Problem 46 - University Physics with Modern Physics

Question

A solid wood door 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high is hinged along one side and has a total mass of 40.0 \(\mathrm{kg}\) . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 \(\mathrm{kg}\) , traveling perpendicular to the door at 12.0 \(\mathrm{m} / \mathrm{s}\) 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

Final angular speed: 0.224 rad/s. The mud's contribution to moment of inertia is insignificant.

Step 1: Identify Known Values

We know the dimensions of the door are 1.00 m by 2.00 m with a mass of 40.0 kg. The mud has a mass of 0.500 kg and impacts the door with a velocity of 12.0 m/s perpendicular to the door's surface. The door pivots freely along its hinge.

Step 2: Calculate Moment of Inertia of the Door

The door is a rectangular plane and its moment of inertia, when pivoted about its edge, is given by the formula \( I_{door} = \frac{1}{3} m L^2 \). For the door, \( m = 40.0 \ \mathrm{kg}\) and \( L = 1.00 \ \mathrm{m}\). Thus, \( I_{door} = \frac{1}{3} \times 40.0 \times (1.00)^2 = 13.33 \ \mathrm{kg} \cdot \mathrm{m}^2\).

Step 3: Calculate Angular Momentum of the Mud

The mud's mass is 0.500 kg and velocity is 12.0 m/s, striking the door at the center of its width. The distance from the hinge to the impact point is 0.5 m. The angular momentum \( L_{mud} = m \cdot v \cdot r = 0.500 \times 12.0 \times 0.5 = 3.0 \ \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\).

Step 4: Apply Conservation of Angular Momentum

Assume no external torques act, so total angular momentum before and after impact remains equal. Before impact, it's only due to the mud, \( L_{mud} = 3.0 \ \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\). After impact, it is the sum of the angular momentum of both the door and mud system: \( I_{total} \cdot \omega_f \), where \( I_{total} = 13.33 + (0.500)(0.5)^2 = 13.395 \ \mathrm{kg} \cdot \mathrm{m}^2 \). Setting equal: \( 3.0 = 13.395 \cdot \omega_f \).

Step 5: Solve for Final Angular Speed

Solving \( 3.0 = 13.395 \cdot \omega_f \), we get the angular speed \( \omega_f = \frac{3.0}{13.395} \approx 0.224 \ \mathrm{rad/s}\).

Step 6: Assess Significance of Mud Contribution

Calculate the mud's moment of inertia: \( I_{mud} = m \cdot r^2 = 0.500 \times (0.5)^2 = 0.125 \ \mathrm{kg} \cdot \mathrm{m}^2\). The total inertia is \( 13.395 \), so the mud's contribution is \( \frac{0.125}{13.395} \approx 0.0093\), about 0.93%. It makes an insignificant contribution.

Key Concepts

Moment of Inertia

The concept of Moment of Inertia is central to understanding rotational motion. It is akin to mass in translational motion and illustrates how an object's mass is distributed relative to an axis of rotation. In simple terms, it measures an object's resistance to changes in its rotation rate.
The formula for the Moment of Inertia (\(I\)) depends on the shape and axis about which it is rotating. For our problem with a rectangular door, the formula is:

• \(I_{door} = \frac{1}{3} m L^2\)

Here, \(m\) is the mass of the door, and \(L\) is the width of the door.
Calculations show that for a 40 kg door, the Moment of Inertia turns out to be 13.33 \(\text{kg} \cdot \text{m}^2\), making it a substantial value for the door's rotational characteristics.
Understanding this concept helps predict how the door will behave when external forces, like the mud in this exercise, act upon it.

Conservation of Angular Momentum

The principle of Conservation of Angular Momentum is a fundamental law in physics. It states that if no external torques act on a system, the total angular momentum remains constant. In our problem, the angular momentum before the mud hits the door must be the same as after.

• Initially, only the mud has angular momentum: \(L_{mud} = m \cdot v \cdot r = 0.500 \times 12.0 \times 0.5 = 3.0 \ \text{kg} \cdot \text{m}^2/\text{s}\).
• After the collision, both the door and the mud share the angular momentum.

The combined system now must still have an angular momentum of 3.0 \(\text{kg} \cdot \text{m}^2\)/s. This calculation provides insights into how the system will move thereafter without external influence.

Angular Speed Calculation

Calculating the Angular Speed (\(\omega\)) after the collision is key to understand the resulting motion. We use the conservation principles to find it. The equation that connects angular momentum and angular speed is:

• \(L_{total} = I_{total} \cdot \omega_f\)

In our example, we calculated that the total moment of inertia, including the door and the mud, is approximately 13.395 \(\text{kg} \cdot \text{m}^2\). Solving for the final angular speed:

• \(\omega_f = \frac{3.0}{13.395} \approx 0.224 \ \text{rad/s}\)

This result reveals how fast the door rotates after being struck by the mud. It’s a modest speed, reflecting both the original strike and the inertia possessed by the door. Importantly, the contribution of the mud to this speed is minimal, highlighting the more significant role of the door's mass distribution.