A wheel is rotating with angular velocity 14 rev/s. What angular acceleration would be needed to bring the wheel to rest in a single rotation?
avg vel = (14 + 0) / 2 = 7 rev/s
stop time = 1/7 s
distance = 1/2 a t^2
... 1 = 1/2 * a * (1/7)^2
To find the angular acceleration needed to bring the wheel to rest in a single rotation, we can use the equation:
ωf^2 = ωi^2 + 2αθ
where ωi is the initial angular velocity, ωf is the final angular velocity, α is the angular acceleration, and θ is the angle through which the object rotates.
In this case, we want the final angular velocity (ωf) to be zero and the initial angular velocity (ωi) to be 14 rev/s. The angle through which the object rotates (θ) is one complete rotation, which is equal to 2π radians.
Plugging these values into the equation, we have:
0 = (14 rev/s)^2 + 2α(2π)
Simplifying, we get:
0 = 196 + 4πα
Rearranging the equation, we can solve for the angular acceleration (α):
4πα = -196
α = -49/π rad/s^2
Note: The negative sign indicates that the wheel needs a deceleration to come to rest.
Therefore, the angular acceleration needed to bring the wheel to rest in a single rotation is approximately -15.57 rad/s^2.