Quick question about moment of inertia

[lc99]

Homework Statement

Homework Equations

I = mr^2 (point mass)

The Attempt at a Solution

I wasn't sure how to calculate the moment of inertia of this triangle if the axis is going through one vertice perpendicularly.
Do i calculate the inerta of 2 sides with Inertia of a rod and the 3 side as point mass?

Or find inertia based on where the center mass is?[/B]

Based on center mass, middle of triangle, the distance r is b/2. m is just m. So the formula is I = mr^2 = m(b/2)^2 =

 

[Doc Al]

Do i calculate the inerta of 2 sides with Inertia of a rod and the 3 side as point mass?

All sides must be treated as rods. For the third one, use the parallel axis theorem.

 

[lc99]

All sides must be treated as rods. For the third one, use the parallel axis theorem.

how can i apply parallel axis theorem is the third side isn't parallel to the axis?

 

[Doc Al]

how can i apply parallel axis theorem is the third side isn't parallel to the axis?

To use the parallel axis theorem, the axes must be parallel, not the rod itself. Consider an axis going through the center of that third side.

 

[Voq]

It is stated that axis is passing through one of it vertices. I think you are on good way. Try to make a sketch.

 

[lc99]

All sides must be treated as rods. For the third one, use the parallel axis theorem.

for the third rod, do i just use Ip = Icm + md^2 where Icm is the center of the third rod and i have to figure out what d is?

 

[Doc Al]

for the third rod, do i just use Ip = Icm + md^2 where Icm is the center of the third rod and i have to figure out what d is?

Exactly!

 

[lc99]

Exactly!

Thank you. Just for clarification, Icm would be the center mass from the center of the triangle right?

 

[Doc Al]

Just for clarification, Icm would be the center mass from the center of the triangle right?

No. Icm is the moment of inertia of the third side about its center of mass. (d will be the distance from that side's center of mass to the axis of rotation.)

 

[lc99]

No. Icm is the moment of inertia of the third side about its center of mass. (d will be the distance from that side's center of mass to the axis of rotation.)

Do you mean Icm is the inerta center of mass at axis of rotation? I'm not quite understand Icm in the equal Ip = Icm +md^2. I understand d is the distance between the do parallel axises

(the answer is .5mb^2 but I am not getting that :( )

maybe I am confused on how parallel axis theorem can be applied to FIND the inertia of the third rod so i can add it to the sum of the inertia of the 2 other rods. This is because i thought that PAT is used to find the inertia of something from a new axis if Icm is parallel to this new axis.

 

[Doc Al]

maybe I am confused on how parallel axis theorem can be applied to FIND the inertia of the third rod so i can add it to the sum of the inertia of the 2 other rods. This is because i thought that PAT is used to find the inertia of something from a new axis if Icm is parallel to this new axis.

If you know the moment of inertia of some object about an axis through its center of mass (axis 1), then you can use the parallel axis theorem to calculate its moment of inertia about any other axis parallel to axis 1.

You know, I presume, the moment of inertia of the third side about its center. That's Icm. (Tell me what that is.) Then you can solve for d, using a bit of trig.

 

[Doc Al]

This might clarify things: Parallel Axis Theorem

 

[lc99]

This might clarify things: Parallel Axis Theorem

So, for Ip, I am getting Ip = 2/3mb^2 +md^2

for d, i found d = 3/4b. But, when i plug those in, I am getting 59/48mb^2 instead of .5mb^2

 

[lc99]

So, for Ip, I am getting Ip = 2/3mb^2 +md^2

for d, i found d = 3/4b. But, when i plug those in, I am getting 59/48mb^2 instead of .5mb^2

the formula for the rods I am using is 1/3ML^2

 

[haruspex]

2/3mb^2

Each rod only has a third of the total mass.

I found d = 3/4b

I bet you did not. Try again.

 

[lc99]

Each rod only has a third of the total mass.

I bet you did not. Try again.

I keep on getting that the distance between the axis of rotation (1) to the new axis rotation at the center of mass of the 3rd rod as 3/4b though! :(

Inertia at vertice = 1/3(1/3m)(b/2)^2 + Inertia at new axis?
Inertia at new axis = inertia at vertice + md^2
d = 3/4b

 

[haruspex]

I keep on getting that the distance between the axis of rotation (1) to the new axis rotation at the center of mass of the 3rd rod as 3/4b though! :(

View attachment 219940

Inertia at vertice = 1/3(1/3m)(b/2)^2 + Inertia at new axis?
Inertia at new axis = inertia at vertice + md^2
d = 3/4b

I do not understand that horizontal green line.
The axes are perpendicular to the plane. Your diagram will be clearer if you just draw the plan view of the triangle and do not attempt perspective. Show the axes as mere points, understanding that they go into the page (or out of it).

 

[lc99]

 

[lc99]

I do not understand that horizontal green line.
The axes are perpendicular to the plane. Your diagram will be clearer if you just draw the plan view of the triangle and do not attempt perspective. Show the axes as mere points, understanding that they go into the page (or out of it).

I think I am getting close. I got mb^2

my equation for Inertia at the first axis is = 1/3(1/3m)b^2 + 1/3(1/3m)b^2 + 1/12(1/3m)b^2 + m(bsqrt(3)/2)^2 = mb^2

the answer says it's .5mb^2 though

 

[haruspex]

I think I am getting close. I got mb^2

my equation for Inertia at the first axis is = 1/3(1/3m)b^2 + 1/3(1/3m)b^2 + 1/12(1/3m)b^2 + m(bsqrt(3)/2)^2 = mb^2

the answer says it's .5mb^2 though

The third rod has mass m/3 for the purposes of the parallel axis theorem also.

 

[lc99]

The third rod has mass m/3 for the purposes of the parallel axis theorem also.

So the M in Ip = Icm + Md^2 always makes up the mass of the total mass around Ip? In this case, just rod 3? I was under the impression the M = the total mass of the whole thing which is just m.

Thanks by the way for your guidance. I reallly want to get this right~

 

[haruspex]

So the M in Ip = Icm + Md^2 always makes up the mass of the total mass around Ip? In this case, just rod 3? I was under the impression the M = the total mass of the whole thing which is just m.

Thanks by the way for your guidance. I reallly want to get this right~

Your approach was to deal with each rod separately. Each has mass m/3, and the parallel axis theorem must be applied separately to each.
For the two rods touching the axis, the moment about the centres is 1/12 m/3 b². Applying the theorem adds 1/4 m/3b² to each, giving 1/3 m/3 b².
For the third rod, the mass is m/3 and the theorem gives b²(1/12+3/4)(m/3).