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GR9677 #32
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Mechanics$\Rightarrow$}Moment of Inertia

The inertia through the point A is $I_A=3mr^2$. From geometry, one deduces that the distance between each mass and the centerpoint A is $r\cos(30^\circ)=l/2\Rightarrow r=l/\sqrt{3}$. T he moment of inertia about A is thus $I_A=ml^2$

The inertia about point B can be obtained from the parallel axis theorem ($I_{displaced}=I_{cm}+\sum_i m_i d^2$, where d is the displaced distance from the center of mass). Because $d=l/\sqrt{3}$, one has $I_B = I_A + 3md^2 =2 I_A$. Since the angular velocity is the same for both kinetic energies, recalling the relation for kinetic energy $K_i=I_i \omega_i^2$, one has $K_B/K_A=I_B/I_A=2$, as in choice (B).

Alternate Solutions
 jmason862009-09-26 14:08:10 The pGRE is largely about answering problems quickly. Use MOE (method of elimination), order of mag, and physical intuition whenever possible. KE = $\frac{1}{2}I\omega^2$ so a ratio of KE will just be a ratio of moment of inertia (I) since $\omega$ is fixed. Your physical intuition should tell you that the axis through the center will have a low I and an axis through any of the edges will have a larger I. Therefore, the ratio must be greater than 1. Eliminate (C) (D) (E). Now you just need to know if it is going to be 2 or 3 times more KE (or I). As others have stated below, there are 3 masses that are not POINT masses, so any calculation you do will involve 3m, not 2. Eliminate (B)Reply to this comment yipee2007-09-15 17:39:34 I think the solution here is completely wrong. For moment of inertia about A, Ia=3*(1/3)mr$\^2$, where r is as shown in the above solution. So, then Ia=(1/3)ml$\^2$. Then, moment of inertia about B, Ib=0 + (1/3)ml$\^2$ + (1/3)ml$\^2$=(2/3)ml$\^2$, where the zero comes from the mass at B. So, dividing the Kb by Ka then gives Ib/Ia = 2 as expected.Reply to this comment
Comments
kaic
2013-10-15 11:40:29
If we call the distance from point A to each mass $R$, then we can immediately think of the total inertia as three point masses at distance $R$.
$I_{A} = 3mR^{2}$
For B, one of the masses is at the center of gyration, so it doesn't contribute. We have two point masses at distance $l$.
$I_{B} = 2ml^{2}$
Apply some basic trig to find the relation between $R$ and $l$,
$cos(30) = \frac{l/2}{R} = \frac{\sqrt{3}}{2}$
$R = \frac{l}{\sqrt{3}}$
plug back into $I_{A}$ to find
$I_{A} = ml^2 = \frac{1}{2}I_{B}$
$\frac{I_{B}}{I_{A}} = 2$
as in answer (B)
rachkov
2013-09-01 15:07:10
On the GRE, it is important to go fast. While the original solution is good, here is a faster alternative:

You know that you need to look at the ratio of moments of inertia, mainly: $\frac{I_{B}}{I_{A}}$.
$I_{A}=I_{cm}$, which goes something like $a*md^{2}$, where a is some positive constant you want to know and d is the distance between A and one of the masses.

$I_{B}=I_{cm}+md^{2}=(a+1)md^{2}$ from parallel axis theorem

so the ratio ends up being: $\frac{I_{B}}{I_{A}}=\frac{a+1}{a}$

The only answer satisfying this condition is B)
w24729695
2012-11-08 06:27:53
Ia=3mr^2 ,since 3^0.5=l, we can get Ia=ml^2
now when you move to point B, what you need is to calculate another inertia, thus Ib=0+ml^2+ml^2=2ml^2
2ml^2/ml^2=2
gravity
2010-11-04 03:13:43
I preferred another way.

Obviously,

I$_{B}$ = 2ml$^2$

The only hard part is figuring out the distance from A to each mass. I just guessed .6l.

So,

I$_{A}$ = 3m(.6l)$^2$ = ml$^2$

And the ration between these is 2.
sullx
2009-11-04 21:43:26
For speed, I just though of it as a hoop (and probably because of the hoop hanging from a nail problem a few pages back).

For a hoop $I_{cm} = mr^2$ . Using the parallel axis theorem for the edge of the hoop yields $I_{edge} = 2*mr^2$.

Thus $\frac{I_{edge}}{I_{cm}}$ = 2/1
jmason86
2009-09-26 14:08:10
The pGRE is largely about answering problems quickly. Use MOE (method of elimination), order of mag, and physical intuition whenever possible.

KE = $\frac{1}{2}I\omega^2$ so a ratio of KE will just be a ratio of moment of inertia (I) since $\omega$ is fixed.

Your physical intuition should tell you that the axis through the center will have a low I and an axis through any of the edges will have a larger I. Therefore, the ratio must be greater than 1. Eliminate (C) (D) (E).
Now you just need to know if it is going to be 2 or 3 times more KE (or I). As others have stated below, there are 3 masses that are not POINT masses, so any calculation you do will involve 3m, not 2. Eliminate (B)
 Dx2wUV2009-10-06 13:30:52 If you eliminate B then you eliminate the correct answer, according to ETS
 dogsandfrogs2009-10-07 14:12:42 Eliminate (B)? But B is the answer...
 mrTrig2010-10-23 13:09:01 You probably dont want to eliminate (B) since (B) is the answer.
 KLS2013-10-12 13:33:52 The answer key states that (B) is the correct answer, so you don't want to eliminate it.
 kaic2013-10-15 11:32:28 The correct answer is (B)
 poopdog2016-04-09 23:52:41 This is wrong.
johw
2009-09-23 20:40:40
Is this solution correct?
We use the same solution as yosun for $I_A$
For $I_B$, we recall the original definition of moment of inertia,
$I=\sum_{m_i} r^2$,
Hence,
$I_B = \sum_{1,2} l^2 m = 2ml^2$
Thus, the ratio is just $\frac{I_B}{I_A} = \frac{2ml^2}{ml^2} = 2$.
sonnb
2008-05-29 19:45:36
A more basic solution is as follows: The kinetic energy of the three particles moving about A with angular velocity $\omega$ is $KE_{A}$ = $\3\frac{1}{2}mv^2_A$ = $\frac{3}{2}m\omega^2r^2$ = $\frac{1}{2}m\omega^2l^2$ where $v_A$=$\omega$r and r = l/$\sqrt{3}$.
The kinetic energy of the two particles moving about B with angular velocity $\omega$ is $KE_{B}$ = $\2\frac{1}{2}mv^2_B$ = m$\omega^2l^2$, since $v_B$=$\omega$l. The ratio of these two kinetic energies is 2, answer (B).
 Camoph2010-03-30 17:05:53 How do you obtain r=l/\sqrt{3}??. r(cos30)=l/2??. I don't agree.
 Camoph2010-03-30 17:11:08 How do you obtain r=l/\sqrt{3}??. r(cos30)=l/2??. I don't agree.
Richard
2007-09-26 00:13:07
First, I want to say thanks to the creator of this site. It looks like it takes a lot of work, and it makes a great resource.

The symmetry of this problem makes it simple.
You know that the center of mass is in the center of the triangle and that it is equidistant from the three masses. Thus, $I_{cm}=3mr^2$.
Now, when you are at axis A, you can use the parallel axis theorem: $I=I_{cm}+Mr^2$ where M is the total mass of the system. In this case it is just $M=3m$.
Finally, because $KE\alpha I\omega^2$ you take the ratio $6mr^2/3mr^2=2$.
 flyboy6212010-10-23 08:15:00 This is the best solution. You don't need to calculate r at all.
 anum2010-11-02 01:28:14 isn't A the center of mass?
 anum2010-11-02 01:31:31 oh got it! scratch the upper remark
 anum2010-11-02 01:31:57 oh got it! scratch the upper remark
yipee
2007-09-15 17:39:34
I think the solution here is completely wrong. For moment of inertia about A, Ia=3*(1/3)mr$\^2$, where r is as shown in the above solution. So, then Ia=(1/3)ml$\^2$. Then, moment of inertia about B, Ib=0 + (1/3)ml$\^2$ + (1/3)ml$\^2$=(2/3)ml$\^2$, where the zero comes from the mass at B. So, dividing the Kb by Ka then gives Ib/Ia = 2 as expected.
 Gaffer2007-10-26 16:33:34 I was totally with you while taking the test. However, in looking back, I realized that we both made an improper assumption. We cannot assume that the moment of inertia for the particle at B is zero since it has extension. If the particles were the "very small" point particles we all love so much, that would be valid; since this problem is on the order of magnitude of the tangible, ie with dimensions, the particle at B will contribute to the net inertia. Yosun's method is the safest mathematically.
 errjones2008-10-17 15:49:39 I think the issue here is that you're misunderstanding the parallel axis theorem, which is what I also did at first. Stated in Fowles, it goes "The moment of inertia of a rigid body about any axis is equal to the moment of inertia about a parallel axis passing through the center of mass plus the product of the mass of the body and the square of the distance between the two axes." So, the distance "d" in Yosun's solution is that distance between the two axes, namely A and B. The total mass would also still be 3m.
 QuantumCat2014-09-18 09:26:17 The moment of inertia between two masses is $ml^2$, and since there are two masses, $I=2ml^2$
ec2250
2007-09-07 20:13:34
I don't understand why we need to add all three point masses in the parallel axis theorem. Doesn't the theorem only say we need to add the moment of inertia of the center of mass plus the mass and distance of the other axis? which would only be 1Md/(3)^.5? I'm pretty sure the solution is right, I just want a deeper explanation. Thanks.
jshariff
2007-04-12 17:25:07
Correction to the LaTeX in the above correction =p

$K_i = \frac{1}{2} I_i \omega_i^2$
nitin
2006-10-28 06:24:07
The kinetic energy of rotation is $K_i=\frac{1}{2}I_i(\omega_i)^2 Post A Comment!  Username: Password: Click here to register. This comment is best classified as a: (mouseover) Mouseover the respective type above for an explanation of each type. ## Bare Basic LaTeX Rosetta Stone LaTeX syntax supported through dollar sign wrappers$, ex., $\alpha^2_0$ produces $\alpha^2_0$.
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