EDIT: Funny... I didn't know how to do this question when I first started this post, but the answer came to me while I was parsing all the tex eqns below.
3. The attempt at a solution
Since the wheel is rolling at a steady velocity V, we can consider the inertial frame with origin at the center of the wheel. At a displacement \theta from the vertical, the forces on the pebble are the following:
- a normal force N exerted by the wheel in the outward radial direction
- gravitational force mg in the downward vertical direction
- frictional force \mu N in the counterclockwise tangential direction. This is just N since \mu=1.
Also, up until the point that the pebble slips, we know:
<br />
N - mg cos(\theta) = -m \omega^2 r<br />
or
<br />
N = mg cos(\theta) -m \omega^2 r<br />
In other words, the centripetal force equals N minus the inward radial component of mg.
The pebble slips when the tangential component of force in the clockwise direction exceeds the tangential component of force in the counterclockwise direction. This is because the tangential acceleration (angular acceleration + coriolis term) equals 0 in our inertial frame. So, we want to find \theta such that:
mgsin(\theta) > N, or, plugging in for N from above,
mgsin(\theta) > mg cos(\theta) -m \omega^2 r
mgsin(\theta) - mgcos(\theta) > -m \omega^2 r
g (sin(\theta) - cos(\theta)) > - \omega^2 r
g (\frac{1}{\sqrt{2}} sin(\theta) - \frac{1}{\sqrt{2}} cos(\theta)) > - \frac{1}{\sqrt{2}} \omega^2 r
\frac{1}{\sqrt{2}} cos(\theta) - \frac{1}{\sqrt{2}} sin(\theta) < \frac{\omega^2 r}{g \sqrt{2}}
cos(\pi / 4)cos(\theta) - sin(\pi / 4)sin(\theta) < \frac{\omega^2 r}{g \sqrt{2}}
cos(\theta + \pi / 4) < \frac{\omega^2 r}{g \sqrt{2}}
cos(\theta + \pi / 4) < \frac{v^2}{rg \sqrt{2}}
\theta + \pi / 4 < arccos(\frac{v^2}{rg \sqrt{2}})
\theta < arccos(\frac{v^2}{rg \sqrt{2}}) - \pi / 4