Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 8)
8.
The moment of inertia of a square of side a about its diagonal is
Discussion:
9 comments Page 1 of 1.
Partha said:
5 years ago
It's about a diagonal axis. It divides the Sqr in two triangles. So h=√3a/2, b=a. By using the formula for triangle I=bh^3/12.
So, I= √3a^4/32. As on diagonal, there are two triangles, So, the net moment of Inertia, I=√3a^4/16. So, I think the given options are wrong.
So, I= √3a^4/32. As on diagonal, there are two triangles, So, the net moment of Inertia, I=√3a^4/16. So, I think the given options are wrong.
SOURABH SWARNKAR said:
5 years ago
a4/24 + a4/24 = a4/12 is answer.
Ashutosh Agrawal said:
7 years ago
Perpendicular axis theorem MIz=MIx+MIy=MI about one diagonal + MI about another diagonal. That is 2*MIx=2*MI about one diagonal.
Because it is square MIx=MId.
Therefore MI about diagonal is b^4/12.
Because it is square MIx=MId.
Therefore MI about diagonal is b^4/12.
(1)
SAMIM MOLLA said:
8 years ago
Diagonal of a square is sqrt of 2 *a ( assume one side) height is a/sqrt of 2. Since the square is consist of two triangles. So MI of the one triangle is 1/12 bh^3..here b= sqrt of 2 *a and h= a/sqrt of 2..
So MI OF THE SQUARE ABOUT ITS DIAGONAL IS TWICE THE MI OF ONE TRIANGLE, IT MEANS 1/12 *√ 2* a * (a/√ 2)^3 * 2= a^4/12.
So MI OF THE SQUARE ABOUT ITS DIAGONAL IS TWICE THE MI OF ONE TRIANGLE, IT MEANS 1/12 *√ 2* a * (a/√ 2)^3 * 2= a^4/12.
Sonu said:
8 years ago
The Correct answer is a4/12.
L. Gopi said:
8 years ago
Correct answer is a^4/24. Since it is about the diagonal.
S.Bera said:
8 years ago
But here give about its diagonal.
Ashish said:
9 years ago
You are Correct @Shariful Alam.
Md.shariful alam said:
1 decade ago
We know Moment of inertia of squqre bar = bh3/12.
Here side are equal then b=a & h=a.
So M.I = a4/12.
Here side are equal then b=a & h=a.
So M.I = a4/12.
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