Question

The moment of inertia of a triangle
of base b and altitude h with respect
to an axis through the apex parallel
to the base is​

Answer 1

Answer:

Moment of inertia of triangle having base b and

\small \underline {\sf{height \ h \ when \ axis \ passing \ through \ the \ center \ of \ }}

height h when axis passing through the center of

\small \underline {\sf{gravity \ is \ bh3/12 \ and \ moment \ of \ inertia \ when }}

gravity is bh3/12 and moment of inertia when

\small \underline {\sf{axis \ passing \ through \ base \ is \ bh3/3 \ and \ ratio \ is \ }}

axis passing through base is bh3/3 and ratio is

\small \underline {\sf{ asked \ it \ gives \ 4.}}

asked it gives 4.

Answer 2

Answer:

The moment of inertia of a triangular base  and altitude  with respect to an axis through the apex parallel to the base, is. $=\frac{b h^{3}}{36}$

Explanation:

The parallel axis theorem states that the moment of inertia of a body about an axis parallel to an axis passing through the center of mass is equal to the sum of the moments of inertia of the body about an axis parallel to an axis passing through the center of mass, the product of mass, and the square of the distance between the two axes.

When the axis passes through the center of gravity, the moment of inertia of a triangle with base b and height h is bh3/12,

The moment of inertia when the axis passes through the base is bh3/3, and the ratio

We will use the parallel axis theorem and we will take the centroid as a reference in this case.

$\mathrm{I}_{\mathrm{AA}^{\prime}}=\mathrm{I}_{\mathrm{BB}^{\prime}}+\mathrm{Ad}^{2}$

$\mathrm{I}_{\mathrm{BB}^{\prime}}=\mathrm{IAA}^{\prime}-\mathrm{Ad}^{2}$

$=\left(\frac{1}{12}\right) \mathrm{bh}^{3}-\frac{1}{2} \mathrm{bh}\left(\frac{1}{3}\right) \mathrm{h}^{2}$

$=\frac{b h^{3}}{36}$

The moment of inertia of a triangular base 'b' and altitude 'h' with respect to an axis through the apex parallel to the base, is.  $=\frac{b h^{3}}{36}$