How Is the Moment of Inertia Calculated for a Spherical Shell?

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SUMMARY

The moment of inertia of a spherical shell with radius R and mass m is calculated as (2mR^2)/3. This can be derived using two methods: direct integration or by analyzing the increase in moment of inertia of a solid sphere as its radius changes. The moment of inertia of a solid sphere is given by I = (8/15) π ρ R^5, where ρ is the density. The calculations involve using the mass of the shell, dm = 4 π R^2 ρ dr, and applying the derivative of the moment of inertia with respect to the radius.

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Familiarity with calculus, specifically integration and differentiation
  • Knowledge of density and mass distribution in physics
  • Basic understanding of spherical geometry
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  • Learn about the application of calculus in physics problems
  • Explore the relationship between density and moment of inertia in various materials
  • Investigate the concept of concentric spheres and their contributions to overall moment of inertia
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Lyuokdea
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I'll start off with the question, it's a question where the answer is given and you have to show it, I have been able to figure out the correct calculations, but have no clue what they mean:

Show that the moment of inertia of a spherical shell of radius R and mass m is (2mR^2)/3 This can be done by direct integration or, more easily, by finding the increase in the moment of inertia of a solid sphere when the radius changes. To do this, first show that the moment of inertia of a solid sphere of density \rho is I = (8/15) \pi \rho R^5 Then compute the change dI in I for a change dR, and use the fact that the mass of this shell isdm = 4 \pi R^2 \rho dr.

Obviously the question gives you much of the answer, by shoving numbers together I got

I = \frac{2}{5}mR^2 *(4/3) \pi r^3 \rho
I = \frac{8}{15}m \pi \rho r^5

\frac{dI}{dR} = \frac{8}{3} m \pi \rho r^4

and then basically dividing the first equation by the dI/dR gives you the right answer, but it makes no sense how that works or why you would do that. So I tried the integration route, because I thought that might work better

z = x^2 + y^2

z = r^2


I = mr^2

I = \frac{m}{r}\int_{-R}^{R}r^2 dr

Where \frac{m}{r}[\tex] is the density. This also gives the right answer, but I have a feeling I&#039;m just seeing answer on the page and thinking up a way to get to the answer that really has nothing to do with being right, if somebody could help me conceptualize what I&#039;m doing in both methods and whether either of these are somewhat right, I would really appreciate it.<br /> <br /> P.S. sorry about the bad looking coding, I&#039;m still getting used to tex
 
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If you are looking for a conceptual understanding of the problem, the easiest way to look at it is to think of a spherical shell as a number of concentric spheres. When you add each sphere to the shell, the moment of inertia increases by an amount that is proportional to the radius of the sphere. That is why when you take the derivative of the moment of inertia with respect to the radius, you get the expression that you have in your equations.

In the integration route, you are essentially summing up all of the different moments of inertia for each concentric sphere and adding them together to get the total moment of inertia for the whole shell. In this case, the density of the shell affects the result, which is why you have the expression for the density in your equation.

I hope this helps clear up the issue a bit and gives you a better understanding of the problem. Good luck!
 


First of all, congratulations on figuring out the correct calculations! That is a great accomplishment and shows that you have a good understanding of the topic.

Now, let's break down what is happening in the two methods you used. In the first method, you are using the given formula for the moment of inertia of a solid sphere, which is derived from the moment of inertia of a point mass. This formula takes into account the density of the sphere and its radius. Then, you are using the fact that the mass of the spherical shell is given by dm = 4 \pi R^2 \rho dr, which means that the mass of the shell increases as the radius increases. This is because the shell has a thickness and as the radius increases, the thickness also increases, resulting in more mass. By taking the derivative of the moment of inertia formula with respect to the radius, you are essentially finding the change in moment of inertia for a small change in radius. This change in moment of inertia is then divided by the change in radius, giving you the moment of inertia per unit change in radius. Multiplying this by the change in radius (which is equivalent to the thickness of the shell) gives you the moment of inertia of the spherical shell.

In the second method, you are using the definition of moment of inertia, which is the integral of the mass distribution multiplied by the square of the distance from the axis of rotation. In this case, the axis of rotation is the z-axis and the mass distribution is given by the density of the sphere. By integrating over the entire volume of the sphere, you are essentially adding up the contributions of all the infinitesimal masses that make up the sphere. This gives you the moment of inertia of the solid sphere. Then, by considering a small change in radius, you are essentially finding the change in moment of inertia for a small change in radius. This gives you the same result as the first method.

So, both methods are correct and they essentially use the same concept of finding the change in moment of inertia for a small change in radius. It is just a matter of using different formulas and approaches to get to the same result. As for why you divide the first equation by dI/dR, it is because you are essentially finding the moment of inertia per unit change in radius, so dividing by dI/dR gives you the moment of inertia for the entire shell. I hope this helps to clarify the concepts and reasoning behind the calculations
 

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