Angular acceleration of a rock, in a tire, on a car

1. Feb 6, 2010

deathcap00

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

A rock is wedged into the treads of a 15 in radius tire. The car to which it is attached is driving down the road at 70mph.
What’s the angular acceleration of the rock?

2. Relevant equations

ar=- $$\omega$$02r

3. The attempt at a solution

Not sure where to begin really, how does the car's speed influence the angular acceleration of the rock? What are the basic steps to take to set this problem up?

2. Feb 6, 2010

xcvxcvvc

angular velocity is the rate of change in angle. Angular acceleration is the rate of change in angular velocity. If a tire is traveling at a constant linear velocity, then its angular velocity must be constant.

$$\frac{d\omega}{dt} = \alpha$$. What is the derivative of a constant? If you do not know calculus, what would the rate of change in velocity be if velocity was constant?

3. Feb 6, 2010

deathcap00

Is my angle 2*pi?

do I need to convert the radius of the tire to miles (or the miles/hr to ft/s)?

and the derivative of a constant is zero

4. Feb 6, 2010

xcvxcvvc

The car is spinning at a constant angular velocity to sustain a constant linear velocity. The rock is doing whatever the tire is doing. The rock then has constant angular velocity. With constant angular velocity, angular acceleration is zero.

5. Feb 6, 2010

deathcap00

So my answer is zero then, that would explain why my professor said that he meant to ask for radial acceleration (but said he would accept angular as well though).

Just for the sake of knowing, how would I find the radial acceleration of the rock?

Thanks so much for your help.

6. Feb 6, 2010

xcvxcvvc

We know to relate an angular value to a linear value, we multiply by the radius from the center of rotation. To remember this equation, note that the units make sense.

So we can either divide your linear velocity by radius to find $$\omega$$ and use
$$a_r = \omega^2 r$$
or we can multiply
$$a_r = \omega^2 r$$
by $$\frac{r}{r}$$
and use the above rules to change the equation to
$$a_r = \frac{v^2}{r}$$
where v is the linear velocity. Then we can use that equation. It's just simple plug n' chug.

7. Feb 6, 2010

8. Feb 6, 2010

deathcap00

thank you both very much!