Circular Angular Acceleration problem

In summary, the problem involved a 2.00 foot tire spinning at 60.0 mi/h on a balancing machine with a small stone stuck in the tread. The solution involved finding the angular velocity, radius, and acceleration of the stone using equations for angular velocity, radius, and tangential acceleration. The final answer for the acceleration of the stone was 2350 m/s^2.
  • #1
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Homework Statement



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?

Homework 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

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]
 
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  • #2
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
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:
  • #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
 

1. What is circular angular acceleration?

Circular angular acceleration is a measure of the rate of change of an object's angular velocity, which is the speed at which an object rotates around an axis. It is a vector quantity, meaning it has both magnitude and direction, and is measured in radians per second squared (rad/s^2).

2. How is circular angular acceleration calculated?

Circular angular acceleration can be calculated using the formula α = (ω2 - ω1) / (t2 - t1), where α is the angular acceleration, ω2 and ω1 are the final and initial angular velocities, and t2 and t1 are the final and initial times. Alternatively, it can also be calculated as the product of the moment of inertia (I) and the angular acceleration (α), α = τ / I, where τ is the torque applied to the object.

3. What are some examples of circular angular acceleration?

Circular angular acceleration can be observed in various everyday phenomena, such as the rotation of a merry-go-round, the spinning of a top, the motion of a car around a curve, or the movement of a planet around the sun. It is also a crucial concept in sports like figure skating, gymnastics, and diving, where athletes perform circular motions with their bodies.

4. How does circular angular acceleration differ from linear acceleration?

Circular angular acceleration and linear acceleration are both measures of how fast an object's velocity changes, but they differ in the type of motion they describe. Linear acceleration describes the change in an object's velocity along a straight line, while circular angular acceleration describes the change in an object's velocity as it rotates around an axis.

5. What factors affect circular angular acceleration?

The magnitude of circular angular acceleration is influenced by several factors, such as the mass and shape of the object, the distance from the axis of rotation, and the torque applied to the object. The direction of circular angular acceleration is determined by the direction of the applied torque and follows the right-hand rule, where the thumb points in the direction of the torque and the fingers curl in the direction of the angular acceleration.

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