---
title: "Engine crank 200 rpm, rod and piston acceleration"
subject: "Applied Mechanics (Dynamics)"
chapter: 4
key: "semester_3/applied_mechanics_dynamics:engine-crank-200rpm-acceleration"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/engine-crank-200rpm-acceleration
site: BCT Engineering
---

# Engine crank 200 rpm, rod and piston acceleration

Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies

- Asked in: 2072 Chaitra, 8 marks
- Video: [Velocity and Acceleration diagram|Slider Crank Chain|velocity and acceleration analysis of mechanism](https://www.youtube.com/watch?v=D5jUncwm1gY) — Pravinkumar Suthar
- 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-200rpm-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4

## Question

Crank AB of the engine system has a constant clockwise angular velocity of 200 rpm, which makes the angle $60^\circ$ with horizontal level. For the crank position shown in figure below. Determine the angular acceleration of the connecting rod BD and the acceleration of point D. [Figure: crank AB, $r = 0.15$ m, at $60^\circ$ to the horizontal; connecting rod BD of length $L = 0.5$ m with centre G; piston P at D.]

## Short answer

From crank speed 200 rpm, omega_AB = 20.94 rad/s cw, geometry gives beta = 15.06°. Using relative acceleration, angular acceleration of BD = 115.2 rad/s^2 counter-clockwise; a_D = 23.0 m/s^2 to the left.

## Answer

#### Geometry
Take $A$ as origin, $x$ horizontal towards the piston, $y$ upward. The crank turns clockwise, so $\omega_{AB} = -20.94$ rad/s ($200$ rpm $= 200\times 2\pi/60$). The piston $D$ moves only horizontally, so $y_D = 0$:

$$
\sin\beta = \frac{AB\sin 60^\circ}{BD} = \frac{0.15\sin 60^\circ}{0.5} = 0.2598 \;\Rightarrow\; \beta = 15.06^\circ
$$

$$
\vec r_{B/A} = (0.0750\,\hat i + 0.1299\,\hat j)\ \text{m}, \qquad \vec r_{D/B} = (0.4828\,\hat i - 0.1299\,\hat j)\ \text{m}
$$

#### Velocities
Crank pin $B$ moves on a circle about $A$:

$$
\vec v_B = \vec\omega_{AB}\times\vec r_{B/A} = (2.72\,\hat i -1.57\,\hat j)\ \text{m/s}, \qquad v_B = \omega_{AB}\,AB = 3.14\ \text{m/s}
$$

For the connecting rod, $\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B}$ with $\vec v_D = v_D\hat i$:

$$
\begin{aligned}
\hat j:&\quad 0 = -1.57 + \omega_{BD}(0.4828) \;\Rightarrow\; \omega_{BD} = 3.25\ \text{rad/s}\\
\hat i:&\quad v_D = 2.72 + \omega_{BD}(0.1299) = 3.14\ \text{m/s}
\end{aligned}
$$

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

$$
\vec a_B = -\omega_{AB}^2\,\vec r_{B/A} = (-32.9\,\hat i -57.0\,\hat j)\ \text{m/s}^2 \quad (a_B = 65.8\ \text{m/s}^2)
$$

For the rod, with $\omega_{BD} = 3.25$ rad/s (counter-clockwise) and $\vec a_D = a_D\hat i$:

$$
\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\,\vec r_{D/B}
$$

$$
\omega_{BD}^2\,\vec r_{D/B} = (5.1\,\hat i -1.4\,\hat j)\ \text{m/s}^2, \qquad \hat k\times\vec r_{D/B} = (0.1299\,\hat i + 0.4828\,\hat j)\ \text{m}
$$

Equating components:

$$
\begin{aligned}
\hat j:&\quad 0 = -57.0 +1.4 + \alpha_{BD}(0.4828) \;\Rightarrow\; \alpha_{BD} = 115.2\ \text{rad/s}^2\\
\hat i:&\quad a_D = -32.9 -5.1 + \alpha_{BD}(0.1299) = -23.0\ \text{m/s}^2
\end{aligned}
$$

**Answer: $\alpha_{BD} \approx 115.2$ rad/s² (counter-clockwise), $a_D \approx 23.0$ 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 w_AB for 200 rpm?
   A. 200 rad/s
   B. 20.94 rad/s
   C. 3.33 rad/s
   D. 125.7 rad/s
   Answer: B. w = 200 x 2 pi / 60 = 20.94 rad/s clockwise. (Key point: Crank angular velocity)

2. How is the rod angle beta found?
   A. sin(beta) = AB sin 60 deg / BD, since D has y = 0
   B. tan(beta) = AB / BD
   C. cos(beta) = AB / BD
   D. beta = 60 deg
   Answer: A. The piston lies on the horizontal axis, so the vertical drops of crank and rod match: sin(beta) = 0.15 sin60 / 0.5 = 0.2598, beta = 15.06 deg. (Key point: Rod angle)

3. How is w_BD obtained and what is its value?
   A. Set the j component of vD to zero, giving 3.25 rad/s ccw
   B. Set the i component of vD to zero, giving 20.94 rad/s
   C. w_BD = w_AB
   D. Using aB, giving 115.2 rad/s
   Answer: A. 0 = -1.57 + w_BD(0.4828) gives w_BD = 3.25 rad/s; then vD = 3.14 m/s. (Key point: Angular velocity of rod)

4. What are alpha_BD and aD?
   A. alpha_BD = 3.25 rad/s^2, aD = 3.14 m/s^2
   B. alpha_BD = 115.2 rad/s^2 ccw, aD = 23.0 m/s^2 towards the crank
   C. alpha_BD = 0, aD = 65.8 m/s^2
   D. alpha_BD = 115.2 rad/s^2 cw, aD = 23.0 m/s^2 to the right
   Answer: B. The j equation gives alpha = 55.6/0.4828 = 115.2 rad/s^2 and the i equation gives aD = -32.9 - 5.1 + 115.2(0.1299) = -23.0 m/s^2. (Key point: Angular acceleration)

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