Why do we use negative mass in this question?

Question

So there is a problem based upon the Center of mass and Moment of Inertia of continuous rigid bodies. So let's say we have a sphere of radius R and its mass is uniform throughout the system of particles. Now we cut out a smaller sphere of radius R/2 from the above sphere of radius R... Now my question is, Why can't we use another way of calculating the center of mass or moment of inertia of uniform bodies instead of using the concept of negative mass?

Answer 1

You can calculate the moment of inertia of a spherical shell from first principles if you like, but it is simpler to use the linearity of moments of inertia. So if MOI of small sphere + MOI of shell = MOI of large sphere then MOI of shell = MOI of large sphere - MOI of small sphere.

In the same way you could calculate the volume of the spherical shell from first principles, but it is simpler to say

$\displaystyle \text{Volume of shell} = \text{Volume of large sphere} - \text{Volume of small sphere} \\ \displaystyle= \frac 4 3 \pi R^3 - \frac 4 3 \pi \left( \frac R 2 \right)^3 = \frac 7 6 \pi R^3$

Answer 2

It is actually like this... I am doing it for center of mass right now :

*ρ(Volumetric density) = M/(4/3πr³)

M = (4/3πr³)ρ

m(R/2 sphere's mass) = ρ[4/3π(r/2)³]*

Now by symmetry, we can say that the center of mass does not lie on the x-axis..

So the y-coordinates of center of mass is :

y = {[ρ*(4/3πr³)0]+[-ρ(4/3π(r/2)³)r]} / {[ρ(4/3πr³)]+[-ρ*(4/3π(r/2)³)]}

We cancel out all the ρ variables and we equate the remaining equation We get it as -R/(4R-1) approx ( i may have done a few errors ) Here is the attached image which I created...

Answer 3

Yes there is a way to define the MMOI out of an arbitrary shape and it involves the integrals

$$ \begin{aligned}m & =\int_{V}\rho{\rm d}V\\ \vec{{\rm CM}} & =\frac{1}{m}\int_{V}\vec{r}\rho{\rm d}V\\ {\rm I}_{0} & =m\left(\frac{\int_{V}\left(\vec{r}\cdot\vec{r}-\vec{r}\odot\vec{r}\right)\rho{\rm d}V}{\int_{V}\rho{\rm d}V}\right) \end{aligned} $$

where $\vec{r}$ is the location of particle inside the volume, $\cdot$ is the inner product, and $\odot$ is the outer product.

But when you have composite shapes with elemental geometries, it is much easier to combine them using the concept of positive or negative volumes.

For example if $V_1$ is positive (part of solid) and $V_2$ is negative (missing material), then the above integral combine as follows:

$$\begin{aligned}m & =\int_{V_{1}}\rho{\rm d}V-\int_{V_{2}}\rho{\rm d}V\\ \vec{{\rm CM}} & =\frac{1}{m}\left(\int_{V_{1}}\vec{r}\rho{\rm d}V-\int_{V_{2}}\vec{r}\rho{\rm d}V\right)\\ {\rm I}_{0} & =m\left(\frac{\int_{V_{1}}\left(\vec{r}\cdot\vec{r}-\vec{r}\odot\vec{r}\right)\rho{\rm d}V-\int_{V_{2}}\left(\vec{r}\cdot\vec{r}-\vec{r}\odot\vec{r}\right)\rho{\rm d}V}{\int_{V_{1}}\vec{r}\rho{\rm d}V-\int_{V_{2}}\vec{r}\rho{\rm d}V}\right) \end{aligned}$$

Answer 4

Its a mathematical method of using negative mass, the other way is slicing the rjejcted piece , calculating the piece's mass and remaining mass of original object, and then recalculating the center of mass.

In both cases,if we take origin to be at center of original sphere, we get:

$$\vec{R_{cm}}=\frac{M(0)-\frac{M}{4}(\vec R/2)}{M-\frac{M}{4}}$$

Interpretations:

Negative mass method: Simply, the equation is written, as we write $R_{cm}$ equation. The negative sign comes as $M/4$ mass is negative. It is possible only in idealistic world (atleast not in our real world so far..), and is generally easier.

Normal method: The rejected mass, is subtracted instead of being added, since its no longer a part of system. That's why negative sign comes.

This method is more realistic and practical according to our real world.

Hope it helped!