A Ferris wheel rotating at 20 rad/s slows down with a constant angular acceleration of magnitude 5.0 rad/s².How many revolutions does it make while slowing down before coming to rest? A)40 B)20 C)6.4 D)3.2

Question

Quizplus · Accepted Answer

The initial angular velocity of the Ferris wheel is 20 rad/s and the angular acceleration is -5.0 rad/s^2 (negative because it's slowing down). We can use the equation:
ω^2 = ω0^2 + 2αθ
where ω0 is the initial angular velocity (20 rad/s), α is the angular acceleration (-5.0 rad/s^2), and θ is the angle of rotation (in radians). We want to find θ when ω = 0 (the Ferris wheel comes to rest).

Substituting the known values, we get:
0^2 = (20 rad/s)^2 + 2(-5.0 rad/s^2)θ
Solving for θ, we get:
θ = (20 rad/s)^2 / (2(5.0 rad/s^2)) = 40 rad
Since one revolution is equal to 2π radians, the number of revolutions is:
θ / 2π = 40 rad / (2π rad/rev) ≈ 6.4 revolutions. 
Therefore, the answer is C.

A Ferris Wheel Rotating at 20 Rad/s Slows Down with a Constant