# Velocity at the top of wheel

1. Oct 23, 2008

### Naldo6

1. The problem statement, all variables and given/known data

A cyclist accelerates from rest at a rate of 1.10 m/s2. How fast will a point on the rim of the tire (diameter = 79.7 cm) at the top be moving after 3.10 s?

2. The attempt at a solution:

i just tried this one...

a_avereage= Δw/Δt so a_avereage= ( w_2 - w_1 ) / Δt

if we set for w_2 we get w_2 = W_1 + (a_avereage)(Δt)
with w_1=0 the response for w_2 is 1.358 rad/s but that was wrong

can anyone help me how to ge the velocity of the wheel at the top t 3.10 s?

2. Oct 23, 2008

### Dick

The first thing you need to figure out is how fast the bike is moving after 3.10 s. What's that? That's the same speed that the hub of the wheel is moving at. Now think about it. A point on the rim of the wheel that is in contact with the road is not moving relative to the road (if the bike isn't skidding). Right? The means the rotational speed of a point on the wheel is the same as the linear speed of the bike but in the opposite direction so they can cancel. Try drawing a picture. What happens at the top of the wheel? That's the easy way to think about it. You don't need the diameter of the wheel.

3. Oct 23, 2008

### Naldo6

i dont undertand whau told me....

i need to know how to calculate that velocity?

4. Oct 23, 2008

### Dick

Yes, you do, whether you do it the way I suggested or not. The bike accelerates at 1.10m/s^2 for 3.10s. What's its velocity?

5. Oct 23, 2008

### Naldo6

ok, but how i calculate that velocity?... i tried in the above way but that was wrong.

6. Oct 23, 2008

### Dick

It's a linear acceleration problem so far. Forget the rotational formulas.

7. Oct 23, 2008

### Naldo6

ok so v= v_i + at
where v_i = 0
so v= (1.10m/s^2)(3.10s)=3.41 m/s right?...

8. Oct 23, 2008

### Dick

That's a good start. Yes. So what's the rotational speed of points on the rim of the wheel? How fast is the rim of the tire spinning? In m/s, not radian/s.

9. Oct 23, 2008

### Naldo6

is at the top 2 times the velocity and at the bottom zero?

10. Oct 23, 2008

### Dick

Sure it is. The bottom of the wheel is at rest, so the top of the wheel is moving at 2*3.41m/s forward. You've got the picture, right? See? No diameter needed.

11. Oct 23, 2008

### Naldo6

ty a lot.....

can u help m ewith other problems i have doubt?...

i will post in new topics....

12. Oct 23, 2008

### Dick

I'll try. You're welcome.