15. A wheel rotates witha constant angular acceleration of 3.5 rad/s². (a) If the angular speed of the wheel is 2 rad/s at t= 0, through what angular displacement does the wheel rotate in 2s?
b) Through how many revolutions has the wheel turned during this time interval?
c) What is the angular speed of the wheel at = 2 s?
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.