A motorcycle is required to travel at a constant speed over an...

At the highest point of the bridge, the normal force $N$ is given by the equation of motion:

$mg - N = \dfrac{mv^2}{R}$

For contact to be maintained, $N \ge 0$, which implies:

$mg \ge \dfrac{mv^2}{R} \Rightarrow v \le \sqrt{Rg}$

Thus, the maximum velocity at the highest point is v_\max = \sqrt{Rg}.

If the motorcycle travels at a speed v = \dfrac{v_\max}{\sqrt{2}} = \sqrt{\dfrac{Rg}{2}}, the normal force at an angular position $\theta$ from the highest point is:

$N = mg \cos\theta - \dfrac{mv^2}{R}$

The motorcycle loses contact when $N = 0$:

$mg \cos\theta = \dfrac{m}{R} \left(\dfrac{Rg}{2}\right) = \dfrac{mg}{2}$

$\cos\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{3}$

The distance along the bridge from the highest point is $s = R\theta = \dfrac{\pi R}{3}$.

To avoid losing contact anywhere on the bridge, the normal force must be non-negative at all points. The minimum normal force occurs at the ends of the bridge, where the angle from the highest point is maximum, \theta_\max = \dfrac{L}{2R}.

For $N \ge 0$ at the ends:

mg \cos\theta_\max - \dfrac{mv^2}{R} \ge 0

$v^2 \le Rg \cos$\left(\dfrac{L}{2R}\right)

Thus, the maximum uniform speed to maintain contact everywhere is $v =$\sqrt{gR \cos\left(\dfrac{L}{2R}\right)}.