Can I Find the Moment of Inertia with Only 8 Equal Parts in My Homework Problem?

[LoveBoy]

Homework Statement

http://i.imgur.com/7ZRsVj5.png

Homework Equations

I can't apply correct method !

The Attempt at a Solution

In bigger figure, i cut into 8 equal parts.So , total moment of inertia=8 I
But it's wrong !
I know it's wrong because i can't able to find angle between any two sides when i cut into 8 equal parts.
So,just as a guessing approach , i supposed each part is equal to I . Therefore,corresponding to total 8 I.

 

[SteamKing]

Homework Statement

http://i.imgur.com/7ZRsVj5.png

Homework Equations

I can't apply correct method !

The Attempt at a Solution

In bigger figure, i cut into 8 equal parts.So , total moment of inertia=8 I
But it's wrong !
I know it's wrong because i can't able to find angle between any two sides when i cut into 8 equal parts.
So,just as a guessing approach , i supposed each part is equal to I . Therefore,corresponding to total 8 I.

You don't have to cut anything, but you should think how the inertia of the sample triangle changes when the dimensions 'a' are multiplied by the same scale factor.

 

[jbriggs444]

You cut the square into eight parts because the area of [a triangular half of] the square is eight times the area of the triangle? How did you manage this feat? Where do you make the cuts to get eight 1x1 triangular half-squares out of a triangular half square that is ##2\sqrt{2}## on a side?

Consider, as SteamKing hints, cutting the square into four pieces, each of which is an isoceles right triangle and each of which has a side other than the hypotenuse on the axis of rotation.

 

[LoveBoy]

As per your given hint, if i double the sides of isosceles triangle , then i would get hypotenuse equals the side of square .
But I'm confused in the thing that if we double the sides , is moment of inertia becomes 4 times ?

 

[Nathanael]

But I'm confused in the thing that if we double the sides , is moment of inertia becomes 4 times ?

The moment of inertia typically looks like kML², right?
For uniform density distributions, similar geometries turn out to have the same coefficient k.

Start with a triangle of side length a and mass m with moment of inertia I₁ = km₁a².
Now scale it up to side length 2a; the moment of inertia becomes I₂ = km₂(2a)².

If m₂ = m₁ then I₂ = 4I₁. Is the "if" part true though?

 

[LoveBoy]

If m2 = m1 then I2 = 4I1. Is the "if" part true though?

I don't think so because if we double the length, mass also differs.

 

[LoveBoy]

@Nathanael

 

[Nathanael]

I don't think so because if we double the length, mass also differs.

Right, because the problem statement said that the triangles are of the same density and thickness. More area with the same density means more mass. What is the exact relationship though? If you double the legs what happens to the mass?
(Then, what is the total effect on the moment of inertia?)

 

[LoveBoy]

If you double the legs what happens to the mass?
(Then, what is the total effect on the moment of inertia?)

If we double the legs keeping density and thickness same,then mass becomes 4 times.
And if mass becomes 4 times ,then leg becomes 2 times, then moment of inertia becomes 16 times.
So, total moment of inertia=4*16 I = 64I

Therefore, 64 I is our answer !

 

[LoveBoy]

Thank you for your help ! @Nathanael , @jbriggs444 and @SteamKing