Moment of inertia of rectangle prism

 Moment of inertia of rectangle prism

A rectangular prism is a three-dimensional object with three distinct dimensions: length ($L$), width ($W$), and height ($H$). Its moment of inertia describes the resistance to rotational motion about a particular axis. The calculation depends on the axis of rotation and the mass distribution.

The moment of inertia of a rectangular prism depends on the axis about which it is being calculated. The formula differs based on the axis of rotation. Let’s assume the prism has dimensions:

• Length ($L$)
• Width ($W$)
• Height ($H$)
• Mass ($M$)

Common Cases:

1. About an axis passing through the center of mass and parallel to the length ($L$):

   $$I_{\text{center}, L} = \frac{1}{12} M \left(W^2 + H^2\right)$$
   

2. About an axis passing through the center of mass and parallel to the width ($W$):

   $$I_{\text{center}, W} = \frac{1}{12} M \left(L^2 + H^2\right)$$
   

3. About an axis passing through the center of mass and parallel to the height ($H$):

   $$I_{\text{center}, H} = \frac{1}{12} M \left(L^2 + W^2\right)$$
   

4. About an axis along one edge (e.g., along the height or width): If the axis is at one edge instead of passing through the center, the parallel axis theorem is used:

   $$I_{\text{edge}} = I_{\text{center}} + M \cdot d^2$$
   

   where $d$ is the distance between the edge and the center of mass.

Let’s work through an example of calculating the moment of inertia of a rectangular prism.

Example:

A rectangular prism has the following dimensions:

• Length ($L$) = 4 m
• Width ($W$) = 2 m
• Height ($H$) = 1 m
• Mass ($M$) = 10 kg

We will calculate the moment of inertia about:

1. An axis through the center of mass parallel to the height ($H$)
2. An axis along one of the height edges using the parallel axis theorem

1. Axis through the center of mass parallel to height ($H$):

The formula is:

$$I_{\text{center}, H} = \frac{1}{12} M \left(L^2 + W^2\right)$$

Substitute the values:

$$I_{\text{center}, H} = \frac{1}{12} (10) \left(4^2 + 2^2\right)$$
$$I_{\text{center}, H} = \frac{1}{12} (10) \left(16 + 4\right)$$
$$I_{\text{center}, H} = \frac{1}{12} (10) (20)$$
$$I_{\text{center}, H} = \frac{200}{12} = 16.67 \, \text{kg·m}^2$$

2. Axis along one of the height edges:

Using the parallel axis theorem,

$$I_{\text{edge}} = I_{\text{center}, H} + M \cdot d^2$$

Here, $d$ is the distance from the center of mass to the edge along the width and length.

$$d = \sqrt{\left(\frac{L}{2}\right)^2 + \left(\frac{W}{2}\right)^2}$$Substitute the values:$$d = \sqrt{\left(\frac{4}{2}\right)^2 + \left(\frac{2}{2}\right)^2}$$
$$d = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \, \text{m}$$
$$d \approx 2.236 \, \text{m}$$

Now calculate $I_{\text{edge}}$:

$$I_{\text{edge}} = 16.67 + 10 \cdot (2.236)^2$$
$$I_{\text{edge}} = 16.67 + 10 \cdot 5$$
$$I_{\text{edge}} = 16.67 + 50 = 66.67 \, \text{kg·m}^2$$

Results:

1. Moment of inertia through the center of mass ($H$): $16.67 \, \text{kg·m}^2$
   
2. Moment of inertia along one edge ($H$): $66.67 \, \text{kg·m}^2$