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Units And Measurement

Question
CBSEENPH11017478

What is the moment of inertia of disc of mass m and radius r about its axis?

Solution
               
Consider, M is the mass of a uniform circular disc of radius R with centre O. 
Let, O be the centre of the ring. And YY' is the length of the axis. 
So the axis is represented by YOY'. 
Now, 
Surface area of the disc = straight pi space straight R squared
Mass space per space unit space are space of space the space disc space equals space fraction numerator straight M over denominator straight pi space straight R squared end fraction space  
Consider a small element of the disc, which would be a circular strip of radius x and width dx, say. 
Length of this element = 2straight pi
Surface are of this element = 2 space straight pi space straight x space dx
Therefore, 
Mass space of space this space element space equals space fraction numerator straight M over denominator straight pi space straight R squared end fraction space left parenthesis 2 πx space dx right parenthesis space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 2 straight M space straight x space dx over denominator straight R squared end fraction space

Moment space of space inertia space of space this space element space
of space the space disc space about space YOY apostrophe space equals space mass space straight x space left parenthesis distance right parenthesis squared

space space space space space space space space equals space fraction numerator 2 straight M space straight x space dx over denominator straight R squared end fraction space straight x space left parenthesis straight x right parenthesis squared space equals space fraction numerator 2 straight M space straight x cubed space dx over denominator straight R squared end fraction

The space small space element space may space lie space anywhere space from space
the space centre space of space disc space left parenthesis straight x space equals space 0 right parenthesis thin space to space left parenthesis straight x space equals space straight R right parenthesis.

Therefore comma space

straight I space equals space integral subscript straight x space equals space 0 end subscript superscript straight x space equals space straight R end superscript space fraction numerator 2 straight M space straight x cubed space dx over denominator straight R squared end fraction space equals space fraction numerator 2 straight M over denominator straight R squared end fraction space open square brackets straight x to the power of 4 over 4 close square brackets subscript straight x space equals space 0 end subscript superscript straight x space equals space straight R end superscript space

straight I space equals space 2 over 4 straight M over straight R squared space left parenthesis space straight R to the power of 4 space minus space 0 right parenthesis space

straight I thin space equals space 1 half space MR squared
Therefore, the moment of inertia of disc about its axis is 1/2 mr2 .