---
title: "Slider crank velocity by instantaneous centre method"
subject: "Theory of Mechanism and Machine I"
chapter: 7
key: "semester_6/theory_of_mechanism_and_machine_i:slider-crank-velocity-instantaneous-centre"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/slider-crank-velocity-instantaneous-centre
site: BCT Engineering
---

# Slider crank velocity by instantaneous centre method

Theory of Mechanism and Machine I · Chapter 7: Kinematic Analysis of Mechanisms

- Asked in: year not given, 8 marks
- Video: [HOW TO LOCATE THE INSTANTANEOUS CENTERS II SLIDER CRANK MECHANISM](https://www.youtube.com/watch?v=_46knsBl7nA) — TECH CAD
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/slider-crank-velocity-instantaneous-centre
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-7

## Question

In a slider crank mechanism the crank OA is 150 mm long and rotates at 300 rpm anticlockwise. The connecting rod AB is 600 mm long. When the crank has turned through 40 degrees from the inner dead centre, locate the instantaneous centre of the connecting rod and find, by the instantaneous centre method, (a) the velocity of the slider B, (b) the angular velocity of the connecting rod AB, and (c) the velocity of the midpoint G of the connecting rod.

## Short answer

Locate IC of connecting rod at intersection of OA extended and perpendicular to stroke at B; v_A=4.712 m/s. Then ω_AB = 6.096 rad/s clockwise, v_B = 3.617 m/s, and velocity of midpoint G = 3.781 m/s.

## Answer

#### Configuration
Take O as origin and the line of stroke along the $x$-axis.

$$
\omega_{OA} = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s}, \qquad v_A = \omega_{OA}\cdot OA = 31.416 \times 0.150 = 4.712\ \text{m/s}
$$

$v_A$ is perpendicular to OA. Obliquity of the rod: $\sin\phi = \dfrac{OA\sin 40^\circ}{AB} = 0.1607 \Rightarrow \phi = 9.25^\circ$.

Coordinates: A $= (114.9,\ 96.4)$ mm, B $= (707.1,\ 0)$ mm.

#### Locating the instantaneous centre $I_{13}$ (connecting rod)
- The velocity of A is perpendicular to OA, so $I$ lies on the line OA produced.
- The velocity of B is along the line of stroke, so $I$ lies on the perpendicular to the line of stroke at B.

The intersection gives $I$ at $(707.1,\ 593.3)$ mm.

```text
           I  (instantaneous centre)
          /|
        /  |
      A/   |        IA = 773.1 mm
     /\    |        IB = 593.3 mm
   O/__\___|__B ->
```

$$
IA = \sqrt{(707.1 - 114.9)^2 + (593.3 - 96.4)^2} = 773.1\ \text{mm}, \qquad IB = 593.3\ \text{mm}
$$

#### (b) Angular velocity of the rod
The rod rotates about $I$, so $v_A = \omega_{AB}\cdot IA$:

$$
\omega_{AB} = \frac{v_A}{IA} = \frac{4.712}{773.1/1000} = 6.096\ \text{rad/s}
$$

(clockwise: A lies to the lower left of $I$ and moves to the upper left.)

#### (a) Velocity of slider
$$
v_B = \omega_{AB}\cdot IB = 6.096 \times \frac{593.3}{1000} = 3.617\ \text{m/s}
$$

(towards the crank end, towards O.)

Check with the analytic result $v_B = \omega r\left[\sin\theta + \dfrac{r\sin2\theta}{2\sqrt{l^2 - r^2\sin^2\theta}}\right] = 3.617$ m/s. It agrees.

#### (c) Velocity of midpoint G
Distance from I to G (G is the midpoint of AB): $IG = 620.4$ mm.

$$
v_G = \omega_{AB}\cdot IG = 6.096 \times \frac{620.4}{1000} = 3.781\ \text{m/s}
$$

perpendicular to IG.

**Answer: $v_B = 3.617$ m/s, $\omega_{AB} = 6.096$ rad/s (clockwise), $v_G = 3.781$ m/s.**

## Check yourself

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

1. How is the IC of the connecting rod found?
   A. Intersection of OA produced and the perpendicular to the line of stroke at B
   B. Intersection of OB and the perpendicular at A
   C. Midpoint of AB
   D. At the crank pivot O
   Answer: A. v_A is perpendicular to OA, so I lies on OA produced; v_B is along the stroke, so I lies on the perpendicular to the stroke at B. (Key point: Locating the IC)

2. What is v_A for a 150 mm crank at 300 rpm?
   A. 9.42 m/s
   B. 2.36 m/s
   C. 47.1 m/s
   D. 4.712 m/s
   Answer: D. omega = 31.416 rad/s, so v_A = 31.416 x 0.150 = 4.712 m/s. (Key point: Crank pin velocity)

3. How is omega_AB found by the IC method?
   A. v_B divided by AB
   B. v_A divided by IA
   C. v_A divided by OA
   D. v_A times IB
   Answer: B. The rod rotates about I, so omega_AB = v_A/IA = 4.712/0.7731 = 6.096 rad/s. (Key point: Angular velocity of rod)

4. How is the slider velocity obtained and what is it?
   A. v_B = omega_AB x IA = 4.71 m/s
   B. v_B = omega_OA x OB = 6.5 m/s
   C. v_B = omega_AB x IB = 3.617 m/s
   D. v_B = v_A cos 40 = 3.61 m/s by projection onto OA
   Answer: C. B is at distance IB = 593.3 mm from I, so v_B = 6.096 x 0.5933 = 3.617 m/s. (Key point: Slider velocity)

## Related in Theory of Mechanism and Machine I

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