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Moment of inertia of a square along an axis inclined at an angle

Homework Statement
Question-:Find moment of inertia of a square of mass M and side length 'a' along an axis lying in the same plane inclined at an angle 15° to one of its sides and passing through vertex.

Help please.
Homework Equations
I=Ma^2+Mr^2
Iz=Ix+Iy
My attempt-:I extended the axis and made a triangle by joining other adjacent vertex to the line such that its angles are 15°,75° and 90°.I found the distance between the centre of square and upper vertex of triangle by using law of sines.And then i found out inertia along upper vertex of triangle I=Ma^2/6+Mr^2.And then i used perpendicular axis theorem but unfortunately the answer did not match.
 

BvU

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Hello lover, ##\qquad## :welcome: ##\qquad## !

Make a sketch to show you understood the problem statement (and post it to help us understand it 😉 )

Then sketch what you attempted (and post it to help us understand it :wink: )

Then try the simplest formula ##\ I = \int r^2 \, {\sf d}m \ \ ## for moment of inertia (a relevant equation I am missing in your list)
 
Hello lover, ##\qquad## :welcome: ##\qquad## !

Make a sketch to show you understood the problem statement (and post it to help us understand it 😉 )

Then sketch what you attempted (and post it to help us understand it :wink: )

Then try the simplest formula ##\ I = \int r^2 \, {\sf d}m \ \ ## for moment of inertia (a relevant equation I am missing in your list)
Sir i want to solve it without integration.My teacher had said that there is a way to solve without integration.

And regarding the question can i post a picture of it as i would have to learn LaTeX first.
 
245096

I think the question is like this.Now how can i solve it without integration.
 
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jbriggs444

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I think the question is like this.Now how can i solve it without integration.
That is an adequate picture of your understanding of the problem. Now you speak about having carved out a right triangle from this. Something like:

square.png

That seems like a good start. Can you fill in more details about what you did with that triangle and how you handled the rest of the square?

I would be thinking about carving a similar triangle from the opposite side of the square and using the parallel axis theorem to track how the moment of inertia changes as that triangle is merged with the first one. The result should be an easy to evaluate parallelogram.
 

haruspex

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I would be thinking about carving a similar triangle from the opposite side of the square and using the parallel axis theorem to track how the moment of inertia changes as that triangle is merged with the first one.
There's quite a sneaky method. Consider first the moment about a line at that angle through the centre of the square, and about another line through the centre orthogonal to the first.
By symmetry, these two moments must be the same. By considering the contribution from a small element, can you see what they must add up to?
 
There's quite a sneaky method. Consider first the moment about a line at that angle through the centre of the square, and about another line through the centre orthogonal to the first.
By symmetry, these two moments must be the same. By considering the contribution from a small element, can you see what they must add up to?
Ok what am i able to make from this is that the lines are not in the plane of square, am i right?So i think they will contribute to Ma^2/6.
 

haruspex

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what am i able to make from this is that the lines are not in the plane of square, am i right?
No, I mean lines in the plane of the square. Sorry, I should have been more precise.
 
No, I mean lines in the plane of the square. Sorry, I should have been more precise.
Ok no problem,Will they contribute to Ma^2/6.
 

haruspex

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Ok no problem,Will they contribute to Ma^2/6.
In total or each, and how do you arrive at that? Remember, these lines are still at angle θ to the sides of the square.
 
In total or each, and how do you arrive at that? Remember, these lines are still at angle θ to the sides of the square.
In total.I got this by perpendicular axis theorem the two lines in xy plane are perpendicular and therefore the z componet will be passing perpendicular to the plane of square through its centre.Therefore they will contribute to Ma^2/6.
 

haruspex

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In total.I got this by perpendicular axis theorem the two lines in xy plane are perpendicular and therefore the z componet will be passing perpendicular to the plane of square through its centre.Therefore they will contribute to Ma^2/6.
Very good.
So what is the moment about one of the lines, and then what is the moment about the line through the corner?
 
Very good.
So what is the moment about one of the lines, and then what is the moment about the line through the corner?
Umm..Ma^2/12

Very good.
So what is the moment about one of the lines, and then what is the moment about the line through the corner?
Umm..Ma^2/12 Is it like this?
245121
 

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