A solid aluminum sphere of radius R has moment of inertia I about an axis through its center. What is the
moment of inertia about a central axis of a solid
aluminum sphere of radius 2R?
1. 21
2. 41
3. 87
4. 161
5. 321​

5. 32I

Explanation:

The moment of inertia of a solid sphere about its central axis is given by

I =            ------------------(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 =       --------------(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 =

=> M = p()              --------------------(*)

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' =       ------------------(iii)

Where;

M' = pV'

p = density of aluminum

V' = volume of the sphere =

=> M' = p()

Rewriting gives;

M' = p()

M' = p()

M' = 8p()

From equation (*), this can be written as

M' = 8M

Now substitute all necessary values into equation (ii)

I' =    

I' =    

I' =    

I' =  

I' =  

I' =  

Comparing with equation (ii)

I' =  

Therefore, the moment of inertia about a central axis of a solid

aluminum sphere of radius 2R is 32I

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

if they have a finite resultant