NLM CYU 31

The motorcyclist moves on a horizontal unbanked circular track. The only horizontal force providing acceleration is the friction exerted by the track on the tyres. This friction force $\vec{f}$ must provide both the centripetal acceleration ($a_c$) needed to turn and the tangential acceleration ($a_t$) needed to increase speed.

The total acceleration $a$ is the vector sum of the tangential and centripetal components:

$$ a = \sqrt{a_t^2 + a_c^2} = \sqrt{\left(\frac{dv}{dt}\right)^2 + \left(\frac{v^2}{R}\right)^2} $$

To prevent slipping, the required frictional force $ma$ must not exceed the maximum static friction:

$$ m\sqrt{a_t^2 + \left(\frac{v^2}{R}\right)^2} \le \mu mg $$ $$ a_t^2 + \frac{v^4}{R^2} \le \mu^2 g^2 $$

To cover the distance in the minimum amount of travel, the motorcyclist should utilize the maximum available friction at all times. Thus, we set the condition to equality:

$$ a_t = \sqrt{\mu^2 g^2 – \frac{v^4}{R^2}} $$

Using the kinematic relation $a_t = v \frac{dv}{ds}$, where $s$ is the distance covered:

$$ v \frac{dv}{ds} = \sqrt{\mu^2 g^2 – \frac{v^4}{R^2}} $$

Rearranging for integration:

$$ ds = \frac{v \, dv}{\sqrt{\mu^2 g^2 – \frac{v^4}{R^2}}} $$

The maximum allowable speed $v_{max}$ is reached when all available friction is used for turning, meaning tangential acceleration $a_t$ becomes zero.

$$ 0 = \mu^2 g^2 – \frac{v_{max}^4}{R^2} \implies v_{max}^2 = \mu g R $$

We integrate from rest ($v=0$) to $v_{max}$:

$$ S = \int_{0}^{v_{max}} \frac{v \, dv}{\sqrt{\mu^2 g^2 – \frac{v^4}{R^2}}} $$

Let substitution $u = \frac{v^2}{R}$. Then $du = \frac{2v \, dv}{R} \implies v \, dv = \frac{R}{2} du$.

The limits change from $v=0 \to u=0$ and $v=\sqrt{\mu g R} \to u = \mu g$.

$$ S = \int_{0}^{\mu g} \frac{\frac{R}{2} du}{\sqrt{\mu^2 g^2 – u^2}} = \frac{R}{2} \int_{0}^{\mu g} \frac{du}{\sqrt{(\mu g)^2 – u^2}} $$

This is a standard integral of the form $\int \frac{dx}{\sqrt{a^2 – x^2}} = \sin^{-1}(\frac{x}{a})$:

$$ S = \frac{R}{2} \left[ \sin^{-1}\left(\frac{u}{\mu g}\right) \right]_{0}^{\mu g} $$ $$ S = \frac{R}{2} \left( \sin^{-1}(1) – \sin^{-1}(0) \right) = \frac{R}{2} \left( \frac{\pi}{2} – 0 \right) $$ $$ S = \frac{\pi R}{4} $$

Given $R = 28$ m and using $\pi = 22/7$:

$$ S = \frac{1}{4} \times \frac{22}{7} \times 28 $$ $$ S = \frac{1}{4} \times 22 \times 4 $$ $$ S = 22 \text{ m} $$