Question

The motorcyclist crosses the turn at a speed V = 40 km / h, which is less than the allowed speed of the sliding threshold. At what angle should the motorcyclist lean from the vertical position, when it is known that the turning radius is R = 40 m?

Answer 1

$$\tan{\theta}=\frac{v^2}{rg}\\\theta=\arctan{\frac{v^2}{rg}} \\\theta=\arctan{\frac{40^2}{40\cdot 9.8\cdot3.6^2}}=17.5\degree$$