# Turning wheel, simply angular momentum

1. Mar 2, 2013

### Ground State

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

A bicycle with 0.8 m diameter tires is coasting on a level road at 5.6 m/s. A small dot has been painted on the rear tire. What is:
a) the angular speed of the tire
b) the speed of the dot when it is 0.8 m above the road in m/s?
c) the speed of the dot when it is 0.4 m above the road in m/s?

3. The attempt at a solution

a)
ω=v/r
ω=5.6/0.4
ω=14

for (b) and (c), can someone help me understand why the answer is not 5.6 m/s for both of them?

2. Mar 2, 2013

### rcgldr

In this case, they want the speed of the dot relative to the road (relatvie to any point on the road).

3. Mar 2, 2013

### Ground State

But when it is 0.8 m above the road, and hence, the full diameter of the wheel, is the speed not tangentially in the same direction as the road, and therefore 5.6 m/s? (It isn't, I know this answer is wrong but am not sure why). Is it because it will have the cumulative linear speed of the moving car and angular speed of the turning wheel? Does that mean that the answer for (c) will simply be answer from (a)?

4. Mar 2, 2013

### ehild

See picture. The centre of the tyre moves with 5.6 m/s to the right. The speed of the perimeter of the tyre is also 5.6 m/s, but the velocity of the dot with respect to the CM changes direction as the wheel rolls.

The velocity of the dot with respect to the road is the vector sum of the velocity of CM and the velocity of the dot with respect to the CM . And the speed of the dot is the magnitude of velocity.

ehild

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5. Mar 2, 2013

### Ground State

So am I understanding your explanation correctly by submitting the following?

(b)
velocity with respect to the road = velocity of CM + velocity of the dot with respect to the CM
velocity with respect to the road = 5.6 m/s + 5.6 m/s
velocity with respect to the road = 11.2 m/s

(c) (arranging vectors tip to tail and taking resultant)
= (5.6 m/s)^2 + (5.6 m/s)^2
= 7.92 m/s

6. Mar 2, 2013

### ehild

It is correct now. And what is the speed of the dot when it is at the bottom of the wheel?

ehild