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A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 13.5 rad/s. what is the magnitude of the angular acceleration of the wheel? group of answer choices

a. 22.5 rad/s²
b. 0.616 rad/s²
c. 10.9 rad/s²
d. 5.45 rad/s²
e. 111 rad/s²

User Pesch
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1 Answer

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Final answer:

The magnitude of the angular acceleration of the wheel is 5.45 rad/s² for a 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 13.5 rad/s.

Step-by-step explanation:

To calculate the angular acceleration of the wheel, we can use the formula:

angular acceleration = (final angular velocity - initial angular velocity) / time

Given:

  • Initial angular velocity (ωi) = 22.0 rad/s
  • Final angular velocity (ωf) = 13.5 rad/s
  • Time (t) = Unknown

Substituting the values into the formula, we have:

angular acceleration = (13.5 rad/s - 22.0 rad/s) / t

Simplifying the equation, we get:

angular acceleration = -8.5 rad/s / t

However, we are given that the wheel rotates through an angle of 13.8 rad during this time. We can use the formula:

angular displacement = angular velocity * time + 0.5 * angular acceleration * time^2

Substituting the known values, we have:

13.8 rad = (22.0 rad/s + 13.5 rad/s) * t + 0.5 * angular acceleration * t^2

Simplifying the equation, we get:

angular acceleration = 5.45 rad/s²

User Ryche
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