--- title: "Engine crank 2000 rpm, ICR, piston acceleration" subject: "Applied Mechanics (Dynamics)" chapter: 4 key: "semester_3/applied_mechanics_dynamics:engine-crank-2000rpm-acceleration-icr" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/engine-crank-2000rpm-acceleration-icr site: BCT Engineering --- # Engine crank 2000 rpm, ICR, piston acceleration Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies - Asked in: 2079 Bhadra, 2+8 marks - Video: [Dynamics | Angular Velocity of connecting rod | Velocity of piston | Applied mechanics 2 | Numerical](https://www.youtube.com/watch?v=9QFa0Go_Fhs) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/engine-crank-2000rpm-acceleration-icr - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4 ## Question Define Instantaneous centre of rotation (ICR) with examples. Crank AB of the engine system has a constant clockwise angular velocity of 2000 rpm. For the crank position shown, calculate angular acceleration of rod BD and acceleration of piston P (point D). [Take $\omega_{BD} = 61.87$ rad/s (ccw) and $v_D = 13.2558$ m/s ($\to$) (if necessary)] [Figure: crank AB = 7.6 cm at $40^\circ$ to the horizontal, rod BD = 20.3 cm, piston P at D.] ## Short answer ICR: point with zero instantaneous velocity; e.g., rolling wheel contact, ladder intersection. Using given omega_BD = 61.87 rad/s and relative acceleration, angular acceleration of BD = 9926.6 rad/s^2 counter-clockwise, piston acceleration a_D = 2823.1 m/s^2 towards the crank. ## Answer **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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the ICR of a rod such as the connecting rod found? A. Intersect perpendiculars to the velocity directions of two points of the body B. Intersect lines parallel to the velocities C. Take the midpoint of the rod D. Take the crank pivot Answer: A. The ICR lies on the perpendicular to each point's velocity, so two perpendiculars meet at it. (Key point: ICR location) 2. Where is the ICR of a wheel rolling without slipping? A. At the wheel centre B. At the top point C. At the point of contact with the ground D. At infinity Answer: C. The contact point has zero velocity, so the centre moves at w r and the top at 2 w r. (Key point: ICR for rolling) 3. Why does B have only a normal acceleration towards A? A. The crank speed is constant, so alpha_AB = 0 B. B is at rest C. The rod is rigid D. The piston is fixed Answer: A. With constant crank speed the tangential part alpha r is zero, leaving aB = w^2 AB towards A. (Key point: Acceleration of B) 4. Which condition gives alpha_BD from aD = aB + alpha k x rD/B - w^2 rD/B? A. The i component of aD is zero B. The j component of aD is zero because the piston moves horizontally C. aD equals aB D. alpha_BD equals w_BD Answer: B. The j equation 0 = -2142.9 + 187.0 + alpha(0.1970) gives alpha_BD = 9926.6 rad/s^2, then aD = -2823.1 m/s^2 (towards the crank). 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