GREPhysics.NET: GR9677 #32

GR9677 #32

Problem

GREPhysics.NET Official Solution

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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

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) Reply to this comment
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. Reply to this comment

Comments

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

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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)

Dx2wUV
2009-10-06 13:30:52

If you eliminate B then you eliminate the correct answer, according to ETS

dogsandfrogs
2009-10-07 14:12:42

Eliminate (B)? But B is the answer...

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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 $\frac{I_B}{I_A} = \frac{2ml^2}{ml^2} = 2$ .

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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).

Camoph
2010-03-30 17:05:53

How do you obtain r=l/\sqrt{3}??. r(cos30)=l/2??. I don't agree.

Camoph
2010-03-30 17:11:08

How do you obtain r=l/\sqrt{3}??. r(cos30)=l/2??. I don't agree.

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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 $6mr^2/3mr^2=2$ .

Reply to this comment

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.

Gaffer
2007-10-26 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.

errjones
2008-10-17 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.

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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.

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jshariff
2007-04-12 17:25:07

Correction to the LaTeX in the above correction =p

$K_i = \frac{1}{2} I_i \omega_i^2$

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nitin
2006-10-28 06:24:07

The kinetic energy of rotation is $K_i=\frac{1}{2}I_i(\omega_i)^2

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