---
title: "Accelerations of points on rolling wheel"
subject: "Engineering Mechanics"
chapter: 5
key: "semester_3/engineering_mechanics:rolling-wheel-acceleration-points"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rolling-wheel-acceleration-points
site: BCT Engineering
---

# Accelerations of points on rolling wheel

Engineering Mechanics · Chapter 5: Kinematics of Rigid Bodies

- Asked in: year not given, 8 marks
- Video: [Rigid Bodies Relative Motion Analysis: Acceleration Dynamics (step by step)](https://www.youtube.com/watch?v=ee_LrUTVE5o) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rolling-wheel-acceleration-points
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-5

## Question

A wheel of radius 0.5 m rolls without slipping on a horizontal road. At a certain instant the centre O has a velocity of 6 m/s and an acceleration of 2 m/s², both in the forward direction. Find the angular velocity and angular acceleration of the wheel, and the accelerations of (a) the point of contact C with the road, (b) the top point T, and (c) the point F on the horizontal diameter at the front of the wheel.

## Short answer

Use rolling condition: ω = v_O/r = 12 rad/s clockwise, α = a_O/r = 4 rad/s² clockwise. Apply relative acceleration a_P = a_O + a_t + a_n. a_C = 72.0 m/s² upward; a_T = 72.1 m/s² at 86.8° below horizontal; a_F = 70.0 m/s² backward, 1.6° below horizontal.

## Answer

**Given:** $r = 0.5$ m, $v_O = 6$ m/s, $a_O = 2$ m/s² (both forward, +x); rolling without slipping.

#### Angular motion
The wheel rolls forward, so the rotation is clockwise.

$$
\omega = \frac{v_O}{r} = \frac{6}{0.5} = 12.0\ \text{rad/s} \qquad \alpha = \frac{a_O}{r} = \frac{2}{0.5} = 4.0\ \text{rad/s}^2
$$

#### Relative acceleration equation
For any point P: $\mathbf{a}_P = \mathbf{a}_O + \mathbf{a}_{P/O}$, where $\mathbf{a}_{P/O}$ has a tangential part $\alpha r$ (perpendicular to OP, in the sense of $\alpha$) and a normal part $\omega^2 r$ directed from P toward O.

$$
\omega^2 r = 12^2 \times 0.5 = 72.0\ \text{m/s}^2 \qquad \alpha r = 4 \times 0.5 = 2.0\ \text{m/s}^2
$$

For a clockwise rotation, the tangential part points backward at the bottom point, forward at the top point and downward at the front point.

#### (a) Contact point C (directly below O)
- $a_O = 2$ m/s² forward
- Tangential: $\alpha r = 2$ m/s² backward
- Normal: $\omega^2 r = 72$ m/s² upward (toward O)

$$
a_{Cx} = 2 - 2 = 0 \qquad a_{Cy} = 72.0\ \text{m/s}^2 \Rightarrow a_C = 72.0\ \text{m/s}^2\ \text{upward}
$$

#### (b) Top point T
- Tangential: 2 m/s² forward; normal: 72 m/s² downward

$$
a_{Tx} = 2 + 2 = 4.0,\quad a_{Ty} = -72.0 \Rightarrow a_T = 72.11\ \text{m/s}^2
$$

directed $\tan^{-1}(72/4) = 86.8^\circ$ below the horizontal (forward).

#### (c) Front point F (level with O)
- Tangential: 2 m/s² downward; normal: 72 m/s² backward (toward O)

$$
a_{Fx} = 2 - 72 = -70.0,\quad a_{Fy} = -2.0 \Rightarrow a_F = 70.03\ \text{m/s}^2
$$

directed backward at $\tan^{-1}(2/70) = 1.6^\circ$ below the horizontal.

**Answer: ω = 12 rad/s clockwise, α = 4 rad/s² clockwise; a_C = 72.0 m/s² (vertically up); a_T = 72.1 m/s² (86.8° below the horizontal); a_F = 70.0 m/s² (backward, 1.6° below the horizontal).**

## Check yourself

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

1. A wheel rolls with v_O = 6 m/s, a_O = 2 m/s^2 and r = 0.5 m. Then omega and alpha are:
   A. 12 rad/s and 4 rad/s^2
   B. 3 rad/s and 1 rad/s^2
   C. 12 rad/s and 2 rad/s^2
   D. 6 rad/s and 2 rad/s^2
   Answer: A. omega = v/r and alpha = a/r for rolling without slipping. (Key point: Omega and alpha)

2. For point P on the wheel, a_P/O has components:
   A. Tangential alpha r and normal omega^2 r directed from P toward O
   B. Tangential omega^2 r and normal alpha r
   C. Only alpha r
   D. Both directed away from O
   Answer: A. These are 2 and 72 m/s^2 here. (Key point: Relative acceleration parts)

3. The acceleration of the contact point C is:
   A. 0
   B. 72 m/s^2 upward (the horizontal parts cancel)
   C. 74 m/s^2 forward
   D. 2 m/s^2 backward
   Answer: B. a_Cx = 2 - 2 = 0 and the normal part 72 is upward toward O, so the contact point is not at rest in acceleration. (Key point: Contact point)

4. The acceleration of the top point T is:
   A. Zero
   B. 72.11 m/s^2, with 4 m/s^2 horizontal and 72 m/s^2 downward
   C. 72 m/s^2 upward
   D. 4 m/s^2 only
   Answer: B. Tangential part 2 adds to a_O = 2 forward, and the normal part is toward O (downward). (Key point: Top point)

## Related in Engineering Mechanics

- [Rolling cylinder attached to a spring](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arolling-cylinder-spring-energy) — Engineering Mechanics, chapter 7
- [Cylinder rolling down an incline](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Acylinder-rolling-down-incline) — Engineering Mechanics, chapter 6
- [Locating the instantaneous centre](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Ainstantaneous-centre-locating-methods) — Engineering Mechanics, chapter 5
- [Rod released from horizontal position](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-release-from-horizontal-reaction) — Engineering Mechanics, chapter 6
- [Sliding ladder by instantaneous centre](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Asliding-ladder-instantaneous-centre) — Engineering Mechanics, chapter 5
- [Velocity analysis of slider-crank mechanism](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Acrank-slider-velocity-numerical) — Engineering Mechanics, chapter 5

## Related in other subjects

- [Sphere rolling down incline, friction and velocity](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Asphere-rolling-incline-80kn) — Applied Mechanics (Dynamics), chapter 6
- [Velocities in plane motion by instantaneous centre](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aladder-sliding-wheel-rolling-velocities) — Theory of Mechanism and Machine I, chapter 7
- [General plane motion; angular acceleration of arm DE](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Ageneral-plane-motion-arm-de-acceleration) — Applied Mechanics (Dynamics), chapter 4
- [Coriolis acceleration; rod AB and BD velocity](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acoriolis-define-rod-ab-bd-velocity) — Applied Mechanics (Dynamics), chapter 4