Question

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the cirucmferences of the discs coincide. The centre of mass of the new disc is `alpha/R` from the center of the bigger disc. The value of `alpha` is
A. `1//3`
B. `1//2`
C. `1//6`
D. `1//4`

Answer 1

Correct Answer - A
Since mass `prop` area

Let mass of the bigger disc `= 4 M`
`:.` mass of the smaller disc = M
`:.` mass of the remaining posrtion `= 4M - M = 3M`
Now put the cut disc at its place again, centre of mass of the whole disc will be at centre O.
Centre of mass of the smaller disc is at its centre that is at B.
Suppose CM of the remaining portion is at A and `AO` is X, Let O as origin
`:. 3 M(x) = RM rArr x = (R )/(3)`
This suggests that centre of mass of remaining disc will shift from the centre of original disc by a distance of `(1//3)R` towards left.
`alpha(1)/(3)`.