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Circular Angular Acceleration problem

  1. Feb 25, 2008 #1
    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 = [tex]\frac{change in theta}{change in time}[/tex]

    r = diameter/2

    [tex]\alpha[/tex] = [tex]\frac{\omega - \omega_{0}} {time}[/tex]

    a[tex]_{t}[/tex] = r*\alpha
    3. The attempt at a solution

    [tex]\omega[/tex] = 840.3 rev/min = 88 radians/sec.

    radius = .305m.

    [tex]\alpha[/tex] = [tex]\frac{88 radians}{1 second}[/tex] = 88 radians/second[tex]^{2}[/tex].

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

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 25, 2008 #2


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    Homework Helper

    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.
  4. Feb 26, 2008 #3
    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[tex]^{2}[/tex] * .305 = 2631.92radians/seconds[tex]^{2}[/tex]. Look right?
    Last edited: Feb 26, 2008
  5. Feb 2, 2010 #4
    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
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