Speed of Point at Front Edge of 6.0cm Wheel Rolling at 20m/s

[G01]

A 6.0cm diameter wheel is rolling along at 20 m/s What is the speed of a point at the front edge of the wheel?

Im kinda lost here. Tha answer given in the back of the book is 28.3 m/s.

I have no idea how to do this problem, though I know i must be missing something simple. I found /omega = 66.6 rad/s After this I have no idea where to go. I am not asking for anyone to do the problem for me. I'm just asking for a hint to help me figure out where to go next. Thanks alot. I would show more work if i could but I just can't figure out where to go or what I'm missing. I know it must be something stupid. Thanks.

 

[LeonhardEuler]

I'm not sure how you found \omega, but it doesn't seem right. The way to go about it is by finding the circumference of the wheel; this gives the length of a revolution, so if you divide the foward velocity by this you will have the number of revolutions per second. Multiply by 2\pi to get\omega. I think you may have just messed up the units.

Once you have that you need to find the total speed. Since the point is at the front edge of the tire the rotational component of the velocity is directed straight downward. There is also another comonent of the velocity that comes from the fact that the bike is moving foward at 20m/s. This component is horizontal. So you have a vertical component and a horizontal component perpendicular to each other: use the pythagorean theorem to find the length of the vector obtained by adding these.

By the way, it is not necessary to find \omega. If you think about it, you know that the rotational component of the velocity on the rim is 20m/s because the bottom of the tire is not moving. This means that the two components cancel.