Find the inertia of a sphere radius R with rotating axis through the center

[annamal]

Homework Statement

Find the moment of inertia of a sphere with axis through the center

Relevant Equations

$I = \int{r^2dm}$

$$I = \int{r^2dm}$$
$$dm = \sigma dV$$
$$dV = 4\pi r^2dr$$
$$\sigma = \frac{M}{\frac{4}{3}\pi*R^3}$$
$$I = \sigma 4 \pi \int_0^R{r^4 dr} = \frac{3*MR^2}{5},$$
which is not the correct moment of inertia of a sphere

 

[haruspex]

$$I = \int{r^2dm}$$

Where r is what, exactly?

 

[annamal]

Where r is what, exactly?

The distance from center of rotation.

 

[haruspex]

The distance from center of rotation.

But that is not how you used it later. You substituted the distance from the centre of the sphere.

 

[annamal]

But that is not how you used it later. You substituted the distance from the centre of the sphere.

Note the problem says sphere has a rotating axis through the center so distance from center of sphere to outside would be 0 to R

 

[haruspex]

Note the problem says sphere has a rotating axis through the center so distance from center of sphere to outside would be 0 to R

But the distance from an arbitrary point within the sphere to its centre is not (in general) the distance from the point to a given axis through the centre.

 

[bob012345]

Note the problem says sphere has a rotating axis through the center so distance from center of sphere to outside would be 0 to R

I think it would help you to sketch a picture of the situation.

 

[PhDeezNutz]

OP as both posters before me mentioned: $r$ is the distance to the axis.

Are you familiar with spherical coordinates? (If not then get familiar with them ASAP). Be mindful of the different conventions (math vs physics).

$\theta$ is the angle from the z-axis going from $0$ to $\pi$. Draw an arbitrary radial vector and find the component that is perpendicular to the z-axis.

Because there is now angular dependence your new $dV$ will have to reflect that.

 

[annamal]

But the distance from an arbitrary point within the sphere to its centre is not (in general) the distance from the point to a given axis through the centre.

I don't understand that. Can you draw something to illustrate that?

 

[bob012345]

I don't understand that. Can you draw something to illustrate that?

We were hoping that you yourself would draw a picture to clarify it in your mind but I will give you a word picture to get started.

Imagine a sphere. Draw a pole through the center in some direction. Now, pick an arbitrary point inside the sphere somewhere. Now draw a line segment from the center of the sphere to that point. Draw another line segment from that point to the pole which is the minimum distance between the point and the pole, i.e. not generally through the center. If your point was an arbitrary point, they should be different lengths.

 

[annamal]

We were hoping that you yourself would draw a picture to clarify it in your mind but I will give you a word picture to get started.

Imagine a sphere. Draw a pole through the center in some direction. Now, pick an arbitrary point inside the sphere somewhere. Now draw a line segment from the center of the sphere to that point. Draw another line segment from that point to the pole which is the minimum distance between the point and the pole, i.e. not generally through the center. If your point was an arbitrary point, they should be different lengths.

Yes I see that.
$dV = 4\pi y^2 dy$
so $I = \int{r^2dm} = I = \int_{-R}^{R}{R^2 - y^2}*\sigma 4\pi y^2 dy$

Is that correct?

 

[PhDeezNutz]

@annamal

 

[haruspex]

Yes I see that.
$dV = 4\pi y^2 dy$
so $I = \int{r^2dm} = I = \int_{-R}^{R}{R^2 - y^2}*\sigma 4\pi y^2 dy$

Is that correct?

Your diagram should also illustrate the elements you are summing.
In principle, this is a volume integral. You can reduce it to a double integral using rotational symmetry about the sphere's axis of rotation, but you still have two largely independent variables of integration: the height y from the equatorial plane and the distance of the mass element from the axis. They interact only in that the height affects the range of the radius.

Edit:
From $dV = 4\pi y^2 dy$, it looks like you are using a spherical shell as your volume element. That cannot work because not all parts of it are the same distance from the rotational axis. Start with a ring element, or a cylindrical shell.

 

[bob012345]

Yes I see that.
$dV = 4\pi y^2 dy$
so $I = \int{r^2dm} = I = \int_{-R}^{R}{R^2 - y^2}*\sigma 4\pi y^2 dy$

Is that correct?

Closer but not exactly correct yet. How did you get $dV = 4\pi y^2 dy$?

 

[annamal]

Closer but not exactly correct yet. How did you get $dV = 4\pi y^2 dy$?

$V = \frac 4 3 \pi r^3$, so $dV = 4\pi y^2 dy$
What is wrong with that?

 

[annamal]

@annamal

View attachment 299205

r (distance indicated) = $\sqrt{r^2 - y^2}$
Thus
$I = \int{r^2dm} = I = \int_{-R}^{R}{R^2 - y^2}*\sigma 4\pi y^2 dy$

 

[haruspex]

$V = \frac 4 3 \pi r^3$, so $dV = 4\pi y^2 dy$
What is wrong with that?

First, what follows from $V = \frac 4 3 \pi r^3$ is that $dV = 4\pi r^2 dr$.
Secondly, as I pointed out in post #13, $4\pi r^2 dr$ would be a spherical shell element. That is of no help since different parts of it are at different distances from the axis of rotation.

Btw, I have been assuming this is for the moment of inertia of a solid sphere. Please clarify whether it is that or a spherical shell. Your answer is wrong either way.

 

[PhDeezNutz]

r (distance indicated) = $\sqrt{r^2 - y^2}$
Thus
$I = \int{r^2dm} = I = \int_{-R}^{R}{R^2 - y^2}*\sigma 4\pi y^2 dy$

Please have a look at this page http://hyperphysics.phy-astr.gsu.edu/hbase/sphc.html

For an entire sphere $\theta$ ranges from $0$ to $\pi$ and $\phi$ ranges from $0$ to $2 \pi$

Are you familiar with multi-variable calculus? Is it a pre-req or co-req for your class? If not then I think it would be very difficult to confirm the moment of inertia for a sphere.