--- title: "Coriolis meaning; engine crank 2000 rpm" subject: "Applied Mechanics (Dynamics)" chapter: 4 key: "semester_3/applied_mechanics_dynamics:engine-crank-2000rpm-coriolis" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/engine-crank-2000rpm-coriolis site: BCT Engineering --- # Coriolis meaning; engine crank 2000 rpm Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies - Asked in: 2074 Asoj, 8 marks - 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-coriolis - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4 ## Question What is the meaning of coriolis's acceleration in plane motion of rigid body? Crank AB of the engine system shown in figure below, has a constant clockwise angular velocity of 2000 rev/min. For the crank position as shown in figure below, determine the angular acceleration of the connecting rod 'BD' and the acceleration of point 'D'. Given that the value of $\omega_{BD} = 61.9$ rad/sec and the angle made by rod BD with horizontal $\beta = 13.9$. [Figure: crank AB = 75 mm at $40^\circ$ to the horizontal, rod BD of length $l = 200$ mm, piston at D.] ## Short answer Coriolis acceleration is 2 omega x v_rel, occurring when a point slides relative to a rotating body. For the rigid connecting rod it is zero. Using omega_BD = 61.9 rad/s, angular acceleration of BD = 9943.0 rad/s^2 ccw, a_D = 2784.6 m/s^2 towards crank. ## Answer **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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the Coriolis acceleration of a point sliding on a rotating body? A. w^2 r B. 2 w x v_rel, perpendicular to the relative velocity C. alpha x r D. a_rel only Answer: B. The Coriolis term 2 w x v_rel arises when a point moves relative to a rotating frame; its magnitude is 2 w v_rel. (Key point: Meaning of Coriolis acceleration) 2. Why is the Coriolis term zero for points B and D on the connecting rod? A. The rod does not rotate B. B and D belong to the same rigid body, so there is no relative sliding C. The crank speed is constant D. D moves in a straight line Answer: B. Coriolis acceleration needs relative motion on the rotating body; two points fixed in the same rigid rod have none. (Key point: Coriolis in engine) 3. What is the acceleration of B, given constant crank speed? A. Zero B. Normal only, w_AB^2 times AB, directed towards A C. Tangential only D. alpha_BD times BD Answer: B. With alpha_AB = 0, aB = -w_AB^2 rB/A, about 3289.9 m/s^2 towards A. (Key point: Crank pin acceleration) 4. Which results are obtained for the rod and piston? A. alpha_BD = 9943 rad/s^2 ccw; aD = 2784.6 m/s^2 towards the crank B. alpha_BD = 61.9 rad/s^2; aD = 3289.9 m/s^2 C. alpha_BD = 0; aD = 743.7 m/s^2 D. alpha_BD = 9943 rad/s^2 cw; aD to the right Answer: A. The j equation 0 = -2114.7 + 184.7 + alpha(0.1941) gives alpha = 9943 rad/s^2, then aD = -2784.6 m/s^2 (left). 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