# Moment of inertia rigid body problem

• i_hate_math
In summary, the Homework Statement discusses a rigid body that falls from rest. The angular speed is found when the plane of the body is vertical. The moment of inertia is found to be 1/3mL^2+mL^2 for the two rods on the RHS of the "H."

## Homework Statement

A rigid body is made of three identical thin rods, each with length L = 0.530 m, fastened together in the form of a letter H, as suggested by the figure here. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of the H is vertical?

## Homework Equations

This is really a conservation of energy question but the hard bit is to find out the moment of inertia of this object.
K+P=K+P
K=0.5I*w^2

## The Attempt at a Solution

==> 2mgL=(1/2)I*w^2 +(1/2)mgL
and from this equation w can be easily found. However, i canot seem to get the correct Rotational Inertia. Can anyone show me how to find the moment of inertia for this system??

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What have you tried? What equations are relevant for dealing with the moment of inertia of thin rods?

Which rods contribute to the MoI about that axis? Find the MoI of each.

gneill said:
What have you tried? What equations are relevant for dealing with the moment of inertia of thin rods?
That is where i got stuck. For some reason i believe the Moment of inertia for this system is 1/3mL^2+mL^2, but it doesn't lead to the correct answer

haruspex said:
Which rods contribute to the MoI about that axis? Find the MoI of each.
I know that only the two rods on the RHS of the 'H' contribute to the moment of inertia. But I have no clue how to find the MoI. Is it the parallel axis theorem?

The parallel axis theorem is involved for one of the rods (which one?).

What is the MoI of a thin rod about its center of mass?

gneill said:
The parallel axis theorem is involved for one of the rods (which one?).

What is the MoI of a thin rod about its center of mass?
Depending on your starting formulae, the parallel axis theorem may apply to both rods.

haruspex said:
Depending on your starting formulae, the parallel axis theorem may apply to both rods.
True, but I've found that the formula most often given for a thin rod is the one for it rotating about a line perpendicular to the rod and passing though its center of mass.

That would be the y-axis or z-axis in the figure.

gneill said:
The parallel axis theorem is involved for one of the rods (which one?).

What is the MoI of a thin rod about its center of mass?
The one parallel to rotation axis i suppose?

i_hate_math said:
The one parallel to rotation axis i suppose?
gneill meant the other one (see post #8), but arguably it can be used for both.

haruspex said:
Depending on your starting formulae, the parallel axis theorem may apply to both rods.
I know the outer rod has I=(1/3)mL^2 +mL^2, but how do i find the inner perpendicular rod's?

i_hate_math said:
I know the outer rod has I=(1/3)mL^2 +mL^2, but how do i find the inner perpendicular rod's?
Ah, the dangers of quoting formulae without understanding when they apply.
The 1/3 formula you quote is for a rod of what length rotating about what axis?

haruspex said:
Ah, the dangers of quoting formulae without understanding when they apply.
The 1/3 formula you quote is for a rod of what length rotating about what axis?
TBH i have no idea. Only these cases are included in the textbook and i got that one from the internet. So I=1/2mL^2 ?

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i_hate_math said:
TBH i have no idea. Only these cases are included in the textbook and i got that one from the internet. So I=1/2mL^2 ?
so after a quick research, 1/3mL^2 is rotation about an axis thru one end of the rod and 1/2mL^2 + mL^2 is the MoI for the parallel rod(by parallel axis theorem). Does that look right?

i_hate_math said:
TBH i have no idea. Only these cases are included in the textbook and i got that one from the internet. So I=1/2mL^2 ?
None in that image show a factor 1/3, so where did that come from?

i_hate_math said:
1/3mL^2 is rotation about an axis thru one end of the rod
An axis perpendicular to the rod and through one end, yes.
i_hate_math said:
1/2mL^2 + mL^2 is the MoI for the parallel rod(by parallel axis theorem).
No. Where did the 1/2 term come from?

haruspex said:
An axis perpendicular to the rod and through one end, yes.

No. Where did the 1/2 term come from?
thats the MoI of the rod thru a parallel axis(ie Icom) and I=0.5mL^2+mL^2 by parallel axis theorem

i_hate_math said:
thats the MoI of the rod thru a parallel axis(ie Icom)
The "i.e." is not appropriate. The direction of the axis and the point it passes through are independent.
To apply the parallel axis theorem you first need to identify an axis parallel to the given axis and passing through the mass centre of the body.
What does that parallel axis look like in this case, from the viewpoint of the rod?

haruspex said:
The "i.e." is not appropriate. The direction of the axis and the point it passes through are independent.
To apply the parallel axis theorem you first need to identify an axis parallel to the given axis and passing through the mass centre of the body.
What does that parallel axis look like in this case, from the viewpoint of the rod?
Is the axis pointing thru the rod? Thats parallel tocthe original axis

i_hate_math said:
Is the axis pointing thru the rod? Thats parallel tocthe original axis
Along the rod, yes. What do you think its moment of inertia would be about such an axis? (You won't find this one in any table, it's too simple!)

haruspex said:
Along the rod, yes. What do you think its moment of inertia would be about such an axis? (You won't find this one in any table, it's too simple!)
Simply m*L^2?

i_hate_math said:
Simply m*L^2?
No, it's even simpler.
Start with a point particle, mass m. What is its moment of inertia about a point distance L away? Based on that, what is its moment of inertia about its own position?

haruspex said:
No, it's even simpler.
Start with a point particle, mass m. What is its moment of inertia about a point distance L away? Based on that, what is its moment of inertia about its own position?
for one particle, MoI is mL^2, and for all the particles on this rod, the sum of all mL^2 would be ML^2 where M is the mass of the rod? and for the rod at the original axis, sinece its a "thin" rod, L can be ignored i guess

i_hate_math said:
for one particle, MoI is mL^2
Yes, but you need to answer my next question too:
haruspex said:
Based on that, what is its moment of inertia about its own position?
(What is L then?)

haruspex said:
Yes, but you need to answer my next question too:

(What is L then?)
L is the length of the rod?

i_hate_math said:
L is the length of the rod?
I asked about the moment of inertia of a point particle, not a rod, about a point at distance L from it. You correctly answered mL2. Now I am asking you about a special case of this: when the point particle is at the axis point. What is the value of L in that case? What moment of inertia do you get?

haruspex said:
I asked about the moment of inertia of a point particle, not a rod, about a point at distance L from it. You correctly answered mL2. Now I am asking you about a special case of this: when the point particle is at the axis point. What is the value of L in that case? What moment of inertia do you get?
Then it would be the radius of the rod, and in this case, as the question has indicated, a "thin" rod has negligible radius and hence L=0

i_hate_math said:
Then it would be the radius of the rod, and in this case, as the question has indicated, a "thin" rod has negligible radius and hence L=0
Right, so what is the moment of inertia?

0 for this rod, and the total would be 1/3mL^2 by itself?

i_hate_math said:
0 for this rod, and the total would be 1/3mL^2 by itself?
Oops no, 1/3mL^2 +0+mL^2 by the parallel axis theorem

i_hate_math said:
0 for this rod, and the total would be 1/3mL^2 by itself?
We are not quite ready to go back to considering the rod. I am still asking about the point particle.

Edit: cancel that, see below.

i_hate_math said:
0 for this rod
Yes, the moment of inertia of a thin rod about itself as axis is 0.
Now use the parallel axis theorem to find its moment of inertia about the rod parallel to it in the H figure.

haruspex said:
Yes, the moment of inertia of a thin rod about itself as axis is 0.
Now use the parallel axis theorem to find its moment of inertia about the rod parallel to it in the H figure.
The parallel axis thrm states that I=Icom+mh^2
here it's I=0+mL^2=mL^2

i_hate_math said:
The parallel axis thrm states that I=Icom+mh^2
here it's I=0+mL^2=mL^2
Yes!
Do you see why this is? Every point of the rod is at distance L from the axis, so each little mass dm contributes L2.dm to the total.

haruspex said:
Yes!
Do you see why this is? Every point of the rod is at distance L from the axis, so each little mass dm contributes L2.dm to the total.
Yeah because its a rigid body of many particles. And adding them up, I-total=1/3mL^2+0+mL^2=4/3mL^2