Question

Consider a Planar Motion Mechanism (PMM) test of a ship model in a towing tank. The transverse motion of the model from the centerline of the tank is described by \( y = a_0 \cos(\omega t) \), where \( \omega \) is the angular frequency. The carriage speed is 3 m/s, \( \omega = \frac{\sqrt{3}}{2} \) rad/s, and the maximum drift angle during the test is \( 30^\circ \). The amplitude of oscillation \( a_0 \) lies between ________.

• A. \( 0.2 { m and } 0.4 { m} \)
• B. \( 0.5 { m and } 0.7 { m} \)
• C. \( 1.0 { m and } 1.2 { m} \)
• D. \( 1.9 { m and } 2.1 { m} \)

Hint:

When dealing with PMM tests, remember that the drift angle is the arctangent of the ratio of lateral to forward velocity. Use max values to find amplitude constraints.

Answer 1

The Correct Option is D

Step 1: Use the formula for drift angle \(\delta\): \[ \tan \delta = \frac{{lateral velocity}}{{forward velocity}} = \frac{a_0 \omega \sin(\omega t)}{U} \] The maximum drift angle occurs when \( \sin(\omega t) = 1 \), so: \[ \tan(30^\circ) = \frac{a_0 \cdot \omega}{3} \quad \Rightarrow \quad \frac{1}{\sqrt{3}} = \frac{a_0 \cdot \frac{\sqrt{3}}{2}}{3} \] Step 2: Solve for \( a_0 \): \[ \frac{1}{\sqrt{3}} = \frac{a_0 \cdot \frac{\sqrt{3}}{2}}{3} \quad \Rightarrow \quad \frac{1}{\sqrt{3}} = \frac{a_0 \cdot \sqrt{3}}{6} \quad \Rightarrow \quad 6 = a_0 \cdot \sqrt{3} \cdot \sqrt{3} \quad \Rightarrow \quad 6 = a_0 \cdot 3 \quad \Rightarrow \quad a_0 = 2 \] So the amplitude \( a_0 \) lies between 1.9 m and 2.1 m.