Heel angle of a two-wheeler on a curve

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

   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.