Partial Sphere Moment (Part 2)

Classical Mechanics Level 3

A partial sphere, described below, has radius $R$ and mass $M$ .

$\large{x = r \, cos \theta \, sin \phi \\ y = r \, sin \theta \, sin \phi \\ z = r \, cos \phi \\ 0 \leq r \leq R \\ 0 \leq \theta \leq 2 \pi \\ \frac{\pi}{4} \leq \phi \leq \frac{3 \pi}{4}}$

The object's moment of inertia with respect to the $z$ -axis can be expressed as:

$\large{I = \frac{A}{B} \, M \, R^2}$

If $A$ and $B$ are positive co-prime integers, what is $A + B$ ?

The answer is 3.

1 solution

Otto Bretscher
Nov 24, 2018

This works out very nicely! Thank you for posting!

Assuming a constant density $\rho$ , the moment of inertia comes out to be $I_z= \int_W \rho (x^2+y^2) dV= \rho\int_{0}^{2\pi}\int_{\pi/4}^{3\pi/4}\int_{0}^{R}r^4\sin^3\phi dr d\phi d\theta=\rho \times 2\pi\times \frac{5}{3\sqrt{2}}\times \frac {R^5}{5}$ and the mass is $M=\int_W \rho dV= \rho \int_{0}^{2\pi}\int_{\pi/4}^{3\pi/4}\int_{0}^{R}r^2\sin\phi dr d\phi d\theta=\rho \times 2\pi\times \sqrt{2}\times \frac {R^3}{3}$ so that $\frac{I_z}{M}=\frac{1}{2}R^2$ and the answer is $1+2=\boxed{3}$ .

I will add this to me repertoire of homework problems in calculus, if I may.

Feel free, and thanks for the solution. I was amused when I found that this yielded the same answer as for a disk.

Steven Chase - 2 years, 6 months ago

Partial Sphere Moment (Part 2)

The answer is 3.

1 solution

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