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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? 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 Answered by Thiyagarajan K | 25 Sep, 2021, 02:01: PM
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