Uniform Solid Sphere Moment of Inertia Calculation

In summary: Where M is the mass of the cylindrical shell and R is the radius of the cylindrical shell.In summary, to find the moment of inertia of a uniform solid sphere of mass m and radius a about an axis through its centre, a volume element approach can be used. However, using cylindrical shells instead of spherical shells makes it easier to calculate the moment of inertia. The moment of inertia of a cylindrical shell is MR^2/2, where M is the mass of the shell and R is the radius.
  • #1
Mirumbe
4
0
Find the moment of inertia of a uniform solid sphere of mass,m and radius,a about an axis through its centre.
I have tried to solve it but I get the different answer, I don't know where I have done mistake. Please! check and correct my solution below:-
Consider a volume element, dv of the sphere; this has mass,mdv/(4/3)∏a^3.
Then
I = ∫(mdv/(4/3)∏a^3)r^2
Where r is the distance of the volume element, dv from the axis, and
I is the moment of inertia
I = m/(4/3)∏a^3∫r^2dv
We know, the volume of the sphere,v is:
v = (4/3)∏r^3
dv = 4∏r^2dr
Thus,
I = 3m/(4/3)∏a^3∫r^2(4∏r^2)dr
= 3m/a^3∫r^4dr under limit [0,a]
= 3m/a^3[r^5/5] under [o,a]
Hence,
I = (3/5)ma^2 --------------WRONG!

The correct answer is: I = (2/5)mr^2
 
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  • #2
For starters you seem to mixing r and a.
 
  • #3
Mirumbe said:
I = ∫(mdv/(4/3)∏a^3)r^2
Where r is the distance of the volume element, dv from the axis, and
I is the moment of inertia
I = m/(4/3)∏a^3∫r^2dv
Here r is the distance to the axis.
We know, the volume of the sphere,v is:
v = (4/3)∏r^3
dv = 4∏r^2dr
Here r is the distance to the center.
 
  • #4
My intention was to find the volume element, dv of the sphere about the axis. But, since the axis was through its centre that's why I said
v = (4/3)∏r^3
Then, after differentiating w.r.t variable, I got the volume element, dv
dv = 4∏r^2dr
So, if I am wrong here,what should I do, so as to get the volume element,dv?
 
  • #5
I'd use cylinder coordinates [itex]\rho,\varphi,z[/itex]. The sphere is given by [itex]\rho \in [0,R][/itex], [itex]\varphi \in [0,2 \pi[[/itex], and [itex]z \in [-\sqrt{R^2-\rho^2},+\sqrt{R^2-\rho^2}[/itex]. Let [itex]n=\text{\const}[/itex] be the mass density. Then the moment of inertia is

[tex]\Theta=n \int_0^{R} \mathrm{d} \rho \int_0^{2 \pi} \mathrm{d} \varphi \int_{-\sqrt{R^2-\rho^2}}^{\sqrt{R^2-\rho^2}} \mathrm{d} z \; \rho^3.
[/tex]

There one factor of [itex]\rho[/itex] comes from the volume element [itex]\mathrm{d}^3 x=\rho \mathrm{d} \rho \mathrm{d} \varphi \mathrm{d} z[/itex].

The integrals over [itex]z[/itex] and [itex]\varphi[/itex] are trivial. You find

[tex]\Theta=4 \pi n \int_0^R \mathrm{d} \rho \; \rho^3 \sqrt{R-\rho^2}.[/tex]

This remaining integral becomes

[tex]\Theta=\frac{8 \Pi}{15} n R^5 = \frac{8 \pi}{15} \frac{3m}{4 \pi R^3} R^5=\frac{2}{5} m R^2.[/tex]
 
  • #6
Thanks Vanhee71 for your solution although it is the new idea to me.
 
  • #7
Unfortunately, vanhees71 just did the problem for you which deprives you of figuring it out on your own.

The key point is to choose a volume element for which you can easily write down the moment of inertia. You chose spherical shells. (And in the process, mixed up the radius of those shells with the distance of a mass element to the axis of rotation.) You could do the problem that way, but you'd have to know the moment of inertia of a spherical shell. The better way is use cylindrical shells, which are easy. What's the moment of inertia of a cylindrical shell?
 
  • #8
The moment of inertia of the Cylindrical shell is:
I = MR^2
 

1. What is a uniform solid sphere?

A uniform solid sphere is a three-dimensional shape with a perfectly round surface and equal mass distribution throughout. It is also known as a perfect sphere or a symmetrical sphere.

2. How is the mass of a uniform solid sphere distributed?

The mass of a uniform solid sphere is evenly distributed throughout the entire volume of the sphere. This means that every point on the surface of the sphere has the same amount of mass.

3. What are the properties of a uniform solid sphere?

A uniform solid sphere has a symmetrical shape, equal mass distribution, and a constant density. It also has a smooth surface and a single radius that extends from the center to any point on the surface.

4. What are the applications of a uniform solid sphere?

Uniform solid spheres have various applications in physics and engineering, such as in modeling planetary bodies, calculating rotational inertia, and understanding the behavior of fluids. They are also used in sports equipment, like golf balls and baseballs.

5. How is the volume of a uniform solid sphere calculated?

The volume of a uniform solid sphere is calculated using the formula V = (4/3)πr^3, where r is the radius of the sphere. This formula can also be used to find the mass of a uniform solid sphere, as long as the density is known.

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