A uniform disc of radius R has a hole cut out which has a radius r. The centre of hole is at a distance `(R )/(2)` from the centre of disc. The position of centre of mass is shifted through a distance x from 'O' find x. If in this question the values of R = 6 m and that if r = 1 m, calculate the value of shift from 'O' and state whether it is towards left or right of O.

Options

A. `(R^(2)r)/(2(R^(2)-r^(2))),(3)/(37)m` towards left of 'O'
B. `(Rr^(2))/(2(R^(2)+r^(2))),(3)/(37)m` towards right of 'O'
C. `(Rr^(2))/(2(R^(2)-r^(2))),(3)/(35)m` towards left of 'O'
D. `(Rr^(2))/(2(R^(2)+r^(2))),(3)/(35)m` towards right of 'O'

Answer

C

Explanation

Mass of complete disc = `piR^(2)trho`
(t = thickness)
Mass of scooped out discc `=pir^(2)trho`
Mass of remaining disc `=pi(R^(2)-r^(2))trho`
Then, `pi(R^(2)-r^(2))trhoxx x=pir^(2)trho(R)/(2)`
or `x=(r^(2)R)/(2(R^(2)-r^(2)))`

Now `x=(r^(2)R)/(2(R^(2)-r^(2)))`
Here `r=1m`
`R=6m`
`therefore x=((1)^(2)xx6)/(2(36-1))=(3)/(35)m`
towards left of centre.