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
Jun 17, 2018

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!!!