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

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

Thanks for your help in advance!!

1 Answer
Narad T.
Jun 17, 2018

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

Explanation:

Mass of cylinder is =M=M

Radius of cylinder is =R=R

Moment of inertia of cylinder is I_C=1/2MR^2IC=12MR2

Radius of hole is =a=a

Mass of the part removed is =m=m

The moment of inertia of the removed part is I_h=1/2ma^2Ih=12ma2

Let the length of the cylinder be =L=L

Volume of the cylinder is V_C=pir^2LVC=πr2L

The density of the cylinder is rho=M/V_c=M/(piR^2L)ρ=MVc=MπR2L

The volume of the "hole" v_h=pia^2Lvh=πa2L

The mass of the "hole" is

m=v_h*rho=(pia^2L)*M/(piR^2L)=a^2N/(R^2)m=vh⋅ρ=(πa2L)⋅MπR2L=a2NR2

Therefore,

I_h=1/2*a^2M/(R^2)*a^2=1/2a^4/R^2MIh=12⋅a2MR2⋅a2=12a4R2M

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=12MR2−12a4R2M=12MR2(1−a4R4)

I hope that this will help!!!

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