Answer:
5. 32I
Explanation:
The moment of inertia of a solid sphere about its central axis is given by
I = [tex]\frac{2}{5} MR^2[/tex] ------------------(i)
Where;
M = mass of the sphere
R = radius of the sphere.
From the question;
Case 1: The aluminum sphere has a radius R and moment of inertia I.
This means that we can substitute these values of R and I into equation (i) and get;
I = [tex]\frac{2}{5} MR^2[/tex] --------------(ii)
M is the mass of the aluminum sphere and is given by;
M = pV
Where;
p = density of aluminum
V = Volume of the sphere = [tex]\frac{4}{3} \pi R^3[/tex]
=> M = p([tex]\frac{4}{3} \pi R^3[/tex]) --------------------(*)
Case 2: An aluminum sphere with a radius of 2R instead.
Let the moment of inertia in this case be I' and mass be M'
Substituting R = 2R, M = M' and I = I' into equation (i) gives
I' = [tex]\frac{2}{5} M'(2R)^2[/tex] ------------------(iii)
Where;
M' = pV'
p = density of aluminum
V' = volume of the sphere = [tex]\frac{4}{3} \pi (2R)^3[/tex]
=> M' = p([tex]\frac{4}{3} \pi (2R)^3[/tex])
Rewriting gives;
M' = p([tex]\frac{4}{3} \pi (2)^3(R)^3[/tex])
M' = p([tex]\frac{4}{3} \pi8(R)^3[/tex])
M' = 8p([tex]\frac{4}{3} \pi R^3[/tex])
From equation (*), this can be written as
M' = 8M
Now substitute all necessary values into equation (ii)
I' = [tex]\frac{2}{5} M'(2R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(2R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(2)^2(R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(4)(R)^2[/tex]
I' = [tex]\frac{2}{5} (32M)(R)^2[/tex]
I' = [tex]32[\frac{2}{5}MR^2][/tex]
Comparing with equation (ii)
I' = [tex]32[I][/tex]
Therefore, the moment of inertia about a central axis of a solid
aluminum sphere of radius 2R is 32I
