We analyze the motion in an inertial frame of reference attached to the center of the wheel. In this frame, the pebble undergoes uniform circular motion with speed v. The acceleration is purely centripetal, directed towards the center, with magnitude $a_c = v^2/R$.

Let $\theta$ be the angle of the pebble's position measured from the top vertical. The forces acting on the pebble are gravity ($mg$), the normal force ($N$), and static friction ($f_s$). We apply Newton's second law in the radial (outward) and tangential directions.

In the radial direction:

$$N - mg\cos\theta = -ma_c = -m\frac{v^2}{R}$$ $$N = m(g\cos\theta - \frac{v^2}{R})$$

The pebble remains in contact as long as $N > 0$.

In the tangential direction, the tangential acceleration is zero:

$$mg\sin\theta - f_s = 0$$ $$f_s = mg\sin\theta$$

The pebble starts to slip when the required static friction equals the maximum possible static friction, $f_s = \mu N$.

$$mg\sin\theta = \mu m(g\cos\theta - \frac{v^2}{R})$$ $$\sin\theta - \mu\cos\theta = -\frac{\mu v^2}{gR}$$

To solve this trigonometric equation, we let $\mu = \tan\phi$.

$$\frac{\sin(\theta-\phi)}{\cos\phi} = -\frac{\mu v^2}{gR}$$ $$\sin(\theta-\phi) = -\frac{\mu v^2}{gR}\cos\phi = -\frac{\mu v^2}{gR\sqrt{1+\mu^2}}$$

The smallest positive angle $\theta$ that satisfies this is:

$$\theta = \phi + \arcsin\left(-\frac{\mu v^2}{gR\sqrt{1+\mu^2}}\right) = \arctan\mu - \arcsin\left(\frac{\mu v^2}{gR\sqrt{1+\mu^2}}\right)$$

The time $t$ is related to the angle $\theta$ by $\theta = \omega t = (v/R)t$.

$$t = \frac{R}{v}\theta$$