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Let's work through the problem step-by-step:

### Problem Setup and Equation of Motion

The problem involves a rigid door with a torsion spring and a torsional damper. The door has a moment of inertia \( J = 19 \, \text{kgm}^2 \). The system needs to be critically damped, and the undamped natural frequency should not exceed 1.15 rad/s.

1. **Equation of Motion:**

   The equation of motion for a rotational system with a torsion spring (constant \( k \)) and a damper (constant \( c \)) is given by:

   \[
   J\ddot{\theta} + c\dot{\theta} + k\theta = 0
   \]

### Part (a): Spring Stiffness and Damping Coefficient

To find \( k \) and \( c \) for a critically damped system:

- **Natural Frequency (\(\omega_n\))**:
  
  For critical damping, the damping coefficient \( c \) is related to \( k \) and \( J \) by the equation \( c = 2\sqrt{kJ} \). The natural frequency is given by:

  \[
  \omega_n = \sqrt{\frac{k}{J}}
  \]

  Given that \( \omega_n \leq 1.15 \, \text{rad/s} \):

  \[
  1.15 \geq \sqrt{\frac{k}{19}}
  \]

  \[
  1.15^2 \geq \frac{k}{19}
  \]

  \[
  1.3225 \times 19 \geq k
  \]

  \[
  k \leq 25.1275 \, \text{Nm/rad}
  \]

- **Damping Coefficient (\(c\))**:

  For critical damping:

  \[
  c = 2\sqrt{kJ}
  \]

  Using the maximum value of \( k = 25.1275 \, \text{Nm/rad} \):

  \[
  c = 2\sqrt{25.1275 \times 19}
  \]

  \[
  c \approx 2 \times 21.85
  \]

  \[
  c \approx 43.7 \, \text{Nms/rad}
  \]

### Part (b): Initial Angular Velocity

For the initial angular velocity required to open the door to an angle of \( 10^\circ \) from the closed position:

- Convert the angle to radians: \( 10^\circ = \frac{\pi}{18} \, \text{radians} \).

  For a critically damped system, the response is of the form:

  \[
  \theta(t) = (\theta_0 + (c/2J) \theta_0 t)e^{-\frac{c}{2J} t}
  \]

  The initial velocity can be found from the initial conditions. However, without knowing the specific dynamics (e.g., if the door is released from rest or if an initial push is applied), we can't provide a precise value for \( \dot{\theta}(0) \). 

### Part (c): Modified Damping Ratio

If a mass of 4 kg is added at a distance of 0.5 m from the hinge:

1. **New Moment of Inertia (J'):**

   \[
   J' = J + mr^2
   \]

   Where \( m = 4 \, \text{kg} \) and \( r = 0.5 \, \text{m} \):

   \[
   J' = 19 + 4 \times 0.5^2
   \]

   \[
   J' = 19 + 1 = 20 \, \text{kgm}^2
   \]

2. **New Damping Ratio (\(\zeta'\)):**

   \[
   \zeta' = \frac{c}{2\sqrt{J'k}}
   \]

   Given \( \zeta' = 0.9747 \):

   \[
   0.9747 = \frac{43.7}{2\sqrt{20 \times 25.1275}}
   \]

   This equation can be used to confirm the damping coefficient \( c \).

Would you like more detailed steps or have specific questions?

Here are some related questions you might explore:

1. How does the addition of mass affect the system's natural frequency?
2. What would happen if the damping coefficient \( c \) is reduced?
3. How can you measure the damping ratio in a physical system?
4. What are the practical implications of critical damping in real-world systems?
5. How does the position of the added mass influence the moment of inertia and system response?

**Tip:** In problems involving angular motion, always check if the angles are in degrees or radians, especially when calculating trigonometric functions or angular velocities.

Explore critical damping in rotational systems using a torsion spring and damper. Learn how to determine spring stiffness, damping coefficient, and initial angular velocity. Includes detailed calculations and formulas. Suitable for advanced college-level study.

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