Moment of Inertia of a sphere about an axis

[phantomvommand]

Homework Statement

Consider a sphere with an axis passing through its centre. I = 2/5 MR^2 when the sphere is rotated through this axis. Consider a new axis which is parallel to the original axis, but at a distance d away. What is the new I about this new axis?

Relevant Equations

Parallel axis theorem

I = 2/5M R^2 + Md^2

This is analagous to Earth's movement about the Sun. Is the moment of inertia of Earth about the centre of mass of the Earth Sun system = 2/5MR^2 + Md^2, where:

M = Mass of earth,
R = Radius of Earth,
d = distance from Earth to centre of mass of earth-sun system.

 

[etotheipi]

It is the moment of inertia of the Earth about an axis passing through the barycentre of the Earth-Sun system, however you need to be careful how you use that information.

In particular, the Earth is rotating about its own axis at a different rate to that at which its centre of mass is rotating around the barycentre. In other words, the Earth is not performing rigid body rotation about an axis through the barycentre of the Earth-Sun system.

If you want the angular momentum of the Earth about that axis, for instance, it's going to be $L_z = Md^2 \omega_{ES} + \frac{2}{5} MR^2 \omega_{E}$, where $\omega_{ES} = 2\pi / (\text{1 year})$ and $\omega_E = 2\pi /(\text{1 day})$. To write that we note that the angular momentum of a rigid body about a certain axis equals the angular momentum of its centre of mass about that axis, plus the angular momentum of the body about the parallel axis through its centre of mass.

 

[phantomvommand]

It is the moment of inertia of the Earth about an axis passing through the barycentre of the Earth-Sun system, however you need to be careful how you use that information.

In particular, the Earth is rotating about its own axis at a different rate to that at which its centre of mass is rotating around the barycentre. In other words, the Earth is not performing rigid body rotation about an axis through the barycentre of the Earth-Sun system.

If you want the angular momentum of the Earth about that axis, for instance, it's going to be $L_z = Md^2 \omega_{ES} + \frac{2}{5} MR^2 \omega_{E}$, where $\omega_{ES} = 2\pi / (\text{1 year})$ and $\omega_E \approx 2\pi /(\text{1 day})$. To write that we note that the angular momentum of a rigid body about a certain axis equals the angular momentum of its centre of mass about that axis, plus the angular momentum of the body about the parallel axis through its centre of mass.

Thanks for the reply. I think you've mentioned an important detail that I am confused about. In $L_z = Md^2 \omega_{ES} + \frac{2}{5} MR^2 \omega_{E}$, I get that 2/5MR^2 w(earth) is the angular momentum due to Earth's own 'spin'. As for the other term, shouldn't it be (2/5MR^2 + Md^2) w(earth-sun)?

 

[etotheipi]

Thanks for the reply. I think you've mentioned an important detail that I am confused about. In $L_z = Md^2 \omega_{ES} + \frac{2}{5} MR^2 \omega_{E}$, I get that 2/5MR^2 w(earth) is the angular momentum due to Earth's own 'spin'. As for the other term, shouldn't it be (2/5MR^2 + Md^2) w(earth-sun)?

Not quite, no. The original is correct, but why? I'll try to explain the general theory, because this sort of thing becomes a lot easier once you have the framework. Imagine you have a general system of $N$ particles with position vectors $\mathbf{x}_{a}$ with respect to some origin $O$. The angular momentum of this system about $O$ is nothing but the sum of the angular momenta of each particle,\begin{align*}

\mathbf{L}_{O} = \sum_{a} \mathbf{l}_{O} &= \sum_a m_a \mathbf{x}_a \times \frac{d\mathbf{x}_a}{dt}
\end{align*}Let's now consider the centre of mass $C$, at position vector $\mathbf{X}$, of this system. By definition, if $M := \sum_a m_a$ is the total mass of all the particles, $\mathbf{X}$ satisfies$$M\mathbf{X} = \sum_a m_a \mathbf{x}_a$$Furthermore, we can write the position vector of any particle as the sum of the position vector of the centre of mass, and the vector from the centre of mass to that particular particle which we'll denote by $\mathbf{x}_a'$, i.e.$$\mathbf{x}_a = \mathbf{X} + \mathbf{x}_a'$$With these things in mind, let's return to our original expression,\begin{align*}

\mathbf{L}_{O} &= \sum_a m_a (\mathbf{X} + \mathbf{x}_a') \times \frac{d}{dt} \left( \mathbf{X} + \mathbf{x}_a' \right) \\

&= \sum_a m_a \mathbf{X} \times \frac{d\mathbf{X}}{dt} + \sum_a m_a \mathbf{X} \times \frac{d\mathbf{x}_a'}{dt} + \sum_a m_a \mathbf{x}_a' \times \frac{d\mathbf{X}}{dt} + \sum_a m_a \mathbf{x}_a' \times \frac{d\mathbf{x}_a'}{dt} \\

&= M \mathbf{X} \times \frac{d\mathbf{X}}{dt} + \mathbf{X} \times \sum_a m_a \frac{d\mathbf{x}_a'}{dt} + \left( \sum_a m_a \mathbf{x}_a' \right) \times \frac{d\mathbf{X}}{dt} + \sum_a m_a \mathbf{x}_a' \times \frac{d\mathbf{x}_a'}{dt}

\end{align*}Now using the definition of the centre of mass we know that $\sum_a m_a \mathbf{x}_a' = 0$, and differentiating that we also know $\sum_a m_a \frac{d \mathbf{x}_a'}{dt} = 0$. So the middle two terms vanish, and we're just left with\begin{align*}

\mathbf{L}_{O} &= M \mathbf{X} \times \frac{d\mathbf{X}}{dt} + \sum_a m_a \mathbf{x}_a' \times \frac{d\mathbf{x}_a'}{dt} \\

&:= \mathbf{L}_{\mathrm{CM}} + \mathbf{L}^*

\end{align*}Notice, then, that the angular momentum of a system of particles about any point is the sum of two terms: the angular momentum about the point $O$ of a hypothetical particle of mass $M$ situated at the centre of mass of the system, and the angular momentum of the system about the centre of mass $C$.

That is the answer to your question in post #3. In your example, the system is nothing but the set of particles which make up the Earth. The first term in that expression is the angular momentum of a hypothetical particle of mass $M_E$ situated at the centre of mass of the Earth, which has moment of inertia $M_E d^2$ and not $M_E d^2 + \frac{2}{5}M_E R^2$ about the axis through the barycentre.

Let me know if that answers your question, or also if you want to know a bit more about the moment of inertia as a tensor and related stuff.

 

[phantomvommand]

Not quite, no. The original is correct, but why? I'll try to explain the general theory, because this sort of thing becomes a lot easier once you have the framework. Imagine you have a general system of $N$ particles with position vectors $\mathbf{x}_{a}$ with respect to some origin $O$. The angular momentum of this system about $O$ is nothing but the sum of the angular momenta of each particle,\begin{align*}

\mathbf{L}_{O} = \sum_{a} \mathbf{l}_{O} &= \sum_a m_a \mathbf{x}_a \times \frac{d\mathbf{x}_a}{dt}
\end{align*}Let's now consider the centre of mass $C$, at position vector $\mathbf{X}$, of this system. By definition, if $M := \sum_a m_a$ is the total mass of all the particles, $\mathbf{X}$ satisfies$$M\mathbf{X} = \sum_a m_a \mathbf{x}_a$$Furthermore, we can write the position vector of any particle as the sum of the position vector of the centre of mass, and the vector from the centre of mass to that particular particle which we'll denote by $\mathbf{x}_a'$, i.e.$$\mathbf{x}_a = \mathbf{X} + \mathbf{x}_a'$$With these things in mind, let's return to our original expression,\begin{align*}

\mathbf{L}_{O} &= \sum_a m_a (\mathbf{X} + \mathbf{x}_a') \times \frac{d}{dt} \left( \mathbf{X} + \mathbf{x}_a' \right) \\

&= \sum_a m_a \mathbf{X} \times \frac{d\mathbf{X}}{dt} + \sum_a m_a \mathbf{X} \times \frac{d\mathbf{x}_a'}{dt} + \sum_a m_a \mathbf{x}_a' \times \frac{d\mathbf{X}}{dt} + \sum_a m_a \mathbf{x}_a' \times \frac{d\mathbf{x}_a'}{dt} \\

&= M \mathbf{X} \times \frac{d\mathbf{X}}{dt} + \mathbf{X} \times \sum_a m_a \frac{d\mathbf{x}_a'}{dt} + \left( \sum_a m_a \mathbf{x}_a' \right) \times \frac{d\mathbf{X}}{dt} + \sum_a m_a \mathbf{x}_a' \times \frac{d\mathbf{x}_a'}{dt}

\end{align*}Now using the definition of the centre of mass we know that $\sum_a m_a \mathbf{x}_a' = 0$, and differentiating that we also know $\sum_a m_a \frac{d \mathbf{x}_a'}{dt} = 0$. So the middle two terms vanish, and we're just left with\begin{align*}

\mathbf{L}_{O} &= M \mathbf{X} \times \frac{d\mathbf{X}}{dt} + \sum_a m_a \mathbf{x}_a' \times \frac{d\mathbf{x}_a'}{dt} \\

&:= \mathbf{L}_{\mathrm{CM}} + \mathbf{L}^*

\end{align*}Notice, then, that the angular momentum of a system of particles about any point is the sum of two terms: the angular momentum about the point $O$ of a hypothetical particle of mass $M$ situated at the centre of mass of the system, and the angular momentum of the system about the centre of mass $C$.

That is the answer to your question in post #3. In your example, the system is nothing but the set of particles which make up the Earth. The first term in that expression is the angular momentum of a hypothetical particle of mass $M_E$ situated at the centre of mass of the Earth, which has moment of inertia $M_E d^2$ and not $M_E d^2 + \frac{2}{5}M_E R^2$ about the axis through the barycentre.

Let me know if that answers your question, or also if you want to know a bit more about the moment of inertia as a tensor and related stuff.

This has been very helpful, thanks so much!