\vec a_P = \vec a_{P'} + \vec a_{rel} + 2\vec\omega\times\vec v_{rel}$$

where $\vec a_{P'}$ is the acceleration of the coincident point of the rotating body and $\vec a_{rel}$ is the acceleration of $P$ relative to the body. The term $\vec a_c = 2\vec\omega\times\vec v_{rel}$ is the Coriolis acceleration. Its magnitude is $2\omega v_{rel}$ and its direction is that of $\vec v_{rel}$ rotated through $90^\circ$ in the sense of $\omega$. It exists only when the point slides on the rotating body (for example a collar sliding on a rotating arm, or a slider in a slotted rotating link); it is zero for two points fixed in the same rigid body, such as the crank pin and piston pin of a connecting rod.

In the engine below, points $B$ and $D$ both belong to the rigid rod $BD$, so there is no relative sliding and the Coriolis term is zero; the relative acceleration has only the $\alpha r$ and $\omega^2 r$ parts.

Geometry

Use $\omega_{BD}=61.9$ rad/s (ccw) as given.

Take $A$ as origin, $x$ horizontal towards the piston, $y$ upward. The crank turns clockwise, so $\omega_{AB} = -209.44$ rad/s ($2000$ rpm $= 2000\times 2\pi/60$). The piston $D$ moves only horizontally, so $y_D = 0$:

$$\sin\beta = \frac{AB\sin 40^\circ}{BD} = \frac{0.075\sin 40^\circ}{0.2} = 0.2410 \;\Rightarrow\; \beta = 13.95^\circ$$

This agrees with the given $\beta = 13.9^\circ$.

$$\vec r_{B/A} = (0.0575\,\hat i + 0.0482\,\hat j)\ \text{m}, \qquad \vec r_{D/B} = (0.1941\,\hat i - 0.0482\,\hat j)\ \text{m}$$

Accelerations

The crank speed is constant ($\alpha_{AB}=0$), so $B$ has only a normal acceleration towards $A$:

$$\vec a_B = -\omega_{AB}^2\,\vec r_{B/A} = (-2520.2\,\hat i -2114.7\,\hat j)\ \text{m/s}^2 \quad (a_B = 3289.9\ \text{m/s}^2)$$

For the rod, with $\omega_{BD} = 61.90$ rad/s (counter-clockwise) and $\vec a_D = a_D\hat i$:

$$\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\,\vec r_{D/B}$$ $$\omega_{BD}^2\,\vec r_{D/B} = (743.7\,\hat i -184.7\,\hat j)\ \text{m/s}^2, \qquad \hat k\times\vec r_{D/B} = (0.0482\,\hat i + 0.1941\,\hat j)\ \text{m}$$

Equating components:

$$\begin{aligned} \hat j:&\quad 0 = -2114.7 +184.7 + \alpha_{BD}(0.1941) \;\Rightarrow\; \alpha_{BD} = 9943.0\ \text{rad/s}^2\\ \hat i:&\quad a_D = -2520.2 -743.7 + \alpha_{BD}(0.0482) = -2784.6\ \text{m/s}^2 \end{aligned}$$

Answer: $\alpha_{BD} \approx 9943.0$ rad/s² (counter-clockwise), $a_D \approx 2784.6$ m/s² (towards the crank, i.e. to the left).