---
title: "Piston velocity and acceleration, equivalent system"
subject: "Theory of Mechanism and Machine II"
chapter: 1
key: "semester_7/theory_of_mechanism_and_machine_ii:piston-velocity-acceleration-derivation"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/piston-velocity-acceleration-derivation
site: BCT Engineering
---

# Piston velocity and acceleration, equivalent system

Theory of Mechanism and Machine II · Chapter 1: Engine Force Analysis

- Asked in: year not given, 3+5 marks
- Video: [Analytical Method for Velocity & Acceleration of a Piston](https://www.youtube.com/watch?v=FeJzfjsEjAo) — Sachendra Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/piston-velocity-acceleration-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-1

## Question

(a) State the conditions for a connecting rod to be replaced by a dynamically equivalent system of two point masses.
(b) Derive expressions for the velocity and acceleration of the piston of a reciprocating engine in terms of crank angle $\theta$, crank radius $r$, connecting rod length $l$ and obliquity ratio $n = l/r$.

## Short answer

For dynamic equivalence, total mass of two point masses equals rod mass, their center of gravity coincides with rod’s, and moment of inertia about CG matches. Piston velocity ≈ rω[sinθ + sin2θ/(2n)], acceleration ≈ rω²[cosθ + cos2θ/n] where n = l/r.

## Answer

#### (a) Conditions for a dynamically equivalent system
A connecting rod of mass $m$ can be replaced by two point masses (one at the small end, one at the big end) if:

1. The sum of the two masses equals the mass of the rod: $m_1 + m_2 = m$.
2. The centre of gravity of the two masses coincides with the centre of gravity of the rod: $m_1 l_1 = m_2 l_2$.
3. The moment of inertia of the two masses about the centre of gravity equals that of the rod: $m_1 l_1^2 + m_2 l_2^2 = m k_G^2$ (so $l_1 l_2 = k_G^2$).

Here $l_1, l_2$ are the distances of the two masses from the centre of gravity and $k_G$ is the radius of gyration of the rod about its centre of gravity. If the third condition is not met, a correcting couple is needed.

#### (b) Piston velocity and acceleration
```text
        P            x measured from IDC
   O----+----------+------> x
   crank O, crank pin A, piston P
   OA = r, AP = l, angle AOP = theta,
   angle OPA = phi
```
Let the piston move a distance $x$ from the inner dead centre (IDC). From the geometry,

$$
x = r(1-\cos\theta) + l(1-\cos\phi)
$$

Also $l\sin\phi = r\sin\theta$, so $\sin\phi = \dfrac{\sin\theta}{n}$ and $\cos\phi = \dfrac{\sqrt{n^2-\sin^2\theta}}{n}$. Hence

$$
x = r\left[(1-\cos\theta) + n - \sqrt{n^2-\sin^2\theta}\right]
$$

**Velocity.** With $\theta = \omega t$ and $\omega$ constant, $v = \dfrac{dx}{dt} = \omega\dfrac{dx}{d\theta}$:

$$
v = r\omega\left[\sin\theta + \frac{\sin 2\theta}{2\sqrt{n^2-\sin^2\theta}}\right]
$$

For $n \ge 4$, $\sin^2\theta$ is small compared with $n^2$, so $\sqrt{n^2-\sin^2\theta}\approx n$:

$$
v \approx r\omega\left[\sin\theta + \frac{\sin 2\theta}{2n}\right]
$$

**Acceleration.** Differentiating once more, $a = \omega\dfrac{dv}{d\theta}$:

$$
a \approx r\omega^2\left[\cos\theta + \frac{\cos 2\theta}{n}\right]
$$

**Special positions**

| Position | $\theta$ | $a$ |
|---|---|---|
| IDC | $0^\circ$ | $r\omega^2\left(1+\frac{1}{n}\right)$ |
| ODC | $180^\circ$ | $-r\omega^2\left(1-\frac{1}{n}\right)$ |

The velocity is zero at both dead centres and the acceleration is greatest at IDC. The inertia force of the reciprocating parts is $F_I = m\,a$ acting opposite to the acceleration.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which set of conditions must be met for a connecting rod to be replaced by a dynamically equivalent system of two point masses?
   A. Total mass equal to rod mass; centre of gravity unchanged; moment of inertia about the centre of gravity unchanged
   B. Total mass equal to rod mass; both masses equal; moment of inertia unchanged
   C. Centre of gravity unchanged and the masses placed at the rod's ends only; mass need not be conserved
   D. Total mass equal to rod mass; the masses placed at equal distances from the centre of gravity; no condition on inertia
   Answer: A. The three conditions are m1+m2=m, m1 l1 = m2 l2 (same CG) and m1 l1^2 + m2 l2^2 = m kG^2 (same moment of inertia). (Key point: Equivalent system conditions)

2. In terms of the distances l1, l2 of the two masses from the rod's centre of gravity and the radius of gyration kG, the inertia condition reduces to:
   A. l1 + l2 = kG
   B. l1 l2 = kG^2
   C. l1^2 + l2^2 = kG^2
   D. l1 = l2 = kG
   Answer: B. Combining m1 l1 = m2 l2 with m1 l1^2 + m2 l2^2 = m kG^2 gives l1 l2 = kG^2. (Key point: Third condition)

3. Which relation between the crank angle theta and the rod angle phi is used to get the piston displacement x?
   A. sin(phi) = n sin(theta)
   B. sin(phi) = sin(theta)/n
   C. cos(phi) = sin(theta)/n
   D. tan(phi) = n tan(theta)
   Answer: B. The vertical heights of the crank pin match: l sin(phi) = r sin(theta), so sin(phi) = sin(theta)/n. (Key point: Piston displacement geometry)

4. For n >= 4, the piston velocity is approximately:
   A. r*w*[sin(theta) + sin(2 theta)/(2n)]
   B. r*w*[cos(theta) + cos(2 theta)/n]
   C. r*w*[sin(theta) + sin(2 theta)/n]
   D. r*w*[sin(theta) - sin(2 theta)/(2n)]
   Answer: A. Replacing sqrt(n^2 - sin^2 theta) by n in the exact expression gives v = r w [sin(theta) + sin(2 theta)/(2n)]. (Key point: Approximate velocity)

5. What is the piston acceleration at the inner dead centre (theta = 0)?
   A. Zero
   B. r w^2 (1 - 1/n)
   C. r w^2 (1 + 1/n)
   D. - r w^2 (1 + 1/n)
   Answer: C. At theta = 0, cos(theta)=1 and cos(2 theta)=1 so a = r w^2 (1 + 1/n), the greatest value; velocity is zero there. (Key point: Acceleration at dead centres)

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