A wheel initially has an angular velocity of 20.0 rad/s, but it is slowing down at a rate of 2.0 rad/s2. By the time it stops, how many revolutions will it have turned through?

To find the number of revolutions the wheel will have turned through by the time it stops, we can use the formula:

θ = ωi * t + (1/2) * α * t^2

Where: θ is the angle turned in radians, ωi is the initial angular velocity in rad/s, α is the angular acceleration in rad/s^2, and t is the time in seconds.

In this case, the initial angular velocity (ωi) is 20.0 rad/s, and the angular acceleration (α) is -2.0 rad/s^2 (negative because the wheel is slowing down). We need to find the time it takes for the wheel to stop, so we can set ωf (final angular velocity) to 0 and solve for t:

ωf = ωi + α * t 0 = 20.0 - 2.0 * t 2.0 * t = 20.0 t = 10.0 s

Now that we have the time it takes for the wheel to stop, we can substitute the values into the formula to find the angle turned (θ):

θ = ωi * t + (1/2) * α * t^2 θ = 20.0 * 10.0 + (1/2) * (-2.0) * (10.0)^2 θ = 200.0 - 100.0 θ = 100.0 rad

To convert this angle to revolutions, we can use the conversion factor:

1 revolution = 2π radians

So, the number of revolutions the wheel will have turned through by the time it stops is:

Number of revolutions = θ / (2π) Number of revolutions = 100.0 / (2π) Number of revolutions ≈ 15.92 revolutions

Therefore, the wheel will have turned through approximately 15.92 revolutions by the time it stops.