Answer :
(a) The wheel had been turning for 8.04 s before the start of the 6.00 s interval.
(b) The angular velocity of the wheel at the start of the 6.00 s interval was 19.2 rad/s.
(a) To find the time the wheel had been turning before the 6.00 s interval, we can use the formula for angular displacement with constant angular acceleration:
θ = ω_initial * t + 0.5 * α * [tex]t^2[/tex]
Given that θ = 67.8 rad and α = 3.20 rad/s², and assuming the initial angular velocity ω_initial is 0 (as the wheel starts from rest), we can solve for t:
67.8 = 0 * t + 0.5 * 3.20 * [tex]t^2[/tex]
67.8 = [tex]1.6t^2[/tex]
[tex]t^2[/tex] = 67.8 / 1.6
[tex]t^2[/tex] ≈ 42.375
t ≈ √42.375 ≈ 6.51 s
However, this time represents the total time elapsed since the wheel started turning until the end of the 6.00 s interval. To find the time before the 6.00 s interval, we subtract 6.00 s from this total:
Time before 6.00 s interval = 6.51 s - 6.00 s = 0.51 s
So, the wheel had been turning for approximately 0.51 s before the start of the 6.00 s interval.
(b) To determine the angular velocity of the wheel at the start of the 6.00 s interval, we use the equation:
ω_final = ω_initial + α * t
Given that
ω_initial = 0, α = 3.20 rad/s², and t = 6.00 s, we can find ω_final:
ω_final = 0 + 3.20 * 6.00
ω_final = 19.2 rad/s
Therefore, the angular velocity of the wheel at the start of the 6.00 s interval is 19.2 rad/s.
