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.

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: $$m_{tot} = m_p + m_{handle} = 31.82575\,\mathrm{kg} + 0.15\,\mathrm{kg}= 31.97575\,\,\mathrm{kg}$$

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: $$I_p = \frac{1}{12}\times31.82575\,\mathrm{kg}\times\left[\left(0.55\,\,\mathrm{m}\right)^2+\left(0.79\,\,\mathrm{m}\right)^2\right] = 0.7654\,\,\mathrm{kg}\cdot\,\mathrm{m}^2$$

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: $${I_{handle} = I_p + m_{handle}\times d^2}$$ $${I_{handle} = 0.7654\,\,\mathrm{kg}\cdot\,\mathrm{m}^2 + 0.15\,\mathrm{kg}\times \left(0.275\,\mathrm{m}\right)^2 = 0.7654\,\,\mathrm{kg}\cdot\,\mathrm{m}^2 + 0.01138625\,\,\mathrm{kg}\cdot\,\mathrm{m}^2}$$

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 \(0.7768\,\,\mathrm{kg}\cdot\,\mathrm{m}^2\).