How do I use the parallel axis theorem for moment of inertia in this homework?

[reminiscent]

Homework Statement

Homework Equations

I = (1/3)ML²
Parallel axis theorem: I_new = I_old + mr²

The Attempt at a Solution

I already solved part a) and I got (1/12)ML². For b), I get that I have to use the parallel axis theorem but I don't know how to go from there. We have not gone over this concept yet in class (not even moment of inertia) so bear with me...

 

[Simon Bridge]

... start from the equation for the parallel axis theorem. You can look it up: the meanings of the symbols I mean.
I think you were supposed to use it for part (a) so how did you do that?

 

[drvrm]

already solved part a) and I got (1/12)ML2. For b), I get that I have to use the parallel axis theorem but I don't know how to go from there. We have not gone over this concept yet in class (not even moment of inertia) so bear with me...

how did you do the part (a)-- if you have doneit correctly then (b) should not be a problem?

Homework Equations

I = (1/3)ML2

the above is not relevant -did you use it?

 

[Simon Bridge]

$I=\frac{1}{3}ML^2$ is the moment of inertia of a uniform rod mass m and length L about one end.
Could be forgiven for thinking it relevant...

It would probably be clearer to start from the moment of inertial about the centre of mass though.
https://en.wikipedia.org/wiki/List_of_moments_of_inertia

While I'm at it: how to use parallel axis theorem:
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html

 

[reminiscent]

how did you do the part (a)-- if you have doneit correctly then (b) should not be a problem?
the above is not relevant -did you use it?

$I=\frac{1}{3}ML^2$ is the moment of inertia of a uniform rod mass m and length L about one end.
Could be forgiven for thinking it relevant...

It would probably be clearer to start from the moment of inertial about the centre of mass though.
https://en.wikipedia.org/wiki/List_of_moments_of_inertia

While I'm at it: how to use parallel axis theorem:
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html

All I did for part a) was that since the rod is bent at its center, that means both segments of the rod have masses of M/2 and lengths of L/2. So for one rod, I = (1/24)ML². For 2 rods, then it's just 1/2(ML²)

 

[SammyS]

All I did for part a) was that since the rod is bent at its center, that means both segments of the rod have masses of M/2 and lengths of L/2. So for one rod, I = (1/24)ML². For 2 rods, then it's just 1/2(ML²)

Is that a typo ?

I thought that you got 1/12 ML² , (which is correct.)

 

[reminiscent]

Is that a typo ?

I thought that you got 1/12 ML² , (which is correct.)

Yes, sorry, that was a typo. That is what I got. But how am I able to use the parallel axis theorem for b)? We did not go over it in class completely yet...

 

[SammyS]

Yes, sorry, that was a typo. That is what I got. But how am I able to use the parallel axis theorem for b)? We did not go over it in class completely yet...

What is the location of the center of mass ?

 

[reminiscent]

What is the location of the center of mass ?

The location of the center of mass for each segment of the rod is the midpoint of each segment, correct?

 

[reminiscent]

What is the location of the center of mass ?

I don't understand the axis that it is being rotated. Is it the midpoint between the 2 ENDS of each segment of the rod? So does it boil down to a trig problem?

 

[SammyS]

The location of the center of mass for each segment of the rod is the midpoint of each segment, correct?

So then where is the center of mass of the whole bent rod?

 

[reminiscent]

So then where is the center of mass of the whole bent rod?

The midpoint of the rod, at the bent center?

 

[SammyS]

The midpoint of the rod, at the bent center?

No, not at the bend.

Make a sketch.

 

[reminiscent]

No, not at the bend.

Make a sketch.

Is it the midpoint of the ends of the two segments if they were connected?

 

[SammyS]

Is it the midpoint of the ends of the two segments if they were connected?

No. That's the pivot point for part (b).

It should be mid-way between CM of each segment, Right?

 

[reminiscent]

No. That's the pivot point for part (b).

It should be mid-way between CM of each segment, Right?

So like 1/4 of the way of each segment?

 

[SammyS]

So like 1/4 of the way of each segment?

No.

Start with the following:

The location of the center of mass for each segment of the rod is the midpoint of each segment, correct?

That's a correct statement.

The C.M. for the bent rod is mid-way between the center of mass for each of the two segments

 

[Arkun]

I have the same problem as OP has, and while I know where the C.M. of the bent rod is, how can you use the center of mass of the two segments of the rod to algebraically obtain the center of mass of the whole rod?

What I did was determine the distance of the center of mass from the rotational axis and use that as the L for the moment of inertia equation.

My attempt at finding $I_{CM}$:

$I_{CM} = m*(\sqrt{((L / 2) * cos30)^{2} + ((L / 2) * sin30)^{2}})^{2}$

This is for part (a) of the problem.

 

[SammyS]

I have the same problem as OP has, and while I know where the C.M. of the bent rod is, how can you use the center of mass of the two segments of the rod to algebraically obtain the center of mass of the whole rod?

If you don't understand that, then it's doubtful that do know the correct location of the center of mass of the whole rod .

What I did was determine the distance of the center of mass from the rotational axis and use that as the L for the moment of inertia equation.

My attempt at finding $I_{CM}$:

$I_{CM} = m*(\sqrt{((L / 2) * cos30)^{2} + ((L / 2) * sin30)^{2}})^{2}$

This is for part (a) of the problem.

Simplifying what you have for $\ I_\text{CM}\$ gives:

$\displaystyle \ m\cdot\left(\sqrt{((L / 2)^2 \cdot \cos^2(30^\circ) + ((L / 2)^2 \cdot \sin^2(30^\circ)}\right)^{2} \$

$\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 \cdot( \cos^2(30^\circ) + \sin^2(30^\circ))}\right)^{2} \$

$\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 }\right)^{2} =mL^2/4\$​

It's not at all clear, how that is the moment of inertia about any point.

 

[Arkun]

Isn't the center of mass of each segment of the rod at their midpoint? Which means that the distance between the center of mass for each segment of the rod and the axis of rotation is $L / 4$?

 

[Arkun]

If you don't understand that, then it's doubtful that do know the correct location of the center of mass of the whole rod .
Simplifying what you have for $\ I_\text{CM}\$ gives:

$\displaystyle \ m\cdot\left(\sqrt{((L / 2)^2 \cdot \cos^2(30^\circ) + ((L / 2)^2 \cdot \sin^2(30^\circ)}\right)^{2} \$

$\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 \cdot( \cos^2(30^\circ) + \sin^2(30^\circ))}\right)^{2} \$

$\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 }\right)^{2} =mL^2/4\$​

It's not at all clear, how that is the moment of inertia about any point.

Sorry about that. Yeah, I had no idea what I was doing with that. I guess I still don't know how to figure out the center of mass for this bent rod and use it in the appropriate moment of inertia equation.

In my second attempt, I applied the parallel axis theorem to each segment and then added their resulting $I_{NEW}$s and I got the correct answer: $\frac{1}{12} * m * L^2$.

My work:
$I_{NEW} = I_{OLD} + m * d^2$
$I_{NEW} = ((1/12) * (m/2) * (L/2)^2) + (m/2) * (L/4)^2$
$I_{NEW} = (1/96) * m * L^2 + (1/32) * m * L^2$
$I_{NEW} = (1/24) * m * L^2$
$2 * I_{NEW} = (1/12) * m * L^2$

 

[SammyS]

Sorry about that. Yeah, I had no idea what I was doing with that. I guess I still don't know how to figure out the center of mass for this bent rod and use it in the appropriate moment of inertia equation.

In my second attempt, I applied the parallel axis theorem to each segment and then added their resulting $I_{NEW}$s and I got the correct answer: $\frac{1}{12} * m * L^2$.

My work:
$I_{NEW} = I_{OLD} + m * d^2$
$I_{NEW} = ((1/12) * (m/2) * (L/2)^2) + (m/2) * (L/4)^2$
$I_{NEW} = (1/96) * m * L^2 + (1/32) * m * L^2$
$I_{NEW} = (1/24) * m * L^2$
$2 * I_{NEW} = (1/12) * m * L^2$

Yes. Of course, that's what OP got, but didn't really explain it.

 

[Simon Bridge]

If we label the endpoint $P$ and $Q$, and the bend $R$ then we can say (from post #1) $|PR|=|RQ|=L/2$ (may need to sketch as I describe.)
Initially the axis of rotation is point $R$ ... which nobody seems to have had trouble with.

The axis of rotation is shifted to a new point exactly half way along the segment $\overline{PQ}$ ... call that point $S$.
The angle $\angle PSR$ is a right angle. Since we also know the other angles it is tempting to try trig and/or pythagoras ... but is there any need?

If the com of the segment $\overline{PR}$ and $\overline{RQ}$ are $A$ and $B$ respectively, then we notice that $\triangle PAS$ and $\triangle SBQ$ are iscoseles triangles with $|PA|=|AS|=|SB|=|BQ|$ and we know that $|PA|=|QB|=L/4$

Each segment therefore has $I=\frac{1}{96}ML^2$ about com, which is a distance $L/4$ from the center of rotation, and there's two of them... so:

$$I = 2\left(\frac{ML^2}{96} + M\left(\frac{L}{4}\right)^2\right)$$ ... and simplify.

Which I think Arkud noticed earlier. If I read the above correctly, that was the method used: well done.

 

[Simon Bridge]

How about the other approach - where the moment of inertia about the center of mass of the bent rod is calculated and the parallel axis theorem is applied to that?
This is how it would normally done IRL: you have a thingy that may get rotated about an arbitrary axis, so you work out the moment of inertia about the com, with the position of the com, and put that information in a set of tables or in the documentation that you sell with the thingy. Some other engineer can. then. work out the details for their particular application.

Using the above labelled diagram, the combined COM is (from symmetry) half way along the segment $\overline{AB}$ call it $C$ the distance we need is:
$\qquad \qquad d=|AC|=|CB|=|AR|\sin(\theta /2)$
$\qquad \qquad \qquad \qquad$ ... where $\theta$ is the angle of the bend from the problem statement, and $|AR|=L/4$
Notice how the angle given was chosen to have nice sines and cosines?

There are a number of other ways of finding $d$.
You end up with $I_{C} = 2M\left( \frac{1}{96}L^2 + d^2 \right)$

The distance from C to S can also be found by pythagoras ... $|CS|=|RC|=\frac{1}{2}|RS|$.
You should be able to work through the steps and check that it gives the same answer as above.

 

[SammyS]

@Simon Bridge ,

You are using M as the mass of each segment, while in the OP, M is the mass of the entire rod.

 

[Simon Bridge]

I like to leave something for the student to find so they don't just copy ;)
That makes the denominator 2x48=96 ... I should go back and edit it now.

 

[SammyS]

I like to leave something for the student to find so they don't just copy ;)
That makes the denominator 2x48=96 ... I should go back and edit it now.

I would be inclined to simply replace M with (M/2) .

 

[Simon Bridge]

You are right:... it may be clearer to have left the (M/2) and (L/2) explicit in the formula as in...

$$I = 2\left[\frac{ml^2}{12} + md^2\right] = 2\left[\frac{(M/2)(L/2)^2}{12} + (M/2)(L/4)^2\right]$$ ... may make the process clearer for people coming in cold.

Now both of them are up so if the earlier one messes up the next one will get 'em.
As always I make no guarantees about the correctness of the actual formulas - it is the understanding that counts and students should repeat my calculation to find errors.