7
$\begingroup$

My physics textbook said that the moment of inertia of a sphere is $\frac25mr^2$ where $m$ and $r$ are the mass and the radius of the sphere respectively. I wanted to verify the result by finding it myself, however, I always end up with $\frac35mr^2$ instead of $\frac25mr^2$. What is wrong with my method?

My approach involved cutting up a sphere into an infinite amount of hollow sphere shells, each of which having volume $4\pi r^2dr$. Then, calling $d$ the density of the sphere and $R$ the radius, the moment of inertia should be: $$\int_{0}^{R}4\pi r^2\cdot d\cdot r^2\cdot dr$$ Calculating this integral gives the following: $$4\pi d\frac{R^5}5$$ Using $m=d\cdot\frac43\pi r^3$, we can simplify this to: $$\frac35mR^2$$

This is clearly not the correct answer. Where have I gone wrong in my method? Is cutting the sphere into hollow shells conceptually wrong in the first place?

$\endgroup$

3 Answers 3

7
$\begingroup$

It did not came clear from the other answers why your approach is wrong. Remember that you calculate the moment of inertia for rotation around an axis not around a point. So if you choose for example the vertical axis, you notice that the points on spherical shell are at a constant distance from the center of the sphere, but they are at different distances from the vertical axis. You can use either point like masses, or cylinders along the vertical direction. You need to compute $$\frac{m}{\frac{4}{3}\pi r^3}\iiint_S(x^2+y^2)dxdydz$$ What you did was instead computing the integral of $x^2+y^2+z^2$.

$\endgroup$
2
$\begingroup$

There is a good trick with the sphere, because it is symmetrical. By definition, $$ I_x + I_y + I_z = \sum_i m_i(2x_i^2+2y_i^2+2z_i^2) = \sum_i 2m_i r_i^2 $$ where the sum is over all the point masses we imagine splitting the object into; $(x_i^2 + y_i^2)$ comes from $I_z$, $(x_i^2 + z_i^2)$ comes from $I_Y$ and $(y_i^2 + z_i^2)$ comes from $I_x$ - when you add them up you get the summand above. Now the $r_i$ in that formula is exactly the $r$ you were using when integrating - the distance from the center (and your mistake was using that instead of, say, $x^2+y^2$).

So it turns out that your integral is useful, after all; it tells us that (you didn't have a factor of $2$ in your integral, that's why I include that now) $$ I_x + I_y + I_z = 2\cdot \frac{3}{5}mR^2 $$ But since all the $I$'s are the same by symmetry, $$ I_x = I_y = I_z = \frac{1}{3}\cdot\frac{6}{5}mR^2 = \frac{2}{5}mR^2. $$

Similarly, for a hollow spherical shell, $r_i = R$ for all points, so there's no need to integrate and $$ I_x = I_y = I_z = \frac{1}{3}\cdot 2mR^2. $$

$\endgroup$
1
$\begingroup$

You have to use the moment of inertia of the spherical shells in your derivation, which is

$$ dI = \frac{2}{3}r^2 \ dm = \frac{2}{3} r^2 d \ (4\pi r^2 \ dr) $$

Integrating this will give the correct answer. Remember, you're adding up the spherical shells, not individual point masses, so this changes the calculation.

$\endgroup$
5
  • $\begingroup$ How would you find the moment of inertia for the spherical shell though? $\endgroup$
    – u8y7541
    Nov 24, 2017 at 19:55
  • $\begingroup$ I'd just look them up from a table, but you can also obtain it from rotating a circular ring around its planar axis. Of course, this requires you to know the moment of inertia of a ring around its horizontal axis, which is $\frac{1}{2}mr^2$ $\endgroup$
    – Dylan
    Nov 24, 2017 at 20:00
  • $\begingroup$ Another way is to disregard spherical shells entirely, and obtaining a sphere by adding up a stack of solid circular disks (whose radii vary along their height) $\endgroup$
    – Dylan
    Nov 24, 2017 at 20:03
  • $\begingroup$ @u8y7541 Or thin hollow cylindrical shells (co-axial) $\endgroup$ Nov 24, 2017 at 20:04
  • $\begingroup$ @NickPavlov Yup, that too! Either way, you'll need to do 3 levels of integration to get from a point mass to a 3D solid $\endgroup$
    – Dylan
    Nov 24, 2017 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.