Why Does a Pebble Fly Off a Rolling Wheel if Its Speed Exceeds sqrt(Rg)?

[asdf60]

Here is the question:

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 of the wheel if V > sqrt(Rg).

What does this mean. In particular, what does instantaneously at rest mean??

 

[Doc Al]

It means that when the pebble is placed on the wheel, its speed is exactly the same as the speed of the top of the wheel. (So, with respect to the wheel, it's at rest.)

 

[code.master]

Like Doc Al says, it means that the wheel and pebble are happily in motion together. Carefully explore this problem. Remember that the velocity at the top of the wheel is not V.

also try to graph the motion of the pebble vs the ground, vs the wheel, and the wheel vs the ground. that's how my teacher started me to understand this problem. we did the problem with a meter stick and pennies, sliding off of a giant snowball...

what you are showing is that the velocity of the pebble is so high that the pebble doesn't say on the wheel at all, and that the velocity at this point can be characterized by the velocity of the center of the wheel relative to the ground, which is a function only of R and g.