How can the pebble stay on the wheel as it rolls?

[cacofolius]

Homework Statement

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.

Show that the pebble will immediately fly off the wheel if V> sqrt(Rg)

The Attempt at a Solution

Hi, I know this problem's been asked before, but since they're old posts, and didn't quite understood the clues given there, I decided to ask one more time.

I understand that if the centripetal acceleration equals or is less than gravity, then the pebble will not immediately fly off, but in this case the tangential velocity at the top will be 2v (taken from the floor), and thus the distance will be 2R. Therefore

(2v)^2/(2R)=g and V=sqrt(Rg/2)

I feel I'm missing something, any hints will be appreciated

 

[TSny]

From the point of view of the reference frame at rest with respect to the ground, a point on the rim of the wheel travels along a cycloid. So, the radius that you would need to use is the radius of curvature of a cycloid at the highest point.

You can avoid all that by going to the frame of reference moving with the axle of the wheel.

 

[cacofolius]

Oh, then its more simple. Thanks!