To solve this problem, we can use the following kinematic equation relating angular displacement, initial angular velocity, angular acceleration, and time:
θ = ω₀t + (1/2)αt²
where θ is the angular displacement, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time.
a) We are given ω₀ = 2 rad/s, α = 3.5 rad/s², and t = 2 s. Plugging in these values, we can solve for θ:
θ = (2 rad/s)(2 s) + (1/2)(3.5 rad/s²)(2 s)²
= 4 rad + (1/2)(3.5 rad/s²)(4 s²)
= 4 rad + 7 rad
= 11 rad
Therefore, the wheel rotates through an angular displacement of 11 radians in 2 seconds.
b) To find the number of revolutions, we can divide the angular displacement by 2π (since 1 revolution is equal to 2π radians):
Number of revolutions = 11 rad / (2π rad/rev)
≈ 1.75 revolutions
Therefore, the wheel has turned approximately 1.75 revolutions during this time interval.
c) To find the angular speed at t = 2 s, we can use the equation:
ω = ω₀ + αt
Given ω₀ = 2 rad/s, α = 3.5 rad/s², and t = 2 s, we can compute:
ω = 2 rad/s + (3.5 rad/s²)(2 s)
= 2 rad/s + 7 rad/s
= 9 rad/s
Therefore, the angular speed of the wheel at t = 2 s is 9 rad/s.