What is the moment of inertia of the plate about z axis?

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Homework Help Overview

The discussion revolves around calculating the moment of inertia of an isosceles triangular plate about the z-axis. The original poster presents an initial attempt at the solution, referencing geometric considerations and the parallel axes theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods for determining the moment of inertia, including integration and geometric reasoning. Questions arise regarding the shape of the plate and the implications of its configuration on the moment of inertia calculation.

Discussion Status

The discussion is active, with participants offering various perspectives on the problem. Some suggest reconsidering the relationship between the triangular plate and a square, while others emphasize the importance of integrating to find the moment of inertia. There is no explicit consensus on the correct approach or final answer.

Contextual Notes

Participants note the specific characteristics of the triangular plate, including its mass, base length, and apex angle. There is an ongoing examination of assumptions related to the geometry and mass distribution of the plate.

zorro
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Homework Statement



What is the moment of inertia of the plate about z axis?


Homework Equations





The Attempt at a Solution



Consider the isosceles triangle to be a part of a square of side l/root(2)
Its mass will be 2M
We know that its moment of inertia about the centre perpendicular to the plane is 2M(l^2/2)/6=Ml^2/6
Applying parallel axes theorem, moment of inertia about z axis will be Ml^2/6 + 2M(l^2/4)
which is 2/3 (Ml^2)
Moment of inertia of the triangle will be half of it i.e. Ml^2/3
but the answer is Ml^2/6
Please explain if there is any other easier approach.
So the moment of inertia of the triangle will be half of it i.e. Ml^2/12
 

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What is the moment of inertia of the plate about z axis?

What is the shape of the plate and the point of suspension?
 
I am sorry I did not post the full question. Here it is.

The figure shows an isosceles triangular plate of mass M and base L. The angle at the apex is 90°. The apex lies at the origin and the base is parallel to X-axis
 
The moment of inertia of the triangle is not half that of the square. See the picture: the points of the upper triangle are farther than those of the lower triangle.

It is easy to get the moment of inertia by integrating (x^2+y^2)dm for the triangle.

ehild
 

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The base of the triangle is fixed ( it is of length l). Since the apex angle is 90 degrees and it is isosceles, You get a square when it is rotated about l. So there is no chance of upper triangle being bigger or smaller.
 
How is the moment of inertia defined for an extended body?

ehild
 
here the extended body is a square. So its moment of inertia is same as that of a square.
 
You did not understand me. Forget for the moment that this triangle is half of a square. It is just a set of elementary masses, organized in a certain shape. How would you get the moment of inertia of a set of point masses?

ehild
 
By using the equation ∫x^2 dm
 
  • #10
What do you mean on x? If x is the horizontal axis, a rod along the y-axis has zero moment of inertia?



ehild
 

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