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
[tex]I=\frac{2}{5}MR^2[/tex]
where
M is the mass of the sphere
R is the radius of the sphere
The mass of the sphere can be rewritten as
[tex]M=\rho V[/tex]
where
[tex]\rho[/tex] is the density
[tex]V=\frac{4}{3}\pi R^3[/tex] is the volume of the sphere
So the moment of inertia becomes
[tex]I=\frac{2}{5}(\frac{4}{3}\pi \rho R^3)R^2 = \frac{8}{15}\pi \rho R^5[/tex]
Calling R the radius of sphere 1, the moment of inertia of sphere 1 is
[tex]I_1=\frac{8}{15}\pi \rho R^5[/tex]
where [tex]\rho[/tex] 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
[tex]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[/tex]
So, the moment of inertia of sphere 2 is 4 times the moment of inertia of sphere 1.
