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Moment of intertia for a system

  • Thread starter dragon695
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Figure 1

Homework Statement


Three identical marbles are on a turntable disk of radius [tex]R[/tex]. The mass of the disk is [tex]{m}_{d}[/tex]. The mass of each marble is [tex]{m}_{b}[/tex]. What is the moment of inertia of the disk and the marbles around an axis of rotation at the edge of the disk (the dot in Figure 1)?
  • [tex]\left({m}_{d} + 4{m}_{b}\right){R}^{2}[/tex]
  • [tex]\left(\frac{5}{2}{m}_{d} + 3{m}_{b}\right){R}^{2}[/tex]
  • [tex]\left(\frac{3}{2}{m}_{d} + 5{m}_{b}\right){R}^{2}[/tex]
  • [tex]\left(\frac{1}{2}{m}_{d} + 3{m}_{b}\right){R}^{2}[/tex]
  • [tex]\left(\frac{1}{2}{m}_{d} + 2{m}_{b}\right){R}^{2}[/tex]

Homework Equations


  • Solid Disk Inertial Moment: [tex]{I}_{disk} = \frac{1}{2}m{r}^{2}[/tex]
  • Solid Sphere Inertial Moment: [tex]{I}_{solid\ sphere} = \frac{2}{5}m{r}^{2}[/tex]
  • Inertial Moment Additivity: [tex]{I}_{sys} = {I}_{1} + {I}_{2} + \ldots + {I}_{n}[/tex]
  • Parallel Axis Theorem: [tex]I = {I}_{cm} + m{d}^{2}[/tex]

The Attempt at a Solution


I initially looked at this problem and I was not sure what assumptions I could make. I guessed that the marbles were not attached, but I am not sure what that means. The only experiment I did with the inertial moment of a system was dropping a hollow cylinder onto a rotating disk. While not attached, the cylinder had enough mass that it did not rotate independent of the disk and thus the system could be treated as a whole such that the principle of inertial moment additivity could be used. However, what am I supposed to assume here? Well, since this was a test question, I was forced to go with my instinct, which was to treat it like the experiment I did earlier. But what about the radius of the sphere? As you'll see below, I accounted for it, but why is the problem not accounting for it?

So, with that in mind, I started by listing what moments of inertia made up the system in terms of the knowns:

Let [tex]d[/tex] be the disk and [tex]b1[/tex], [tex]b2[/tex], and [tex]b3[/tex] be the marbles. Then we know that:

[tex]{I}_{sys} = {I}_{d} + {I}_{b1} + {I}_{b2} + {I}_{b3}[/tex]

I then found the inertial moments of each object independent of each other in terms of the knowns (but here is where I had to account for the marble's radius):

For the moments of inertia for each object, we simply substitute our variables into the given equation for the object type:

[tex]{I}_{dcm} = \frac{1}{2}{m}_{d}{R}^{2}[/tex]

[tex]{I}_{b1cm} = \frac{2}{5}{m}_{b}{r}_{b}^{2}[/tex]

[tex]{I}_{b2cm} = \frac{2}{5}{m}_{b}{r}_{b}^{2}[/tex]

[tex]{I}_{b3cm} = \frac{2}{5}{m}_{b}{r}_{b}^{2}[/tex]

My next step was to apply the parallel axis theorem to find the axis-adjusted moments of inertia for each object:

Since the axis is moved to one end of the disk and all distances from the new axis can be expressed in terms of the disk's radius, the axis-adjusted moments of inertia are as follows:

[tex]{I}_{d} = {I}_{dcm} + {m}_{d}{R}^{2} = \frac{1}{2}{m}_{d}{R}^{2} + {m}_{d}{R}^{2} = \left(\frac{1}{2}{m}_{d} + \frac{2}{2}{m}_{d}\right){R}^{2} = \frac{3}{2}{m}_{d}{R}^{2}[/tex]

[tex]{I}_{b1} = {I}_{b1cm} + {m}_{b}{0}^{2} = \frac{2}{5}{m}_{b}{r}_{b}^{2}[/tex]

[tex]{I}_{b2} = {I}_{b2cm} + {m}_{b}{R}^{2} = \frac{2}{5}{m}_{b}{r}_{b}^{2} + {m}_{b}{R}^{2}[/tex]

[tex]{I}_{b3} = {I}_{b3cm} + {m}_{b}{\left(2R\right)}^{2} = \frac{2}{5}{m}_{b}{r}_{b}^{2} + 4{m}_{b}{R}^{2}[/tex]

With those equations in hand, it was easy to substitute back into the equation for the moment of inertia of the system and perform some algebraic simplification. The result was:


[tex]{I}_{sys} = \frac{3}{2}{m}_{d}{R}^{2} + \frac{2}{5}{m}_{b}{r}_{b}^{2} + \frac{2}{5}{m}_{b}{r}_{b}^{2} + {m}_{b}{R}^{2} + \frac{2}{5}{m}_{b}{r}_{b}^{2} + 4{m}_{b}{R}^{2}[/tex]

[tex]{I}_{sys} = \frac{3}{2}{m}_{d}{R}^{2} + 3\left(\frac{2}{5}{m}_{b}{r}_{b}^{2}\right) + 5{m}_{b}{R}^{2}[/tex]

[tex]{I}_{sys} = \frac{3}{2}{m}_{d}{R}^{2} + \frac{6}{5}{m}_{b}{r}_{b}^{2} + 5{m}_{b}{R}^{2}[/tex]

[tex]{I}_{sys} = \frac{3}{2}{m}_{d}{R}^{2} + 5{m}_{b}{R}^{2} + \frac{6}{5}{m}_{b}{r}_{b}^{2}[/tex]

[tex]{I}_{sys} = \left(\frac{3}{2}{m}_{d} + 5{m}_{b}\right){R}^{2} + \frac{6}{5}{m}_{b}{r}_{b}^{2}[/tex]

As you can see, I arrive at a solution which is close to one of the options presented in the multiple choice answer block. On the test, I was forced to select option c, but that is not why am posting this. I'm just not satisfied with the answer. Am I making incorrect assumptions? Is the question itself making incorrect assumptions? Is it possible that I am procedurally incorrect? How can they just discount the size of the marbles? I understand that we are already in unreal terrirtory since marbles and turntables are not of uniform density and mass. Thanks in advance for helping to clear my mind of what has been a nagging issue for me since I took the test.
 

Answers and Replies

  • #2
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The problem figure suggests that r_b<<R so the extent of the marble can be neglected. I think a better problem formulation would have made that explicit, but since there is no mention of the marble's size in the multiple choice answers it is pretty clear that is the intention...

To convince yourself that this assumption quickly becomes a good one for r_b<R, consider what the total inertia is for r_b=frac*R where frac takes values of (say) 0.1, 0.05 etc. The relative contribution from r_b goes as (r_b/R)^2...
 
  • #3
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The problem figure suggests that r_b<<R so the extent of the marble can be neglected. I think a better problem formulation would have made that explicit, but since there is no mention of the marble's size in the multiple choice answers it is pretty clear that is the intention...

To convince yourself that this assumption quickly becomes a good one for r_b<R, consider what the total inertia is for r_b=frac*R where frac takes values of (say) 0.1, 0.05 etc. The relative contribution from r_b goes as (r_b/R)^2...
I should have known it would be something like this, but I still wonder. what if you had marbles made of iridium or some other extremely dense material? If you were to say that the radius of the marble was 1/16 of the turntable, it should still have non-insignificant impact on the moment of inertia then? This is what frustrates me the most about Physics and, to a large extent, the idiotic questions that Mastering Physics comes up with. Why be so convoluted about it? I don't understand the purpose of modeling something mathematically where we are free to just throw away inconvienent things to make it "simple."
 
  • #4
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I should have known it would be something like this, but I still wonder. what if you had marbles made of iridium or some other extremely dense material? If you were to say that the radius of the marble was 1/16 of the turntable, it should still have non-insignificant impact on the moment of inertia then? This is what frustrates me the most about Physics and, to a large extent, the idiotic questions that Mastering Physics comes up with. Why be so convoluted about it? I don't understand the purpose of modeling something mathematically where we are free to just throw away inconvienent things to make it "simple."
The problem can be solved at many levels of detail and determining a reasonable level is often part of a real-world problem. If the turntable is made out of balsa wood and the marbles are lead the balance shifts and the dimensions of the marbles clearly become more important. The power of solving the more complicated problem (as you did) is that you can ask these questions in a quantifiable way: you can determine the tolerance you can accept and see what kind of marbles that would allow for.

The purpose of making the problem "simpler" (even if you can solve it in a more complicated formulation) is usually to promote understanding of a particular concept. When I find a solution as you did I will consider what happens to it in extreme cases, and what possible effects I get by varying the various problem parameters. It is not only a good way to find possible issues with the solution, but it allows you to get a more generalized idea of the physics that the problem is trying to illustrate.

For the practical case of solving this particular problem, the clue to the level of detail is given to by the answers...
 
  • #5
52
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Also, if the radius of the marbles is 1/16 of that of the turntable (and they are made of similar materials), the relative impact of considering the finite size of the marbles is less than one in a thousand. Is that significant? Maybe, maybe not - depends on the application, but I think it would have to be a rather sensitive setup for it to make a difference.
 

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