---
title: "Crank OA 12 rad/s, piston B, constrained motion"
subject: "Applied Mechanics (Dynamics)"
chapter: 4
key: "semester_3/applied_mechanics_dynamics:crank-oa-12rads-piston-constrained-motion"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/crank-oa-12rads-piston-constrained-motion
site: BCT Engineering
---

# Crank OA 12 rad/s, piston B, constrained motion

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

- Asked in: 2078 Kartik, 6+2 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/crank-oa-12rads-piston-constrained-motion
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4

## Question

If crank OA rotates with an angular velocity 12 rad/s, determine the velocity of piston B, velocity of midpoint of AB and the angular velocity of rod AB at the instant shown. Define constrained motion with examples. [Figure: crank OA = 0.3 m (vertical), rod AB = 0.6 m at $30^\circ$ to the horizontal, B sliding on a vertical guide; 12 rad/s at O.]

## Short answer

By locating the instantaneous centre, v_B = 6.24 m/s (vertical), midpoint velocity = 3.60 m/s, ω_AB = 12.00 rad/s. Constrained motion is motion restricted by guides or links, e.g., a piston in a cylinder or a pendulum bob.

## Answer

Assumed layout: $OA$ is vertical, so $v_A = \omega_{OA}\cdot OA$ is horizontal. Rod $AB$ makes $30^\circ$ with the horizontal, and $B$ slides in a vertical guide, so $v_B$ is vertical. Use the instantaneous centre $I$ of rod $AB$.

#### Locating the ICR
$I$ lies on the perpendicular to $v_A$ through $A$ (the vertical line through $A$) and on the perpendicular to $v_B$ through $B$ (the horizontal line through $B$):

```text
   A *   --> vA (horizontal)
     |\
     | \  AB
   y |  \
     |   \
   I *----* B   (vB vertical)
        x
```

Horizontal and vertical distances of $B$ from $A$:

$$
x = AB\cos30^\circ = 0.6\cos30^\circ = 0.5196\ \text{m}, \qquad y = AB\sin30^\circ = 0.6\sin30^\circ = 0.300\ \text{m}
$$

So $IA = y = 0.300$ m and $IB = x = 0.5196$ m.

#### Velocities
$$
v_A = \omega_{OA}\,OA = 12\times0.3 = 3.60\ \text{m/s}
$$

$$
\omega_{AB} = \frac{v_A}{IA} = \frac{3.60}{0.300} = 12.00\ \text{rad/s}
$$

$$
v_B = \omega_{AB}\,(IB) = 12.00\times 0.5196 = 6.24\ \text{m/s (vertical)}
$$

The midpoint $M$ of $AB$ is at $IM = \sqrt{(x/2)^2 + (y/2)^2} = \sqrt{(0.2598)^2 + (0.150)^2} = 0.300$ m, so

$$
v_M = \omega_{AB}\,(IM) = 12.00\times 0.300 = 3.60\ \text{m/s}
$$

(perpendicular to $IM$).

**Answer: $v_B = 6.24$ m/s, $v_M = 3.60$ m/s, $\omega_{AB} = 12.00$ rad/s.**

**Constrained motion** is motion in which the path or position of a body is restricted by guides, pins, links or contact, so its coordinates are related by geometric equations and it cannot move freely in all directions. The restricted motion lets us find unknown velocities and accelerations from the geometry.

**Examples**
- A piston in a cylinder: it can move only along the cylinder axis.
- A pendulum bob on a rod: it is forced to move on a circular arc.
- A collar sliding on a fixed rod, or a block on a slotted guide.
- A wheel rolling without slipping: the centre speed and angular speed are related by $v = \omega r$.

## Check yourself

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

1. Where is the ICR of rod AB found?
   A. Where the perpendiculars to vA (through A) and to vB (through B) meet
   B. At the midpoint of AB
   C. At O
   D. At B
   Answer: A. The ICR lies on the perpendicular to each velocity; vA is horizontal and vB vertical, so the vertical through A meets the horizontal through B. (Key point: Locating the ICR)

2. How is w_AB computed and what is its value?
   A. w_AB = vA / IA = 3.6 / 0.300 = 12 rad/s
   B. w_AB = vA / AB = 3.6 / 0.6 = 6 rad/s
   C. w_AB = 12 x 0.3 = 3.6 rad/s
   D. w_AB = vB / IA = 20.8 rad/s
   Answer: A. vA = 12 x 0.3 = 3.6 m/s and IA = AB sin30 = 0.300 m, so w_AB = 12 rad/s. (Key point: Angular velocity of AB)

3. What is vB of the piston?
   A. 3.60 m/s
   B. 6.24 m/s vertical (w_AB x IB, with IB = 0.5196 m)
   C. 7.20 m/s
   D. 4.32 m/s
   Answer: B. vB = 12 x 0.5196 = 6.24 m/s. (Key point: Piston velocity)

4. What is constrained motion?
   A. Motion that is free in all directions
   B. Motion whose path or position is restricted by guides, pins, links or contact
   C. Motion at constant speed
   D. Motion without rotation
   Answer: B. Constraints relate the coordinates geometrically, e.g. a piston in a cylinder, helping solve for unknown velocities. (Key point: Constrained motion)

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