Problem 73 - University Physics with Modern Physics

Question

A sheet of plywood \(1.3 \mathrm{~cm}\) thick is used to make a cabinet door \(55 \mathrm{~cm}\) wide by \(79 \mathrm{~cm}\) tall, with hinges mounted on the vertical edge. A small 150 - \(\mathrm{g}\) handle is mounted \(45 \mathrm{~cm}\) from the lower hinge at the same height as that hinge. If the density of the plywood is \(550 \mathrm{~kg} / \mathrm{m}^{3},\) what is the moment of inertia of the door about the hinges? Neglect the contribution of hinge components to the moment of inertia.

Step-by-Step Solution

Answer

The moment of inertia of the door about the hinges is approximately \(0.7768\,\,\mathrm{kg}\cdot\,\mathrm{m}^2\).

Step 1: Determine the mass of the entire door using the given density

First, let's find the mass of the entire door, including the plywood and handle. The volume of the plywood can be found by multiplying its dimensions: $$V_p = 55\,\text{cm}\times 79\,\text{cm}\times 1.3\,\text{cm} = \:57865\,\text{cm}^3 = 0.057865\,\text{m}^3$$ where we have converted the dimensions to meters for consistency. Now, with the given density of the plywood, we can find the mass of the plywood: $$m_p = \rho V_p = 550\,\mathrm{kg/m}^3\times 0.057865\,\mathrm{m}^3 =\: 31.82575\,\mathrm{kg}$$ The total mass of the door, including the handle, is: $$m_{tot} = m_p + m_{handle} = 31.82575\,\mathrm{kg} + 0.15\,\mathrm{kg}= 31.97575\,\,\mathrm{kg}$$

Step 2: Find the moment of inertia of the door without the handle

The moment of inertia of the plywood can be found using the formula for the moment of inertia of a rectangle rotating about one side (since the hinges are mounted along the vertical edge). The moment of inertia of the plywood about the side can be calculated using the formula: $$I_p = \frac{1}{12}m_p(b^2 + h^2)$$ where \(b\) is the width of plywood and \(h\) is the height of plywood. Plugging in the values, we get: $$I_p = \frac{1}{12}\times31.82575\,\mathrm{kg}\times\left[\left(0.55\,\,\mathrm{m}\right)^2+\left(0.79\,\,\mathrm{m}\right)^2\right] = 0.7654\,\,\mathrm{kg}\cdot\,\mathrm{m}^2$$

Step 3: Use the parallel axis theorem to find the moment of inertia of the door with the handle

The parallel axis theorem states that the moment of inertia of a body about an axis parallel to and a distance \(d\) away from an axis through the body's center of mass is \(I=I_{cm}+md^2\), where \(I_{cm}\) is the moment of inertia of the body about the axis through the center of mass and \(m\) is the body's mass. Since the handle is positioned at the same height as the bottom hinge, the distance between the handle and the plywood's center of mass is: $${d = \frac{55}{2}~\text{cm} = 27.5~\text{cm} =\:0.275~\text{m}}$$ Now, using the parallel axis theorem to find the moment of inertia of the door with the handle, we get: $${I_{handle} = I_p + m_{handle}\times d^2}$$ $${I_{handle} = 0.7654\,\,\mathrm{kg}\cdot\,\mathrm{m}^2 + 0.15\,\mathrm{kg}\times \left(0.275\,\mathrm{m}\right)^2 = 0.7654\,\,\mathrm{kg}\cdot\,\mathrm{m}^2 + 0.01138625\,\,\mathrm{kg}\cdot\,\mathrm{m}^2}$$

Step 4: Calculate the moment of inertia of the door about the hinges

Adding the terms, we get the final moment of inertia of the entire door, $$I_{door} = I_{handle} = 0.7654\,\,\mathrm{kg}\cdot\,\mathrm{m}^2 + 0.01138625\,\,\mathrm{kg}\cdot\,\mathrm{m}^2 = 0.77678625\,\,\mathrm{kg}\cdot\,\mathrm{m}^2$$ So, the moment of inertia of the door about the hinges is approximately \(0.7768\,\,\mathrm{kg}\cdot\,\mathrm{m}^2\).

Key Concepts

Parallel Axis Theorem

The Parallel Axis Theorem is a crucial concept in physics for calculating the moment of inertia of an object around an axis that isn't its center of mass axis. Imagine a door with hinges at its edge. The theorem helps find the moment of inertia when the rotation axis is parallel to the center of mass axis.
The formula is:

• \[I = I_{cm} + md^2\]

where \(I_{cm}\) is the moment of inertia through the center of mass, \(m\) is the mass, and \(d\) is the distance between the two axes.
This theorem allows us to include additional elements, like handles, and understand how they affect the door's inertia.

Density of Materials

Density is a fundamental property that relates the mass of a material to its volume. It's denoted by \(\rho\) and defined as:

• \[\rho = \frac{m}{V}\]

where \(m\) is the mass and \(V\) is the volume.
For plywood, a given density tells us how massive a specific volume will be. This is essential for calculating the mass before determining the moment of inertia. A denser material will generally result in a higher moment of inertia if the shape remains constant.

Mass Calculation

Calculating mass is one of the first steps in solving physics problems involving inertia. Using the formula:

• \[m = \rho V\]

we can determine the mass from density and volume.
For the cabinet door:

• Find the volume by multiplying its dimensions.
• Convert dimensions to meters to keep measurements consistent.
• Use given density to find mass.

Including additional components, like the handle, completes our mass calculation, incorporating all elements affecting inertia.

Geometric Dimensions

Understanding geometric dimensions is vital for calculating the moment of inertia. These dimensions define the shape and volume of the object. For the door, we consider:

• The height (79 cm), width (55 cm), and thickness (1.3 cm).

Knowing these helps compute volume, which is essential for mass calculation.
Additionally, the position of components like handles in relation to the axis of rotation can affect calculations. Therefore, accurately measuring and converting geometric dimensions is a key part of the process.

Physics Problem Solving

Tackling physics problems successfully requires a systematic approach:

• Understand the problem and identify known values and unknowns.
• Use relevant formulas, like those for mass, volume, and inertia.
• Apply concepts such as the Parallel Axis Theorem when necessary.
• Check calculations for consistency with units and conversions.

This structured method ensures each step logically leads to the next, providing clarity and accuracy in finding solutions. In the case of our cabinet door, combining these principles helps in precisely calculating the moment of inertia.