# Circular Angular Acceleration problem

1. Feb 25, 2008

### the7joker7

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

A tire 2.00 feet in diameter is placed on a balancing machine, where it is spun so that its tread is moving at a constant speed of 60.0 mi/h. A small stone is stuck in the tread of the tire. What is the acceleration of the stone as the tire is being balanced?

2. Relevant equations

\omega = $$\frac{change in theta}{change in time}$$

r = diameter/2

$$\alpha$$ = $$\frac{\omega - \omega_{0}} {time}$$

a$$_{t}$$ = r*\alpha
3. The attempt at a solution

$$\omega$$ = 840.3 rev/min = 88 radians/sec.

$$\alpha$$ = $$\frac{88 radians}{1 second}$$ = 88 radians/second$$^{2}$$.

a$$_{t}$$ = .305 *88 = 26.84 m/s$$^{2}$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 25, 2008

### Dick

You got omega (I'll call it w) just fine somehow. I'm not sure how because it's not clear what you did. But acceleration isn't w*r, that's velocity v. 26.8m/sec=60mi/hr. You are back where you started. Acceleration is v^2/r=(w*r)^2/r=w^2*r.

3. Feb 26, 2008

### the7joker7

Well, going from 60mi/hr to radians/second is pretty basic math as long as you know the transformation values.

Thanks for the help!

EDIT: I got 88$$^{2}$$ * .305 = 2631.92radians/seconds$$^{2}$$. Look right?

Last edited: Feb 26, 2008
4. Feb 2, 2010

### cherriesx11x

Use the equation centripetal acceleration = v^2/r. The radius s 1.00 ft, or .305 m. V is 60 mi/h, or 26.8 m/s. Then the acceleration would be (26.8)^2/0.305, which is 2350 m/s^2