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

• Physics lover
In summary: The moment about one of the lines is ##\ I = \int r^2 \, {\sf d}m \ \ ## and the moment about the line through the corner is ##\ I = \int r^2 \, {\sf d}m \ \cos(\theta) \ \ .The moment about one of the lines is ##\ I = \int r^2 \, {\sf d}m \ \ ## and the moment about the line through the corner is ##\ I = \int r^2 \, {\sf d}m \ \cos(\theta) \ \ .
Physics lover
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.

Relevant 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.

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 )

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)

BvU said:

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 )

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.

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

BvU
Physics lover said:
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:

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.

jbriggs444 said:
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?

haruspex said:
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.

Physics lover said:
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.

haruspex said:
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.

Physics lover said:
Ok no problem,Will they contribute to Ma^2/6.
Sir am i correct.

Physics lover said:
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.

haruspex said:
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.

Physics lover said:
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?

haruspex said:
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

haruspex said:
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?

Physics lover said:
Umm..Ma^2/12Umm..Ma^2/12 Is it like this?
View attachment 245121
Looks right, but you can evaluate the trig function to simplify the answer.

haruspex said:
Looks right, but you can evaluate the trig function to simplify the answer.
Ok thanks for the help.

## 1. What is moment of inertia and why is it important?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is important because it helps us understand how an object will behave when subjected to rotational forces.

## 2. How is moment of inertia calculated for a square along an axis inclined at an angle?

The moment of inertia for a square along an axis inclined at an angle can be calculated using the formula I = (mL^2)/6, where m is the mass of the square and L is the length of one side. The angle of inclination also needs to be taken into account in the calculation.

## 3. How does the moment of inertia change as the angle of inclination changes?

The moment of inertia will increase as the angle of inclination increases. This is because the square will have a greater distance from the axis of rotation, resulting in a larger moment of inertia.

## 4. How does a change in the shape of the square affect its moment of inertia?

A change in the shape of the square, such as increasing or decreasing its side length, will also affect the moment of inertia. The moment of inertia will increase as the square's size increases, and decrease as its size decreases.

## 5. Can the moment of inertia of a square along an axis inclined at an angle be negative?

No, the moment of inertia cannot be negative. It is always a positive value. If the calculated moment of inertia is negative, it is most likely due to an error in the calculation.

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