# Moment of inertia of a sphere

1. Feb 3, 2008

### Oerg

1. The problem statement, all variables and given/known data

Ok, so i was trying to derive the moment of inertia about an axis that runs through the centre of mass of the sphere and i couldnt get the answer of $$I_{CM}=\frac{2}{5}MR^2$$

3. The attempt at a solution

The moment of inertia is defined as

$$\int r^2dm$$

and since

$$dm=\rho \times 4\pi r^2\times dr$$

so

$$I_{CM}=\int_{0}^{R}4\rho r^4 \pi dr$$
$$I_{CM}=\frac{4\rho\pi R^5}{5}$$

and expanding rho gives

$$I_{CM}=\frac{3}{5}R^2M$$

which is not correct.

2. Feb 3, 2008

### mjsd

the typical way to derive the moment of inertia of a sphere is to use the result for the moment of inertia of a thin disk (axis through center). The sphere is just a collection of thin disks with different radius.....

3. Feb 3, 2008

### Oerg

Hmm, then is there something wrong with my working? I would like to know in case I have some conceptual misunderstanding.

4. Feb 3, 2008

### mjsd

looks like you were trying to add up contribution from many many thin spherical shell (of radius r) but u haven't used the moment of inertia for shell. remember the "r^2" in the moment of inertia formula represents the perpendicular distance from axis of rotation to the point mass with mass m

personally, i would do this problem by
first MoI of thin hood
then MoI of thin disk
then MoI of sphere

or the hood->shell->sphere path

5. Feb 3, 2008

### Oerg

oh man pelase help me im going crazy

so i decided to try solving using the disk method

$$\int r^2 dm$$

so

$$dm= \rho \times \pi r^2 \times dr$$

and

$$I_{CM}=\int_{0}^{R} \frac{M}{V} \pi r^4 dr$$

and the moment of inertia of half a sphere is then

$$I_{CM}\frac{=3MR^2}{20}$$

and this moment of inertia of half a sphere obviously doesn't give the moment of inertia of a sphere HELP!

6. Feb 3, 2008

### mjsd

Assuming you know how to derive the disk, let's use that result

here I sliced the sphere into many disks with different radius

see attachment

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Last edited: Feb 4, 2008
7. Feb 3, 2008

### rock.freak667

I found another post here on how to derive it but it is suggested that you try it yourself a bunch of times...

8. Feb 3, 2008

### D H

Staff Emeritus
This $r$ refers to the distance from the rotational axis ...
and this $r$ refers to the distance from the center of the sphere. You got an erroneous result by erroneously equating two very different measures that happen to have the same symbol.

9. Feb 3, 2008

### D H

Staff Emeritus
Rather than integrating over a series of spherical shells I suggest you look at integrating over a series of cylindrical shells. The height (and hence mass) of these shells will vary with the radial distance from the axis of rotation. It will help to remember the formula for the length of a chord length of a circle as a function of the radius and chord angle ...

10. Feb 4, 2008

### Gear300

The $$dm=\rho \times 4\pi r^2\times dr$$ doesn't work. If you were to cut the sphere in half and try to straighten out one of the halves, it wont work out too well and thats probably the source of the mistake.

11. Feb 6, 2008

### Oerg

thanks for the replies, i finally figured how to derive it on my own. The moment of inertia of a disk is

$$I_{CM}=\int_{0}^{R} r^2 \times \rho \times 2 \pi r \times dr$$

$$I_{CM}=\frac{MR^2}{2}$$

Integrating these differential moment of inertia gives us the moment of inertia of a sphere
$$dI=\frac{r^2}{2} dm$$

$$dm=\rho \times \pi r^2 \times dz$$

So the integral becomes

$$I=\int_{-R}^{ R} \frac{\rho \pi}{2} (R^2-z^2)^2dz$$

$$I=\frac{2}{5} MR^2$$

Last edited: Feb 6, 2008