Question

How do I find the moment of inertia of a hollow cylinder?

The answer is A, but I'm unsure how they got the answer.

Thanks for your help in advance!!

Answer 1

The answer is `option (A)`, Please see the explanation below.

Explanation:

Mass of cylinder is `=M`

Radius of cylinder is `=R`

Moment of inertia of cylinder is `I_C=1/2MR^2`

Radius of hole is `=a`

Mass of the part removed is `=m`

The moment of inertia of the removed part is `I_h=1/2ma^2`

Let the length of the cylinder be `=L`

Volume of the cylinder is `V_C=pir^2L`

The density of the cylinder is `rho=M/V_c=M/(piR^2L)`

The volume of the "hole" `v_h=pia^2L`

The mass of the "hole" is

`m=v_h*rho=(pia^2L)*M/(piR^2L)=a^2N/(R^2)`

Therefore,

`I_h=1/2*a^2M/(R^2)*a^2=1/2a^4/R^2M`

Therefore,

The new moment of inertia is

`I=1/2MR^2-1/2a^4/R^2M=1/2MR^2(1-a^4/R^4)`

I hope that this will help!!!