Calculate the moment of inertia of a uniform triangular lamina of mass

In summary, the homework statement is trying to calculate the moment of inertia of a uniform triangular lamina of mass m in the shape of an isosceles triangle with base 2b and height h, about its axis of symmetry. The attempt at a solution involves trying various integrals, but the author eventually finds an algebraic method that yields the same result.
  • #1
mattgad
15
0

Homework Statement



Calculate the moment of inertia of a uniform triangular lamina of mass m in the shape of an isosceles triangle with base 2b and height h, about its axis of symmetry.

The Attempt at a Solution



I've tried various things for this and never get the correct answer, 1/2*m*b^2.
I'm beginning to think this may involve a double integral.

Thanks.
 
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  • #2
I don't think ½mb² is the right result.

Here is a similar example I did earlier:
https://www.physicsforums.com/showthread.php?t=278184

In this case I think you would attack the sum of the x²*dm by observing that you can construct m in terms of x as something like h*(1-x/b) so that you arrive at an integral over an expression something like (hx² -x³/b)*dx.

At the end you will be able note that the area of the lamina triangle times the implied density ρ yields you an M total mass in the product that defines your moment.
 
  • #3
I have coded this problem as a double integral in Maple.

> x(y):=b*(1-y/h);
> rho:=M/(b*h);
> dJ:=int(rho*z^2,z=0..x(y));
> J:=2*int(dJ,y=0..h);

In the first line, the right boundary is defined.
In the second line, the mass density is expressed.
In the third line, the integration in the x-direction is performed from the axis of symmetry to the right edge
In the fourth line, the integration is performed in the y-direction from bottom to top. The result is M*b^2/6. It is reasonable that h should not be in the result. The altitude should not affect this function, only the base width which describes how far the mass is distributed off the axis of rotation.
 
  • #4
Happily algebraic methods arrive at the same result.
 
  • #5
I did actually mean to put mb^2 / 6 in my first post. Thanks for replies. Last night I managed to get it myself as well after spotting errors in my work. Thanks.
 
  • #6
how about the inertia product of this problem?
 

1. What is a uniform triangular lamina?

A uniform triangular lamina is a two-dimensional shape with three equal sides and three equal angles. It is often used in geometry and physics as a simplified representation of various objects, such as a cone or a pyramid.

2. What is the moment of inertia?

The moment of inertia is a measure of an object's resistance to rotational motion. It is calculated by multiplying the mass of the object by the square of its distance from the axis of rotation. In simpler terms, it represents how difficult it is to change the rotational motion of an object.

3. How do you calculate the moment of inertia of a uniform triangular lamina?

To calculate the moment of inertia of a uniform triangular lamina, you will need to know its mass, length of its sides, and the distance from the axis of rotation to its center of mass. Then, you can use the formula I = (1/12) * m * a^2, where m is the mass of the lamina and a is the length of its sides.

4. Why is the moment of inertia important?

The moment of inertia is important because it helps us understand an object's rotational motion. It is used in various areas of physics, such as mechanics and dynamics, to analyze and predict the behavior of objects when they are rotating. It is also crucial in designing and building structures and machines that involve rotational motion.

5. Can the moment of inertia of a uniform triangular lamina change?

Yes, the moment of inertia of a uniform triangular lamina can change if any of the factors used in the calculation (such as mass or distance) change. For example, if the lamina's mass increases, its moment of inertia will also increase. Additionally, the moment of inertia can also change depending on the axis of rotation. It will be different if the lamina is rotated around its center of mass compared to being rotated around one of its vertices.

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