GSO002 Problem 13

1. Mechanical Energy in the Rotating Frame

Let the $z$ -axis point vertically downward with O as origin. Let $\theta(t)$ be the angle between each rod and the $z$ -axis at any time.

In the rotating frame, the speed of each ball is $v_m' = l\dot{\theta}$ . Since the quadrilateral is a rhombus, sleeve C is always at depth $2l\cos\theta$ , so its speed is $v_M' = 2l\dot{\theta}\sin\theta$ .

Kinetic energy in the rotating frame:

K' = (m + 2M\sin^2\theta)l^2\dot{\theta}^2

The effective potential energy is the sum of gravitational and centrifugal potential energies:

U_{eff}(\theta) = -2(m+M)gl\cos\theta - ml^2\omega^2\sin^2\theta

The mechanical energy $E' = K' + U_{eff}(\theta) = \text{const}$ is conserved in the rotating frame.

2. Steady-State Condition

At steady state $\theta = \theta_0$ ( $\dot{\theta} = 0$ , $\ddot{\theta} = 0$ ), i.e., $U_{eff}'(\theta_0) = 0$ :

\cos\theta_0 = \frac{(m+M)g}{ml\omega^2}

Substituting:

\cos\theta_0 = \frac{(4 + 3) \times 12}{4 \times 1.4 \times 25} = \frac{84}{140} = \frac{3}{5}

So $\cos\theta_0 = 3/5$ and $\sin\theta_0 = 4/5$ .

3. Angular Frequency of Small Oscillations

Differentiating the energy conservation equation with respect to time and linearizing for $\theta = \theta_0 + \Delta\theta$ :

2(m + 2M\sin^2\theta_0)l^2 \Delta\ddot{\theta} = -U_{eff}''(\theta_0) \Delta\theta

Computing $U_{eff}''(\theta_0)$ :

U_{eff}''(\theta) = 2(m+M)gl\cos\theta - 2ml^2\omega^2(\cos^2\theta - \sin^2\theta)

At $\theta_0$ , using $2(m+M)gl\cos\theta_0 = 2ml^2\omega^2\cos^2\theta_0$ :

U_{eff}''(\theta_0) = 2ml^2\omega^2\sin^2\theta_0

The angular frequency of small oscillations:

\omega_{osc}^2 = \frac{m\omega^2\sin^2\theta_0}{m + 2M\sin^2\theta_0}

Numerically:

Numerator: $4 \times 25 \times (4/5)^2 = 100 \times 16/25 = 64$
Denominator: $4 + 2 \times 3 \times (4/5)^2 = 4 + 96/25 = 196/25$

\omega_{osc}^2 = \frac{64}{196/25} = \frac{1600}{196} = \frac{400}{49} \implies \omega_{osc} = \frac{20}{7} \text{ rad/s}

Period:

T = \frac{2\pi}{\omega_{osc}} = \frac{7}{10}\pi \text{ [s]}

With $A = 7$ , $B = 10$ (coprime): $100A + B = 710$ .

Final Answer: 710

GSO002 Problem 13

Watt Governor: Non-Linear Small Oscillations in a Rotating System

Problem Statement

Constraints

Input Format

Solution

1. Mechanical Energy in the Rotating Frame

2. Steady-State Condition

3. Angular Frequency of Small Oscillations