A flywheel rotating at 12 rev/s is brought to rest in 6 s. The magnitude of the average angular acceleration in rad/s² of the wheel during this process is: A) 1/ $\pi$ B) 2 C) 4 D) 4 $\pi$ E) 72

Question

Quizplus · Accepted Answer

We can use the equation:
Δω = αt
where Δω is the change in angular velocity, α is the angular acceleration, and t is the time interval.
We want to find α, so we rearrange the equation to solve for α:
α = Δω / t

At the start, the rotational speed is:
ωi = 12 rev/s
We convert this to rad/s by multiplying by 2π:
ωi = 12 rev/s * 2π rad/rev = 24π rad/s

At the end, the rotational speed is zero:
ωf = 0

The change in angular velocity is:
Δω = ωf - ωi = -24π rad/s

The time interval is given as t = 6 s.

Plugging in the values, we get:
α = Δω / t = (-24π rad/s) / (6 s) = -4π rad/s^2

The magnitude of the average angular acceleration is just the absolute value of α, which is:
|α| = 4π rad/s^2 = 4π/π rad/s^2 = 4 rad/s^2

Therefore, the answer is D) 4.

A Flywheel Rotating at 12 Rev/s Is Brought to Rest π\piπ

A Flywheel Rotating at 12 Rev/s Is Brought to Rest $\pi$