Answer:
Δθ = 15747.37 rad.
Explanation:
- The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
- Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:
[tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]
- Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:
[tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]
[tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]
- The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:
[tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]
- Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:
[tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]
- Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:
[tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]
- The total angular displacement is just the sum of (3), (4) and (6):
- Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
- ⇒ Δθ = 15747.37 rad.
