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CBSE Class 11-science Answered

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?

Asked by adipadmakarri | 25 Sep, 2021, 11:18: AM
Expert Answer
As per the given statement " peripheries of discs are touching each other ",
it is assumed that the discs are positioned as shown in figure.
 
Let A ( R, h/2 ) be the centre of mass of bigger disc and B( R/2, 3h/2) be the centre of
mass of smaller disc as per the assumed Coordinates axes as shown in figure .
 
Where R is radius of bigger disc and h is common thickness of discs .
 
Since both discs have same density and thickness , mass of disc is prportional to radius of disc .
 
Let M be mass of bigger disc and m is mass of smaller disc .
 
( M / m ) = R2 / ( R/2)2 = 4   i.e. M = 4 m 
 
X-coordinate of centre of mass of combined discs,  x  =  [ m / ( m+M) ] (R/2) + [ M / ( m+M) ] (R) 
 
X-coordinate of centre of mass of combined discs,  x = R [ (1/10) + (4/5) ]  = 0.9 R
 
Y-coordinate of centre of mass of combined discs,  y  =  [ m / ( m+M) ] (3h/2) + [ M / ( m+M) ] (h/2) 
 
Y-coordinate of centre of mass of combined discs,  y = h [ (3/10) + (4/10) ]  = 0.7 h
 
Hence , centre of mass of combined discs is at  ( 0.9 R , 0.7 h ) 
 
Distance d between centre of mass of bigger disc and that of combined discs is given as
 
begin mathsize 14px style d space equals space square root of open parentheses 0.9 space minus space 1 close parentheses squared R squared space plus space left parenthesis 0.7 space minus space 0.5 right parenthesis squared h squared end root space equals space square root of 0.01 space R squared space plus space 0.04 space h squared end root end style
Answered by Thiyagarajan K | 25 Sep, 2021, 02:01: PM
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