After 12.6 s, a spinning roulette wheel has slowed down to an angular velocity of 1.32 rad/s. During this time, the wheel has an angular acceleration of -6.07 rad/s2. Determine the angular displacement of the wheel.
alpha = -6.07
initial omega = wi
final omega = 1.32
t = 12.6
1.322 = wi - 6.07 (12.6) solve for wi
then
theta = wi t - (6.07/2) t^2 = wi(12.6) - (3.035)(12.6)^2
To determine the angular displacement of the wheel, we can use the equations of motion for rotational motion.
The equation that relates angular displacement (θ), final angular velocity (ω), initial angular velocity (ω₀), angular acceleration (α), and time (t) is:
θ = ω₀t + (1/2)αt²
Given:
t = 12.6 s
ω = 1.32 rad/s
α = -6.07 rad/s²
First, let's find the initial angular velocity (ω₀). Since the problem statement does not mention the initial angular velocity, we assume it to be zero, as it is often the case when a spinning object is brought to rest before the experiment begins.
ω₀ = 0 rad/s
Now, we can substitute the given values into the equation:
θ = ω₀t + (1/2)αt²
= 0 rad/s * 12.6 s + (1/2) * (-6.07 rad/s²) * (12.6 s)²
Calculating the expression inside the parenthesis:
(1/2) * (-6.07 rad/s²) * (12.6 s)²
= -3.035 rad/s² * (158.76 s²)
= -481.176 rad
Hence, the angular displacement of the wheel after 12.6 seconds is -481.176 rad. The negative sign indicates that the wheel has rotated in the opposite direction of its initial motion.