1.

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the disc coincide. The centre of mass of the new disc is alphaR from the centre of the bigger disc. The value of alpha is :

Answer»

`(1)/(2)`
`(1)/(6)`
`(1)/(4)`
`(1)/(3)`

Solution :Let .O. be the centre of bigger disc of radius `2R=AO` and O. be the centre of SMALLER disc which has been removed from the bigger and has a radius `R=O.A`.

MASS of bigger disc `M=pi(2R)^(2).t.rho=4piR^(2)trho`
Mass of smaller disc `m=piR^(2)trho=piR^(2)trho=(M)/(4)`
Mass of remaining portion = `M-m`
`=M-(M)/(4)=(3)/(4)M`
Let G be the centre of mass of remaining portion and OG be its distance from the centre of mass of the bigger disc. Then taking moments about .O., we get
`(3)/(4)MxxOG=(M)/(4)xxO.O`
or `(3)/(4)OG=(R)/(4)orOG=(R)/(3)`
`ALPHA.R=(R )/(3)`
or `alpha=(1)/(3)`


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