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

AI Thread Summary
The discussion focuses on calculating the moment of inertia of a uniform isosceles triangular lamina about its axis of symmetry. The initial incorrect assumption was that the moment of inertia is 1/2 * m * b². The correct approach involves using a double integral to express mass in terms of x, leading to the correct result of M * b² / 6. The integration process accounts for the mass distribution along the axis, confirming that height does not influence the moment of inertia. The contributor successfully resolved their errors and arrived at the correct solution.
mattgad
Messages
15
Reaction score
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.
 
Physics news on Phys.org
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.
 
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.
 
Happily algebraic methods arrive at the same result.
 
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.
 
how about the inertia product of this problem?
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top