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Units And Measurement

Question
CBSEENPH11017611

From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc as shown in figure. Find the moment of inertia of remaining disc about an axis perpendicular to the plane of the disc and passing through center.


Solution
The moment of inertia of complete disc about a perpendicular axis passing through center O is,
 straight I subscript 1 equals 1 half open parentheses 9 straight M close parentheses straight R squared
space space equals 9 over 2 MR squared
 
The mass of cut out disc if radius R/3 is,
straight m space equals fraction numerator 9 straight M over denominator pi R squared end fraction straight pi open parentheses straight R over 3 close parentheses squared space

space space space space equals straight M 
Now, using the theorem of parallel axis, the moment of inertia of cut out disc about the perpendicular axis passing through center O is,
straight I subscript 2 space equals 1 half straight M open parentheses straight R over 3 close parentheses squared plus straight M open parentheses fraction numerator 2 straight R over denominator 3 end fraction close parentheses squared

space space space space equals 1 half MR squared  

The moment of inertia of residue disc is, 

straight I equals straight I subscript 1 minus straight I subscript 2 space

space equals 9 over 2 MR squared minus 1 half MR squared space space

space equals 4 MR squared