Question

Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.38 x 10-3 rad/s2 for 2.04 x 103 s. For the next 1.48 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.63 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.42 rad/s. Find the total angular displacement of the propeller.

Answer 1

Δθ = 15747.37 rad.

Step-by-step 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:

`\omega_(f1) = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)`

• 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:

`\omega_(f1)^(2) = 2* \alpha *\Delta\theta (2)`

• Solving for Δθ in (2):

`\theta_(1) = (\omega_(f1)^(2))/(2*\alpha ) = ((4.9rad/sec)^(2))/(2*2.38*e-3rad/sec2) = 5044.12 rad (3)`

• 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:

`\theta_(2) =\omega_(f1) * \Delta_(t2) = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)`

• 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:

`\omega_(f2)^(2) - \omega_(o2)^(2) = 2* \alpha *\Delta\theta (5)`

• Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

`\theta_(3) = ((\omega_(f2))^(2)- (\omega_(f1)) ^(2) )/(2*\alpha ) = ((2.42rad/s^(2)) -(4.9rad/sec)^(2))/(2*(-2.63*e-3rad/sec2)) = 3451.25 rad (6)`

• 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.