Moment of Inertia (Triangular Prism)

[derrickb]

Homework Statement

A triangular prism (like a box of toblerone) of mass M, whose ends are equilateral triangles parallel to the xy plane with side 2a, is centered on the origin with its axis along the z axis. Find its moment of inertia for rotation about the z axis. Without doing any integrals write down and explain its two products of inertia for rotation about the z axis.

Homework Equations

I=∫r²dm
ρ=M/V
r²=x²+y²

The Attempt at a Solution

ρ=M/(.5bhH);H is total length along z-axis
ρdV=dM=.5ρdxdydz

I=∫∫∫x²+y²dxdydz

I evaluated this with integration limits
for x: 0 to a
for y: 0 to √3a/2
for z: 0 to h/2

after doing it all out and substituting M/V back in for ρ, I got an answer of 7Ma²/96. I muliplied by 2 since that integral was only for half of the object, so my final answer would be 7Ma²/48. I have an intuition that this is incorrect. Can anyone point me in the right direction? Are my limits of integration correct? Any help would be much appreciated. Thank you

 

[vela]

Homework Statement

A triangular prism (like a box of toblerone) of mass M, whose ends are equilateral triangles parallel to the xy plane with side 2a, is centered on the origin with its axis along the z axis. Find its moment of inertia for rotation about the z axis. Without doing any integrals write down and explain its two products of inertia for rotation about the z axis.

Where exactly is the axis of the prism?

Homework Equations

I=∫r²dm
ρ=M/V
r²=x²+y²

The Attempt at a Solution

ρ=M/(.5bhH);H is total length along z-axis
ρdV=dM=.5ρdxdydz

Where did the 0.5 come from in the last line?

I=∫∫∫x²+y²dxdydz

I evaluated this with integration limits
for x: 0 to a
for y: 0 to √3a/2
for z: 0 to h/2

I don't understand your limits at all. Why are you integrating z from 0 to half the height of the triangle?

after doing it all out and substituting M/V back in for ρ, I got an answer of 7Ma²/96. I muliplied by 2 since that integral was only for half of the object, so my final answer would be 7Ma²/48. I have an intuition that this is incorrect. Can anyone point me in the right direction? Are my limits of integration correct? Any help would be much appreciated. Thank you

 

[derrickb]

Where exactly is the axis of the prism?

Not sure I really understand what you're asking. It says it's along the z-axis

Where did the 0.5 come from in the last line?

ρ=M/V
ρV=M; V=A*H; A=.5bh
ρ(.5bhH)=M
ρdV=dM
I substituted in .5bhH for V but looking back on it I don't think I should have.
would dM=ρdxdydz be correct?

I don't understand your limits at all. Why are you integrating z from 0 to half the height of the triangle?

Sorry that was a typo. "h" should have been "H"

 

[vela]

Not sure I really understand what you're asking. It says it's along the z-axis

I'll assume one of the edges of the prism lies along the z-axis then. Is that right?

ρ=M/V
ρV=M; V=A*H; A=.5bh
ρ(.5bhH)=M
ρdV=dM
I substituted in .5bhH for V but looking back on it I don't think I should have.
would dM=ρdxdydz be correct?

Yes, that's generically correct because in Cartesian coordinates, the volume element is given by dV = dx dy dz.

Sorry that was a typo. "h" should have been "H"

I still don't understand why you're only integrating to H/2. It's not like dividing by 2 and then subsequently multiplying by 2 simplifies the calculation any.

 

[derrickb]

I'll assume one of the edges of the prism lies along the z-axis then. Is that right?

The way I interpret it is that the axis of rotation is along the z-axis, and that the z-axis runs through the center of mass. Basically, the triangle is centered on the z-axis. The wording is very vague, and I am a little confused by what is meant.

I still don't understand why you're only integrating to H/2. It's not like dividing by 2 and then subsequently multiplying by 2 simplifies the calculation any.

So I should integrate from -H/2 to H/2 is what your saying?

 

[vela]

The way I interpret it is that the axis of rotation is along the z-axis, and that the z-axis runs through the center of mass. Basically, the triangle is centered on the z-axis. The wording is very vague, and I am a little confused by what is meant.

That sounds reasonable. So the z-axis lies somewhere in the middle of the prism. Doesn't this mean that x and y should take on both positive and negative values? Have you drawn a sketch?

So I should integrate from -H/2 to H/2 is what your saying?

Yes.

You'll see that the integration over the z-coordinate simply amounts to multiplying by H. Rather than using dV = dx dy dz, you could write dV = H dx dy and forgo integrating over z altogether. The infinitesimal volume element would be a rectangular solid with sides of length dx, dy, and H. You can do this because every piece of that element is the same distance r from the axis of rotation.

 

[derrickb]

That sounds reasonable. So the z-axis lies somewhere in the middle of the prism. Doesn't this mean that x and y should take on both positive and negative values? Have you drawn a sketch?

I have drawn a sketch. I'm going to try and work it out with the z-axis in the center of the triangle, and then with the z-axis as one of the prism edges, and use the parallel-axis theorem. I can take a picture of my work then. I know x and y would have positive and negative values, but if I integrated from them, wouldn't they just cancel out because of the symmetry?

 

[vela]

No, it wouldn't cancel out because r² is positive regardless of the signs of x and y.

 

[haruspex]

So I should integrate from -H/2 to H/2 is what your saying?

Why not 0 to H? It won't make any difference. In fact, forget z and just treat it as a flat triangle.
I suggest using the symmetry of the triangle. Cut it into six identical right angled triangles, each with a corner at the z axis. That will simplify the integration bounds.