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A Uniform Disc of diameter R/2 is put over another uniform disc of radius R of same thickness and Density. The peripheries of the two discs touch each other. The Centre of Mass of the system from centre of the big disc is? (I asked this question before but the answer was wrong. They took the Radius as R/2 while it is the Diameter) Expert Answer Let R be the radius of bigger disc and (R/2) be the diameter of smaller disc .
Let h be the common thickness of both discs. Both discs have same density and same thickness.
Hence mass is proportional to Square of radius

Let M be the mass of bigger disc and m be the mass of smaller disc.

( M / m )  = [ R2 / (R/4)2 ] = 16  ;  Hence M = 16m

X-coordinate of centre of mass , X = [ M / (M+m) ] × R + [ m / (M+m) ] × ( R / 4 )

X-coordinate of centre of mass , X = [ (16/17) + (1/68) ] R =  (65/68) R

Y-coordinate of centre of mass , Y = [ M / (M+m) ] × (h/2) + [ m / (M+m) ] × ( 3h / 2 )

Y-coordinate of centre of mass , Y = [ (16/34) + (3/34) ] h =  (19/34) h

Centre of mass = ( x, Y) = [ (65/68)R , (19/34)h ]
Answered by Thiyagarajan K | 26 Sep, 2021, 09:26: PM
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