Problem 64 - University Physics with Modern Physics

Question

While exploring a castle, Exena the Exterminator is spotted by a dragon that chases her down a hallway. Exena runs into a room and attempts to swing the heavy door shut before the dragon gets her. The door is initially perpendicular to the wall, so it must be turned through \(90^{\circ}\) to close. The door is 3.00 \(\mathrm{m}\) tall and 1.25 \(\mathrm{m}\) wide, and it weighs 750 \(\mathrm{N} .\) You can ignore the friction at the hinges. If Exena applies a force of 220 \(\mathrm{N}\) at the edge of the door and perpendicular to it, how much time does it take her to close the door?

Step-by-Step Solution

Answer

It takes approximately 0.672 seconds to close the door.

Step 1: Understanding the Task

We need to calculate the time it takes for Exena to close a door by applying a force. She applies a perpendicular force at the edge of the door, and the goal is to rotate the door through 90 degrees. We will calculate the moment of inertia, torque, angular acceleration, and finally the time taken to close the door.

Step 2: Calculate the Moment of Inertia

The door behaves as a rectangular body rotating about its hinges. The moment of inertia for a rectangle about an axis through one edge is given by the formula \( I = \frac{1}{3} M L^2 \). We first find the mass, which is \( M = \frac{750}{9.8} = 76.53 \) kg. Hence, the moment of inertia is \[ I = \frac{1}{3} \times 76.53 \times (1.25)^2 = 39.84 \text{ kg} \cdot \text{m}^2 \].

Step 3: Calculate the Torque Applied

Torque \( \tau \) is calculated using the formula \( \tau = F \times r \times \sin(\theta) \). With \( F = 220 \) N, \( r = 1.25 \) m, and since the force is perpendicular, \( \theta = 90^{\circ} \) and \( \sin(90^{\circ}) = 1 \):\[ \tau = 220 \times 1.25 \times 1 = 275 \text{ N} \cdot \text{m} \].

Step 4: Calculate Angular Acceleration

Angular acceleration \( \alpha \) is given by the relation \( \alpha = \frac{\tau}{I} \). Plugging in the values:\[ \alpha = \frac{275}{39.84} \approx 6.90 \text{ rad/s}^2 \].

Step 5: Determine Time to Close Door

To find the time to rotate through \(90^{\circ}\) (or \(\frac{\pi}{2}\) radians), we use the equation for angular motion:\( \theta = \frac{1}{2} \alpha t^2 \). Solving for \(t\):\[ \frac{\pi}{2} = \frac{1}{2} \times 6.90 \times t^2 \]\[ t^2 = \frac{\pi}{2 \times 6.90} \]\[ t = \sqrt{\frac{\pi}{13.8}} \approx 0.672 \text{ seconds} \].

Key Concepts

Moment of Inertia

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 Calculation

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

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

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

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.