91Ó°ÊÓ

The moment of inertia is an important concept when dealing with rotational motion. It is like mass in linear motion but for rotating objects. It determines how difficult it is to change an object's rotational speed. For a door, which can be modeled as a rectangle, the moment of inertia I is calculated using the formula:\[ I = \frac{1}{3} ML^2 \] where \( M \) is the mass of the door and \( L \) is its width (the point from which it rotates).In our problem, we need to first determine the mass of the door, which is given by dividing its weight by the acceleration due to gravity:

• Weight = 750 N
• Gravity, g = 9.8 m/s²
• Mass, M = \( \frac{750}{9.8} \approx 76.53 \text{ kg} \)

With these values, the moment of inertia becomes:\[ I = \frac{1}{3} \times 76.53 \times (1.25)^2 \approx 39.84 \text{ kg} \cdot \text{m}^2 \]This tells us how the distribution of mass affects rotational acceleration.

Torque is the force that causes objects to rotate around an axis. In this scenario, it is the key to understanding how the door starts moving when Exena applies a force. The torque \( \tau \) is calculated using the simple formula:\[ \tau = F \times r \times \sin(\theta) \]where:

• \( F \) is the force applied (220 N in this case)
• \( r \) is the distance from the pivot point to where the force is applied (1.25 m, the width of the door)
• \( \theta \) is the angle between the force direction and the line from the pivot, which is 90 degrees here because the force is perpendicular

Because \( \sin(90^{\circ}) = 1 \), the formula simplifies to:\[ \tau = 220 \times 1.25 \times 1 = 275 \text{ N} \cdot \text{m} \]This value of torque shows how much rotational force is being applied to the door.

Angular acceleration is a measure of how quickly the rotational speed of an object is changing. It is crucial in this problem to understand how fast the door can be closed. The relationship between torque \( \tau \) and angular acceleration \( \alpha \) is given by:\[ \alpha = \frac{\tau}{I} \]where:

• \( \tau \) is the torque applied (275 N·m)
• \( I \) is the moment of inertia (39.84 kg·m²)

Substituting these values gives:\[ \alpha = \frac{275}{39.84} \approx 6.90 \text{ rad/s}^2 \]This angular acceleration tells us how quickly the door begins to move faster from its initial stationary position until it reaches the final angular speed.

Rotational motion describes the behavior of objects that rotate about an axis. The analysis of this motion provides insight into how forces translate into rotational speeds. For a door rotating about its hinges, the key equation relates angular displacement \( \theta \), angular acceleration \( \alpha \), and time \( t \) as follows:\[ \theta = \frac{1}{2} \alpha t^2 \]For Exena's door, we want it to rotate through 90 degrees (or \( \frac{\pi}{2} \) radians). Plugging in known values, the equation becomes:\[ \frac{\pi}{2} = \frac{1}{2} \times 6.90 \times t^2 \]Solving for \( t \) gives:\[ t^2 = \frac{\pi}{13.8} \]\[ t = \sqrt{\frac{\pi}{13.8}} \approx 0.672 \text{ seconds} \]This calculation shows how the applied force translates into time taken to complete the rotation.

Newton's Laws of Motion are fundamental principles that describe the behavior of objects in motion. In this problem, we particularly use the second law, which relates force, mass, and acceleration. But here, rotational versions of these concepts are used.Newton's second law can be adapted to rotational dynamics as:\[ \tau = I \alpha \]where:

• \( \tau \) is torque
• \( I \) is moment of inertia
• \( \alpha \) is angular acceleration

This law helps us understand how Exena's force results in the door's rotation.By calculating the necessary quantities like torque and moment of inertia, we linked linear force concepts to rotational movement. Exena's effort resulted in the door's accelerated rotation, exemplifying Newton's laws in action.