---
title: "Velocity analysis of slider-crank mechanism"
subject: "Engineering Mechanics"
chapter: 5
key: "semester_3/engineering_mechanics:crank-slider-velocity-numerical"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/crank-slider-velocity-numerical
site: BCT Engineering
---

# Velocity analysis of slider-crank mechanism

Engineering Mechanics · Chapter 5: Kinematics of Rigid Bodies

- Asked in: year not given, 8 marks
- Video: [Velocity Diagram | Slider Crank Mechanism | Dr. Veeresh Fuskele](https://www.youtube.com/watch?v=j36wPW4bGIs) — MechTuts
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/crank-slider-velocity-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-5

## Question

In a slider-crank mechanism the crank OA is 200 mm long and rotates counter-clockwise at 300 rpm. The connecting rod AB is 800 mm long and the slider B moves along a horizontal line passing through the crank centre O. When the crank makes 60° with the line of stroke OB (crank pin A above the line), determine the velocity of the slider B and the angular velocity of the connecting rod. Verify using the instantaneous centre method.

## Short answer

ω_crank = 31.42 rad/s, v_A = 6.283 m/s. Using relative velocity, ω_AB = 4.022 rad/s clockwise, v_B = 6.138 m/s toward O. Instantaneous centre method gives same results.

## Answer

**Given:** $OA = r = 0.2$ m, $AB = l = 0.8$ m, $N = 300$ rpm counter-clockwise, $\theta = 60^\circ$.

```text
        A (pin)
       /\
      /  \  AB = 0.8
     /60  \
   O-------B-------> slider (horizontal guide)
```

#### Crank speed

$$
\omega_{OA} = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s}
\qquad
v_A = \omega_{OA}\,r = 6.283\ \text{m/s}
$$

$v_A$ is perpendicular to OA. Its components: $v_{Ax} = -v_A\sin60^\circ = -5.441$, $v_{Ay} = v_A\cos60^\circ = 3.142$ m/s.

#### Geometry of the rod
$\sin\phi = \dfrac{r\sin60^\circ}{l} = \dfrac{0.1732}{0.8} = 0.2165$, so the rod angle $\phi = 12.50^\circ$. Horizontal projection of AB: $d = l\cos\phi = 0.7810$ m.

#### Relative velocity method
$\mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A}$, with $\mathbf{v}_B$ horizontal and $\mathbf{v}_{B/A}$ perpendicular to AB (magnitude $\omega_{AB}\times l$).

Vertical components (B has none): $v_{Ay} = \omega_{AB}\,d$

$$
\omega_{AB} = \frac{3.142}{0.7810} = 4.022\ \text{rad/s (clockwise)}
$$

Horizontal components:

$$
v_B = v_{Ax} - \omega_{AB}(AB\text{ vertical drop}) = -5.441 - (4.022)(0.1732) = -6.138\ \text{m/s}
$$

The negative sign means the slider moves toward O (to the left).

#### Verification by instantaneous centre
I is the intersection of the perpendicular to $v_A$ at A (line OA extended) and the perpendicular to $v_B$ at B (vertical through B).

- $x_B = 0.2\cos60^\circ + 0.7810 = 0.8810$ m
- I lies on line OA at that x: $y_I = x_B\tan60^\circ = 1.5260$ m, so $IB = 1.526$ m
- $IA = \sqrt{(x_B - 0.1)^2 + (y_I - 0.1732)^2} = 1.562$ m

$$
\omega_{AB} = \frac{v_A}{IA} = \frac{6.283}{1.562} = 4.022\ \text{rad/s}
$$

$$
v_B = \omega_{AB}\times IB = 4.022 \times 1.526 = 6.138\ \text{m/s}
$$

The two methods agree.

**Answer: Slider velocity = 6.14 m/s toward the crank centre; ω of connecting rod = 4.02 rad/s clockwise.**

## Check yourself

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

1. The crank rotates at 300 rpm with OA = 0.2 m. The velocity of A is:
   A. 31.4 m/s
   B. 6.283 m/s, perpendicular to OA
   C. 60 m/s
   D. 6.283 m/s along OA
   Answer: B. omega = 2 pi 300/60 = 31.416 rad/s, v_A = omega r = 6.283 m/s perpendicular to OA. (Key point: Crank pin speed)

2. The rod angle is found from:
   A. sin(phi) = r sin60 / l = 0.2165
   B. cos(phi) = r/l
   C. tan(phi) = l/r
   D. phi = 60 deg
   Answer: A. B lies on the line of stroke, so the height of A, r sin60, equals l sin(phi); hence sin(phi) = r sin(theta)/l, phi = 12.5 deg. (Key point: Rod angle)

3. Why does omega_AB equal v_Ay / d?
   A. B has no vertical velocity, so the vertical components of v_A and v_B/A cancel
   B. A has no vertical velocity
   C. d is the rod length
   D. The rod does not rotate
   Answer: A. v_B is horizontal, so v_Ay = omega_AB d, giving 4.02 rad/s clockwise. (Key point: Rod angular velocity)

4. Where is the instantaneous centre of the rod located?
   A. At A
   B. At the intersection of the perpendiculars to v_A at A (line OA extended) and to v_B at B (vertical)
   C. At O
   D. At the midpoint of AB
   Answer: B. It gives omega = v_A/IA = 4.022 rad/s and v_B = omega x IB = 6.14 m/s, agreeing with the relative velocity method. (Key point: Instantaneous centre)

## Related in Engineering Mechanics

- [Acceleration analysis of slider-crank](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Acrank-slider-acceleration-numerical) — Engineering Mechanics, chapter 5

## Related in other subjects

- [Engine crank 1800 rpm, connecting rod velocity](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aengine-crank-1800rpm-rod-velocity) — Applied Mechanics (Dynamics), chapter 4
- [Slider crank velocity by instantaneous centre method](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aslider-crank-velocity-instantaneous-centre) — Theory of Mechanism and Machine I, chapter 7
- [Engine crank 200 rpm, rod and piston acceleration](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aengine-crank-200rpm-acceleration) — Applied Mechanics (Dynamics), chapter 4
- [Static force analysis of slider crank](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Astatic-force-analysis-slider-crank) — Theory of Mechanism and Machine I, chapter 8