Question

You have two steel solid spheres. sphere 2 has twice the radius of sphere 1. part a by what factor does the moment of inertia i2 of sphere 2 exceed the moment of inertia i1 of sphere 1?

Answer 1

Answer:

moment of inertia of sphere 2 is 32 times the moment of inertia of sphere 1

Explanation:

The moment of inertia of a solid sphere about its axis is

$I=\frac{2}{5}MR^2$

where

M is the mass of the sphere

R is the radius of the sphere

The mass of the sphere can be rewritten as

$M=\rho V$

where

$\rho$ is the density

$V=\frac{4}{3}\pi R^3$ is the volume of the sphere

So the moment of inertia becomes

$I=\frac{2}{5}(\frac{4}{3}\pi \rho R^3)R^2 = \frac{8}{15}\pi \rho R^5$

Calling R the radius of sphere 1, the moment of inertia of sphere 1 is

$I_1=\frac{8}{15}\pi \rho R^5$

where $\rho$ is the density of steel, since the sphere is made of steel

Sphere 2 has twice the radius of sphere 1, so

R' = 2R

and so its moment of inertia is

$I_2=\frac{8}{15}\pi \rho R'^5=\frac{8}{15}\pi \rho (2R)^5=32(\frac{8}{15}\pi \rho R^5)=32I_1$

So, the moment of inertia of sphere 2 is 4 times the moment of inertia of sphere 1.