Constant angular acceleration of a rotating wheel

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SUMMARY

The discussion centers on calculating the constant angular acceleration of a rotating wheel that completes 33.5 revolutions in 4.7 seconds, reaching an angular speed of 45.5 rad/s. The initial angular velocity is not provided, which complicates the calculation. Participants suggest using the formula \(wf^2 - wo^2 = 2a(Of - Oo)\) to derive the angular acceleration, emphasizing the need for a standard constant acceleration approach in angular motion.

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  • Understanding of angular kinematics
  • Familiarity with the formula \(wf^2 - wo^2 = 2a(Of - Oo)\)
  • Knowledge of angular velocity and acceleration concepts
  • Basic proficiency in physics problem-solving techniques
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Students studying physics, educators teaching angular motion concepts, and anyone interested in solving problems related to rotational dynamics.

jscherf92
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A rotating wheel requires 4.7 s to rotate
through 33.5 rev. Its angular speed at the
end of the 4.7 s interval is 45.5 rad/s.
What is its constant angular acceleration?
Assume the angular acceleration has the same
sign as the angular velocity.
Answer in units of rad/s2.

wf^2-wo^2=2a(Of-Oo)

Ive got a page of failed equations on my paper, and i need a push in the right direction.
 
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Hi jscherf92! :smile:

The question doesn't give you the initial angular velocity, so I don't think we can assume it's zero.

So use a standard constant acceleration formula with s t a and v (angular version) :wink:
 

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