Find the Moment of Inertia of the Triangle about its base which lies on the X axis. The height and base length are shown in d attatched fig..

To find the moment of inertia of a triangle about its base, we can use a straightforward approach based on the geometry of the triangle. The moment of inertia, often denoted as I, is a measure of an object's resistance to rotational motion about a given axis. For a triangle with its base along the X-axis, we can derive the moment of inertia using integration or by applying a known formula.

Understanding the Triangle's Dimensions

Let’s denote the base length of the triangle as \( b \) and the height as \( h \). The triangle is positioned such that its base lies along the X-axis, extending from \( (0, 0) \) to \( (b, 0) \), and its apex is at the point \( (b/2, h) \).

Using the Formula for Moment of Inertia

The moment of inertia \( I \) of a triangle about its base can be calculated using the formula:

• Formula: \( I = \frac{1}{3} b h^3 \)

This formula arises from the integration of the area elements of the triangle, taking into account their distances from the axis of rotation (the base in this case).

Deriving the Moment of Inertia

To derive this, consider the triangle as being composed of infinitesimally thin horizontal strips. Each strip has a width \( dx \) and a height \( y \), where \( y \) varies linearly from 0 at the base to \( h \) at the apex. The relationship between \( y \) and \( x \) can be expressed as:

• Equation of the line: \( y = \frac{h}{b} x \) for \( 0 \leq x \leq b \)

The area \( dA \) of a thin strip at a distance \( y \) from the base is given by:

• Area of strip: \( dA = y \cdot dx = \frac{h}{b} x \cdot dx \)

The moment of inertia \( dI \) of this strip about the base is:

• Moment of inertia of strip: \( dI = y^2 \cdot dA = y^2 \cdot \frac{h}{b} x \cdot dx \)

Substituting \( y = \frac{h}{b} x \) into the equation gives:

• Substituted equation: \( dI = \left(\frac{h}{b} x\right)^2 \cdot \frac{h}{b} x \cdot dx = \frac{h^3}{b^3} x^3 \cdot dx \)

Integrating to Find Total Moment of Inertia

Now, we integrate \( dI \) from \( 0 \) to \( b \):

• Integration: \( I = \int_0^b \frac{h^3}{b^3} x^3 \, dx \)

Calculating the integral:

• Integral result: \( I = \frac{h^3}{b^3} \cdot \left[\frac{x^4}{4}\right]_0^b = \frac{h^3}{b^3} \cdot \frac{b^4}{4} = \frac{1}{4} \cdot \frac{h^3 b}{b^3} = \frac{1}{4} \cdot \frac{h^3 b}{b^2} = \frac{1}{3} b h^3 \)

Final Result

Thus, the moment of inertia of the triangle about its base is:

• Final Moment of Inertia: \( I = \frac{1}{3} b h^3 \)

This formula is quite useful in various applications, especially in structural engineering and mechanics, where understanding the distribution of mass relative to an axis is crucial for analyzing rotational dynamics.