---
title: "Velocities in plane motion by instantaneous centre"
subject: "Theory of Mechanism and Machine I"
chapter: 7
key: "semester_6/theory_of_mechanism_and_machine_i:ladder-sliding-wheel-rolling-velocities"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/ladder-sliding-wheel-rolling-velocities
site: BCT Engineering
---

# Velocities in plane motion by instantaneous centre

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

- Asked in: year not given, 3+3 marks
- Video: [Instantaneous Center of Zero Velocity (learn to solve any problem step by step)](https://www.youtube.com/watch?v=1aQ9EZGMdDk) — Question Solutions
- 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/ladder-sliding-wheel-rolling-velocities
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-7

## Question

(a) A ladder AB, 4 m long, rests with its foot A on a horizontal floor and its top B against a vertical wall. The ladder makes 60 degrees with the floor and the foot A is pulled away from the wall at 2 m/s. Locate the instantaneous centre and find the angular velocity of the ladder, the velocity of B and the velocity of the midpoint of the ladder. (b) A wheel of radius 0.3 m rolls without slipping on a horizontal road with its centre moving at 6 m/s. Find the velocity of the topmost point of the rim and of the rim point at the same level as the centre.

## Short answer

The instantaneous centre is at the intersection of perpendiculars to A and B velocities, opposite the wall-floor corner. Angular velocity 0.577 rad/s anticlockwise, velocity of B 1.155 m/s downward, midpoint velocity 1.155 m/s. For the wheel, contact point is IC, angular velocity 20 rad/s; topmost point 12.0 m/s forward, side point 8.485 m/s at 45° to horizontal.

## Answer

#### (a) Ladder
The velocity of A is horizontal and the velocity of B is vertical (along the wall). Draw perpendiculars to these velocities at A and B (a vertical line through A and a horizontal line through B). They meet at the instantaneous centre $I$, which is the corner of the rectangle formed by the floor, the wall and the two perpendiculars, opposite the wall-floor corner O.

```text
   wall
   |
   B -------------- I    <- horizontal line through B
   | \              |
   |    \  ladder   |    <- vertical line through A
   |       \        |
   O------------- A ---> 2 m/s     floor
```

Distances from $I$: $IA = L\sin 60^\circ = 3.464$ m (equal to the height of B), $IB = L\cos 60^\circ = 2.0$ m (equal to the distance of A from the wall).

$$
\omega_{AB} = \frac{v_A}{IA} = \frac{2}{3.464} = 0.577\ \text{rad/s (anticlockwise)}
$$

$$
v_B = \omega_{AB}\cdot IB = 0.577 \times 2.0 = 1.155\ \text{m/s (downwards)}
$$

The midpoint M of the ladder is the centre of the rectangle (the diagonals OI and AB bisect each other), so $IM = L/2 = 2$ m.

$$
v_M = \omega_{AB}\cdot IM = 0.577 \times 2 = 1.155\ \text{m/s}
$$

directed perpendicular to $IM$.

#### (b) Rolling wheel
For pure rolling, the point of contact with the road is the instantaneous centre. Angular velocity:

$$
\omega = \frac{v_C}{r} = \frac{6}{0.3} = 20.0\ \text{rad/s}
$$

- Topmost point (distance $2r = 0.6$ m from the IC): $v_{top} = 20 \times 0.6 = 12.0$ m/s, horizontal, in the direction of travel.
- Rim point at the level of the centre (distance $r\sqrt{2}$ from the IC): $v = 20 \times 0.3\sqrt{2} = 8.485$ m/s, at $45^\circ$ to the horizontal.

**Answer: (a) $\omega = 0.577$ rad/s, $v_B = 1.155$ m/s, $v_M = 1.155$ m/s; (b) top $= 12.0$ m/s, side point $= 8.485$ m/s.**

## Check yourself

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

1. Where is the instantaneous centre of the sliding ladder?
   A. At the midpoint of the ladder
   B. At A
   C. At the intersection of the vertical through A and the horizontal through B
   D. At B
   Answer: C. v_A is horizontal and v_B vertical, so perpendiculars at A (vertical) and B (horizontal) meet at I. (Key point: Ladder IC location)

2. With IA = L sin60 = 3.464 m and v_A = 2 m/s, the ladder's angular velocity is:
   A. 0.577 rad/s
   B. 1.732 rad/s
   C. 1.155 rad/s
   D. 0.5 rad/s
   Answer: A. omega = v_A/IA = 2/3.464 = 0.577 rad/s. (Key point: Ladder angular velocity)

3. Why is IM = L/2 for the midpoint M of the ladder?
   A. M is the IC
   B. M is the centre of the circle through A, B and I (AB is a diameter)
   C. IM is perpendicular to AB
   D. The ladder is rolling
   Answer: B. A, B, I lie on a circle with AB as diameter, so M is the centre and IM = L/2 = 2 m. (Key point: Ladder midpoint)

4. For a wheel rolling without slipping, the instantaneous centre is:
   A. The wheel centre
   B. The top of the rim
   C. The rim point level with the centre
   D. The point of contact with the road
   Answer: D. The contact point has zero velocity in pure rolling, so it is the IC. (Key point: Rolling wheel IC)

5. A wheel of radius 0.3 m has its centre moving at 6 m/s. What is the velocity of the top point?
   A. 6 m/s
   B. 12 m/s
   C. 8.485 m/s
   D. 18 m/s
   Answer: B. The top is 2r from the IC: 20 x 0.6 = 12 m/s. (Key point: Top point speed)

## Related in Theory of Mechanism and Machine I

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