Two uniform discs made of same material and thickness of redii `R` and `2 R` are joined as shown. Locate `c.m.`

Explanation

This diagram is symmetrical about the `x`-axis.
`Mass = density xx volume = density xx area xx thickn ess`
Here , density and thickness is same , hence mass `alpha` area
`Disc (1), m_(1) prop pi R^(2) rArr m_(1) = k xx pi R^(2)`
`rArr m_(1) = m, x_(1) = R`
`Disc(2) , m_(2) prop pi( 2R)^(2) rArr m_(2) = k xx 4pi R^(2)`
`rArr m_(2) = 4m , x_(2) = 2 R + 2R = 4R`
`x_(c.m.) = (m_(1)x_(1) + m_(2) x_(2))/(m_(1) + m_(2)) = ( m xx R + 4m xx 4 R)/(m + 4m)`
`= (17 R)/(5) = 3.4 R`
Center of mass from the center of smaller disc, i.e. from `O_(1)` is
`3.4 R - R = 2.4 R`
Center of mass from the contact point of disc , i.e. from `O_(2)` is
`3.4 R - 2 R = 1.4 R`
Center of mass from the center of the larger disc , i.e. from `O_(3)` is
`4 R - 3.4 R = 0.6 R` Note : In general , assume reference axes , as explained previously but in the question if the reference point is given , e.g. locate `c.m.` from center of smaller disc. You can assume center of smaller disc as origin.
`{:(m_1=m,,,,x-co o rdi nate),(,,,,0),(m_2=4m,,,,R+2R=3R):}`
`x_(c.m.) = (m xx 0 + 4m xx 3R)/(m + 4m) = (12)/(R ) = 2.4 R`
`c.m.` from joining point of discs :
`{:(,,,,x-co o rdi nate),(m_(1) = m,,,,-R),(m_2=4m,,,,2R):}`
`x_(c.m.) = ( m xx (-R) + 4m xx 2 R)/(m + 4m) = (7 R)/(5) = 1.4 R`
`c.m.` from center of large disc :
`{:(,,,,x-co o rdi nate),(m_(1) = m,,,,-(2R + R) = -3R),(m_2=4m,,,,0):}`
`x_(c.m.) = (m(-3 R) + 4m xx 0)/(m + 4m)`
`= -(3 R)/(5) = -0.6 R`
The `-ve` sign indicates that `c.m.` will be on the left of `O`. Note that `c.m.` will be nearer to the `c.m.` of larger mass.