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
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
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.