---
title: "Heel angle of a two-wheeler on a curve"
subject: "Theory of Mechanism and Machine II"
chapter: 3
key: "semester_7/theory_of_mechanism_and_machine_ii:two-wheeler-heel-angle-gyroscopic"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/two-wheeler-heel-angle-gyroscopic
site: BCT Engineering
---

# Heel angle of a two-wheeler on a curve

Theory of Mechanism and Machine II · Chapter 3: Gyroscopic Couple

- Asked in: year not given, 8 marks
- Video: [STABILITY OF TWO WHEEL VEHICLE TO TAKE TURN](https://www.youtube.com/watch?v=8CK_hRi1ug0) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/two-wheeler-heel-angle-gyroscopic
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-3

## Question

A motorcycle with its rider has a mass of 250 kg, and their combined centre of gravity is 0.6 m above the ground. Each wheel has a rolling radius of 0.3 m and a moment of inertia of 1.0 kg·m². The engine rotating parts have a moment of inertia of 0.2 kg·m², rotate in the same sense as the wheels, and run at four times the wheel speed. The motorcycle takes a curve of 50 m radius at 54 km/h. Find the angle of heel (lean from the vertical) required for the motorcycle to remain in equilibrium.

## Short answer

The heel angle is found by equating the weight restoring couple (mgh sinθ) to the sum of centrifugal and gyroscopic couples (both proportional to cos θ). With v=15 m/s, R=50 m, ΣIω=140 kg·m²/s, ω_p=0.30 rad/s, tanθ=(675+42)/1471.5=0.4873, giving θ=26.0° towards the curve center.

## Answer

**Data:** $m = 250$ kg, $h = 0.6$ m, $r_w = 0.3$ m, $I_w = 1.0$ kg·m² (each), $I_e = 0.2$ kg·m², $G = 4$, $v = 54$ km/h $= 15.0$ m/s, $R = 50$ m.

#### Spin quantities
$$
\omega_w = \frac{v}{r_w} = \frac{15.0}{0.3} = 50.00\ \text{rad/s}, \qquad \omega_e = 4\omega_w = 200.00\ \text{rad/s}
$$

$$
\sum I\omega = 2I_w\omega_w + I_e\omega_e = 2(1.0)(50.00) + 0.2(200.00) = 140.0\ \text{kg·m}^2/\text{s}
$$

Precession rate about the vertical axis: $\omega_p = \dfrac{v}{R} = \dfrac{15.0}{50} = 0.30$ rad/s.

#### Couples (at heel angle $\theta$)
- Gyroscopic couple: $C_g = \left(\sum I\omega\right)\omega_p\cos\theta = 42.0\cos\theta$ N·m
- Centrifugal couple about the ground contact: $C_c = \dfrac{mv^2}{R}\,h\cos\theta = \dfrac{250 \times 15.0^2}{50}\times 0.6\cos\theta = 675.0\cos\theta$ N·m
- Weight couple (restoring): $C_w = mgh\sin\theta = 250 \times 9.81 \times 0.6\sin\theta = 1471.5\sin\theta$ N·m

```text
   CG *   weight mg (down)
      |\
      | \  centrifugal force mv^2/R (outward)
   h  |  \
      |   \
  ----O----  ground contact, lean angle theta
```

#### Equilibrium
The weight couple balances the sum of the centrifugal and gyroscopic couples:

$$
mgh\sin\theta = C_c + C_g \;\Rightarrow\; 1471.5\sin\theta = (675.0 + 42.0)\cos\theta
$$

$$
\tan\theta = \frac{675.0 + 42.0}{1471.5} = 0.4873
$$

$$
\theta = 26.0^\circ
$$

**Answer:** The motorcycle must be heeled (leaned) by about $26.0^\circ$ towards the centre of the curve.

## Check yourself

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

1. How is the total angular momentum of the spinning parts computed?
   A. Sum of I w for both wheels and the engine rotor, with engine speed 4 times wheel speed
   B. Only the wheels' I w
   C. Sum of I w with engine at wheel speed
   D. Mass times velocity
   Answer: A. Sum I w = 2(1.0)(50) + 0.2(200) = 140 kg.m^2/s. (Key point: Total spin momentum)

2. At heel angle theta the gyroscopic couple is:
   A. (Sum I w) w_p cos(theta)
   B. (Sum I w) w_p sin(theta)
   C. (Sum I w) w_p
   D. (Sum I w) / w_p
   Answer: A. The spin axis is horizontal but the precession axis is vertical, giving the cos(theta) factor: 140 x 0.3 cos theta = 42 cos theta N.m. (Key point: Gyroscopic couple)

3. What balances the centrifugal and gyroscopic couples?
   A. The weight couple m g h sin(theta)
   B. Friction at the road
   C. The engine torque
   D. The wheel reaction only
   Answer: A. m g h sin(theta) = (C_c + C_g) cos(theta), giving tan(theta) = (C_c + C_g)/(m g h). (Key point: Equilibrium condition)

4. The required heel angle is closest to:
   A. 15 deg
   B. 26 deg
   C. 42 deg
   D. 64 deg
   Answer: B. tan(theta) = (675 + 42)/1471.5 = 0.4873, so theta = 26 deg. (Key point: Heel angle)

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