Starting from rest at t = 0 s, a wheel undergoes a constant angular acceleration. When t = 1.6 s, the angular velocity of the wheel is 5.5 rad/s. The acceleration continues until t = 17 s, when the acceleration abruptly changes to 0 rad/s2. Through what angle does the wheel rotate in the interval t = 0 s to t = 37 s?
To solve this problem, we need to use the kinematic equations of rotational motion. Let's break down the problem into two parts:
Part 1: From t = 0 s to t = 17 s
In this interval, the wheel undergoes a constant angular acceleration. We are given that the wheel starts from rest, so its initial angular velocity (ω₀) is 0 rad/s. At t = 1.6 s, the angular velocity (ω) is given as 5.5 rad/s. We need to find the angle of rotation (θ) in this interval.
We can use the equation:
ω = ω₀ + αt,
where ω is the angular velocity, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time.
Rearranging the equation, we have:
α = (ω - ω₀) / t.
Substituting in the given values:
α = (5.5 rad/s - 0 rad/s) / (1.6 s - 0 s)
α = 5.5 rad/s / 1.6 s
α = 3.4375 rad/s²
Next, we can use the equation:
θ = ω₀t + 0.5αt²,
where θ is the angle of rotation.
Substituting the known values:
θ = 0 rad/s * 17 s + 0.5 (3.4375 rad/s²) (17 s)²
θ = 0 + 0.5 (3.4375 rad/s²) * 289 s²
θ = 0 + 1484.375 rad
θ = 1484.375 rad
So, in the interval t = 0 s to t = 17 s, the wheel rotates by 1484.375 rad.
Part 2: After t = 17 s
From t = 17 s to t = 37 s, the wheel has no angular acceleration (α = 0 rad/s²). Therefore, the wheel rotates at a constant velocity (ω) during this interval.
We need to find the angle of rotation (θ) in this interval. Since the angular velocity is constant, we can use the equation:
θ = ωt,
where θ is the angle of rotation.
Substituting the known values:
θ = 5.5 rad/s * (37 s - 17 s)
θ = 5.5 rad/s * 20 s
θ = 110 rad
So, in the interval t = 17 s to t = 37 s, the wheel rotates by 110 rad.
To find the total angle of rotation from t = 0 s to t = 37 s, we sum up the angles from both intervals:
Total angle of rotation = 1484.375 rad + 110 rad
Total angle of rotation = 1594.375 rad
Therefore, the wheel rotates by 1594.375 rad in the interval t = 0 s to t = 37 s.
To find the angle rotated by the wheel in the given time interval, we need to break the problem into two parts:
1. From t = 0 s to t = 17 s (when the acceleration is non-zero).
2. From t = 17 s to t = 37 s (when the acceleration is zero).
Let's solve each part step-by-step:
1. From t = 0 s to t = 17 s:
We can use the formula for angular velocity in terms of angular acceleration and time:
ω = ω0 + αt
Given: ω0 = 0 rad/s (starting from rest), ω = 5.5 rad/s, t = 1.6 s
Using the formula, we can find the angular acceleration:
5.5 rad/s = 0 + α × 1.6 s
α = 5.5 rad/s ÷ 1.6 s
α ≈ 3.44 rad/s^2
Now we can use the formula for angular displacement in terms of initial angular velocity, angular acceleration, and time:
θ = ω0t + 0.5αt^2
θ = 0 × 17 s + 0.5 × 3.44 rad/s^2 × (17 s)^2
θ = 0 + 0.5 × 3.44 rad/s^2 × 289 s^2
θ ≈ 495.52 rad
2. From t = 17 s to t = 37 s:
Since the acceleration abruptly changes to 0 rad/s^2, the wheel continues to rotate with a constant angular velocity.
To find the angle rotated during this time interval, we can use the formula:
θ = ωt
Given: ω = 5.5 rad/s, t = 20 s
θ = 5.5 rad/s × 20 s
θ = 110 rad
Adding the angles from both time intervals:
θ_total = θ1 + θ2
θ_total ≈ 495.52 rad + 110 rad
θ_total ≈ 605.52 rad
Therefore, the wheel rotates approximately 605.52 radians in the interval t = 0 s to t = 37 s.