Question

The trajectory of a model ship during a pure sway PMM test is shown below. The steady forward speed, u is 2.0 m/s. The maximum amplitude of sway motion, \(y_{Max}\) is 0.5 m and its period is 8 s. The magnitude of maximum drift angle, in degrees (round off to the nearest integer), and the magnitude of maximum sway acceleration, in m/s\(^2\) (round off to one decimal place), of the model respectively are 

• A. 11 and 0.3
• B. 13 and 0.5
• C. 15 and 0.2
• D. 9 and 0.1

Hint:

For any simple harmonic motion described by \( x = A \sin(\omega t) \), the maximum velocity is \( A\omega \) and the maximum acceleration is \( A\omega^2 \). This is a fundamental concept in physics and is directly applicable here to the sway motion.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves analyzing the kinematics of a ship model undergoing a pure sway test on a Planar Motion Mechanism (PMM). The model has a steady forward speed (surge) and a superimposed sinusoidal motion in the transverse direction (sway). We need to calculate two quantities: the maximum drift angle and the maximum sway acceleration.
Step 2: Key Formula or Approach:
The sway motion is sinusoidal, so we can describe the sway position \(y(t)\) and sway velocity \(v(t)\) as: \[ y(t) = y_{Max} \sin(\omega t) \] \[ v(t) = \frac{dy}{dt} = y_{Max} \omega \cos(\omega t) \] where \( \omega \) is the angular frequency, \( \omega = \frac{2\pi}{T} \).
1. Drift Angle (\(\beta\)): The drift angle is the angle between the ship's longitudinal axis and its instantaneous velocity vector. It is given by \( \tan(\beta) = \frac{v}{u} \), where \(v\) is the sway velocity and \(u\) is the forward speed. The maximum drift angle occurs when the sway velocity \(v\) is maximum. \[ v_{max} = y_{Max} \omega \] \[ \beta_{max} = \arctan\left(\frac{v_{max}}{u}\right) \] 2. Sway Acceleration (\(a_y\)): The sway acceleration is the second derivative of the sway position: \[ a_y(t) = \frac{d^2y}{dt^2} = -y_{Max} \omega^2 \sin(\omega t) \] The magnitude of the maximum sway acceleration is: \[ (a_y)_{max} = y_{Max} \omega^2 \] Step 3: Detailed Calculation:
Given values:
- Forward speed, \( u = 2.0 \) m/s.
- Max sway amplitude, \( y_{Max} = 0.5 \) m.
- Period of sway motion, \( T = 8 \) s.
First, calculate the angular frequency \(\omega\):
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \approx 0.7854 \text{ rad/s} \] 1. Calculate Maximum Drift Angle:
- Calculate maximum sway velocity:
\[ v_{max} = y_{Max} \omega = 0.5 \times \frac{\pi}{4} = \frac{\pi}{8} \approx 0.3927 \text{ m/s} \] - Calculate the maximum drift angle in radians: \[ \beta_{max} = \arctan\left(\frac{v_{max}}{u}\right) = \arctan\left(\frac{0.3927}{2.0}\right) = \arctan(0.19635) \approx 0.194 \text{ rad} \] - Convert to degrees: \[ \beta_{max} (\text{deg}) = 0.194 \text{ rad} \times \frac{180}{\pi} \approx 11.12^\circ \] - Round to the nearest integer: 11\(^{\circ}\). 2. Calculate Maximum Sway Acceleration:
- Calculate the magnitude of maximum sway acceleration: \[ (a_y)_{max} = y_{Max} \omega^2 = 0.5 \times \left(\frac{\pi}{4}\right)^2 = 0.5 \times \frac{\pi^2}{16} = \frac{\pi^2}{32} \] \[ (a_y)_{max} \approx \frac{9.8696}{32} \approx 0.3084 \text{ m/s}^2 \] - Round to one decimal place: 0.3 m/s\(^2\). Step 4: Final Answer:
The maximum drift angle is 11 degrees, and the maximum sway acceleration is 0.3 m/s\(^2\). This corresponds to option (A).