Engine crank 2000 rpm, ICR, piston acceleration

Instantaneous centre of rotation (ICR) is the point in the plane of motion (on the body or on its imaginary extension) whose velocity is zero at that instant. At that instant the body behaves as if it were rotating about this point, so every point has speed $v = \omega\,r$, where $r$ is its distance from the ICR, and its velocity is perpendicular to the line joining it to the ICR.

Location: draw perpendiculars to the velocity directions of two points of the body; they meet at the ICR. For a body rolling without slipping, the ICR is the contact point.

Examples

• Rolling wheel: the ICR is the point of contact with the ground, so $v_{centre} = \omega r$ and the top point moves at $2\omega r$.
• Ladder sliding on a wall and floor: the ICR is the intersection of the vertical through the foot and the horizontal through the top end.
• Connecting rod of an engine: the ICR lies where the perpendicular to the piston path through the piston meets the perpendicular to the crank-pin velocity (the crank line extended).

Acceleration analysis

The given values $\omega_{BD}=61.87$ rad/s (ccw) and $v_D = 13.26$ m/s are used directly; only $\omega_{BD}$ is needed for the accelerations.

Geometry

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 $y_D = 0$:

$$\sin\beta = \frac{AB\sin 40^\circ}{BD} = \frac{0.076\sin 40^\circ}{0.203} = 0.2406 \;\Rightarrow\; \beta = 13.92^\circ$$ $$\vec r_{B/A} = (0.0582\,\hat i + 0.0489\,\hat j)\ \text{m}, \qquad \vec r_{D/B} = (0.1970\,\hat i - 0.0489\,\hat j)\ \text{m}$$

Accelerations

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

$$\vec a_B = -\omega_{AB}^2\,\vec r_{B/A} = (-2553.8\,\hat i -2142.9\,\hat j)\ \text{m/s}^2 \quad (a_B = 3333.7\ \text{m/s}^2)$$

For the rod, with $\omega_{BD} = 61.87$ rad/s (counter-clockwise) and $\vec a_D = a_D\hat i$:

$$\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\,\vec r_{D/B}$$ $$\omega_{BD}^2\,\vec r_{D/B} = (754.2\,\hat i -187.0\,\hat j)\ \text{m/s}^2, \qquad \hat k\times\vec r_{D/B} = (0.0489\,\hat i + 0.1970\,\hat j)\ \text{m}$$

Equating components:

$$\begin{aligned} \hat j:&\quad 0 = -2142.9 +187.0 + \alpha_{BD}(0.1970) \;\Rightarrow\; \alpha_{BD} = 9926.6\ \text{rad/s}^2\\ \hat i:&\quad a_D = -2553.8 -754.2 + \alpha_{BD}(0.0489) = -2823.1\ \text{m/s}^2 \end{aligned}$$

Answer: $\alpha_{BD} \approx 9926.6$ rad/s² (counter-clockwise), $a_D \approx 2823.1$ m/s² (towards the crank, i.e. to the left).