The given problem involves a wheel that has an initial angular speed of 87 rev/s and experiences a constant angular acceleration of 73 rev/s^2, causing it to spin faster. The goal is to determine the time the wheel was experiencing this angular acceleration and its final angular speed.
To solve this problem, we can use the following equations:
1. Δθ = ωi*t + (1/2)α*t^2
2. ωf = ωi + α*t
where:
Δθ is the change in angle (in radians)
ωi is the initial angular speed (in rad/s)
ωf is the final angular speed (in rad/s)
α is the angular acceleration (in rad/s^2)
t is the time (in seconds)
We are given that Δθ = 56 revolutions. To convert this to radians, we need to multiply it by 2π, since 1 revolution = 2π radians. So, Δθ = 56 * 2π radians.
We are also given that ωi = 87 rev/s and α = 73 rev/s^2. To convert these values to radians, we need to multiply them by 2π, since 1 revolution = 2π radians. So, ωi = 87 * 2π rad/s and α = 73 * 2π rad/s^2.
Now, let's solve for t using equation 1:
56 * 2π = (87 * 2π) * t + (1/2) * (73 * 2π) * t^2
Simplifying the equation:
56 = 87t + 73t^2
Rearranging the equation and setting it equal to zero:
73t^2 + 87t - 56 = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
t = (-87 ± √(87^2 - 4 * 73 * -56)) / (2 * 73)
Simplifying the equation:
t = (-87 ± √(7569 + 13184)) / 146
t = (-87 ± √(20753)) / 146
Calculating the square root:
t = (-87 ± 143.99) / 146
Simplifying further:
t ≈ (-87 + 143.99) / 146 or t ≈ (-87 - 143.99) / 146
t ≈ 0.753 seconds or t ≈ -1.372 seconds
Since time cannot be negative, we discard the negative solution. Therefore, the wheel was experiencing this angular acceleration for approximately 0.753 seconds.
To find the final angular speed, we can use equation 2:
ωf = ωi + α * t
Plugging in the values:
ωf = 87 * 2π + (73 * 2π) * 0.753
Simplifying the equation:
ωf ≈ 546.75 rad/s
Therefore, at the end of this period, the wheel was spinning at a speed of approximately 546.75 rad/s.