Problem 44 - University Physics with Modern Physics

Question

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 kg, traveling perpendicular to the door at 12.0 m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

Step-by-Step Solution

Answer

The final angular speed of the door is 0.225 rad/s; the mud's contribution to the inertia is insignificant.

Step 1: Identify the Moment of Inertia of the Door

The moment of inertia \( I \) for a rectangular door about the hinge is given by the formula \( I = \frac{1}{3} mL^2 \), where \( m \) is the mass of the door and \( L \) is the width of the door. Here, \( m = 40.0 \) kg and \( L = 1.00 \) m, so \( I = \frac{1}{3} \times 40.0 \times (1.00)^2 = 13.33 \) kg \( \text{m}^2 \).

Step 2: Recognize Conservation of Angular Momentum

Initially, the door is at rest, so its angular momentum \( L_i = 0 \). Right after the collision, the mud sticks to the door, turning its linear momentum into angular momentum, which can be expressed as \( L_f = (m_m \cdot v \cdot \frac{L}{2}) \), since the mud strikes at the center. Here, \( m_m = 0.5 \) kg and \( v = 12.0 \) m/s, giving \( L_f = 0.5 \times 12.0 \times \frac{1.00}{2} = 3.0 \) kg \( \text{m}^2/\text{s} \).

Step 3: Calculate the Final Angular Speed

By the conservation of angular momentum, the initial angular momentum \( L_i \) is equal to the final angular momentum \( L_f \). Thus, \( I \cdot \omega = 3.0 \), where \( I = 13.33 \). Solving for \( \omega \), we get \( \omega = \frac{3.0}{13.33} = 0.225 \) radians per second.

Step 4: Assess the Moment Contribution by Mud

The mud's contribution to the final moment of inertia is \( I_{mud} = m_m \times (L/2)^2 = 0.5 \times (0.5)^2 = 0.125 \) kg \( \text{m}^2 \). Compare this with the door's moment of inertia (13.33 kg \( \text{m}^2 \)). The mud makes a minor contribution, which is not significant.

Key Concepts

Moment of Inertia

In understanding the movement of a door struck by mud, we start with the concept of moment of inertia. It's like mass in linear motion but for rotating objects. The moment of inertia \( I \) translates to how difficult it is to change the rotational speed. For us to calculate it, the shape and axis of rotation matter.
For a rectangular door hinged at one side, use the equation:
- \( I = \frac{1}{3} mL^2 \) - where \( m \) is the mass of the door, and \( L \) is its width.In our exercise, a door weighing 40 kg, 1 meter wide, gives:
- \( I = \frac{1}{3} \times 40.0 \times (1.00)^2 = 13.33 \ \text{kg} \ \text{m}^2 \)This formula shows us that width contributes more than height to rotational resistance. Also, check other objects like rods or cylinders. They each have unique formulas for inertia.

Conservation of Angular Momentum

When an object rotates, its angular momentum \( L \) remains constant if no external torque acts. This is the conservation of angular momentum, a key principle in physics. In simpler terms, it means the door and mud must conserve momentum when they interact.
- We start with the door at rest, so its initial angular momentum \( L_i \) is 0.
- The mud, traveling at 12 m/s, transfers its linear momentum to the door as angular momentum:
- \( L_f = m_m \cdot v \cdot \frac{L}{2} \)
- With 0.5 kg of mud, - \( L_f = 0.5 \times 12.0 \times \frac{1.00}{2} = 3.0 \ \text{kg} \ \text{m}^2/\text{s} \)By conservation, the initial \( L_i \)and final \( L_f \) angular momenta must equal. This is how impact velocity and position create rotation.

Angular Speed Calculation

After understanding momentum conservation, we calculate angular speed \( \omega \). This speed tells how fast the door spins post-impact.
The relation of angular momentum \( L \) and angular velocity is:- \( I \cdot \omega = L \)Here, \( I \) stands for moment of inertia and \( \omega \) for angular speed in radians per second.
Plugging known values in, from the conservation equation:- \( 13.33 \cdot \omega = 3.0 \)- Solving, \( \omega = \frac{3.0}{13.33} = 0.225 \) radians/second.This result tells the door's new spinning rate due to the mud.Such calculated outcomes are crucial in designing mechanisms where precise rotational control is necessary. Remember, everything affects speed—mass, distance from the rotation point, and impact strength.