How do I find moments of inertia for different shapes?

[yazan_l]

Hi, can anyone help me understand how to find the moments of inertia for the following:
1- A triangular lamina (isosceles) of mass M, base 2B and height H. about line of symmetry.
2- A uniform lamina of mass M, bounded by the curve with equation y²=4ax and the line x = 4a about the x-axis.

For (1), I managed to get the answer, but I’m not sure if my way is right: it was finding M.I. of a rectangle base 2B, height H, mass 2M about the line of symmetry through the base, and divide it by 2 (as the triangle is the half of the rectangle!) I just don’t know if this method is right or wrong, or whether there is another method that I should had used. Those questions are from the book, and the book doesn’t explain how to find M.I. for such shapes, it only shows: rod, hoop, and discs. But not a triangle or curves!

Thank you very much,
Help is appreciated

 

[arildno]

You are proceeding along a wrong track here, I'm afraid.
Have you learned that in general, an object's moment of inertia I with respect to an axis is given by the integral:
[tex]I=\int_{V}r^{2}dm[/tex]
where V is the volume of the object (in your 2-D case, an area), r the distance of a mass point within the object to the axis, and dm the (infinitesemal) mass of the mass point?

 

[tacman]

To find the moment of inertia for different shapes, you will need to use the formula for moment of inertia, which is I = ∫r²dm. This formula involves integrating over the entire mass of the object, taking into account the distance of each infinitesimal mass element from the axis of rotation.

For the first shape, the triangular lamina, you can use the method of dividing it into smaller shapes with known moments of inertia. In this case, you can divide the triangle into two rectangles, each with a base of B and a height of H. The moment of inertia of a rectangle about its center of mass is given by (1/12)Mh², where M is the mass of the rectangle and h is its height. Since the triangle is half of the rectangle, you can divide the moment of inertia of the rectangle by 2 to get the moment of inertia of the triangle about its line of symmetry.

For the second shape, the uniform lamina bounded by the curve and line, you will need to use the formula for moment of inertia for a continuous distribution of mass. In this case, you will need to use the equation I = ∫y²dm, where y is the distance from the axis of rotation to the infinitesimal mass element and dm is the mass element. You can then solve for dm by expressing it in terms of y and integrating over the entire mass of the lamina, which is given by M. This will give you the moment of inertia of the lamina about the x-axis.

Overall, finding moments of inertia for different shapes involves breaking down the shape into smaller, simpler shapes and using the formula for moment of inertia to calculate the moment of inertia for each shape. You can also refer to tables or online resources for moments of inertia for common shapes, such as triangles and curves.