# Pebble on wheel

1. Apr 29, 2007

### anand

1. The problem statement, all variables and given/known data
A wheel of radius R rolls along the ground with velocity V.
A pebble is carefully released on top of the wheel so that it is instantaneously at rest on the wheel.
(a) Show that the pebble will immediately fly off the wheel if V> sqrt(Rg)
(b) Show that in the case where V< sqrt(Rg) and the coefficient of friction is u=1,the pebble starts to slide when it has rotated through an angle given by theta=arccos[(1/sqrt(2))(V^2/Rg)]-pi/4

3. The attempt at a solution

I've tried resolving the forces along a set of axes parallel and perpendicular to the surface of the sphere,but I can't arrive at the required answer.

2. Apr 29, 2007

### mukundpa

For no slipping the gravitational force must be less then the required centripetal force.
The wheel is rolling, so the path of the point on the rim is not a circle but a cycloid.

3. Apr 29, 2007

4. Mar 6, 2009

### bjj_99

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:
1. a normal force N exerted by the wheel in the outward radial direction
2. gravitational force mg in the downward vertical direction
3. 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:
$$N - mg cos(\theta) = -m \omega^2 r$$
or
$$N = mg cos(\theta) -m \omega^2 r$$

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,
$$mgsin(\theta) > mg cos(\theta) -m \omega^2 r$$
$$mgsin(\theta) - mgcos(\theta) > -m \omega^2 r$$
$$g (sin(\theta) - cos(\theta)) > - \omega^2 r$$
$$g (\frac{1}{\sqrt{2}} sin(\theta) - \frac{1}{\sqrt{2}} cos(\theta)) > - \frac{1}{\sqrt{2}} \omega^2 r$$
$$\frac{1}{\sqrt{2}} cos(\theta) - \frac{1}{\sqrt{2}} sin(\theta) < \frac{\omega^2 r}{g \sqrt{2}}$$
$$cos(\pi / 4)cos(\theta) - sin(\pi / 4)sin(\theta) < \frac{\omega^2 r}{g \sqrt{2}}$$
$$cos(\theta + \pi / 4) < \frac{\omega^2 r}{g \sqrt{2}}$$
$$cos(\theta + \pi / 4) < \frac{v^2}{rg \sqrt{2}}$$
$$\theta + \pi / 4 < arccos(\frac{v^2}{rg \sqrt{2}})$$
$$\theta < arccos(\frac{v^2}{rg \sqrt{2}}) - \pi / 4$$